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First Edition, 2009 ISBN 978 93 80168 42 5 © All rights reserved. Published by: Global Media 1819, Bhagirath Palace, Chandni Chowk, Delhi-110 006 Email: [email protected]
Table of Contents
1. Temperature Measurement
2. Derived Theory
3. Macroscopic Properties
4. Change of State
5. Heat Capacity
6. Heat and Temperature
7. The Reversibility
8. Physical Significance
9. Changing State
1Temperature Measurement
1
Temperature Measurement
Scale of Temperature
In order to indicate the temperature of a body as a number,a ‘scale of temperature’ is adopted and the temperature of thebody is estimated against this scale. The instrument on whichsuch a scale is used, is called a ‘thermometer.’ The temperaturescale is completed in the following steps:
(i) First of all, some measurable physical property of asubstance is chosen which changes with temperature :Such properties are (a) the length of a liquid columnin a glass capillary, (b) the volume of a fixed mass ofgas at constant pressure, (c) the pressure of a fixedmass of gas at constant volume, (d) the electricalresistance of a metallic wire, (e) the e.m.f. of athermocouple, (f) the colour of a lamp filament, etc.The chosen substance and its property are calledrespectively the ‘thermometric substance’ and the‘thermometric property’.
2 Thermal Physics
Suppose we have chosen a thermometric substance whosethermometric property is represented by X. Suppose thethermometer using this substance is in thermal equilibriumwith a system. We arbitrarily define that the magnitude of thisproperty varies linearly with the temperature T of the system.That is
X = aT, ...(i)
where a is some constant. This means that equal changes intemperature correspond to equal changes in X.
(ii) Now, an easily reproducible state of some standard systemis chosen. The temperature corresponding to this state iscalled a ‘fixed-point’: This point is chosen to be the ‘triplepoint’ of water, i.e. the temperature and pressure atwhich ice, liquid-water and water-vapour coexist inequilibrium. (The water vapour-pressure at the triplepoint is 4.58 mm of mercury).
(iii) The chosen fixed point is given an arbitrary numericalvalue for the temperature. The temperature at the triple-point of water is arbitrarily given a value of 273.16 K(kelvin). The kelvin is a unit temperature-interval.
(iv) Finally, the kelvins are numbered by using the chosenthermometric property X.
The temperature is kept at the triple point (by dipping thethermometer in a triple-point cell), and the magnitude of theproperty X at this point (273.16 K) is measured. Let it be Xtr.Then, from equation (i), we have
Xtr = α (273.16). ...(ii)
Then the thermometer is placed in contact of the systemwhose temperature T is to be measured, and the magnitudeof X at this unknown temperature is measured. Let it be XT.Then, again from equation (i), we have
XT = aT. ...(iii)
3Temperature Measurement
Dividing equation (iii) by equation (ii), we have
273.16T
tr
X TX
=
∴ 273.16 .T
tr
XT K
X⎛ ⎞
= ⎜ ⎟⎝ ⎠
Thus T can be calculated from the two measurements Xtr
and XT. This is the temperature defined on the scale using thethermometric property X. To emphasise this, we write Tx forT in the above equation :
273.16 Tx
tr
XT K
X⎛ ⎞
= ⎜ ⎟⎝ ⎠
...(iv)
Let us now define the temperature T on the scales givenby actual thermometers using different thermometricproperties:
Temperature on Liquid-in-glass Scale
In a liquid-in-glass thermometer, the thermometric propertyX is the length of a liquid column, l, (in a capillary tube) whichincreases with temperature. If ltr be the length of the columnat the triple-point, and lT the length at an unknown temperature,then the unknown temperature measured on the length scaleis defined by [from equation (iv)]
1 273.16 .T
tr
lT K
l⎛ ⎞
= ⎜ ⎟⎝ ⎠
Temperature on Constant-volume Gas Scale : At constantvolume, the pressure p of a gas varies with temperature; andthis is the thermometric property used in a constant-volumegas thermometer. If ptr be the pressure at the triple-point, andpT at an unknown temperature, then the unknown temperaturemeasured on the gas pressure scale is defined by
4 Thermal Physics
273.16 .Tp
tr
pT K
p⎛ ⎞
= ⎜ ⎟⎝ ⎠
Temperature on Constant-pressure Gas Scale : At constantpressure, the volume V of a gas varies with temperature andis the thermometric property used in a constant-pressure gasthermometer. As above, the temperature on the gas volumescale is defined by
273.16 .Tv
tr
VT K
V⎛ ⎞
= ⎜ ⎟⎝ ⎠
Temperature on Resistance Scale : In a platinum resistancethermometer, the thermometric property is the electricalresistance R of a platinum wire which varies with temperature.If Rtr be the resistance at the triple point and RT at an unknowntemperature, then the unknown temperature measured on theresistance scale is defined by
273.16 .TR
tr
RT K
R⎛ ⎞
= ⎜ ⎟⎝ ⎠
Similar expressions hold for other thermometric substancesand thermometric properties.
Need of a Standard Scale of Temperature : The differentproperties of matter utilised in different thermometers givedifferent scales of temperature. Now, the difficulty is that ifwe measure the temperature of a body with different kinds ofthermometers, we find differences among their readings. (Allthermometers agree only at the fixed point 273.16 K). Evenwhen different thermometers of the same kind are used, suchas constant-volume gas thermometers filled with different gases,we obtain different temperature readings for the body. Hence,to obtain a completely definite scale of temperature, we selectone particular kind of thermometer as the ‘standard’ andcalibrate all the other thermometers against it.
5Temperature Measurement
The Standard Scale (Ideal Gas Temperature Scale) : Outof the various types of thermometers, the smallest differencein readings is found among the constant-volume gasthermometers filled with different, so-called permanent gases.The difference is only of the order of 0-2 - 0.3%. It is noticeablethat if the amount of the gas filled in the thermometers isreduced i.e. the pressure of the gas is lowered; the differencein readings among different thermometers further decreases.This can be seen from the following experiment:
Suppose an amount of a gas, say O2, is filled in a constant-volume gas thermometer such that the pressure of the gas ptr
is 1000 mm of Hg when the bulb of the thermometer is in atriple-point cell. Now the bulb is placed in condensing steamand the pressure of the gas (pT) at steam temperature ismeasured. The temperature of the steam is calculated by theformula
273.16 ,Tp
tr
PT K
P⎛ ⎞
= ⎜ ⎟⎝ ⎠
where pT is the gas pressure. Now the quantity of the gas inthe thermometer is reduced in steps (so that ptr becomes smallerand smaller) and each time the temperature of steam Tp isdetermined.
6 Thermal Physics
A graph is then plotted between Tp and ptr, which ispractically a straight line. The line is produced back to intersectthe axis where ptr = 0. The whole experiment is then repeatedby filling the thermometer with other gases like air, N2, He,H2 turn by turn. These graphs show that the different gasthermometers do give different temperatures of the steam, butthe difference decreases as the gas pressure is lowered.
As pressure ptr→ 0, ‘all’ gas thermometers give the samevalue of 373.15 K for the steam temperature. The temperaturemeasured by a constant-volume gas thermometer, andextrapolated to a value corresponding to ptr→ 0, is known as‘ideal gas temperature’ and the corresponding scale is knownas ‘ideal or perfect gas temperature scale’. It is defined by therelation
0
lim273.16 .T
trtr
PT K
p P→
⎛ ⎞= ⎜ ⎟⎝ ⎠
The standard thermometer is therefore chosen to be aconstant-volume gas thermometer, using a temperature scaledefined by the above relation.
The International Bureau of Weights and Measures hasaccepted a ‘constant-volume hydrogen thermometer’ as thestandard thermometer when the hydrogen has been filled ata pressure of 1 meter of mercury at 0°C. For measurement oflow temperatures, a constant-volume helium thermometer hasbeen recommended, which can be used upto 1 K. Thus, 1 Kis the lowest ideal gas temperature that can be measured bya gas thermometer. (The temperatures below 1 K remain as yetundefined). The choice for H2 among all the gases is due to thefact that it is (except helium) nearest to an ideal (perfect) gas,and can be obtained in the pure state anywhere in the world.Near 1 K only helium can be used because all gases liquify.
Absolute Scale of Temperature : An absolute scale oftemperature is one which is independent of the properties of
7Temperature Measurement
any particular substance. No scale given by any thermometeris absolute as it depends upon the properties of thethermometric substance. Even the scale defined above is notabsolute, because although it is independent of the propertiesof any one particular gas, it still depends upon the propertiesof gases in general, that is, on the properties of an ideal gas.
Lord Kelvin, from the consideration of the efficiency of anideal heat engine, defined a scale which is absolutelyindependent of the properties of the working substance in theengine. This scale is called the ‘Kelvin’s absolute thermo-dynamical scale of temperature’. It is defined such that if anideal engine absorbs a quantity of heat Q1 at temperature 1θ ,
and rejects a quantity of heat Q2 at θ2 , then
1 1
2 2
θθ
=
It is, however, remarkable that in the temperature-rangein which a gas thermometer can be used, the ideal gas scaleand the Kelvin’s absolute scale are exactly identical in allrespects. Hence, the realization of an ideal gas scale is in factthe realization of Kelvin’s absolute scale. This is why thetemperatures on the ideal gas scale are expressed in K (kelvin).
Relation of Celsius and Fahrenheit Scales with the Kelvin’sScale : The Kelvin’s absolute scale has an absolute zero of0 K and the Kelvin temperature of triple-point of water (Ttr)is 273.16 K.
By experiment, the Kelvin temperature of the ice point (Tice)comes to be 273.15 K and that of the steam point (Tsteam) 373.15 K.
The temperature scales in common use are, however, theCelsius scale and the Fahrenheit scale. Both of these scales havebeen defined in terms of the Kelvin’s scale.
The Celsius temperature scale employs a degree of thesame magnitude as that of the Kelvin’s absolute scale; but its
8 Thermal Physics
zero has been set at 273.15 K. Thus, if the temperature of a bodyon the Celsius scale is t (°C) and that on the Kelvin’s scale isT (K), then
t = T- 273.15.
Thus, the triple point of water on the Celsius scale is
ttr = Ttr- 273.15
= 273.16 - 273.15 = 0.01°C.
Similarly, the temperatures of the ice-point and steam-point are
tice = Tice - 273.15 = 273.15 - 273.15
= 0.00°C
and tsteam = Tsteam - 273.15
= 373.15 - 273.15 = 100.00°C.
A temperature on the Fahrenheit scale (tF) is related to thecorresponding temperature on the Celsius scale (tC) as
tF = 32° + 9
.5 ct
Clearly, the ice-point (0.00 °C), the triple-point of water(0.01°C) and the steam-point (100.00 °C) are 32.0 °F, 32.018 °Fand 212.0 °F respectively on the Fahrenheit scale.
9Derived Theory
2
Derived Theory
Dulong and Petit Law : In 1819, Dulong and Petitestablished empirically that “the atomic specific heat of allsolids is nearly 6 cal/(gm-atom-K) and is independent oftemperature.” This is Dulong and Petit Law.
Derivation from Classical (Kinetic) Theory : The atoms ina metallic solid are arranged in a crystalline array, held inposition by strong electromagnetic forces exerted on oneanother. Thus they are not free to wander about (as they arein a gas). They can simply vibrate about their fixed positionsunder the electromagnetic interaction forces. Thus they havekinetic energy of vibration and potential energy of interaction.
We know from the law of equi-partition of energy that theaverage kinetic energy per atom in each degree of freedom is½kT. Now, in simple oscillatory motion, the average kineticenergy is equal to the average potential energy. Therefore, theaverage potential energy per atom in each degree of freedomwill also be ½ kT. Hence the total energy per atom in eachdegree of freedom will be ½ kT+ ½kT = kT†. As the atom has
10 Thermal Physics
3 degrees of freedom in oscillatory motion, its total energyis 3 kT.
Now, 1 gm-atom of a solid has N atoms in it, where N isAvogadro’s number. Therefore, the total energy of 1 gm-atomof the solid is given by
U = N × 3kT.
But N × k = R, the universal gas constant.
∴ U = 3RT.
The atomic specific heat Cv of a solid at constant volumeis the energy required to raise the temperature of 1 gm-atomof the solid by 1°C (or 1K). Thus
Cv = .dUdT
Putting the value of U from above, we get
CV =d
dT (3RT) = 3R.
We know that R 2 cal/(gm-atom-K).
∴ CV 6 cal/(gm-atom-K).
This is Dulong and Petit Law.
Variation of Specific Heat with Temperature
Applicability of the Law: According to the Dulong andPetit law, the specific heat of all solids must be about 6 cal/(gm-atom-K) and it should not vary with temperature. Actually,this is not true. The observed relationship between the atomicspecific heat of solids and temperature.
The non-metallic solids such as boron, carbon and siliconhave specific heats appreciably different from 6 at ordinarytemperatures. At very high temperatures, however, the specificheats of these solids approach the value given by Dulong andPetit law.
11Derived Theory
The specific heats of metallic solids, however, lie near 6 cal(gm-atom-K) at ordinary temperatures and change little whenthe temperature is raised. Thus, metallic solids obey the Dulongand Petit law at ordinary as well as at high temperatures. But,the law completely fails at very low temperatures. The atomic heatat low temperatures is found to decrease with fall intemperature. Below a certain temperature (which is differentfor different solids), the atomic heat decreases rapidly withdecreasing temperature. It tends to zero at absolute zero forall solids.
The non-agreement of the result of the classical theory ofspecific heats with experiment led to the belief that classicalmechanics breaks down in this case.
Einstein’s Quantum Theory of Specific Heat : Einstein, in1907, applied Planck’s quantum theory to the specific heats ofsolids. According to this theory, an atomic oscillator offrequency ν can only have energy values , 2 , 3 ...h h hν ν ν whereh is Planck’s constant, and not in between. On this basis, theaverage energy of an oscillator per degree of freedom is
( )/ not .1h kT
hkT
e ν
ν−
Einstein assumed that all atoms of a given solid vibrateindependently of each other with the same frequency in each of thethree degrees of freedom. Therefore, the total energy of a gm-atomof the solid consisting of N atoms is
U = /3 .1h kT
hN
e ν
ν−
Therefore, the specific heat is given by
CV =( )
/2
2/3
1
h kT
h kT
he
dU kTN hdT e
ν
ν
ν⎛ ⎞⎜ ⎟⎝ ⎠= ν−
12 Thermal Physics
=( )
2 /
2/3 .
1
h kT
h kT
h eN k
kT e
ν
ν
ν⎛ ⎞⎜ ⎟⎝ ⎠ −
But Nk = R (the universal gas constant).
∴ CV =( )
2 /
2/3 .
1
h kT
h kT
h eR
kT e
ν
ν
ν⎛ ⎞⎜ ⎟⎝ ⎠ −
Let us put ,E
hkν= Θ where EΘ is known as Einstein
characteristic temperature. Then, we have
( )2 /
2/3 .
1
E
E
TE
V T
eC R
T e
Θ
Θ
Θ⎛ ⎞= ⎜ ⎟⎝ ⎠ − …(i)
This is Einsteins’s specific heat formula.
(i) At high temperatures (T >> EΘ ), we have 1E
TΘ
<< so that
/ 1 .E T EeT
Θ Θ≈ + The eq. (i) then takes the form
2
2
13 3 .
E
EV
E
TC R RT
T
Θ+Θ⎛ ⎞≈ ≈⎜ ⎟⎝ ⎠ Θ⎛ ⎞
⎜ ⎟⎝ ⎠…(ii)
This is in agreement with Dulong and Petit law whichagrees with experiment at high temperatures.
(ii) At low temperatures (T << EΘ ), we have 1,E
TΘ
>> so that
/ 1.E TeΘ >> Now, the Einstein’s formula (i) gives
2/3 .E TE
VC R eT
−ΘΘ⎛ ⎞≈ ⎜ ⎟⎝ ⎠...(iii)
This shows that as T → 0, the specific heat CV alsoapproaches zero, a conclusion in general agreement withexperimental results.
13Derived Theory
Limitations : Although Einstein’s theory gives qualitativeaccount of the experimental specific heat curves, there is noquantitative agreement throughout. The theory fails at verylow temperatures where Cv is found to be nearly proportionalto T3 rather than as given by eq. (iii). This disagreement is dueto the neglect of the mutual forces exerted by the atoms uponone another. Taking this into consideration, Debye in 1912developed a specific heat formula which gives excellentagreement with experiment over the whole temperature-range.
Debye’s Theory of Specific Heat of Solids
Debye, presented a theory of the specific heat of solidswhich was a modification of the Einstein’s quantum theory. Heconsidered solid as a continuous elastic body; in which thevibrations of the atoms generate stationary waves. The internalenergy of the solid resides in these elastic stationary waves.
Let us consider a unit volume of a continuous solid inwhich elastic waves are set up. It can be shown that the numberof possible waves of any kind with frequencies between ν andν + d ν is
23
4d
πν ν
ν
where ν is the wave-speed. There are two kinds of elasticwaves that can occur in a solid; longitudinal waves with speedvl and transverse waves with speed vt moreover, there are twoperpendicular directions of polarisation for a transverse wave.Hence the total number of possible elastic stationary waveswith frequencies between and dν ν+ ν is
23 3
1 24 .
tl
dv v
⎛ ⎞π + ν ν⎜ ⎟⎝ ⎠
If V is the volume of a gm-atom of a particular solid, the
number of waves with frequencies between and dν ν+ ν in it
will be
14 Thermal Physics
23 3
1 24 .
tl
V dv v
⎛ ⎞π + ν ν⎜ ⎟
⎝ ⎠
Hence the total number of waves comprising all possiblefrequencies is
23 3 0
1 24 .
mV
tl
V dv v
⎛ ⎞π + ν ν⎜ ⎟
⎝ ⎠∫
The upper limit of integration cannot be infinite, as thiswould make the number of waves and hence the internalenergy of the solid infinite. Debye assumed that in a solid nofrequency can go beyond a definite upper limit vm, which ischaracteristic of the substance. This limit is so chosen that thetotal number of waves is equal to the total number of degreesof freedom of the solid i.e. equal to 3N, where N is Avogadro’snumber. Hence vm can be determined by putting the expression(i) equal to 3N. That is
3N =2
3 3 0
1 24
mV
tl
V dv v
⎛ ⎞π + ν ν⎜ ⎟
⎝ ⎠∫
= 33 3
4 1 2,
3 mtl
Vv v
⎛ ⎞π+ ν⎜ ⎟
⎝ ⎠
and hence
3mν =
3 3
9.
1 24
tl
N
Vv v
⎛ ⎞π +⎜ ⎟
⎝ ⎠
…(i)
Thus the value of vm for a given solid can be calculatedfrom a knowledge of its elastic constants.
Let us now calculate the energy of solid. According toPlanck’s theory, the average energy E per wave is given by
E = / ,1h kT
he ν
ν−
15Derived Theory
where k is Boltzmann’s constant and T is Kelvin tempera-ture. Therefore, the total internal energy of a gm-atom of thesolid is
U = E × total number of waves asgiven by (i).
Thus U =3
3 3 /0
1 24
1mV
h kTtl
h dV
v v e ν
⎛ ⎞ ν νπ +⎜ ⎟ −⎝ ⎠ ∫
Substituting the value of 3 3
1 24
tl
Vv v
⎛ ⎞π +⎜ ⎟
⎝ ⎠from eq. (ii), we
have
U =3
3 /0
9.
1mV
h kTm
N h dv e ν
ν ν−∫
For convenience let us put hkTν
= x, so that kTh
ν = x and
dv = kTh dx. Then
U =3 3
3 0
9;
1mx
xm
N kT x dxkT
v h e⎛ ⎞⎜ ⎟ −⎝ ⎠ ∫
where mm
hx
kTν
=
=3 3
4
09 .
1mx
xm
k x dxNk T
h e⎛ ⎞⎜ ⎟ν −⎝ ⎠
∫
Let us now define a characteristic Debye temperature ,DΘ as
DΘ = , so that .m m Dm
h hx
k kT Tν ν Θ
= =
In view of eq. (ii), DΘ can be independently calculated for
a given solid. Now, the energy expression becomes
U =( )4 3/
3 09
1D T
xD
T x dxR
e
Θ
Θ −∫ [ ]R Nk=Q
16 Thermal Physics
Therefore, the atomic specific heat of the solid, CV, isgiven by
CV =v
UT∂⎛ ⎞⎜ ⎟∂⎝ ⎠
=( )
3 3/
0
19 4
1 / 1D T
Dx
D D
T x dxR
e T e T
Θ⎡ ⎤⎛ ⎞ Θ⎛ ⎞−⎢ ⎥⎜ ⎟⎜ ⎟ ⎝ ⎠Θ − Θ −⎝ ⎠⎢ ⎥⎣ ⎦∫
This is Debye’s specific heat formula.
At high temperatures both and Dhx
kT TΘν⎛ ⎞=⎜ ⎟
⎝ ⎠ are very small
so that /1 , and 1 / .Dx TDe x e TΘ≈ + ≈ +Θ Then Debye’s formula (iii)
gives
CV ≈( )
3/ 2
0
19 4
/D T
D
D D
TR x dx
T T
Θ⎡ ⎤⎛ ⎞ Θ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟Θ Θ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦∫
≈( )3 3/
9 4 13
D
D
TTR⎡ ⎤Θ⎛ ⎞
−⎢ ⎥⎜ ⎟Θ⎢ ⎥⎝ ⎠⎣ ⎦
≈ 3R,
which is the Dulong-Petit value in agreement with experimentat high temperatures.
At very low temperatures / ,D TΘ →∞ so that the second termin the brackets of eq. (iii) becomes negligible, and we canreplace the upper limit of integration in the first term by .∞Thus, we have
CV ≈3 3
09 4 .
1xD
T x dxR
e
∞⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟Θ −⎢ ⎥⎝ ⎠⎣ ⎦
∫
But the integral 3 4
0.
1 15x
x dxe
∞ π=
−∫
17Derived Theory
∴ CV ≈3 4
9 415D
TR⎡ ⎤⎛ ⎞ π⎢ ⎥⎜ ⎟Θ⎢ ⎥⎝ ⎠⎣ ⎦
≈4
33
12.
5 D
RT
πΘ
Since DΘ is a constant for the solid, we obtain
CV ∝ T3.
Thus the specific heat at extremely low temperatures variesas the cube of the Kelvin temperature, a conclusion in excellentagreement with experiment. This is known as “Debye’s T3
law.”
At intermediate temperatures the Debye’s formula hasbeen evaluated numerically. The results have been found inexcellent agreement with experiment for many substances.
PROBLEMS
1. Calculate Einstein temperature for a case for whichν = 2.49 × 1012 sec–1. Given : k = 1. 38 × 10–23 joule/K andh = 6.63 × 10–34 joule-sec.
Solution: The Einstein temperature is given by
EΘ =hkν
=( )( )34 12 1
23
6.63 10 joule - sec 2.49 10 sec
1.38 10 joule/K
− −
−
× ×
×
= 119.6 K.
2. Calculate the Einstein’s frequency for a case for which
Einstein temperature EΘ = 236K. Given : k = 1.38 × 10–23 joule/
K and h = 6.63 × 10–34 joule-sec.
Ans: 4.9 × 1012 sec–1.
18 Thermal Physics
3. Calculate Einstein’s frequency for copper.
Given: EΘ = 230 K, h = 6.63 × 10–34 Js and k = 1.38 × 10–23 J/
K. Show that if T >> 230 K, then the classical result CV = 3Rwill hold good.
Solution: By the definition of Einstein temperature
/ ,E h kΘ = ν the Einstein’s frequency is
( )2312 1
34
230 1.38 10 /4.79 10 .
6.63 10E
K J Kks
h Js
−−
−
× ×Θ ×ν = = = ×
×
If T >> 230 K ( ) ,EΘ then E
TΘ
<< 1 and
/E TeΘ = 1 .E
TΘ
+
Substituting this in Einstein’s specific heat formula, we get
CV =( )
2 /
2/3
1
E
E
TE
T
eR
T e
Θ
Θ
Θ⎛ ⎞⎜ ⎟⎝ ⎠ −
or CV ≈2
2
13 3 .
E
E
E
TR RT
T
Θ+Θ⎛ ⎞ ≈⎜ ⎟Θ⎝ ⎠ ⎛ ⎞
⎜ ⎟⎝ ⎠
19Macroscopic Properties
3
Macroscopic Properties
System and its Surroundings : Any portion of matter whichis considered as separated from its surroundings is called a‘system’. All those things which are outside that system andinfluence its behaviour are known as the ‘surroundings’ of thesystem. For example, let a gas be filled in a cylinder fitted witha piston and heated by a burner. Here the ‘gas is the system’,while the ‘piston and burner are the surroundings’.
The behaviour of the system can be described in terms oftwo types of properties ; the macroscopic and the microscopic.
These are the properties which (i) describe the grosscharacteristics of the system, (ii) can be measured directly inthe laboratory and (iii) can usually be directly experienced byour sense of perception. They are not concerned with thestructure of the system. Thus, in the above example, the chemicalcomposition (if any), pressure, volume, temperature, internalenergy, entropy, etc. are the macroscopic properties of the gas(system).
20 Thermal Physics
Microscopic Properties : These are the properties which(i) describe the internal structure (atoms and molecules) of thesystem, (ii) cannot be measured directly in the laboratory, and(iii) cannot be directly experienced by our sense of perception.In the above example, the masses, velocities, energies, momenta,etc. of the molecules of the gas are the microscopic propertiesof the gas.
Relation between Macroscopic and Microscopic Properties :The macroscopic and microscopic properties are simply thedifferent ways of describing the same system. Hence, theymust be related to each other. For example, the pressure of agas is related to the average rate of change of momentum dueto all the molecular collisions made on a unit area. Higher therate of change of momentum, higher the pressure. Here, thepressure is a macroscopic property, whereas the rate of changeof momentum due to molecular collisions is a microscopicproperty. Similarly, the temperature of a gas, macroscopicproperty, is related to the average kinetic energy of translationof its molecules, which is a microscopic property.
Pressure, Volume and Temperature—Macroscopic orMicroscopic : These properties have no meaning for anindividual molecule, but only for a large collection of molecules.They can be directly experienced by our sense of perceptionand can be measured in the laboratory. Hence, they areintrinsically macroscopic concepts.
The Temperature
Temperature : We can distinguish hot bodies from coldbodies by our sense of touch. For example, by touch, we canroughly arrange bodies in the order of their hotness, decidingthat A is hotter than B, B is hotter than C, and so on. This isknown as our ‘temperature’ sense and it is said that A is at ahigher temperature than B, B is at a higher temperature thanC, and so on. Thus, the temperature of a body is a measureof the hotness or coldness of the body.
21Macroscopic Properties
Thermal Equilibrium
A system is said to be in thermal equilibrium if, any twoof the independent properties describing the state of the systemremain constant as long as the surrounding conditions remainunchanged. For example, if a gas enclosed in a vessel has apressure p and a volume V at the temperature of thesurroundings, the values of p and V will remain constant aslong as the temperature of the surroundings remainsunchanged. The gas is said to be in thermal equilibrium withthe surroundings. The values of pressure and volume areconnected by a functional relationship
( ), 0.f p V =
This is called the ‘equation of constraint’. For a gas obeyingBoyle’s law, pV= constant, it would read
pV- constant = 0.
Let us consider two systems A and B isolated from thesurroundings, but in thermal contact with each other. Theirindependent properties undergo a change until an equilibriumstate of the combined system is reached when these propertiesassume constant values. The systems are now in thermalequilibrium with each other. Thus, thermal equilibrium is thestate attained by two (or more) systems in thermal contact withone another and characterised by constant values of theproperties of the systems.
Zeroth Law of Thermodynamics : This fundamental lawstates that if two systems A and B are separately in thermalequilibrium with a third system C, then A and B are in thermalequilibrium with each other.
Suppose there are three systems A, B and C. Systems A andB are isolated from each other but are in thermal contact withC. Experiments show that both A and B individually attainthermal equilibrium with C. If now A and B are put in thermal
22 Thermal Physics
contact of each other, no further change takes place. That is,A and B are found to be in thermal equilibrium with eachother.
A simple illustration of the law is provided when A andB are two gases enclosed in vessels and C is a mercurythermometer. The zeroth law says that if there is no changein the length of mercury thread when the thermometer C isplaced in thermal contact of A nor when it is placed in thermalcontact of B, then there will be no change if A and B arebrought in thermal contact with each other.
Zeroth Law
The zeroth law can be used to define temperature. It leadsto that all systems in thermal equilibrium with one anotherhave a common property having same value for all of them.This property is ‘temperature’. Thus, the temperature of asystem is the property which determines whether or not thesystem is in thermal equilibrium with other systems. That is,two systems in thermal equilibrium with each other are at thesame temperature. Conversely, if two systems have differenttemperatures, they cannot be in thermal equilibrium with eachother.
PROBLEMS
1. Three systems (gases) 1,2, and 3 have coordinates (p1, V1);(p2 , V2) and (p3, V3). When 1 and 3 are in thermal equilibrium,the equation p1V1-nbp1 -p3V3 = 0 holds. When 2 and 3 are in
thermal equilibrium, the equation p2V2 -p3V3 + 3 3
2
'0
nb p VV
=
holds, where n, b, b’ are constants.
(a) What equation expresses thermal equilibrium between1 and 2 ? (b) What are three functions which are equal to oneanother at thermal equilibrium and each is equal to the empirictemperature T.
23Macroscopic Properties
Solution : (a) The systems 1 and 3 are in thermal equilibrium:
1 1 1 1 3- 0PV nbp p V− =
or 1 1 3 3( ) .P V nb p V− = ...(i)
The systems 2 and 3 are in thermal equilibrium :
3 32 2 3 3
2
'0
nb p vP V p V
V− + =
or 2 2 3 32
'1 0
nbP V P V
V⎛ ⎞
− − =⎜ ⎟⎝ ⎠
or2 2
3 3
2
.'
1
p VP V
nbV
=− (ii)
According to zeroth law, 1 and 2 must also be in thermalequilibrium. To obtain the required equation, we must eliminatep3 and V3 from eq. (i) and (ii), when we get
( ) 2 21 1
2
'1
p Vp V nb
nbV
− =− ...(iii)
(b) Looking at eq. (i), (ii) and (iii), the three functions equalto one another at thermal equilibrium are
( ) 2 21 1 3 3
2
, and '
1
p Vp V nb p V
nbV
−−
Since the systems are gases, each of these is equal to theempiric temperature T (by gas equation pV= RT).
4
Change of State
Clausius-Clapeyron Equation : Clapeyron in 1834, andClausius in 1850, deduced an important equation whichdescribes conditions governing changes of state, such as meltingof solids and boiling of liquids. This is known as, the “Clausius-Clapeyron equation” or the “first latent heat equation.”
Let us consider two isothermals ABCD and EFGH at closetemperatures T and T- d T respectively for unit mass of asubstance below its critical point. Along the sections AB andEF the substance is in the liquid state, along the sections BCand FG the liquid and vapour states coexist in equilibrium,while along CD and GH only the unsaturated vapour stateexists. At B and F the substance is entirely liquid, while at Cand G it is entirely just saturated vapour. Let p and p - d p bethe saturated vapour pressures of the liquid at temperaturesT and T— d T respectively. Let V1 and V2 be the specific volumes(volume per unit mass) of the substance at B and C respectively.
Let us draw two adiabatics through B and C .They intersectthe lower isothermal at points M and N (very close to F and
26 Thermal Physics
G) respectively. The figure BCNM represents the indicatordiagram when the substance is enclosed in the cylinder ofan ideal heat engine and is taken through a reversibleCarnot cycle along BCNMB. The efficiency e of the engine isdefined as
W (work done by the substance during the cyclee = .
Q (heat taken in at temperature T)1...(i)
Now, during a Carnot cycle, the work done is given by thearea BCNMB . As the adiabatics are very small, they can betaken straight and parallel. Thus, the figure BCNMB is aparallelogram, and so
W = area BCNMB
= BC × perpendicular distance between BC and MN
= (V2 - V1) × d p.
During the process BC, the unit mass of the substance ischanged completely from the liquid state at B to the vapourstate at C at constant temperature T. The heat Q1 taken in bythe substance during the process BC is given by
27Change of State
Q1 = L
where L is the latent heat of vaporisation at temperature T.Substituting these values of W and Q1 in eq. (i), we get
2 1
1
(V - V ) pWe = =
Q Ld
In this expression, the numerator (V2 - V1) d p is in workunit, and so the latent heat L should also be put in work unit.If L is given in heat unit it must be converted into work unitby multiplying it by J.
Now, by the definition of thermodynamic scale oftemperature, the efficiency of a Carnot engine working between
temperatures T and T- d T is given by
1T T T
eT T
d d-= - =
Equating the last two expressions for e, we get
2 1(V - V ) p T =
L Td d
or2 1
p L =
T T (V - V )dd
Infinitesimally, 2 1
dp L =
dT T (V - V )
This is Clausius-Clapeyron equation. In this equation, V1
and V2 are the specific volumes (volume/mass) of the substancein the liquid state and the vapour state respectively.
If L is given is cal/gram or in kcal/kg, then
2 1
dp J L =
dT T (V - V )
The above equation also holds to the change from the solidto the liquid state ; but L will then represent the latent heat
28 Thermal Physics
of fusion, V1 the volume occupied by the unit mass of thesubstance in the solid state and V2 that in the liquid state.
Effect of Pressure on Boiling Point of Liquids : If the partBC of the isothermal ABCD at temperature T represents thechange of state of the substance from liquid to vapour, thenT is the boiling point of the substance under the pressure p .Now , the specific volume V2 of a substance in the vapour stateis much larger than the specific volume V1 in the liquid state.Therefore, in case of change of state from liquid to vapourstate, the quantity (V2-V1) is always positive. Since L and T arenecessarily positive, the Clausius-Clapeyron equation showsthat dp/dT is positive. This means that the boiling point of everyliquid rises with increase in pressure .
Effect of Pressure on Melting Point of Solids : If the partBC of the isothermal ABCD represents the change from solidto liquid state, then T is the melting point of the substanceunder the pressure p .For substances like wax and sulphur,which expand on melting, the quantity (V2 - V1) is positive andso dp/dT is positive. The means that the melting point of wax-type substances rises with increase in pressure. But for a substancelike ice, which contracts on melting, the quantity (V2 - V1) isnegative. Therefore, dp/dT is negative which means that themelting point of ice-type substances is lowered with increase inpressure.
Conditions in which Latent Heat become Zero : At thecritical point the distinction between liquid and vapour phasesdisappears (V1 = V2). Hence according to Clausius-Clapeyronequation :
2 1 0( / )
LV V
T dp dT− = =
Hence either L = 0 or dp/dT= ∞ .
Generally the specific volume of a liquid is much smallerthan that of its vapour. Assuming the vapour to obey theperfect gas equation, prove that p = constant e-(L/RT)
29Change of State
Solution : From the Clausius-Clapeyron equation, we have
2 1
L dTdp .
(V V ) T=
-
Ignoring volume V1 of liquid in comparison to volume V2of vapour, we get
L dTdp = .,
V T
where V2 has been replaced by V. From gas equation pV = RT,we may write
2
pLL dT dTdp = =
RT/p T R T
or 2
dp L dT =
P R T
Integrating : we get
e
L 1log p = - + constant
R Tæ ö÷ç ÷ç ÷çè ø
or p = constant L / RTe−
Clausius’ Latent Heat Equation
Let us consider two isothermals ABCD and EFGH at closetemperatures T and T- d T respectively for unit mass of asubstance below its critical point. Along the sections AB andEF the substance is in the liquid state, along sections BC andFG the liquid and vapour states coexist in equilibrium, whilealong CD and GH only the unsaturated vapour state exists. Thedotted curve represents the boundary which separates thedifferent states. Thus, along BF the substance is entirely in theliquid state, and along CG it is entirely in the just saturatedvapour state. Let p and p— d p be the saturated vapour pressuresof the liquid at temperatures T and T- d T respectively. Let V1and V2 be the volumes of the unit mass of the substance at Band C respectively.
30 Thermal Physics
Let the substance be taken round the cycle FBCGF. (Thisis not a true Carnot cycle because BF and CG are not a diabatics).Let s1 and s2 be its specific heats in the liquid and in thesaturated vapour state respectively. If L be the latent heat attemperature T , and dL/dT be the rate of variation of latent heatwith temperature, then the latent heat at temperature(T- d T) will be
dLL- T
dTd
æ ö÷ç ÷ç ÷÷çè ø
As the substance is taken from F to B, its temperature risesby d T and so it takes in a quantity of heat s1 d T. Along BC,the substance changes from liquid to vapour state at constanttemperature T and takes in a quantity of heat L. In passingfrom C to G, the temperature of the substance falls by d T andso it gives out a quantity of heat s2 d T. Finally, in passing alongGF, it changes from vapour to liquid state at constant
temperature (T- d T), and gives out a quantity of heatdL
L- TdT
dæ ö÷ç ÷ç ÷÷çè ø.
Thus , the net quantity of heat taken in during the cycle is givenby
31Change of State
21
dLQ = (s T+ L) - s T + L T
dTd d d d
æ ö÷ç - ÷ç ÷÷çè ø
1 2
dL= + s - s T
dTd
é ùê úê úë û
Now, the external work W done by the substance duringthe cycle is equal to the area of the cycle i.e.
W = area BCGF.
Since the two isothermals are very close together, the figureBCGF is practically a rectangle of area BC × d p . Therefore
W = BC × d P
= (V2 - V1) d p. …(ii)
The substance, after the end of the cycle, returns to itsinitial state. Therefore , its internal energy remains unchanged.Hence, by the first law of thermodynamics , the net heat d Qtaken in is equivalent to the external work W i.e.
d Q = W.
Substituting the values of d Q and W from eq. (i) and (ii),we get
( )1 2 2 1
dL + s - s T = V - V P
dTd d
é ùê úê úë û
or ( ) 1 22 1
1 dL P + s - s
V - V dT Tdd
é ùê ú=ê úë û
But ( )2 1
p LT T V - V
dd
= by Clausius-Clapeyron equation.
1 22 1 2 1
1 dL L + s - s =
(V - V ) dT T (V - V )é ù
\ ê úê úë û
or 1 2
dL L + s - s =
dT T
32 Thermal Physics
or 2 1
dL Ls s
dT T- = -
This is known as “equation of Clausius” or “second latentheat equation.”
Specific Heat of Saturated Water Vapour
From the Clausius equation, the specific heat of saturatedvapour is given by
2 1
dL Ls = s + -
dT T
The specific heat of (liquid) water s1 = 1 cal/(gm-K), andat the normal boiling point of water i.e. at T= 100 + 273 = 373K, the latent heat L = 540 cal/gm.
Putting the values of s1 ,L and T in the above equation, weget
2
dL 540s = 1 + -
dT 373dL
= 1 + - 1.45dT
=dL
- 0.45 + dT
We know that the latent heat of water diminishes with risein temperature i.e. dL/dT is negative. Thus, the specific heat s2
of saturated water vapour at 100 °C is negative.
Explanation of the Negative Specific Heat of SaturatedWater Vapour : The specific heat of a saturated vapour is thequantity of heat required to raise the temperature of 1 gm ofvapour through 1 °C (or 1 K) while at the same time thepressure and volume are varied such that the whole massremains saturated. Thus the negative specific heat of a saturatedvapour means that in order to raise the temperature of thevapour, while keeping it saturated , a certain quantity of heatis to be withdrawn from the vapour. That this is possible can
33Change of State
be explained with the help of Fig. in which ABCD and EFGHrepresent two isothermals of a substance at temperatures Tand T- d T respectively. At the point C the substance is in thesaturated vapour state at temperature T. At the point G it isin the saturated vapour state at temperature T- d T. The curve(dotted) which passes through these points is called the ‘curveof saturation.’
Now, suppose 1 gm of vapour is to be taken from thesaturated state G (at temperature T- d T) to the saturated stateC (at temperature T) along the saturated curve GC. Thus itwould remain saturated throughout the process. The figureshows that the volume V2 of the saturated vapour attemperature T is less than the volume V2’ at temperature(T- d T). Therefore, to take the vapour from G to C , it must becompressed, and so heat will be generated. In the case ofsaturated water vapour this heat is so much that it will raisethe temperature of the vapour above T. Hence some heat willhave to be withdrawn from the vapour so that the temperaturedoes not rise above T. This means that the specific heat of asaturated water vapour will be negative.
Practical Application of Negative Specific Heat : We haveseen that if a saturated water vapour is compressed adiabatically,it does not remain saturated and becomes superheated.Similarly, if the saturated water vapour is expanded adiabatically,it becomes super-saturated. If nuclei are present, the vapourcondenses and forms a fog. The effect is used in observing thetracks of charged particles passing through a Wilson cloudchamber.
Triple Point of a Substance
A substance is found to exist in three states solid, liquidand gas. For each substance there is a set of temperatures andpressures at which any two of the three states of the substancemay coexist in equilibrium.
34 Thermal Physics
Let us consider an enclosure filled with a substance, partlyin the liquid state and partly in the state of saturated vapour.The saturated vapour pressure is found to be a function of thetemperature and increases with increase in temperature. If weplot a graph between saturated vapour pressure and thecorresponding temperature, we obtain a curve OA as shownin Fig. (a). It is called the “curve of vaporisation.” It representsthe boundary between the liquid and the vapour states, thesubstance being liquid at all pressures above the curve for eachtemperature, while below it there exists only vapour.
Similarly, liquid and solid states can coexist in equilibriumat temperatures at which the solid melts under the givenpressures. This pressure-temperature relationship can berepresented by a curve OB or OB’ which is called the “curveof fusion”. It represents the variation of melting point withpressure. The curve OB which slopes to the left is for ice-typesubstances (melting point lowered with increase in pressure).The curve OB’ which slopes to the right is for wax-typesubstances (melting point raised with increase in pressure). Onthe right of the curve the substance is entirely liquid and onthe left entirely solid.
A solid and its vapour may also coexist in equilibrium. Thetemperatures and the corresponding pressures at which thesolid and the vapour states coexist are represented by a curveOC . This curve is called the “curve of sublimation”. Above thiscurve the substance is entirely solid and below it entirelyvapour.
35Change of State
When all the three curves are plotted on the same diagram,they are found to intersect in a common point O . This pointis called the “triple point”. The triple point for a substance maybe defined as the point at which the temperature and pressure are suchthat the solid , liquid and vapour states of the substance may coexistin equilibrium. For water the triple point has coordinates 0.0075°Ctemperature and 4.58 mm (or 0.006 atmosphere) pressure. Thecurves OA , OB and OC, in case of water, are called the steamline, ice line and hoar-frost line respectively.
There is only one triple point for a substance : If thetemperature or the pressure is even very slightly changed fromits value at the triple point, one of the three states will disappear.For example, it the temperature is raised the ice will melt andthe state of the system (water and saturated vapour) will berepresented by a point on the curve OA.
36 Thermal Physics
To prove that there is only one triple point, suppose thatthe three curves intersect at three points O1 , O2 and O3 asshown in Fig. Let us consider the state of the substance withinthe triangle O1 O2 O3 . As it is below the curve of sublimation,the substance must be entirely vapour ; as it is above the curveof vaporisation, the substance must be entirely liquid ; and asit is to the left of the curve of fusion, the substance must beentirely solid. But these are three contradictory conclusions.Hence the triangle O1 O2 O3 cannot exist, and the three curvesmust intersect in a single point.
PROBLEMS
1. Find the change in the melting point of ice (or freezingpoint of water) at 0°C for an increase of pressure by 1atmosphere. Latent heat of ice at 0 °C = 80 cal/gram . At meltingpoint the volumes of 1 gram of water and ice are respectively1.000 cm3 and 1.091 cm3; 1 atmosphere = 1.013 × 106 dyne/cm2
and J = 4.18 × 107 erg/calorie)
Solution : The change in the melting point (dT) of asubstance at absolute temperature T with change in pressure(dp) is given by Clapeyron equation
2 1
dp L =
dT T (V - V )
where V2 and V1 are the specific volumes (volume/mass) ofthe substance in the liquid and solid states respectively, andL is the latent heat of fusion in work unit. If L is given in heatunit, then we shall write
2 1
dp J L =
dT T (V - V )
So that2 1T (V - V ) dp
dT = JL
Here T = 0 + 273 = 273 K, V2 = 1.000 cm3/g, V1 = 1.091 cm3/g,dp = 1 atmosphere = 1.013 × 106 dyne/cm2 and L = 80 cal/g.
37Change of State
3 6 2273 K × (-0.091 cm /g) × (1.013 × 10 dyne/cm )dT =
( . erg / cal) × (80 cal/g)\
´ 74 18 10
= - 0.0075 K(erg = dyne x cm)
= - 0.0075 °C.
The negative sign means that the melting point of ice, orthe freezing point of water, is lowered by increase in pressure.
2. Calculate the change in the melting point of ice peratmosphere increase in pressure. The latent heat of ice is 80kcal / kg and the ratio of the densities of ice and water is 10/11. Given : 1 atmosphere = 105 N/m2 and J = 4.2 joule/cal.
Solution : By Clausius-Clapeyron’s equation, we have
2 1
dp J L =
dT T (V - V )
where L is in heat unit. From this, we have
2 1T (V - V ) dpdT =
JL ...(i)
Here T = 0 + 273 = 273 K, dp = 1 atmos. = 105 N/m2, L =80 kcal / kg and J = 4.2 joule/cal = 4.2 × 103 joule/kcal. Now
o2
o1
sp. volume of water at 0 CV =
V sp. volume of ice at 0 C
o
o
density of ice at 0 C 10= =
density of water at 0 C 11
But V2 = 1.000 cm3/g = 1.000 × 10-3 m3/kg (this value
should be remembered), so V1 = 1110 × (1.000 × 10-3) = 1.100
× l0-3 m3 kg.
38 Thermal Physics
Putting these values in eq. (i), we get
4 2
3 3 5 2
3
273 K × (-0.100 × 10 m /kg) × 10 N/mdT =
( . 10 J/kcal) × (80 kcal/kg)
-
´
= - 0.0081 K = - 0.0081 ºCThe melting point will be lowered by 0.0081 °C .
3. Calculate the lowering of melting point of ice subjectedto a pressure-increase of 50 atmospheres. The density of ice at0 °C is 0.917 gram/cm3 and its latent heat is 334 joule/gram.Take 1 atmosphere = 1.013 x 106 dyne/cm2.
Solution : Ice melts into water at 0 °C (T=273 K) under 1atmosphere pressure. Under a pressure-change dp , the changein melting point, dT, is given by
2 1( )dp LdT T V V
=−
where V2 and V1 are the specific volumes (reciprocal of densities)of water and ice respectively and L is latent heat in work unit.From this, we have
2 1( )T V V dpdT
L−
=
Now, T = 273 K , V2 = 1.000 cm3/g (it should beremembered), V1 = 1/0.917 = 1.091 cm3/g, dp = 50 atmos = 50x (1.013 x 106) = 5.065 x 107 dyne/cm2 and L = 334 joule/gram= 334 x 107 erg/gram.
3 7 2
7
7
7
273 K (1.000 1.091)cm /g (5.065 10 dyne/cm )334 10 crg/g
273 ( 0.091) (5.065 10 )334 10
0.38K 0.38 C.
dT
K
× − × ×∴ =
×
× − × ×=
×= − = − °
4. Calculate the change in the melting point of napthalenefor one atmosphere rise in pressure, given that the meltingpoint is 80 °C, latent heat of fusion is 4563 cal/mole and
39Change of State
increase in volume on fusion is 18.7 cm3/mole. One calorie= 4.18 x 107 erg.
Solution : In usual notations, the change in the meltingpoint of napthalene is given by
2 1( )T V V dpdT
L−
=
Here T = 80 + 273 = 353 K, V2 - V1 = + 18.7 cm3/mole,dp = 1 atmos = 1.013 x 106 dyne/cm2 and L = 4563 cal/mole= 4563 x 4.18 x 107 erg/mole.
.3 6 2
7
353 K x 18.7 cm /mole x (1.013 x 10 dyne/cm )4563 x 4.18 x 10 erg/mole
dT∴ =
6
7
353 x 18.7 x 1.013 x 10 )4563 x 4.18 x 10
K= [Q erg = dyne x cm]
0.035K 0.035 C.= + = + °
dT is positive. It means that the melting point of napthalenerises with increase in pressure.
5. Calculate the change in the melting point of wax subjectedto a pressure of 50 atmospheres from the following data :
Melting point = 64 °C, volume of solid wax at 64 °C =1.161, volume of liquid wax at 64 °C = 1.166 , density of solidwax at 64 °C = 0.96 g/cm3, latent heat of fusion = 97 cal/g, J= 4.18 x 107 erg/cal 1 atmos = 1.013 x 106 dyne/cm2.
Solution : Wax melts at 64 °C (T=337K) under 1 atmospherepressure. According to Clausius-Clapeyron latent heat equation,the change in the melting point of wax (dT) with increase inpressure (dp) is given by
2 1( )T V V dpdT
L−
= ...(i)
where V2 and V1 are volume/mass of liquid and solidwax respectively and L is the latent heat of fusion in terms ofwork unit.
40 Thermal Physics
Since volume is inversely proportional to density, we have
density of liquid wax volume of solid wax=
density of solid wax volume of liquid wax
∴ density of liquid wax = 1.1611.166
x 0.96 (g/cm3)
= 0.956 g/cm3.
Thus, here we have
T= 64 + 273 = 337 K, V2 = 1/0.956 = 1.046 cm3/g, V1 = 1/0.96 = 1.042 cm3/g, dp = 50 -1 = 49 atmos = 49 x 1.013 x 106
dyne/cm2 and L = 97 cal/g = 97 x 4.18 x 107 erg/g.
Substituting these values in eq. (i), we get
3 6 2
7
337 K x (1.046 - 1.042) cm /g x (49 x 1.013 x 10 dyne/cm )97 x 4.18 x 10 erg/g
dT =
6
7
337 x 0.004 x 49 x 1.013 x 1097 x 4.18 x 10
K= [Q erg = dyne x cm]
0.0165K 0.0165 C.= = °
6. Calculate the pressure required to make water freeze (orice melt) at -1 °C. Given : density of ice = 0.917 g/cm3, latentheat of ice = 80 cal/g and J = 4.18 x107 erg/cal.
Solution : Water freezes at 0 °C at a pressure of 1atmosphere. Let dp be the change in pressure required to makethe ice freeze at -1 °C. By Clausius-Clapeyron latent heatequation, we have
2 1( )L
dp dTT V V
=−
where V2 and V1 are specific volumes (volume/mass orreciprocal of density) of water and ice respectively and L is
41Change of State
latent heat in work unit (as erg/gm). If L is in heat unit (ascal/gm), then we must multiply it by J, that is
2 1( )JL
dp dTT V V
=−
Here L = 80 cal/g, T= 0 + 273 = 273 K, dT = -1oC = - 1 K,V2= 1.000 cm3 /g , V1=1/0.917 = 1.091 cm3/g so that V2- V1
= -0.091 cm3/g .
7
3
6 3 6 3
(4.18 10 erg/cal)(80 cal/g)(-1 K)273 K (-0.091 cm /g)
134.6 10 erg/cm =134.6 10 dyne/cm .
dp×
∴ =
= × ×
Now , 1 atmosphere pressure = 1.013 × 106 dyne/cm2 .
6
6
134.6 10133 atmosphere.
1.013 10dp
×∴ = =
×
Thus the pressure must be increased by 133 atmosphere tolower the freezing point of water (or melting point of ice) by1 °C. This means that water will freeze at - 1 °C under apressure of 134 atmosphere.
7. Calculate the specific gravity of solid sulphur from thefollowing data:
Melting point of sulphur = 115 °C ; latent heat of fusionof sulphur = 9.3 cal/g, volume of 1 g of liquid sulphur = 0.513cc, rate of change of melting point with pressure = 0.025 °C peratmosphere. (Atmospheric pressure = 106 dyne/cm2 and J = 4.18x107 erg/cal)
Solution : The Clausius-Clapeyron equation for the melting ofa solid substance is
2 1( )dp JLdT T V V
=− .
42 Thermal Physics
where T is melting point (in Kelvin) of the substance, L is latentheat of fusion in heat unit (cal/gm) and V2 and V1 are specificvolumes (volume/mass) in liquid and solid state of thesubstance respectively. From this, the rate of change of meltingpoint with pressure is
2 1( )T V VdTdp J L
−=
or 2 1
JL dTV V
T dp− = .
Here L = 9.3 cal/g, T = 115 + 273 = 388 K and dTdp
=o0.025 C
1atmos
= 6 2
0.025 K10 dyne/cm
.
7
2 1 6 2
(4.18 × 10 erg/cal)(9.3 cal/g) 0.025 K388 K 10 dyne/cm
V V∴ − = ×
= 0.025 cm3/g. [∴erg = dyne x cm]
Thus, specific volume of solid sulphur is
V1 = V2 – 0.025 = 0.513 – 0.025 = 0.488 cm3/g.
Hence sp. gr. of solid sulphur = 3
1
1 12.05 g/cm
0.488V= = .
7. Calculate the change in the boiling point of water whenthe pressure of steam on its surface is increased from1.0 atmosphere to 3.1 atmospheres. The latent heat of waterat 100 °C is 537 cal/g and the volume of 1 g of steam is,1676 cm3. Take J = 4.2 × 107 erg/cal and 1 atmosphere= 1.0 × 106 dyne/cm2.
Solution : Water boils at 100 °C (T= 100 + 273 = 373 K)under one atmosphere pressure. Let dT be the change in boilingpoint when pressure is increased by dp. From Clausius-Clapeyron equation, we have
43Change of State
2 1( )dp LdT T V V
=−
where V1 and V2 are specific volumes (volume/mass) of waterand steam respectively and L is latent heat in work unit. If Lis given in heat unit (cal/gm), then we shall write
2 1( )dp JLdT T V V
=−
or2 1( )T V V dp
dTJL−
=
Here T = 373 K, V1 = 1 cm3/g (it should be remembered),V2 = 1676 cm3/g, dp= 3.1 - 1.0 = 2.1 atmos = 2.1 x 106 dyne/cm2, J = 4.2 x 107 erg/cal and L = 537 cal/g .
3 6 2
7
373 K x 1675 cm /g x (2.1 x 10 dyne/cm )(4.2 x 10 erg/cal) x (537 cal/g)
dT∴ =
= 58 K = 58 °C.
The boiling point of water will be raised by 58 °C.
8. Calculate the change in the boiling point of water dueto an increase of pressure of 1 cm of mercury. Datas as in thelast problem.
Solution : Refer to last problem.
2 1( )T V V dpdT
JL−
=
Here dp = 1 cm of Hg = 1 x 13.6 x 981 dyne/cm2.
3 2
7
373 K x 1675 cm /g x (1 x 13.6 x 981 dyne/cm )(4.2 x 10 erg/cal) x (537 cal/g)
dT∴ =
= 0.37 K = 0.37 °C.
The boiling point of water will be raised by 0.37 °C .
44 Thermal Physics
8. One gram of water vapour at 100 °C and one atmosphericpressure occupies a volume of 1640 cm3. Calculate the vapourpressure of water at 99 °C. Latent heat of vaporisation = 536cal/g and density of water at 100 °C 1 g/cm3.
Solution : The vapour pressure of water at 100 °C (T = 100+ 273 = 373 K) is 1 atmosphere. Let dp be the change in thevapour pressure when the temperature decreases from 100 °Cto 99 °C, that is, dT= -1°C = - 1 K .
Now, the rate of change of vapour pressure withtemperature is given by the Clausius-Clapeyron equation
2 1( )dp JLdT T V V
=−
where V2 and V1 are the specific volumes in the vapour stateand in the liquid state respectively, and L is the latent heat ofvaporisation in heat unit (cal/g).
Here L = 536 cal/g, V2 = 1640 cm3/g and V1 = 1 cm3/g. Jmay be taken 4.18 x 107erg/cal .
Substituting these values in the above equation, we get
7
3
7
3
(4.18 10 erg/cal) (536 cal/g) (-1 K)
373 K (1640 -1) cm /g
4.18 10 536 erg373 1639 cm
dp× ×
= ××
× ×=
×
= _ 3.66 x 104 dyne/cm2. [Q erg = dyne x cm]
Now, 1 atmosphere pressure is the pressure of 76 cmmercury column, that is, 76 x 13.6 x 981 = 1.01 x 106 dyne/cm2.
4
6
3.66 100.0362 atmos.
1.01 10dp
×= = −
×Q
The minus sign shows that the vapour pressure will decreaseby 0.0362 atmos. Hence the vapour pressure of water at 99°Cwill be 1- 0.362 = 0.964 atmos.
This is equivalent to 733 mm of mercury.
45Change of State
9. Calculate the change in the vapour pressure of water asthe boiling point changes from 100 °C to 102 °C. The specificvolume of steam is 1671 cm2/g and the latent heat of steamis 540 cal/g. J = 4.18 × 107 erg/cal and 1 atmosphere = 1.01 × 106
dyne/cm2.
Solution : The vapour pressure of water at its normalboiling point 100 °C (T=373 K) is 1 atmosphere. Let dp be thechange in vapour pressure as the boiling point changes from100 °C to 102 °C, that is, when dT= 2 °C = 2 K.
From Clausius-Clapeyron equation, we have
2 1( )JL
dp dTT V V
=−
where V2 and V1 are specific volumes in the vapour state andliquid state respectively.
7
3
4 2
4
6
(4.18 x 10 erg/cal) (540 cal/g)(2 K)(373 K) (1671 -1) cm /g
7.25 10 dyne/cm
7.25 100.0718 atmos.
1.01 10
Q dp×
=×
= ×
×= =
×
The vapour pressure will increase by 0.0718 atmosphere orby 0.0718 760 = 54.6 mm of mercury.
10. Calculate under what pressure water would boil at 120°C. one gm of steam occupies a volume of 1677 cm3. Latent heatof steam = 540 cal/g, J = 4.2 × 107 erg/cal, 1 atmosphere pressure= 1.0 × 106 dyne/cm2 .
Solution : Water boils at 100 °C (T = 373 K) at 1 atmospherepressure. Let dp be the change in pressure required to makewater boil at 120 °C, that is, for dT= 20 °C = 20 K.
From Clausius-Clapeyron equation, we have
2 1( )JL
dp dTT V V
=−
46 Thermal Physics
where V2 (specific volume of steam) is 1677 cm3/g and V1
(specific volume of water) is 1 cm3/g . Substituting the values,we get
7
3
7
3
6 2
6
6
(4.2 x 10 erg/cal) (540 cal/g)(20 K)(373 K) (1677 -1) cm /g
erg4.2 10 540 20373 1676 cm
0.725 10 dyne/cm
0.725 100.725 atmosphere.
1.0 10
dp×
=×
× × ×=
×= ×
×= =
×
Thus, in order to make water boil at 120 °C, the pressureshould be increased by 0.725 atmos, that is, it should be increasedto 1.725 atmosphere .
11. Calculate the specific volume of the vapour of carbontetrachloride at the boiling point from the following data.Boiling point = 77 °C at 1 atmosphere, latent heat = 46 cal/g, density of liquid = 1.6 g/cm3, dp/dT = 23 mm of mercury/ °C,J = 4.18 × 107 erg/cal.
Solution : The Clausius-Clapeyron equation for the boilingpoint of a liquid is
2 1( )dp JLdT T V V
=−
.
where T is boiling point (in Kelvin) of the liquid, L is latentheat of vaporisation in heat unit (cal/gm) and V2 and V1 arespecific volumes of vapour and liquid respectively. From this,we have
2 1 ( / )JL
V VT dp dT
− = .
Here L = 46 cal/g, T = 77 + 273 = 350 K
and 23 mm of Hg
CdpdT
=°
47Change of State
= 32.3 cm 13.6 g/cm 980 dyne/g
K= × ×
= 4 23.065 10 dyne/cmK
×
7
2 1 4 2
(4.18 x 10 erg/cal) x (46 cal/g)(3.065 x 10 dyne/cm )
350 K xK
V V∴ − =
= 179.2 cm3/g.
Now, the density of liquid is given 1.6 g/cm3. Therefore,the specific volume (volume/mass) is
31 3
1= 0.6cm /g
1.6 g/cmV = .
∴ V2 = 179.2 + V1 = 179.8 cm3/g .
12. Calculate the latent heat of steam at 100 °C underatmospheric pressure. Given : specific volume at saturatedsteam and water = 1677 cm3/g and 1 cm3/g respectively, dp/dT= 27.1 mm of Hg per °C, J = 4.18 x 107 erg/cal.
Solution : Refer to the last problem.
2 1( ) dpT V VL
J dT−
= .
Here T = 100 + 273 = 373 K, (V2- V1) = (1677-1) = 1676
cm3/g,
dpdT
= 27.1mm of mercury
C°
= 32.71 cm 13.6 g/cm 980 dyne/g
K× ×
= 4 23.61 x 10 dyne/cmK
.3 4 2
7
373 K x 1676 cm /g 3.61 x 10 dyne/cm4.18 x 10 erg/cal K
L∴ = ×
= 540 cal/g.
48 Thermal Physics
13. A Carnot’s engine having 100 g water-steam as workingsubstance has, at the beginning of the stroke ; a volume 104cm3 and pressure 788 mm (B.P. = 101 °C). After a completeisothermal change from water into steam, the volume is 167,404 cm3 and the pressure is then lowered adiabatically to 733.7mm (BP =99 °C). If the engine is working between 99 °C and101 °C, calculate the latent heat of steam.
Solution : Volume of 100 g of water in the beginning is104 cm3.
∴ specific volume, 31
104= 1.04cm /g
100V = .
After the isothermal change, the volume of 100 g of steamis 167,404 cm3.
∴ specific volume, 32
167,404= 1674.04cm /g
100V = .
∴ change in specific volume is
(V2- V1) = 1674.04 -1.04 = 1673 cm3/g.
Now, change in pressure, dp = 788 - 733.7 = 54.3 mm= 5.43 cm
= 5.43 x 13.6 x 980 = 7.24 x 104 dyne/cm2,
change in boiling point, dT= (273 + 101) - (273 + 99) = 2 K,
and mean temperature, 99 101273 373K.
2T
+= + =
According to Clapeyron equation
2 1( )dp JLdT T V V
=−
2 1
3 4 2
7
( )
373 K 1673 cm /g 7.24 10 dyne/cm4.18 10 erg/cal 2 K
540 cal/g.
dpT V VL
J dT−
∴ =
× ×= ×
×=
49Change of State
13. Calculate the specific heat of saturated vapour (steam)at 100 °C if the latent heat L varies with temperature T accordingto the equation L=800 - 0.705 T, and the specific heat of waterat 100 °C is 1.0 cal/(g-°C)
Solution : By the equation of Clausius, the specific heat s2
of a saturated vapour is given by
2 1
dL Ls s
dT T= + −
where s1 is the specific heat in the liquid phase. Here
L = 800 - 0.705 T
so that dLdT
= 0.705.
Also, at T= 100 + 273 = 373 K, we have
L = 800 - 0.705 x 373 = 537.
Putting the values of dL/dT, L/T and s1 in eq. (i), we get
2
5371.0 0.705
3731.0 0.705 1.44
1.145 cal/(g C).
s = − −
= − −= − − °
Thus the specific heat of saturated steam is negative.
14. The latent heat of Water diminishes by 0.64 cal/g foreach degree centigrade rise in temperature in the neighbourhoodof 100 °C and the latent heat of water vapour at 100 °C is 540cal/g. Find the specific heat of saturated steam at 100 °C. Thespecific heat of water at 100 °C is 1.01 cal/g- °C .
Solution : By the equation of Clausius, the specific heat s2
of a saturated vapour is given by
2 1
dL Ls s
dT T= + − .
50 Thermal Physics
where s1 is the specific heat in the liquid state and dL/dT is therate of change of latent heat with temperature.
Here dL/dT = - 0.64 cal/g-°C , T = 100 + 273 = 373 K,L = 540 cal/g and s1=1.01cal/g-°C.
∴ s2 = 1.01 + (-0.64)-540373
= 1.01- 0.64 - 1.447
= - 1.077 cal/(g-°C).
15. Calculate the specific heat of saturated steam at100 °C. Given : L at 90°C = 545.25 cal/g, at 100°C = 539.30 cal/g and at 110°C = 533.17 cal/g. The specific heat of water at100 °C is 1.013 cal/(g-°C).
Solution : By the equation of Clausius, the sp. heat s2 ofsaturated steam is given by
2 1
dL Ls s
dT T= + −
where s1 is the sp. heat of water and L is the latent heat of steamat temperature T.
Here s1 = 1.013 cal/(g-K), L (at 100 °C) = 539.30 cal/g ,
T= 100 + 273 =373 K and 533.17 545.25 12.08
110 90 20dLdT
−= =
−= - 0.604
cal (g-K)
2
539.301.013 0.604
373s∴ = − −
= 1.013 - 0.604 - 1.446 = -1.037 cal/(g-K).
16. The specific heat of a certain liquid is given by
s1 = 0.53 + 0.0004 t
and its latent heat of vaporisation is given by
L = 94 - 0.06 t - 0.0006t2,
51Change of State
where t is temperature in °C. Calculate the specific heatof saturated vapour at 50 °C.
Solution : By the equation of Clausius, the specific heat ofsaturated vapour is given by
2 1
dL Ls s
dT T= + − ...(i)
From the given equations, at 50 °C, we have
s1 = 0.53 + (0.0004 x 50) = 0.55 cal/(g-K),
L = 94 -(0.06 x 50) - (0.0006 x 50 x 50) = 89.5 cal/g,
T = 50 + 273 = 323 K
anddL dLdT dt
= = - 0.06 - 0.0006 (2 t)[dT= dt, Q1 °C = 1 K]
= - 0.06 - 0.0006 x 2 x 50
= - 0.12 cal/(g-K).
Substituting all these values in eq. (i), we get
2
89.50.55 0.12
323s = − −
= 0.55 - 012 - 0.277
= 0.153 cal/(g-K).
The specific heat of the given vapour is thus positive .
17. The vapour pressure p (in mm of Hg) of solid ammoniais given by loge p = 23.03 - (3754/T) while that of liquid ammoniais given by loge p = 19.49 - (3063/T) where T is in Kelvin.Calculate the triple point of ammonia.
Solution : At the triple point the solid, liquid and vapourphases co-exist in equilibrium, i.e. vapour pressure will besame for solid and liquid states. Thus
3754 3063log 23.03 19.49e p
T T= − = −
52 Thermal Physics
or 3754 3063
23.03 19.49T−
− =
691195.2K.
3.54T = =
Substituting this value of T in the given equation, we get
3754log 23.03
195.223.03 19.23 3.8
e p = −
= − =
∴ p = 45 mm of Hg.
18. The coordinates of the triple point of water aret = 0.0075°C and p = 0.006 atmos. Calculate the slope of theice line in atmos/°C.
Solution : At triple point, all the three states of matter arein equilibrium. The slope of the ice line is
2 1( )dp JLdT T V V
=−
where L is latent heat of ice, T is melting point and V2 and V1
are the specific volumes of water and ice respectively. Puttingthe known values :
7
2
(4.18 10 erg/cal) (80 cal/gram)273K ( 0.091 cm /gram)
dpdT
× ×=
× −
= - 13.46 × 107 dyne/(cm2-K)
7
6
13.46 10= = 132.8 atmos/K
1.013 10×
− −×
,
because 1 atmos = 1.013 × 106 dyne/cm2. clearly, the ice lineis almost vertical.
53Heat Capacity
5
Heat Capacity
Kilo-calorie: 1 Kilo-calorie is the quantity of heat requiredto raise the temperature of 1 kilogram of water through 1°C(from 14.5 to 15.5°C).
Calorie: 1 calorie is the quantity of heat required to raisethe temperature of 1 gm of water through 1°C (from 14.5 to15.5°C). Evidently, 1 kilo-calorie = 1000 calories.
Materials differ from one another in the quantity of heatrequired to produce a given rise in temperature in a givenmass. If a quantity of heat ΔQ raises the temperature of a bodythrough ΔT, then the ratio of ΔQ to ΔT is called the ‘heatcapacity’ C of the body.
C= .QT
ΔΔ
If ΔT= 1°, then C = ΔQ . Thus, the heat capacity of a bodyis numerically equal to the quantity of heat required to raiseits temperature by 1°.
54 Thermal Physics
Specific Heat : The heat capacity per unit mass of a bodyis called the ‘specific heat’ of the material of the body. If wedenote it by c, then
heat capacity /.
massQ t Q
cm m T
Δ Δ Δ= = =
Δ ...(i)
If m = 1, ΔT = 1, then c = ΔQ. Thus, the specific heat ofa material is numerically equal to the quantity of heat requiredto raise the temperature of unit mass of that material through1°. It is expressed in cal-gm-1–°C-1 or kilocal-kg -1–°C-1 .
From the definitions of cal and kilo-cal it is clear that 1 cal-gm-1–°C-1 = 1 kilocal-kg-1–°C-1 .
Strictly speaking, the specific heat of a material dependson the location of the temperature-interval in the temperaturescale. The eq. (i) defines the average specific heat of a materialover the temperature-interval ΔT. The true specific heat of amaterial at any temperature is defined from eq. (i) byconsidering an infinitesimal temperature-rise dT, so that
dQ
Cm dT
=
or dQ = mc dT.
Thus the heat required to raise the temperature of a materialof mass m and specific heat c from Ti to Tf is
,f
i
T
TQ m c dT= ∫ ..(ii)
where c is a function of the temperature. At any ordinarytemperature and over small temperature-intervals, c can betaken constant.
Molar Heat Capacity or Gram-molecular Sp. Heat
The amount of heat required to raise the temperature of1 gm-molecule (or 1 mole) of a substance through 1° is called
55Heat Capacity
the ‘molar heat capacity’ of the substance. It is equal to theproduct of the specific heat and the molecular weight of thesubstance.
Specific heat and Molar heat capacity depend also on theconditions under which the heat is added to the substance.Hence we must specify the conditions also, such as specificheat at constant pressure (cp), sp. heat at constant volume (cV),molar heat capacity at constant pressure (Cp), molar heatcapacity at constant volume (Cv), and so on. For solids andliquids the difference between cp and cv or the difference betweenCp and CV is negligible, but certainly not so for gases.
Suppose the specific heat c of a substance is found to varywith temperature in a parabolic fashion, that is
c = A + BT2
where A and B are constants and T is the Celsius tempe-rature.
(i) Calculate the quantity of required to raise m gm of thesubstance from a temperature T1 to T2
(ii) Compare the mean specific heat of the substance in thetemperature-range T=0 to T=T with the true specificheat at T/2
The true specific heat of a substance is given by
c = A + BT2. ...(i)
The heat required to raise the temperature of a mass m ofthe substance from temperature T1 to T2 is given by
( )2 2
1 1
2T T
T TO m c dT m A BT dT= = −∫ ∫
( ) ( )2
1
33 3
2 1 2 13 3
T
T
T Bm AT B m A T T T T⎡ ⎤ ⎡ ⎤= + = − + −⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
( ) ( )2 22 1 2 1 T .
3B
m A T T T⎡ ⎤= + + −⎢ ⎥⎣ ⎦
56 Thermal Physics
(i) The mean specific heat in temperature-range T= 0 toT= T is given by
( )2
0 0
3 2
0
1 1
1.
3 3
T T
mean
T
c cdt A BT dtT T
T BTAT B A
T
= = +
⎡ ⎤= + = +⎢ ⎥
⎣ ⎦
∫ ∫
The true specific heat at T/2 is obtained by putting T= T/2in eq. (i). Thus
cT/2 = A+B(T/2)2 = A+2
.4
BT
∴2 2 2
/2 .3 4 12mean T
BT BT BTc c A A
⎛ ⎞ ⎛ ⎞− = + − + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Thus, the mean specific heat exceeds that at T/2 by BT2/12.
Specific Heat of Water and Climate
The specific heat of water is about five times higher thanthat of clay or sand. Therefore, if same quantity of heat begiven to (or taken from) the same mass of water and earth, thewater will have a much smaller rise (or fall) in temperaturethan the earth. This has a very important effect on the climateof islands and coastal areas which have a sea or ocean nearthem.
During a hot day the temperature of the sea rises moreslowly than that of the land, and likewise falls more slowlywhen heat is radiated during the night. Therefore, during theday the cooler sea keeps the land near it cool, while during thenight it tends to keep the land warm. This transfer of heat takesplace through the movement of air. Thus, the land near the seaundergoes a smaller day and night temperature-variation thanthe land elsewhere. Similarly, the land near the sea remainscooler during the summer and warmer during the winter thanthe land far from the sea.
57Heat and Temperature
6
Heat and Temperature
The temperature of a body is a measure of its degree ofhotness or coldness. A body appearing hotter to our sense oftouch than the other is said to be at a higher temperature thanthe other.
When two bodies A and B, the body A being at a highertemperature, are placed in contact, then after some time theyboth acquire the same temperature which is somewherebetween the two initial temperatures. This means thatsomething from A has been transferred to B. This ‘something’is called heat. Thus, ‘heat is that which is transferred from onebody to the other, because of a temperature difference betweenthem’. In fact heat is a form of energy which is transferred fromone body to the other because of a temperature differencebetween them.
There is a distinction between the temperature of a bodyand the heat that it contains. The heat that a body containsdepends upon its mass as well as upon its temperature. Thesparks from a blacksmith’s hammer are white hot (i.e. at a veryhigh temperature) but they do not burn the hand since their
58 Thermal Physics
mass is very small, and therefore they contain little heat. Onthe other hand, a jug of hot water, though at a much lowertemperature than the spark, causes a severe burn because itcontains more heat.
Heat and temperature can be compared with water andwater-level, if two vessels filled with water upto differentheights are connected, water flows from the higher to thelower level irrespective of the quantities of water in the twovessels. Similarly, heat flows from a body at higher temperatureto a body at lower temperature, irrespective of the quantitiesof heat in the two bodies.
Hot and Cold Bodies Placed in Contact : When a hot bodyand a cold body are put in mutual contact, heat flows from thehot to the cold body until they attain a common temperature.This means that the temperature of the hot body falls and thatof the cold body rises. These temperature-changes are notnecessarily equal because the masses (and also the specificheats) of the two bodies may be different.
Suppose 10 gm of hot water at 50°C is added to 20 gm ofcold water at 20°C and acquire a common temperature T °C.Thus
heat lost by hot water = heat gained by cold water
or 10 × 1 × (50 - T) = 20 × 1 × (T - 20)
∴ T = 30°C.
Thus, the temperature of hot water falls by 20°C and thatof cold water rises by 10°C.
We cannot say that the temperature has passed from hotto cold water. It is the ‘heat’ that has passed.
Equivalence of Heat and Work
Heat, a form of Energy: Formerly it was believed that heatwas a fluid called ‘caloric’ which was filled in the spaces between
59Heat and Temperature
the molecules of matter, and a body at higher temperaturecontained more caloric than that at a lower temperature. Thistheory, called ‘caloric theory of heat’, could, however, notexplain many phenomena concerning heat and hence it wasrejected.
Today, it is an established fact that heat is a form of energyrather than a substance. We know that in a situation involvingfriction, heat is always produced. When the brakes of a movingcar are applied, the brake-shoes and drums become hot. Whenair is pumped into a cycle tyre, the pump becomes hot. Thematch-stick on being rubbed becomes hot to burn. In all thesesituations, and in many others, mechanical energy (work)disappears and heat is produced.
Rumford’s Experiments : The first evidence of relationshipbetween work and heat came from Count Rumford’sexperiments. He noticed, while supervising the boring ofcannons, that large amount of heat was generated during theboring process, and that the generation of heat continued evenwhen the boring tool became so dull that it was no longercutting the metal. If heat were a substance (caloric) it wouldnot be possible to evolve it indefinitely from the metal. Rumford,therefore, hinted that the mechanical work done by the boreragainst the friction of the metal of cannons created the heat.
Davy’s Experiments : Davy liquefied two pieces of ice byrubbing them together. It was well known that a considerableamount of heat must be supplied to ice to melt it. Since therewas no external source of heat, it was considered that themechanical work done in rubbing the ice pieces against eachother created the necessary heat.
Joule’s Experiments : Joule performed accurate quantitativeexperiments to establish relation between work and heat. Hetried to find the amount of heat produced by a certain amountof work done. He rotated a brass paddle-wheel in watercontained in a copper calorimeter, with the expenditure of a
60 Thermal Physics
measured amount of mechanical work, and measured the risein temperature of the water. He found that the amount of heat(thermal energy) produced was always proportional to theamount of work done.
Joule performed may more experiments to measure theheat produced by measured amounts of mechanical work. Inone experiment, he stirred mercury contained in an iron vesselby an iron paddle and measured the heat produced. In anotherexperiment he rubbed two iron rings against each other undermercury. He also measured the heat produced when electricalcurrent was flown in a metallic wire. All these experimentsgave practically the same result. Thus, Joule established thata given amount of mechanical work done in any way alwaysproduces the same amount of heat; and that the disappearanceof that very amount of heat always gives the same amount ofmechanical work. This establishes the fact that ‘heat is a formof energy.
Mechanical Equivalent of Heat : Whenever mechanicalwork is transformed into heat, or heat into mechanical work,there is a constant ratio between the work and the amount ofheat. This ratio is called “mechanical equivalent of heat” andis denoted by J . Thus, if W be the amount of work done andQ the amount of heat produced, we have
WJ
Q=
or .W JQ=
If Q = 1 unit then J = W. Therefore, J is numerically equalto the mechanical work required to produce one unit of heat.
The value of J is 4.186 joule/kilo-calorie (or 4.186 joule/calorie or 4.186 x 107 erg/calorie). This means that 4.186 jouleof mechanical work will produce 1 kilo-calorie of heat. But1 kilo-calorie of heat is defined as the heat which raises thetemperature of 1 kg of water from 14.5 to 15.5 °C. Hence, we
61Heat and Temperature
may say that 4.186 joule of mechanical work, when transformedinto heat, will increase the temperature of 1 kg of water from14.5 to 15.5 °C.
It is clear from the definition of J that it is not a physicalquantity. It is only a conversion factor used to convert energyfrom heat units (kilo-calorie or calorie) to mechanical units(joule or erg) and vice-versa.
Thermodynamic System : A thermodynamic system is onewhich may interact with its surroundings in at least two distinctways and one of these is necessarily a transfer of heat in orout of the system. The other (or others) may be some othermeans of transfer of energy, say by performance of mechanicalwork by or on the system, or through electromagneticinteraction such as magnetisation. A gas contained in a cylinder,a vapour in contact with its liquid, a stretched wire, a pieceof magnetic material are examples of thermodynamic system.
Thermodynamic Equilibrium : When there are nounbalanced forces between the system and its surroundings,the system is said to be in ‘mechanical equilibrium’. If thesystem has no tendency to undergo a change in internalstructure and also has no tendency to transfer matter from onepart of the system to another, it is said to be in ‘chemicalequilibrium.’ Again, if all parts of the system are at the sametemperature which is equal to the temperature of thesurroundings, the system is said to be in ‘thermal equilibrium’.When the system is under all the three types of equilibrium,it is said to be in ‘thermodynamic equilibrium’. In this conditionany change in the state of the system or of the surroundingswould not occur.
If, however, the conditions for any one of the three typesof equilibrium are not satisfied, the system is said to be in a‘non-equilibrium state’. When this is the case, the phenomenalike acceleration, eddies, chemical reaction, heat-transfer withsurroundings, etc. would take place.
62 Thermal Physics
State Variables : The state of a thermodynamic system inequilibrium can be completely specified in terms of certainmeasurable (macroscopic) quantities. In the case of a gas, forexample, these quantities are pressure p, volume V andtemperature T. They are functions of the state of the systemalone and return to the same values whenever the systemreturns to the same equilibrium state. They are known as ‘statefunctions’ or ‘state variables’ or ‘thermodynamic coordinates’of the system.
Equation of State : The equation of state of a thermodynamicsystem in equilibrium is a functional relationship among thestate variables of that system. The general form of the equationof state of a gas, for example, may be written as
f (p,V,T) = 0.
It can be solved with respect to any one of the variables.Its three equivalent forms are
p = p(V,T)
V = V(p,T)
T = T(p,V).
This implies that any two of the three variables p , V andT are enough to specify the state of a gas and to fix the valueof the third variable.
If, however, the system is in a non-equilibrium state thenthis state cannot be described in terms of thermodynamiccoordinates.
Dependence of Temperature on Pressure and Volume of aSystem : If the pressure and volume of a system inthermodynamic equilibrium are given, then its temperature isalso fixed and is determined by the equation of state. If, however,the system is in a non-equilibrium state, then no equation ofstate exists and the temperature cannot be uniquely determined.
63Heat and Temperature
Temperature of an Isolated System : When a system is notinfluenced in any way by its surroundings, it is said to be‘isolated’. The temperature of an isolated system is conservedprovided no internal changes like acceleration, eddies, chemicalreaction, heat-transfers between different parts are taking placein the system.
Relation between Isothermal Bulk Modulus, VolumeExpansion and Pressure Expansion : Let p, V and T be the statevariables of a system in thermodynamic equilibrium. Thegeneral equation of state is
f (p,V,T) = 0.
We can solve it for p and V, to get
p = p (V,T)
and V = V(p,T).
From partial differentiation, we get
T V
p pdp dV dT
V T∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ ...(i)
and .pT
V VdV dp dT
p T
⎛ ⎞∂ ∂⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠...(ii)
If the pressure is increased on the system at constanttemperature (dT = 0), we get from eq. (i)
T
pdp dV
V∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠
If we consider the relative change in volume, we can definethe isothermal bulk modulus of elasticity as
/TT
dp pE V
dV V V∂⎛ ⎞= − = − ⎜ ⎟∂⎝ ⎠ ...(iii)
64 Thermal Physics
If the system is heated at constant pressure (dp = 0), weget from eq. (ii)
.p
VdV dT
T∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠ ...(iv)
The coefficient of volume expansion can, therefore, bedefined by
/ 1.
p
dV V VdT V T
α ∂⎛ ⎞= = ⎜ ⎟∂⎝ ⎠ ...(iv)
If the system is heated at constant volume (dV= 0), we getfrom eq. (i)
.V
pdp dT
t∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠
The coefficient of pressure expansion can, therefore, bedefined by
/ 1
V
dp p pdT p T
β∂⎛ ⎞= = ⎜ ⎟∂⎝ ⎠ . ...(v)
Now, by the reciprocity theorem for the variables p, V andT, we have
1pT V
p V TV T p
⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞ = −⎜ ⎟⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠
Making substitutions from eq. (iii), (iv) and (v), we get
( ) 1 1TEV
V pα
β⎛ ⎞⎛ ⎞− = −⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
orTE
p=αβ .
This is the required relation.
Change in Pressure for a Change in Temperature atConstant Volume
65Heat and Temperature
By the reciprocity theorem for the variables p, V and T ofa gas, we have
1pT V
p V TV T p
⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞ = −⎜ ⎟⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠
or .pV T
p p VT V T∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠
Now ET = – VT
pV∂⎛ ⎞
⎜ ⎟∂⎝ ⎠ and 1V
α =p
VT∂⎛ ⎞
⎜ ⎟∂⎝ ⎠
∴ TV
pE
Tα
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠ .
The change in pressure dp corresponding to a change intemperature dT at constant volume is therefore given by
dp = ET α dT
= (104N/m2) (0.0037/K) (0.1 K)
= 3.7 N/m2.
Equation of State of a Stretched Wire
Let us consider a stretched wire, and let its state dependon its length L, temperature T and tension F. Let us write
F = f (L,T).
On differentiation:
dF = .T L
F FdL dT
L T∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
If the wire undergoes an infinitesimal change from onestate of equilibrium to another, then the differentials dF, dL anddT are perfect differentials.
Of the three state variables F, L and T, only two areindependent. Choosing F and T as independent variables, the
66 Thermal Physics
above equation must be true for all sets of values dF and dT.Then, if dF = 0 and dT ≠ 0, we have
L T
F FdT dL
T L∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
or .L T F
F F VT L T∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
...(i)
Now, 1
F
LL T
λ∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠ (coefficient of linear expansion) and
2T
L FY
r L∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠π (isothermal Young’s modulus), where 2rπ is
the area of cross-section of the wire. Making these substitutionsin eq. (i), we get
2 .L
Fr Y
Lπ λ∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠
External Work and Internal Work
When a system undergoes a displacement under the actionof a force, ‘work’ is said to be done, its magnitude being equalto the product of the force and the component of thedisplacement parallel to the force. If the system ‘as a whole’exerts a force on its surroundings and a displacement takesplace, then the work done by the system is called ‘externalwork’. For example, when a gas contained in a cylinder expandsand pushes out the piston, external work is done by the gason the piston.
When the work is done by one part of a system on anotherpart of the same system, then it is called ‘internal work’ . Forexample, the molecules of an actual gas attract one another.Therefore, when a gas expands, work is done against the mutualattraction between its molecules. This is internal work.
67Heat and Temperature
Internal work has no place in thermodynamics. Only theexternal work which involves an interaction between a systemand its surroundings is important in thermodynamics, and isa macroscopic concept.
If W is the (external) work done by a system against itssurroundings, then - W will stand for the work done on thesystem by its surroundings. If in a process the volume of asystem increases, then the work is done by the system. Ifvolume decreases, then the work is done on the system.
Performance of External Work Versus
A system which can interact with its surroundings canpass from one state to another by two different processes : bydoing work and by transfer of heat.
The water, its container and the paddle constitute thesystem, and the weight constitutes the surroundings. There isno temperature difference between the system and thesurroundings. When the weight is let fall a certain distance,the paddle rotates and the water is churned. Thus the state ofthe system changes as a result of the work done by thesurroundings (falling weight) on the system. “We may thereforedefine work as the energy which is transferred between asystem and its surroundings when no temperature differenceexists between them”.
The system has some water in a container, which is incontact with a Bunsen flame which constitutes the surroundings.The temperature of the flame is higher than that of the water,so that heat flows from the flame into the water. Certainly nowork has been done. Thus, in this case, the state of the system(water + container) changes because heat is transferred fromthe surroundings (flame) into the system. “We may thereforedefine heat as the energy which is transferred between a systemand its surroundings only by virtue of a temperature-differencebetween them”.
68 Thermal Physics
Naturally, a decision as to whether a particular change ofstate is due to the work done or due to the transfer of heatcannot be taken until we decide, which is the system and whatare the surroundings’. This may be clarified by an example. Afalling weight which turns a generator G, which in turn sendsan electric current through a resistor R immersed in a watercontainer.
Suppose the water plus the resistor plus the generator ischosen as the system, and the weight as the surroundings.Now, the temperature of the surroundings does not differfrom that of the composite system. Hence during the fallingof weight the surroundings do work on the system, there beingno transfer of heat between the two.
Again, let us choose the water along as the system, and theresistor plus the generator plus the weight as the surroundings.Now there is a temperature-difference between thesurroundings (resistor) and the system (water). Hence heatwill flow from the resistor into the system during the fallingof the weight, there being no work done by the surroundingson the system Thus, in this case, there is a transfer of heatbetween the system and the surroundings.
Heat given without Temperature-rise
If a substance in a condensed state, say ice, is put in contactof a flame, the heat from the flame will flow into the ice byvirtue of a temperature-difference between them. But it wouldnot cause the temperature of ice to rise. Instead, the condensedstate begins to change into the liquid state (ice melts) at constanttemperature. Thus, the addition of heat without temperature-rise is possible, and this does not contradict the concept of heatas energy transferred by virtue of a temperature-difference.This also does not violate energy conservation because thegiven heat is stored as potential energy of the molecules of thesubstance.
69Heat and Temperature
Heat, a Form of Stored Energy
The heat added to the melting ice is used in displacing themolecules from their positions and making their order irregular.That is, it is stored as the internal potential energy of themolecules. Thus, heat can also be considered as a form ofpotential energy and this again does not contradict the conceptof heat as energy transferred by virtue of a temperature-difference.
Changing Temperature without Heat-transfer : In alldynamic processes involving friction (or resistance), heat isalways produced causing temperature-rise, although no heatis transferred. For example, when water is churned by a rotatingpaddle wheel or current is sent through a wire, the temperatureof water (or of wire) rises.
Dependence of Work Done on the Path
Let a gas be contained in a cylinder with insulating wallsand conducting base, and fitted with a movable piston whichcarries weights upon it. Let a heat-reservoir of adjustabletemperature T be available on which the cylinder maybe placed, when desired. Let the gas be the system, andthe piston (with weights) or the heat-reservoir be thesurroundings. Work can be done on the system (or by thesystem) by increasing (or decreasing) weights on the piston.Similarly, heat can be made to flow into the system (or out ofthe system) by placing the cylinder on the heat-reservoir andadjusting the temperature of the reservoir to be higher (orlower) than that of the system.
Suppose initially the system (gas) is in equilibrium with itssurroundings, having a pressure pi, and volume Vi. It is thenmade to interact with the surroundings with some ‘quasi-static’ process in which it expands and reaches a finalequilibrium state with pressure pf and volume Vf. Let uscompute the external work done by the gas during this process.
70 Thermal Physics
Let p be the instantaneous pressure of the gas whileexpanding against the piston. The instantaneous force on thepiston is pA, where A is the face-area of the piston. The workdone by the gas in displacing out the piston through aninfinitesimal distance ds is
dW = p A ds (work= force × distance)
= p dV,
where dV (= A ds) is the infinitesimal increase in the volumeof the gas. The total work done by the gas in expanding fromthe initial volume Vi to the final volume Vf would be
.iV
ViW dW p dV= =∫ ∫
The value of this integral can be obtained graphically byplotting the p-V curve of the gas. The nature of the curvedepends upon the conditions under which the gas has
expanded. It is clear that the integral fV
Vip dV∫ is the area under
the p-V curve between the initial and final states. This area is,
71Heat and Temperature
therefore, the measure of the work done by the gas. Now, there aremany processes (paths) by which the system (gas) can be takenfrom an initial state i (with coordinates pt, Vi, Ti) to a final statef (with coordinates pf, Vf, Tf). Let us represent these two statesby points i and f on a p-V diagram Now, one path from i tof is the path iaf.
The pressure is kept constant from i to a and then thevolume is kept constant from a to f. In this case, the work doneis equal to the area under the line ia. Another path is ibf, inwhich case the work done is the area under the line bf whichis much smaller than the area under the line ia. The otherpossible paths are the continuous curve from i to f, and theseries of short zig-zag paths from i to f. The areas under thesepaths are different from each other and also different from thefirst two paths i af and ibf. Thus, we see that the work doneby or on a system depends not only on the initial and finalstates but also on the intermediate states, i.e. on the path ofthe process.
Dependence of Heat-transfer on Path : Similarly, the amount
of heat flowed ( dQ∫ ) into the system (or out of the system),when it passes from state i (at temp. Ti) to state f (at temp. Tf)also depends on how the system is heated. For example, wecan heat the gas at constant pressure pi until the temperaturerises to Tf , and then lower the pressure to pf, keeping thetemperature constant at Tf . Alternatively, we can first lowerthe pressure to pf , and then heat it to the temperature Tf,keeping the pressure constant at pf. Each process would give
a different value of dQ∫ . Thus, we see that the heat flowed
72 Thermal Physics
into (or out of) a system depends not only on the initial andfinal states but also on the intermediate states, i.e. on the pathof the process.
Both work and heat do depend on the path of the processi.e. neither one can be conserved alone.
Cyclic Process
When a system in a given state goes through a series ofdifferent processes which bring it back to its initial state, thesystem is said to have undergone a ‘cycle’.
Suppose a gas goes from state i to state f along path iaf andreturns to state i along path fbi. During the first process (iaf)the gas expands. The work is done by the gas and is equal tothe area between the curve iaf and the V-axis. During thesecond process (fbi) the gas is compressed. Now the work isdone on the gas and is equal to the area between the curve fbiand the V- axis. The net work done by the gas in the cycle isequal to the difference between the two areas, that is, equalto the area enclosed by the cycle.
When the p-V cycle is traced clockwise, the net work isdone by the system and is said to be positive. When the cycleis traced anti-clockwise, the net work is done on the system,and is said to be negative.
73Heat and Temperature
Let W1 and W2 represent the work in cycle 1 and 2respectively in
W1 is positive (cycle 1 is clockwise) and W2 is negative(cycle 2 is anti-clockwise). Since area of cycle 1 is smaller,W1 < W2. Hence the net work is negative.
A similar consideration the net work is positive, while thenet work is zero.
Internal Energy of a System
A system may seem to have no apparent mechanical energybut may still be capable of doing work. It is therefore said topossess ‘internal energy’. For example, a mixture of hydrogenand oxygen does not possess any external kinetic or potentialenergy but still it can do work on explosion. This is due to theinternal energy of the system.
The internal energy is associated with microscopicmechanical energy. For example, the internal energy of anactual gas is the sum of the kinetic and potential energies ofits molecules. The molecules of an ideal gas do not possesspotential energy (because they do not attract one another), andhence, the internal energy of an ideal gas is only the kineticenergy of its molecules. This is, however, not an operationaldefinition, as microscopic mechanical energy cannot bemeasured directly.
Measurement of Internal Energy
Let us formulate a macroscopic definition of internal energywhich provides a means of measuring it. Let a thermodynamicsystem interact with its surroundings, and pass from an initialequilibrium state i to a final equilibrium state f through acertain process (path). Let Q be the heat absorbed by thesystem and W the work done by the system during this process.The quantity (Q - W) can be computed. It is experimentallyfound that if the system be carried from the state i to f through
74 Thermal Physics
different paths, the quantity (Q - W) is found to be the same,although Q and W individually are different for different paths.Thus, when a thermodynamic system passes from state i tostate f, the quantity (Q - W), depends only upon the initial andfinal (equilibrium) states and not at all on the path taken bythe system between these states. This quantity is defined as the‘change in the internal energy of the system. ’Thus, if Ui andUf be the internal energies of the system in the initial and thefinal states respectively, we have
Uf-Ui = ΔU = Q-W. …(i)
We conclude that there is a function U of thethermodynamic coordinates of the system whose final valueminus initial value is definite and equals Q — W in the process,by whatever path the system went from the initial to the finalstate. This function is called ‘internal energy function .’ It isa point function.
Eq. (i) can be used to measure the change in the internalenergy of the system. If some arbitrary value is assumed forthe internal energy in some standard reference state, its valuein any other state can be computed from the above equation.In practice, however, only the change in internal energy isimportant.
The internal energy of a system in a given state is a functionof that state only and does not at all depend upon the way bywhich that state has been acquired. Hence, we cannot tellwhether the internal energy of the system has been acquiredby heat-transfer or by performance of mechanical work or byboth. In other words, it is impossible to divide the internalenergy into a thermal part and a mechanical part.
First Law of Thermodynamics
The first law of thermodynamics is simply the principle ofconservation of energy applied to a thermodynamic system.In one form it is stated as follows :
75Heat and Temperature
“Whenever other forms of energy are converted into heator vice-versa, there is a fixed ratio between the quantities ofenergy and of heat thus converted “.
Mathematically, the change in the internal energy of asystem in passing from an initial state i to a final state f is
Uf-Ui = ΔU = Q-W,
where Q is the amount of heat absorbed by the systemand W, the external work done by the system. This equationis taken as mathematical form of the first law ofthermodynamics.
If the system undergoes only an infinitesimal change instate absorbing an infinitesimal amount of heat dQ and doingan infinitesimal amount of work dW, then the infinitesimalchange in internal energy dU would be given by
dU = dQ-dW
or dQ = dU + dW.
This is the differential form of the first law ofthermodynamics. (Q, U and W must all be expressed in thesame unit).
Significance : The mathematical formulation of thefirst law contains three related ideas : (i) heat is a form ofenergy in transit, (ii) energy is conserved in thermodynamicsystem, (iii) every thermodynamic system in a equilibriumstate possesses internal energy, which is a function of the stateonly.
Thus, the first law establishes an exact relation betweenheat and other forms of energy. For example, it tells us thata certain quantity of heat will produce definite amount ofwork, and vice-versa. It follows directly from this law that itis impossible to derive any work without expenditure of anequivalent amount of energy in some other form.
76 Thermal Physics
Cycle of Processes
When a closed system is taken from an initial to a final stateby one or more processes, and then back to the initial state bysome other one or more processes, the net change in internalenergy of the system is zero (ΔU = 0). This is because theinternal energy of the system depends only on the state. Hence,
by the first law of thermodynamics, ΔU = Q - W , we have
0 = Q-W
or Q = W .
In the differential form we may write
dQ∫ = dW∫ .
Thus, for a closed system undergoing a cycle of processes,the cyclic integral of heat is equal to the cyclic integral of work.This is also taken as a statement of the first law ofthermodynamics.
Isobaric Process : A process taking place at constant pressureis called an ‘isobaric’ process. For example, the boiling of waterto steam or the freezing of water to ice taking place at aconstant pressure (and also at a constant temperature) areisobaric processes.
Boiling of Water : Let us consider the boiling of water at100 °C and atmospheric pressure. Let m be the mass of water,Vl its volume in the liquid state , Vv in the vapour state andL the latent heat of vaporisation. The heat absorbed by themass m during the change of state is
Q = mL.
The work done by the mass in expanding from volume Vl
to volume Vv against the constant external pressure p is
( ) .v v
l l
V V
v lV VW p dV p dV p V V= = = −∫ ∫
77Heat and Temperature
If ΔU be the change in the internal energy, then from thefirst law of thermodynamics, we have
ΔU = Q-W
or ΔU = mL-p (Vv-Vl).
This is the expression for the change in the internal energyof the system.
We know that at 1 atmospheric pressure, 1.000 gm of waterhaving a volume 1.000 cm3 becomes 1671 cm3 of steam whenboiled. The heat of vaporisation of water at 1 atmosphere is539 cal/gm. Thus, the heat spent is
Q = mL = 1.000 × 539 = 539 cal. The external work doneagainst the atmosphere is
W = p(Vv - Vl)
= (1.013 × 106) × (1671 - 1.000)
= 169.2 × l07 erg.
We know that 1 calorie = 4.18 × 107 erg.
W =7
7
169.2 1040.5 cal.
4.18 10×
=×
Substituting these values in the above expression, we get
ΔU = 539 cal - 40.5 cal = 498.5 cal,
which is positive. Thus, the internal energy of the systemincreases by 498.5 cal during the process.
Of the 539 cal of heat needed to boil 1 gm of water, 40.5cal are used in doing external work and 498.5 cal are addedto the internal energy of the system. This energy is in fact theinternal work done in overcoming the strong attractionsbetween the H2O molecules in the liquid state.
Freezing of Water : When a given quantity of water freezesinto ice at 0 °C under atmospheric pressure, it gives out a
78 Thermal Physics
definite quantity of (latent) heat. At the same time, it doeswork against the atmospheric pressure. We know that iceoccupies a greater volume than an equal mass of water (i.e.water expands on freezing).
At 1 atmospheric pressure, 1.000 gm of water, having avolume of 1.000 cm becomes 1.091 cm3 of ice on freezing. Theheat of fusion of water at 1 atmosphere is 80.0 cal/gm. Thus,the heat given out is
Q = - mL = - 1 x 80 = - 80 cal
and the external work done is
W = p(Vice-Vwater)
= (1.013xl06) × (1.091-1.000)
= 0.0922 × 106 erg
=6
7
0.0922×10= + 0.0022 cal.
4.18 ×l0
∴ by the first law of thermodynamics, the change in internalenergy is
ΔU = Q – W
= - 80-0.0022 = - 80.0022cal,
which is negative. Thus, the internal energy of the systemdecreases by an amount greater than the heat extracted from it.
Isochoric Process : A process taking place at constantvolume (ΔV = 0) is called an ‘isochoric’ process. In such aprocess, the work done on or by the system is zero (W = 0).Hence, by the first law of thermodynamics, ΔU=Q- W, wehave
ΔU = Q.
Thus, in an isochoric process, the heat added to (or takenfrom) the system becomes entirely the increase (or decrease)in the internal energy of the system.
79Heat and Temperature
Adiabatic Process : When a system passes from an initialstate i to a final state f through a process such that no heat flowsinto or out of the system, then the process is called an ‘adiabaticprocess’. Such a process can occur when the system is perfectlyinsulated from the surroundings, or when the process is veryrapid so that there is little time for the heat to flow into or outof the system.
Let a system pass from an initial state i to a final state fby absorbing heat Q from outside and by doing external workW . If Ui and Uf be the internal energies of the system in theinitial and the final states respectively then, from the first lawof thermodynamics, the change in energy is given by
ΔU = Uf-Ui = Q-W. ...(i)
If the process be adiabatic then Q = 0 (because no heat entersor leaves the system in adiabatic process). Then the change ininternal energy is
ΔU = -W, ...(ii)
which is negative. Thus, if an amount of work is done by thesystem in an adiabatic process (i.e. W is positive), the internalenergy of the system decreases by that amount. That is, thesystem does adiabatic work at the cost of its own internalenergy. If, on the other hand, the work is done on the systemin the adiabatic process, the energy of the system increases bythat amount.
Let us take a few examples. When a gas is suddenlycompressed (adiabatic process), the work done on the gas isadded to its internal energy so that its temperature rises. Thisis why the bicycle pump becomes hot when the air in it iscompressed into the cycle tube. Similarly, when a gas is suddenlyexpanded, the work done by it against the surroundings isdrawn from its internal energy which therefore decreases andthe temperature of the gas falls. When a motor-car tyre bursts,
80 Thermal Physics
the sudden expansion of its air into the atmosphere is adiabaticand the tyre is cooled.
The eq. (ii) shows that during an adiabatic process thework done is exactly equal to the change in the internal energyof the system. But we know that the change in the internalenergy of a system depends only on the initial and the finalstates of the system, and not at all on the path connecting them.Hence, in the adiabatic process, the work done will also dependonly on the initial and the final states i.e. it will be same forall adiabatic paths connecting the two states.
Free Expansion : If a system, say a gas, expands in sucha way that no heat enters or leaves the system (adiabaticprocess) and also no work is done by or on the system, thenthe expansion is called the ‘free expansion’.
Let us consider an asbestos-covered vessel with rigid wallsand divided into two parts, one containing a gas and the otherevacuated. When the partition is suddenly broken, the gasrushes into the vacuum and expands freely. Uf and Uf be theinitial and final internal energies of the gas, we have, by thefirst law of thermodynamics
Uf - Ui = Q - W.
Since the vessel is heat-insulated and the process is sudden,no heat enters or saves the system i.e. Q = 0.
Again, since the walls of the vessel are rigid and the gasexpands into the vacuum, no external work is done i.e. W=0.Hence, from the last expression, we have
Uf - Ui = 0
or Ui = Uf.
Thus, the initial and the final internal energies are equalin free expansion.
81Heat and Temperature
PROBLEMS
1. A mass of 5 kg falling vertically through a height of 50meter rotates a paddle wheel which in turn churns 0.5 kg ofwater initially at 15°C. What is the final temperature of water ?sp. heat of water is 1.0 k- cal/(kg- °C) and g = 9.8N/kg.
Solution : The falling mass loses potential energy whichis used in rotating the wheel and is finally converted into heat.Now, the work done in rotating the wheel is
W= mgh = 5 × 9.8×50 = 2.45×103 joule.
The heat equivalent of this work is
( )32.45 10
cal J=4.18 joule/cal4.18
WQ
J×
= =
= 586 cal = 0.586 k-cal.
This heat is taken by the water. Let ΔT be the rise intemperature. Then
Q = mass of water × specific heat × rise in temperature
or 0.586 = 0.5 × 1.0 × ΔT
∴ ΔT =0.586
0.5 × 1.0 = 1.17°C.
2. The height of the Niagara falls is 50 meter. Calculate thedifference between the temperature of water at the top and atthe bottom of the fall, if
J = 4.2 Joule/calorie
Solution. Suppose m kg of water fall in one second. Thepotential energy lost in one second is
W = mgh
= m × 9.8 × 50 joule.
This lost energy is converted into heat. If Q be the heatproduced, then
82 Thermal Physics
( 9.8 50) joule4.2 joule/calorie
mWQ
J× ×
= =
= 117 m cal = 0.117 m k-cal.
If this heat causes a temp-rise ΔT in water, then
Q = mass × sp. heat of water x temp-rise
or 0.117 m k-cal = m kg× 1.0 k-cal/(kg-°C) x ΔT
0.117 k-cal kg×l.0k-cal/(kg-°C)
mT
mΔ =
= 0.117°C
3. A 2.0-g bullet moving with a velocity of 200 meter/secis brought to a sudden stoppage by an obstacle. The total heatproduced goes to the bullet. Calculate the rise in temperatureof the bullet. Sp. heat for the bullet may be taken d0.03 cal/(g-°C). (J = 4.2 × 107 erg/cal)
Solution. The kinetic energy in the moving bullet is
W =12
mv2 = 12
×2.0g × (200,00cm)2
= 4.0× 108 erg.
By sudden stoppage of the bullet, the whole of the energyis converted into heat. The heat equivalent of this energy is
8
7
4.0 10=9.52cal.
4.2 ×10W
QJ
×= =
ΔT be the rise in temperature of the bullet due to this heat,then
Q = m x c × ΔT (c is sp. heat)
or 9.52 = 2.0 × 0.03 × ΔT
∴9.52
1592.0 0.03
T CΔ = = °×
83Heat and Temperature
4. A lead bullet at 100 °C strikes a steel plate and melts.What was its minimum speed ? Sp. heat of lead = 0.03 cal/(g-°C), latent heat = 5 cal/g and melting point = 327 °C. Theheat produced is shared equally between the plate and thebullet. (J= 4.2 × 107 erg/cal)
Solution. Let m gm be the mass of the bullet and v cm/secthe velocity with which it strikes the plate. Its kinetic energy
is 12
mv2 erg.
By sudden stoppage of the bullet, the whole of the energyis converted into heat. According to relation W = JQ , the heatgenerated is
212 calories.
mvWQ
J J= =
This heat is shared equally between the plate and thebullet. Hence the heat shared by the bullet is
214
mvJ calories.
This heat is used up in raising the temperature of the bulletto its melting point and then in melting the bullet.
Now, the heat required to raise the temperatureof the bullet, from 100 °C to 327 °C (melting point) is mcΔT = m × 0.03 x (327 -100) cal and the heat required to meltit is mL = m × 5 cal. Therefore, the total heat required to meltthe bullet is
Q = m × 0.03 × (327 – 100) + m × 5
= m (0.03 × 227 + 5) cal = 11.81 m cal.
Clearly, this must be equal to the heat shared by thebullet. Thus,
11.81 m =21
4mv
J
84 Thermal Physics
or v2 = 4J× 11.81
= 4×(4.2× 107)× 11.81 = 19.84×108
∴ v = √(19.84× 108) = 4.4 × 104 cm/sec.
5. A piece of ice of mass 50 kg is pushed with a velocityof 5 meter per sec along a horizontal surface. As a result offriction between the piece and the surface, the piece stops aftertravelling 25 meter. How much ice has been melted? (J = 4.2joule/cal, latent heat of ice is 80 cal/g.)
Solution : The initial kinetic energy in the piece is
W = 12
mv2 = 12
× 50 kg × (5 meter/sec)2 = 625 joule.
The energy becomes zero when the piece stops. This lostenergy has been converted into heat, which is
625149 cal=
4.2W
QJ
= =
Let m gm of ice melt due to this heat. Then
Q = mL
or 149 = m x 80
∴149
1.8680
m = = g.
Here we have assumed that all the heat produced goes tomelt the ice and that the ice melts after it is brought to rest.
6. An athlete consumes 4000 kilo-calories per day throughhis diet. Compute his power in watt.
Solution : W = JQ, where J = 4.18 joule/cal.
The work equivalent to 4000 kilo-calories (= 4000 × 1000cal) per day is
W = 4.18 × (4000 × 1000) joule per day
85Heat and Temperature
=4.18×(4000×l000)
= 193.5 joule per sec.24 x 60 x 60
Now, the power is the rate of doing work, and 1 watt =1 joule per sec.
∴ power = 193.5 watt.
7. A metallic ball falls from a height of 10 meter on theground and rebounds to a height of 0.50 meter. Has its internalenergy changed ? Compute the rise in temperature of the ball.Take the specific heat of the ball as 0.12 cal-g-1-°C-1, g = 980dyne/gram and J = 4.2 x 107 erg/cal.
Solution. Let m gram be the mass of the ball. The energyreleased when it falls by 10 meter (=1000 cm) is
mgh = m × 980 × 1000 erg.
A part of this is used to rebound the ball to a height of 0.50meter (=50 cm) and the rest is added to the internal energy ofthe ball.
Energy used to raise the ball = m × 980 × 50 erg.
∴ energy added to the ball
= (m × 980 × 1000) - (m × 980 × 50)
= m × 980 x 950 erg = 7
× 980 x 950 cal.
4.2 ×l0m
Let it be Q . Then if ΔT be the temperature-rise due to this,we have
Q = m × c × ΔT
or 7
× 980 x 950 = ×0.12×
4.2 ×l0m
m T
∴ ΔT= 7
980×950 = 0.18
4.2 ×l0 × 0.12C°
86 Thermal Physics
8. 0.004 kg air is heated at constant volume from 2°C to6°C. Find the change in its internal energy. The sp. heat of airat constant volume is 0.172 cal/(g-°C) and J = 4.18 joule/cal.
Solution : By the first law of thermodynamics, the changein the internal energy of a system is given by
ΔU = Q-W,
where Q is the heat absorbed by the system and W is theexternal work done by the system.
Here the air is heated at ‘constant’ volume. Hence W = 0.
∴ ΔU = Q.
The heat absorbed by the air is
Q = mass x sp. heat × temp-rise
= 0.004 kg ×0.172 kcal/(kg-°C) x 4°C
= 2.752 x 10-3 kcal
= 2.752 cal
∴ ΔU = Q= 2.752 cal
= 2.752 × 4.18 = 11.5 J.
9. 50 cal per gram heat is added to a system and the processtakes place at constant volume. What will be the change inthe specific internal energy in joule ? J = 4.18 joule/cal.
Solution : At constant volume, the change in internal energyequals heat added:
ΔU = Q.
Here Q = 50 cal/g. Therefore, change in internal energyper gram (i.e. specific internal energy) is
ΔU = 50 cal/g
= 50 × 4.18 = 209 joule/gram.
87Heat and Temperature
10. A system absorbs 1000 cal of heat and does 1675 jouleof external work. The internal energy of the system increasesby 2505 joule. Compute the value of J .
Solution : By the first law of thermodynamics, we have
ΔU = Q-W,
where Q is the heat absorbed and W the external work done bythe system. Here ΔU = + 2505 joule, Q = + 1000 cal = + 1000 Jjoule, and W = + 1675 joule. Thus
2505 = 1000 J-1675 .
∴ J = 2505 1675 41804.18
1000 1000+
= = joule/cal
11. At atmospheric pressure 1.0 g of water, having volume10 cm3, becomes 1671 cm3 of steam when boiled. Calculate theexternal work and the change in the internal energy. The heatof vaporisation of water is 539 cal/g. (J = 4.18 × 107 erg/cal and1 atmospheric pressure = 1.013 × 106 dyne/cm2)
Solution : The external work against the atmosphericpressure p is
W = p dV
= (1.013 × 106 dyne/cm2)
(1671 cm3 - 1.0 cm3)
= 169.2 × 107 erg.
The heat taken by the water to become steam is
Q = mL = 1.0 × 539 = 539 cal.
In work unit, it is
Q = 539 × (4.18 × 107)
= 2253 × 107 erg.
By the first law of thermodynamics, the change in internalenergy is
88 Thermal Physics
ΔU = Q-W
= 2253 ×l07-169.2 ×l07
= 2083.8 ×l07 erg
= 2083.8 joule.
12. When water is boiled under 2 atmosphere pressure, theheat of vaporisation is 2.20 × 106 joule/kg and the boiling pointis 120°C. At this pressure, 1 kg of water has a volume of 10-
3 m3 and 1 kg of steam a volume of 0.824 m3 . Compute the workdone and the increase in internal energy when 1 kg of wateris converted into steam at 120°C (1 atmosphere = 1.013 × 105
N/m2).
Solution : The heat taken in boiling 1 kg of water to steamat 120°C is
Q = mL = 1 kg × (2.20 × 106 J/kg)
= 2.20 × 106 joule.
During this process the volume increases from 0.001 m3
to 0.824m3 at 2 atmosphere pressure. The external workdone against the atmosphere is
W = pdV
= (2 × 1.013 × 105 nt/m2) × (0.823 m3)
= 0.167 × 106 joule.
By the first law of thermodynamics, the increase in internalenergy is
ΔU = Q -W = 2.20 × 106- 0.167 × 106
= 2.033×106 joule.
13. When a system is taken from state i to state f alongthe path iaf, it is found that the heat Q absorbed by the systemis 50 cal and the work W done by the system is equal to 20cal. Along the path ibf ; Q = 36 cal.
89Heat and Temperature
(i) What is W along the path ibf ?
(ii) If W = - 13 cal for the curved return path fi, what isQ for this path ?
(iii) Take Ui = 10 cal. What is Uf ?
(iv) If Ub = 22 cal, what are Q for the processes bf and ib ?
Solution : According to the first law of thermodynamics,the change in internal energy of a system in going from thestate i to f is
ΔU = Uf-Ui = Q-W, ...(i)
where Q is heat absorbed by the system. For the path iaf;we have Q = + 50 cal, W = + 20cal.
∴ ΔU = Q- W = 50 - 20 = 30 cal.
Since internal energy U is a state function,
ΔU between i and f is always 30 cal, whatever path beadopted,
(i) For the path ibf; Q = 36 cal (given) and ΔU = 30 cal.
∴ W = Q -ΔU=36-30 = 6cal.
(ii) For the return curved path fi ; W = - 13 cal. (given) andΔU= - 30 cal.
∴ Q = ΔU+W= - 30 - 13 = -43 cal.
(iii) Ui = 10 cal (given). Now,
ΔU = Uf-Ui = 30 cal.
∴ Uf = 30 + Ui = 30+10 = 40 cal.
90 Thermal Physics
(iv) For the process bf, the volume remains constant so thatwork done is zero. Hence from eq. (i)
Q = ΔU=Uf-Ub.
Now, Uf= 40 cal (obtained above) and Ub = 22 cal (given).
∴ Q = 40 - 22 = 18 cal.
For the path ibf , Q = 36 cal (given), while for the path bf,Q = 18 cal (just determined). Hence for the path ib, we have
Q = 36 -18 = 18 cal.
14.When a system is taken from state a to state b alongthe path acb, 80 joules of heat flow into the system, and thesystem does 30 joules of work,
(i) How much heat flows into the system along path adb, if the work done is 10 joule ?
(ii) When the system is returned from b to a along the curvedpath, the work done on the system is 20 joule. Does thesystem absorb or liberate heat and how much ?
(iii) If Ua = 0 and Ud = 40 joule, find the heat absorbedin the processes ad and db .
Solution : According to the first law of thermodynamics,the change in internal energy of a system in going from thestate a to the state b is
ΔU = Ub-Ua = Q-W, ...(i)
where Q is heat absorbed by the system and W is work doneby the system.
91Heat and Temperature
For the path acb , Q = 80 joule and W= 30 joule.
∴ ΔU = Ub - Ua = 80 - 30 = 50 joule.
(i) For the path adb , W= 10 joule and ΔU=Uh-Ua = 50 joule.Therefore, from eq. (i), we get
Q = ΔU + W = 50 + 10 = 60 joule.
(ii) For the curved path ba , W = - 20 joule and ΔU = Ua
- Ub = - 50 joule.
Therefore, from eq. (i), we get
Q = ΔU+W = -50-20 = - 70 joule.
Minus sign means that heat is liberated from the system.
(iii) For the path ad, the work would be same as foradb (because no work is done along db ) i.e. W= 10 jouleand ΔU = Ud - Ua = 40 joule (given). Thus,
from eq. (i), we get
Q = ΔU + W= 40 + 10 = 50 joule.
For the path db , W=0 (volume is constant) and ΔU=Ub-Ud=(Ub-Ua) - (Ud - Ua) = 50 - 40 = 10 joule. Thus, from eq. (i),we get
Q = AU+W= 10 joule.
15. A system is taken from a state i to a state f along thepath iaf, where path ia is isobaric and path af is isochoric. Theheat absorbed by the system is 36 cal and the work done bythe system is 6 cal. (i) If Ui = 10 cal and Ua = 22 cal, find theheat absorbed along ia and af. (ii) Will the heat rejected forany return path fi be just 36 cal ? If not, on what factor willit depend ?
Ans. (i) 18 cal, 18 cal; (ii) no ; it will depend on the actualpath fi which has been adopted.
92 Thermal Physics
16. A gas-filled cylinder fitted with a piston is immersedin ice (0 °C). The piston is rapidly pushed down to compressthe gas which is therefore heated. It is then left for some timeso that the gas is once again at 0 °C. The piston is now slowlyraised up to the initial position. If 100 g of ice is melted duringthis cyclic process, compute the work done on the gas. Representthe cycle on a p-V diagram.
Solution : Let A be the Initial state of the gas characterisedby thermodynamic coordinates (p1, V1). It is first compressedadiabatically to B whose coordinates are (p2 V2). It is then leftto reach the state C when its volume remains constant at V2,the temperature falls to 0 °C, and pressure falls to p3(>p1)-Finally, it is expanded isothermally to reach the initial state A,its pressure falling to p1 and volume increasing to V1.
Let us write the first law of thermodynamics for the entireprocess ;
ΔU = Q – W.
Since the gas returns to the initial state, its internal energyis restored to the initial value i.e. ΔU = 0.
∴ W = Q.
The heat Q is used to melt 100 gm of ice.
∴ Q = mL = 100 × 80 = 8000 cal.
∴ work done on the gas during this cycle
W = Q = 8000 cal.
93Heat and Temperature
17. A given system is subjected to three different processesas shown in the table below. Supply the missing information.All the data are in joule.
Calculate the unknown values with the help of the firstlaw of thermodynamics. Which of these processes is adiabatic?
Process Q W Ut UfΔU = Uf – Ui
1 35 -15 … -10 …
2 -15 … … 60 -20
3 … -20 80 … 20
Ans. (1) Ui = -60, ΔU =50,(2) W = 5, Ui= 80, (3) Q = 0,Uf = 100.Process 3 is adiabatic because for it Q = 0.
7
The Reversibility
Quasi-static Process : A process during which a system isnever more than infinitesimally far away from an equilibrium stateis a ‘ quasi-static ‘process. For example, a quasi-static expansionof a gas may be obtained by an infinitesimal change in pressure,or a quasi-static flow of heat may be obtained by an in-finitesimalchange in temperature. In fact, quasi-static process is an idealprocess which can never be exactly obtained, but can only beapproximated under some circumstances.
Reversible Process
A reversible process is one which can be reversed in such away that all changes taking place in the direct process are exactlyrepeated in the inverse order and opposite sense, and no changesare left in any of the bodies taking part in the process or in thesurroundings . For example, if an amount of heat is suppliedto a system and an amount of work is obtained from it in thedirect process; the same amount of heat should be obtainableby doing the same amount of work on the system in the reverseprocess.
96 Thermal Physics
Conditions of Reversibility : A process can be reversibleonly when it satisfies two conditions :
(i) Dissipative forces such as friction , viscosity, inelasticity,electrical resistance, magnetic hysteresis , etc. must becompletely absent: Suppose a gas is contained in a cylinderfitted with a piston and placed in contact with aconstant-temperature source. The piston is loaded sothat the pressure exerted by the piston on the gasexactly balances the pressure of the gas on the piston.If the load on the piston is now decreased, the gas willexpand, doing external work in pushing up the pistonand also in overcoming the friction between the pistonand the walls of the cylinder. The heat necessary forthis work is taken from the source. If now the load onthe piston is increased, the gas will be compressed. Thework used in pushing up the piston during theexpansion is now recovered. On the contrary, morework has to be done against the friction. The expansionis therefore irreversible. Similarly, other dissipativeeffects like inelasticity, electrical resistance , etc. makethe process irreversible.
(ii) The process must be quasi-static : When the gas expands,an amount of work is done by the gas to give kineticenergy to the piston. This work cannot be recoveredduring the reverse process, but on the contrary, morework is to be done to give kinetic energy to the piston.Hence in order to make the expansion of the gasreversible; the pressure of the gas on the piston shouldbe only infinitesimally different from the pressureexerted by the piston on the gas. Under this conditionthe expansion or compression will take place infinitelyslowly so that no kinetic energy will be produced.These conditions are never realised in practice. Hence,a reversible process is only an ideal conception.
97The Reversibility
Irreversible Process
Any process which is not reversible exactly is an irreversibleprocess. All practical processes such as free expansion, Joule-Thomson expansion, electrical heating of wire, diffusion ofliquids or gases, etc. are irreversible. All natural processes suchas conduction, radiation, radioactive decay, etc. are alsoirreversible. Thus , irreversibility is a rule.
Isothermal Expansion of a Gas : Let us imagine a gascontained in a cylinder having perfectly insulating walls buta perfectly conducting base, and fitted with a frictionless piston.The cylinder is placed on a heat-reservoir maintained at aconstant temperature, which is the same as the temperatureof the gas. The piston is loaded, so that the pressure exertedby the piston on the gas exactly balances the pressure of thegas on the piston. Suppose the load on the piston is decreasedby an infinitesimally small amount. The gas will expand, doingexternal work in pushing up the piston and its temperaturewill tend to fall. It will thus very slightly deviate fromequilibrium, but an amount of heat equivalent to the workdone will immediately flow from the heat-reservoir to the gaswhich will again be at the temperature of the reservoir andattain equilibrium. Thus, the infinitely-slow isothermalexpansion of a gas in the absence of any friction is an exampleof a reversible process. The conditions described above are,however, ideal. In practice, a very slow isothermal expansionis “approximately” reversible.
Adiabatic Compression of a Gas : Again, if the cylindercontaining the gas is a perfect insulator (including its base) andthe gas is compressed infinitesimally slowly, the compressionis reversible. In practice, it can be made only approximatelyreversible.
Diffusion of Gases : When two or more gases diffuse intoone another, there is a change in chemical composition and theprocess is irreversible.
98 Thermal Physics
Transfer of Heat from Hot to Cold Body : This process isirreversible, because heat cannot be transferred back from thecold to the hot body without leaving any change elsewhere.
Joule Expansion (or Free Expansion) of a Perfect Gas: Inthis process, the (perfect) gas changes from a volume Vi to alarger volume Vf without any change in temperature. To revertthe gas to its initial state, it would have to be compressed toVi by some external device. The external work so done wouldbe converted into heat. To ensure that the gas retains its initialtemperature and no changes are left in the surroundings, theheat produced would have to be extracted from the gas andconverted completely into work. Since this last step isimpossible, the process is irreversible.
Joule-Thomson Effect : The Joule-Thomson expansion of agas is an irreversible process. The reason is same as for freeexpansion. (Any heat cannot be completely converted into work.)
Transfer of Heat by Radiation : Heat coming from a hotbody by radiation cannot be radiated back to the hot bodywithout leaving any change elsewhere. Hence the process isirreversible.
Electrical Heating of a Wire : The electrical energydissipated as heat in the wire cannot be fully converted intoelectrical energy and so the process is irreversible.
Very Slow Extension or Contraction of Spring : In thisprocess, if carried extremely slowly, the spring passes throughstates of thermodynamic equilibrium, which may be traversedjust as well in one direction as in the opposite direction. Theprocess is therefore approximately reversible.
Dissipation of Mechanical Energy to Heat through Friction:Suppose a body moves on a surface from an initial position,spends its mechanical energy to overcome friction betweenitself and the surface, and again returns to its initial position.The energy spent is dissipated as heat. Now, if the body be
99The Reversibility
allowed to go round its path in the reverse direction, its energyspent previously cannot be recovered. On the contrary, thebody will have to further spend its energy against the frictionin the reverse path. Hence the process is irreversible.
Complete Conversion of Work into Heat : We can converta given quantity of work ‘completely’ into heat. For example,when we rub two stones together under water, the work doneagainst friction is converted into heat which is communicatedto the surrounding water. Since the state of the stones is thesame at the end of the process as at the beginning, the netresult of the process is merely the conversion of mechanicalwork into heat with 100 per cent efficiency (W = Q). Thisconversion can be continued indefinitely.
The reverse process is, however, not possible. We cannotmake a device by which a given amount of heat can be‘completely’ converted into work. At first thought, theisothermal expansion of an ideal gas can be considered as aprocess in which heat is converted completely into work . Inthis case D U = 0 (since the temperature remains constant) andso the heat absorbed by the gas is equal to the work done bythe gas during the expansion (Q = W). But here the state ofthe gas changes . Its volume increases and pressure decreasesuntil atmospheric pressure is reached at which the processstops. Thus the conversion of heat into work cannot becontinued indefinitely.
Further, any device which converts heat completely intowork would lead to a decrease in entropy of the universe,which is impossible.
Heat Engine : Any “cyclic” device by which heat is convertedinto mechanical work is called a heat engine.
There are three main parts in an engine: a hot body called‘source’, a working substance, and a cold body called ‘sink’.The working substance takes in heat from the source, converts
100 Thermal Physics
a part of it into useful work, and gives out the rest to the sink.This series of processes is called a ‘cycle ‘ since the workingsubstance returns to its original state. This is shown in Fig. Byrepeating the same cycle over and over again, work can becontinuously obtained.
Suppose the working substance takes in an amount of heatQ1 from the source, and gives out an amount Q2 to the sink.Suppose W is the amount of work obtained. The net amountof heat absorbed by the substance is Q1 – Q2, which has beenactually converted into work. Applying the first law ofthermodynamics to one complete cycle , we get
Q1 –Q2 = w.
Thermal Efficiency: The ‘thermal efficiency‘ e of an engineis defined as the ratio of the work obtained to the heat takenin from the source , that is ,
1 2
1 1
W Q –Qe = =
Q Q
2
1
Qe = 1-
Q
This equation indicates that the efficiency of the heat enginewill be unity (100%) when Q2 = 0 (no heat is given out to the
101The Reversibility
sink). This is, however, not possible in practice. This meansthat the engine cannot convert all the heat taken in from thesource into work.
We cannot define the efficiency as W/Q2 , because in thatcase we shall have
1 2 1
2 2 2
Q – Q QWe = = = = –1 ,
Q Q Q
so that the condition for the ideal value of efficiency (i.e. fore = 1) would be Q1= 2Q2 which is absurd.
Indicator Diagram : If the pressure of a system undergoinga process is plotted against its volume, the resulting curve iscalled the ‘indicator diagram’ of the system. This diagramenables us to represent the working of an engine and to calculatethe useful work obtained from the engine.
Let a given mass of a working substance, say a gas, becontained in a cylinder closed by a frictionless piston whichis loaded so that the pressure exerted by the piston on the gasexactly balances the pressure of the gas on the piston whenthe cylinder is placed in contact with a heat source. Let p andV be the pressure and volume of the gas and A the area ofcross-section of the piston.
Suppose the load on the piston is decreased by an infinitely”small amount. The gas will expand very slowly, pushingup the piston through an infinitely small distance ds and reacha new equilibrium state. The external work done by the gaswill be
dW = force exerted on the piston × displacement
= pressure × area of piston face x displacement
= p × A × ds .
The pressure p has been assumed constant during thesmall change. But A x ds = dV is the increase in volume.
102 Thermal Physics
\ dW = pdV.
This work is drawn from the hot source. This process maybe carried further by decreasing the load on the piston ininfinitely small steps. Clearly, the external work done for anexpansion from an initial state (p1, V1) to a final state (p2 , V2)will be
2
1
V
V
W = p dVò
Let us choose the coordinate axes the pressure and volumeaxes. Let the points A and B (Fig.) represent the initial state(p1, V1) and the final state (p2 , V2) of the gas, and the curveAB represent the expansion of the gas.
Let us consider a point M . Let p be the pressure and Vthe volume at this point. Let (V+dV) be the volume at aninfinitely close point N . Then MN represents expansion of thegas through a volume dV. The strip MNRS will be a rectangleof area p dV . But this is external work done by the gas whichis thus represented by the area of the strip. The whole areaABCD between the curve AB and the volume-axis is clearly
2
1
V
V
p dVò which represents the external work done by the gas in
expanding from V1 to V2 . Thus , the area bounded between
103The Reversibility
the p- V curve and the volume axis gives the work donedirectly . This result is very useful in the study of heat engines.
Reversible Engine: In an engine the working substancegoes through a cycle of processes. It takes in heat from a hotbody, converts a part of it into work and gives out the rest toa cold body, returning to its initial state. During this cycle theconditions of the hot and cold bodies and of the surroundingschange. If this cycle can be traversed in the reverse order suchthat all the parts of the engine completely recover their originalconditions and no changes are left in the surroundings, thecycle is a “reversible cycle”, and the engine is a “reversibleengine”.
Such an engine can be realised if (i) the working parts ofthe engine are free from friction, (ii) the pressure andtemperature of the working substance never differ appreciablyfrom its surroundings at any stage of the cycle, so that all theprocesses involved in the cycle are quasi-static.
These conditions can never be realised in practice. Hencea reversible engine is an ideal conception. (Carnot has presentedan imaginary picture of such an engine.)
Carnot’s Heat Engine
Carnot developed a plan of an idealised heat engine, freefrom all the imperfectness of an actual engine. As shown inFig., the Carnot’s engine consists of four components :
(i) A cylinder with perfectly heat-insulating walls but perfectlyconducting base, and closed with a tight-fitting perfectlyinsulating and frictionless piston. A fixed mass of a gas(working substance) is filled in the cylinder, and someweights are placed on the piston of the cylinder.
(ii) A hot body of infinitely large heat capacity at a constanttemperature T1 , serving as ‘source’.
104 Thermal Physics
(iii) A cold body of infinitely large heat capacity at a constanttemperature T2 , serving as ‘sink’.
(iv) A perfectly heat-insulating stand.
The cylinder may be placed on any of the three bodies(ii), (iii) and (iv) and may be moved from one to the otherwithout doing any work.
The working substance is imagined to go through a cycleof four processes, known as the “Carnot’s cycle”. Suppose itis in an initial state represented by the point A on the p-Vdiagram (Fig.).
Process 1 : The cylinder is put on the source and theweights on the piston are removed in infinitely small steps. The
105The Reversibility
working substance thus expands infinitely slowly, doing workin raising the piston. During this process the substance takesin heat from the source by conduction through the base. Theexpansion is isothermal. It is continued until the staterepresented by the point B is reached. The curve AB representsthe isothermal expansion of the substance at the constanttemperature T1 (of the source).
Process 2 : The cylinder is now put on the heat-insulatingstand and, by further removing the weights on the piston , thesubstance is further expanded. This expansion is adiabaticbecause now no heat can leave or enter the substance throughthe (insulating) cylinder. The substance does work in raisingthe piston and its temperature falls. The expansion is continueduntil the temperature falls to T2 which is the temperature ofthe sink. It is represented by the curve BC.
Process 3: The cylinder is now put on the sink. Weightsare now placed on the piston in infinitely small steps.The substance is thus compressed isothermally until thestate represented by the point D is reached. The curve CDrepresents the isothermal compression at the constanttemperature T2 (of the sink). During this process work is doneon the substance by the piston and a quantity of heat is givenout to the sink.
Process 4 : The cylinder is once more put on the heat-insulating stand and, by further placing the weights on thepiston, the substance is compressed adiabatically. A furtheramount of work is done on the substance and its temperaturerises. The process is continued until the temperature rises oncemore to T1 and the state A is recovered. The curve DA representsthe adiabatic compression.
The external works done by the substance during theprocesses AB and BC are represented by the areas ABB’ A’ andBCC B’ respectively. Similarly, the external works done on thesubstance during the processes CD and DA are represented by
106 Thermal Physics
the areas CDD’ C and DAA’ D’ respectively. The differencebetween these two parts of areas is
area (ABB’ A’ + BCC’ B’) – area (CDD’ C’ + DAA’ D’)
= area ABCD.
Thus, the area ABCD represents the net external workdone W by the substance in one complete cycle. The net amountof heat absorbed by the substance in the cycle is Q1 — Q2 whereQ1 is the heat taken in from the source during the process AB,and Q2 the heat given out to the sink during the process CD.Since the initial and final states of the substance are the same;there is no net change in its internal energy. Hence , by thefirst law of thermodynamics,
W = Q1–Q2
for the cycle. The result of the cycle is that heat has beenconverted into work by the system. Any required amount ofwork can be obtained by simply repeating the cycle.
It is not necessary to start with the cycle at the point A .Since the cycle is reversible at each step, we can choose anypoint in the cycle as the starting point. The net external workobtained will always be the same.
Efficiency: The efficiency, e, of an engine is defined as theratio of net work done by the engine during one cycle to theheat taken in from the source in one cycle. Thus
1 2 2
1 1 1
Q –Q QWe = = 1 –
Q Q Q ...(i)
The expression is independent of the nature of the workingsubstance.
Let us now calculate the efficiency taking 1 mole of an idealgas as the working substance in terms of T1 and T2 . Let(pa ,Va), (pb ,Vb), (pc ,Vc) and (pd ,Vd) be the coordinates of thepoints A, B, C and D respectively. The heat Q1 taken in by the
107The Reversibility
gas will be used in increasing the internal energy of the gasin expanding isothermally along AB and in the external workdone by the gas. But since the gas is ideal, the internal energyis wholly kinetic depending on the temperature only, and soit remains unchanged during the isothermal expansion AB.Therefore, the heat Q1 will be equivalent only to the externalwork done by the gas in expanding from A to B at temperatureT1 i.e.
[ ] logb b b
aa a
V V V b1 1 1 e 1 eVV V
a
VdVQ = p dV =RT = RT log V = RT
V Vò ò ...(ii)
Similarly, Q2 will be equivalent to the external work done onthe gas during compression from C to D at temperature T2 i.e.
[ ]d cV V c
2 2 2 e 1 edVc Vdd
VdVQ = – p dV =RT = RT log V = RT ,
V V
Vc
Vlogò ò ...(iii)
minus sign indicating that work is done on the gas.
The work done by the gas during the adiabatic expansionfrom B to C is exactly equal to the work done on the gas duringthe adiabatic compression from D to A , so the net work duringadiabatic processes is zero.
From (ii) and (iii), we have
be
a1 1
c2 2e
d
VVQ T
= = VQ TV
log
log ...(iv)
The points B and C, and similarly the points D and A lieon the same adiabatic. Therefore, we have, from Poisson’s law
2 cbT1 V = T V –1 –1 ,g g ...(v)
and a 2 dT1 V = T V –1 –1 .g g ...(vi)
Dividing eq. (v) by eq. (vi), we get
108 Thermal Physics
b c
a d
V V =
V V.
Using this result in equation (iv), we get
1 1
2 2
Q T =
Q T.
Hence the efficiency, given by eq. (i), is
1 2
2 1
Q Te = 1 – = 1 – .
Q T
Thus , the efficiency of Carnot’s reversible engine isindependent of the working substance and depends only onthe absolute temperatures of the sink and the source.
For the engine to have 100 per cent efficiency (e = 1), T2must be zero. Since we cannot obtain a sink at absolute zero,an engine with 100 per cent efficiency is a practical impossibility.
Efficiency of Carnot’s Engine in Terms of AdiabaticExpansion Ratio : We can obtain an alternative expression forthe efficiency. From eq. (v), we have
–1
b2
1 c
VT=
T V
gæ ö÷ç ÷ç ÷ç ÷÷çè ø
Now, = p (adiabatic expansion ratio).
∴–1
.1
2
T 1T p
gæ ö÷ç ÷= ç ÷ç ÷÷çè ø
Making this substitution in the above expression, we get
–gæ ö÷ç ÷ç ÷ç ÷÷çè ø
1
e = – .p1
1
Carnot Engine is not a Practical Possibility : Carnot enginehas two ideal features :
109The Reversibility
(i) The source and the sink are bodies of infinitely large heatcapacities. Therefore, the working substance takes in allthe heat at a constant temperature (of the source) andgives up all the heat at another constant temperature(of the sink). Thus it utilises the full temperaturedifference available.
(ii) Each process in Carnot’s cycle is completely reversible.This is because of three reasons: (a) The workingsubstance is contained in a cylinder with perfectlyinsulating walls and fitted with a perfectly insulatingfrictionless piston. Thus the dissipative forces are absent,(b) The base of the cylinder is perfectly conducting . Sowhen it is placed on the hot or the cold body, theworking substance differs in temperature with the bodyonly by an infinitesimal amount, (c) The gas is expandedand compressed infinitely slowly , so the substance differsin pressure with its surroundings only by aninfinitesimal amount.
These features cannot be actually realised. Hence it is notpossible to obtain a Carnot engine in practice.
The way to Increase Efficiency of Carnot Engine : Theefficiency of a Carnot engine operating between temperaturesT1 and T2 is given by
2
1
Te = – .
T
To increase e , we must decrease T2/T1 . This can be done eitherby decreasing T2 or by increasing T1 . Since T2 < T1 , a decreasein T2 will be more effective than an equal increase in T1 .
The working substance takes in heat from the body attemperature T1 and, after doing work, gives out the balanceto the body at temperature T2 . This continues until T1 dropsto Tf and T2 rises to Tf . Thus the total heat taken in from thehot body is
110 Thermal Physics
Q1 = Cp(T1 – Tf) ...(i)
and that given out to the cold body is
Q2 = Cp(Tf–T2). ...(ii)
The balance Q1 - Q2 is converted into work W. Thus
W = Q1– Q2
= Cp(Tl -Tf) – Cp(Tf - T2)
= Cp(T1 + T2 - 2Tf).
This is the required expression.
The heat Q1 is taken in at a mean temperature 1 fT +T
2and
the heat Q2 is given out at a mean temperature 2 fT +T
2. Carnot
engine is the maximum efficient.
So for maximum work, we must have
1 f1
2 f2
(T +T ) /Q = .
(T +T ) /Q
2
2
From eq. (i) and (ii), this becomes
1 1f f
2 2f f
T –T T +T = .
T –T T +T
On simplifying:
1fT = T T2 .
Two Carnot engines are operating in series. The first engineabsorbs a quantity of heat Q1 at a temperature T1 and afterdoing work W1, rejects the remaining heat Q2 at a lowertemperature T2. The second engine absorbs the heat Q2 attemperature T2 (rejected by first) and after doing work W2,rejects the remaining heat Q3 at a still lower temperature T3.Compute the efficiency of the combination.
111The Reversibility
The total work by the combination in one cycle is
W1 + W2 = (Q1– Q2) + (Q2– Q3) = Ql–Q3 .
The heat absorbed by the combination is Q1. Therefore, theefficiency of the combination is
total work doneheat absorbed
e =
31
1 1
QQ – Q3= = 1 – .
Q Q
31
1 1
QQ – Q3= = 1 – .
Q Q31
1 1
QQ – Q3= = 1 – .
Q Q
But3 3
1 1
Q T= .
Q T
∴ 3
1
1 .T
eT
= −
This is the same as the efficiency of a single engine operatingbetween T1 and T3.
Carnot’s Ideal Refrigerator
Any device for removing heat from a cold place and addingit to a hotter place is called a “refrigerator”. It is essentially aheat engine running backwards. In a heat engine the workingsubstance takes in heat from a body at a higher temperature,converts a part of it into mechanical work, and gives out therest to a body at a lower temperature.
In refrigerator, a working substance takes in heat from abody at a lower temperature, has a net amount of work doneon it by an external agent, and gives out a larger amount ofheat to a hot body (Fig.). Thus it continually transfers heatfrom a cold to a hot body at the expense of mechanical energysupplied to it by an external agent. The working substance iscalled a “refrigerant”.
112 Thermal Physics
The engine employing the Carnot cycle may be adoptedas a refrigerator. Each step in the cycle is reversible, thereforeit is possible to reverse the entire cycle.
Let Q2 be the heat removed from the cold body attemperature T2 , W the net work done on the refrigerant, andQ1 the heat delivered to the hot body at temperature T1 . Then,we have
Q1 = Q2+ W
or W = Q1– Q2
=1
22
QQ – 1 .
Q
æ ö÷ç ÷ç ÷ç ÷÷çè ø
As in Carnot’s engine, if we use an ideal gas as a workingsubstance, we can show that
1 1
2 2
Q T= .
Q T
\ 21 1 2
22 2
T T – TW= Q –1 Q .
T T
æ ö÷ç ÷=ç ÷ç ÷ç ÷è ø
This is the expression for the work that must be suppliedto run the refrigerator.
113The Reversibility
Coefficient Performance
This purpose of the refrigerator is to remove as much heat(Q2) as possible from the cold body with the expenditure ofas little work (W) as possible. Therefore, a measure of theperformance of the refrigerator is expressed by the “coefficientof performance” K which is defined as the ratio of the heattaken in from the cold body to the work needed to run therefrigerator. That is
2 2
11
2
Q QK = .
QW Q – Q –Q
2
1
1= =
But1 1
2 2
Q T=
Q T
∴2
1 2
1
11
2
TK .
T T – T–T
= =
This is the expression for the coefficient of performance.A good refrigerator should have a high coefficient ofperformance, typically 5 or 6. Thicker and high-qualityinsulation tends to increase the coefficient of performance.
Relation between the Efficiency e of a Carnot Engine andthe Coefficient of Performance K of a Carnot Refrigerator : Leta Carnot engine and a Carnot refrigerator work between thesame temperatures T1 and T2 . Then, we have
1
1
T – Te =
T2
and1
TK =
T – T2
2
NOW1 1
T TK = = = .
T – T T – T e2 1
2 2
11 1+ +
∴1
e =K+1
This is the required relation.
114 Thermal Physics
Unattainability of Absolute Zero : A refrigerator is a devicefor producing cold. An analysis of Carnot’s ideal refrigeratorshows that its performance coefficient is
1 2
TK = .
T –T2
Obviously, K becomes vanishingly small as the temperatureT2 of the reservoir from which heat is taken approaches absolutezero. This means that as the reservoir becomes more and morecold, the refrigerator finds it more and more difficult to run.Therefore, it is impossible to take the reservoir down toT2 = 0 .
Heat Pump : A heat pump transfers heat from some coldplace to a hotter place at the expense of energy suppliedexternally by a motor. Thus it does not violate any law ofthermodynamics.
Opening the door of Refrigerator : A room cannot be cooledby having opened the door of the refrigerator placed in it. Onthe contrary, the room will warm up. The reason is as follows :
The refrigerator removes heat from its interior and expelsit into the surrounding air, thus warming the air. For doingthis, additional energy is supplied to the refrigerator by anelectric motor. The heat expelled into the air is the sum of theenergy from the motor and that removed from the interior ofthe refrigerator. In other words, the refrigerator adds moreheat into the room than it removes from its interior. On openingits door it will run continuously and hence add even more heatto the room than when its door is closed.
Purpose of the Second Law of Thermodynamics :Limitations of the First Law: The first law of thermodynamicsstates the equivalence of mechanical work and heat, when oneis completely converted into other (W = Q). Thus , it is theprinciple of conservation of energy applied to a thermodynamicsystem.
115The Reversibility
If, however, we propose to extract a certain quantity ofheat from a body and convert it completely into work, the firstlaw would not be violated. But, in actual practice this is foundto be impossible. If this were possible, we could drive shipsacross an ocean by extracting heat from the water of the ocean.Thus , the first law simply tells that if a process takes place,energy will remain conserved. It does not tell us whether theprocess is possible or not. Similarly, if a hot body and a coldbody are brought in contact, the first law is not violated whetherthe heat flows from the hot to the cold body or vice versa. Byexperience we know that heat never flows from cold to hotbody. The purpose of the second law is to incorporate suchexperimental facts into thermodynamics.
There are two equivalent statements of the second law ofthermodynamics.
Kelvin-Planck Statement
In a heat engine, a working substance takes in heat froma hot body, converts a part of it into mechanical work, andgives out the rest to a cold body. No engine has ever beendesigned which can operate in a cycle by taking heat from abody and converting all of it into work ; some heat mustalways be given to a colder body. This experience led Kelvinand Planck to state the following:
It is impossible to construct a device which, operating ina cycle, will take heat from a body and convert it’ completely‘into work, without leaving any change anywhere.
Clausius Statement
In a refrigerator, a working substance takes in heat froma cold body, has a net amount of work done on it by an externalagent (electric supply), and gives out a larger amount of heatto a hot body. It thus transfers heat from a cold body to a hotbody with the aid of external supply. No refrigerator has ever
116 Thermal Physics
been designed which can run without supply of external energy.This experience led Clausius to state the following :
“It is impossible to construct a device which ,operating in a cycle, will transfer heat from a coldbody to a hot body without expenditure of workby an external energy source . In other words, heatcannot flow spontaneously from a colder to a hotterbody.”
Equivalence of the two Statements : We can show thatthese two statements of the second law are equivalent.
Let us suppose that there is a refrigerator R (Fig.) whichtransfers an amount of heat Q2 from a cold body to a hot bodywithout having any supply of external energy. It is thus againstthe Clausius statement. Now, suppose an engine E workingbetween the same hot and cold bodies takes in heat Q1 fromthe hot body, converts a part W (= Q1 - Q2) into work, and givesup the remaining heat Q2 to the cold body. The engine E alonedoes not violate the law. But if the refrigerator R and theengine E are combined together, they form a device that takesin heat Q1 - Q2 from the hot body and converts all into workwithout giving up any amount to the cold body. This is clearlyagainst the Kelvin-Planck statement.
Similarly, let us suppose that there is an engine E (Fig.)which takes in an amount of heat Q1 from a hot body andconverts it completely into work W (= Q1), without giving anyheat to the cold body. It is against the Kelvin-Planck statement.Now, suppose a refrigerator R working between the same hotand cold bodies takes in heat Q2 from the cold body, has workW(= Q1) done upon it by an external agent, and gives out heatQ1 + Q2 to the hot body. The refrigerator R alone does notviolate the law. But both E and R together form a device whichtransfers an amount of heat Q2 from a cold body to a hot bodywith no external energy source. This is clearly against theClausius statement.
117The Reversibility
The second law of thermodynamics supplements thefirst law. The first law simply tells us that any device cannotdeliver more energy than it receives. It does not speak regardingany limitation , or any condition necessary for the deliver ofenergy.
The second law, however, does it . For example, heat takenin by a substance cannot be all delivered as work, or heatcannot flow spontaneously from a colder to a hotter body.These phenomena are not disallowed by the first law, but theyare disallowed by the second law.
Production of 8000 K Temperature by Sun-rays : It is notpossible to produce a temperature of 8000 K by focussing sun-rays . The reason is that the creation of a temperature of 8000K by transferring heat from a colder body (sun at 6000 K) bymeans of a lens is a violation of the second law ofthermodynamics.
Driving a Ship by Extracting Heat from Ocean : It is anattractive idea to drive ship on the energy drawn from theinternal energy of water. At the start of its cycle the engine ofthe ship will draw some heat Q1 from the water, convert a partof it into work, but where it would reject the rest ? By thesecond law, it must reject some heat into a colder reservoir,but none is available at hand. Theoretically, it is possible if we
118 Thermal Physics
can arrange some conveyance to the cold upper atmospherebut practical difficulties would make it almost impossible.
From Carnot’s cycle also we see that the efficiency is
Te
T= - 2
1
1
Thus , e = 0 if T2 = T1 i.e. without a temperature-differencethe conversion of thermal energy into mechanical work isimpossible.
Heat Conduction is Irreversible : Suppose there are twobodies 1 and 2 at temperatures T1 and T2 where T1 > T2. Whenthey are brought into contact, heat flows by conduction from1 to 2 till they reach a common temperature. Heat cannot flowin the reverse direction, from 2 to 1, because heat-flow by itselffrom cold to hot body is not allowed by the second law ofthermodynamics. Thus heat-conduction is an irreversiblephenomenon.
Atomic Power Plant: An atomic energy power plant doesnot violate any law of thermodynamics.
Carnot’s Theorem : It states that no engine working betweentwo given temperatures can be more efficient than a reversible engineworking between the same two temperatures, and that all reversibleengines working between the same two temperatures have the sameefficiency, whatever the working substance.
Let us consider two engines A and B working between agiven pair of source and sink. Let A be irreversible and Breversible. Let the quantities of working substance used in thetwo engines be such that they perform equal quantities ofwork per cycle. Let the engine A take in heat Q1 from thesource, perform work W, and give out heat Q1 - W to the sink.Its efficiency will be W/Q1’x. Similarly, let the engine B take inheat Q1’ from the source, perform work W, and give out heatQ1’ - W to the sink. Its efficiency will be W/Q1’ .
119The Reversibility
Let us suppose that the irreversible engine A is more efficientthan the reversible engine B i.e.
Q '1 1
W W >
Q
or .1 1Q < Q ...(i)
Now, let us imagine the two engines to be so coupled thatthe engine A works directly and it drives the engine B reversiblyas shown in Fig. B now acts as a refrigerator, taking in heatQ1’-W from the sink, having work IV done on it, and givingout heat Q1’ to the source. The couple in this way forms a self-acting device, since all the work needed to run the refrigeratorB is supplied by the engine A. The net heat taken in from thesink is thus
(Q1’ - W) - (Q1 - 2) = Q1’ - Q1
which is positive since Q1 < Q1’ from (i). This is also thenet heat given to the source. Thus the couple is transferringin each cycle a quantity of heat Q1’ - Q1 from the cold sink tothe hot source, without the aid of any external agent. But thisis against the second law of thermodynamics, hence impossible.Hence our supposition that the irreversible engine A is moreefficient than the reversible engine B is wrong. Thus the firstpart of Carnot’s theorem is proved.
120 Thermal Physics
The second part of the theorem follows as a corollary. Letus consider two reversible engines A and B working betweenthe same source and sink. If we suppose that A drives Bbackward, then A cannot be more efficient than B . Similarly,if we suppose that B drives A backward, then B cannot be moreefficient than A . Hence the two engines are equally efficient.Thus, the efficiency of a reversible engine depends only on thetemperatures of the source and the sink, and is completelyindependent of the nature and properties of the workingsubstance.
Imagine an engine I and a Carnot engine R operatingbetween the same two reservoirs. Suppose that they absorbfrom the hotter reservoir different amounts of heat, do differentamounts of work, but reject to the cooler reservoir equal amountof heat. Prove Carnot’s theorem, assuming the efficiency of Ito be greater than that of R and coupling the two engines.
Suppose the engine I takes in heat Q1 from the source, performswork W, and gives out heat Q2 to the sink. Its efficiency is
'1 2 2
11 1 1
W' Q ' - Q Qe = = = 1 -
Q Q ' Q
121The Reversibility
Suppose the Carnot (reversible) engine R takes in heat Q1
from the source, performs work W and gives out heat Q2 (sameas I) to the sink. Its efficiency is
'1 2 2
R1 1 1
W' Q ' - Q Qe = = = 1 -
Q Q ' Q
Let us assume that I is more efficient than R, that is,
Q QQ Q '
- > -2 2
1 1
1 1
or Q1 > Q1’ ...(i)
Now, let us imagine the two engines to be so coupled thatI works directly and it drives R reversibly as shown in theFigure above. Now R acts as refrigerator taking in heat Q2 fromthe sink, having work W done on it, and giving out heat Q1’to the source. The couple formed by I and R takes in net heatQ1 - Q1’ from the source, performs net work W- W’, and givesout no net heat to the sink. Now
Q1 - Q1’ = (W + Q2)- (W’ + Q2)
= W - W’
and is positive by eq. (i). This means that the couple convertsthe heat taken in from the source completely into work, withoutgiving out any heat to the sink. This is against the Kelvin-Planck statement of the second law. Hence our assumptionthat I is more efficient than R is wrong i.e. no engine can bemore efficient than a reversible engine working between thesame temperatures. This is Carnot’s theorem.
Absolute Scale of Temperature : A temperature scale whichis independent of the properties of any particular substance is calledan absolute scale of temperature. No scale furnished by anythermometer is absolute as it depends upon the properties ofthe thermometric substance.
According to Carnot’s theorem, the efficiency of a reversibleengine is independent of the working substance and depends
122 Thermal Physics
only on the two temperatures between which it is working.Taking this hint, Lord Kelvin defined a temperature scale whichdoes not depend upon the properties of any particularsubstance. This is the “Kelvin’s absolute thermodynamic scaleof temperature”.
Let a reversible engine take in heat Q1 from a source attemperature t1 and give out heat Q2 to a sink at temperaturet2, where the temperatures t1 and t2 have been measured on
any scale. The efficiency of this engine is e = 1 -Q
'Q
2
1which
depends on t1 and t2 only. We may thus write
f (t ,t )21 2
1
Qe = 1 -
Q
where f ’ is an unknown function. From this, we may also saythat Q1/Q2 must be a function of t1 and t2 only. Thus
1 2
Qf' (t , t )
Q1
2
= ...(i)
where f ’ is some other unknown function.
Similarly, for a reversible engine taking in heat Q2 attemperature t2 and giving out heat Q3 at temperature t3, wehave
2 3
Qf' (t , t )
Q2
3
= ...(ii)
where the function f’ remains unchanged.
The heat Q2 given out by the first engine is taken in by thesecond. Thus both engines, working together, form a thirdengine which takes in heat Q1 at temperature t1 and gives outheat Q3 at temperature t3, where
1 3
Qf' (t , t )
Q1
3
= ...(iii)
123The Reversibility
Since 1Q Q / Q
Q Q / Q1 3
3 2 3
= , we have from eq. (i), (ii) and (iii)
2
f'(t ,t )f '(t ,t )
f ' (t ,t )1 3
1 23
= ...(iv)
This equation does not contain t3 on the left hand side. Itmeans that the function f ’ must be such that t3 cancels out inthe right-hand side also. Hence we choose f ’ in the followingform :
(t )f '(t ,t )
(t )ff
= 11 3
3
and(t )
f '(t ,t )(t )
ff
= 22 3
3
where f is some other function. If we substitute these values
of f’(t1, t3) and f’ (t2 , t3) in the right hand-side of eq. (iv), weshall have
(t )f '(t ,t ) .
(t )ff
= 11 2
2...(v)
Now, from eq. (i) and (v), we get
(t )QQ (t )
ff
11
2 2
=
We know that Q1 > Q2. Hence the function f (t1) > f (t2)
when t1 > t2. It means that the function f (t) increases as the
temperature rises. Hence it can be used to measuretemperatures.
Let us suppose that f (t) denotes a temperature q on a
new scale. Then we may write.
orQQ
1 1
2 2
=
124 Thermal Physics
=1 1
2 2...(vi)
This equation defines the Kelvin’s absolute thermodynamicscale of temperature. That is, the ratio of any two temperaturesmeasured on this absolute scale is equal to the ratio of the quantitiesof heat taken in and given out by a Carnot’s reversible engine workingbetween these temperatures. It is independent of the propertiesof any particular substance.
In order to complete the definition of the Kelvin’s absolutescale, we assign the arbitrary value of 273-16 K to thetemperature of the triple point of water, i.e.
t rq = 27316 K.
Now, for a Carnot’s engine working between a source at
temperature 8 and a sink at temperature t rq , we have, from
eq. (vi),
t r t r
=
or t rt r
q q´ =
ort r
Q. K.
Qq= ´ 273 16
Comparing this with the corresponding equation for theideal (perfect) gas temperature T, which is T =
T
t r t r
Lim p.
p p
æ ö÷ç ÷ç ÷ç ÷®ç ÷ç ÷è ø273 16
0 K we see that on the Kelvin’s absolute scale,
Q acts as a thermometric property.
Absolute Zero : We can find out the zero of the absolutescale. The last expression shows that the heat Q transferredisothermally between two given adiabatics decreases as thetemperature q decreases. Conversely, the smaller the value of
125The Reversibility
Q, the lower is the corresponding temperature q . The smallestpossible value of Q is zero. The corresponding value of q(= 0 K) is called ‘absolute zero’. Thus , the absolute zero is thetemperature at which a reversible isothermal process takes place‘without any transfer of heat’. This means that at absolute zerothe isothermal and adiabatic processes are identical.
Comparison of Absolute Scale and Ideal Gas Scale : Letus consider a reversible engine with one mole of an ideal gasas the working substance and performing a Carnot cycle ABCD(Fig.). Let (pa , Va), (pb , Vb), (pc , Vc) and (pd , Vd) be thecoordinates of the points A, B, C and D respectively. Let T1 andT2 be the temperatures of the source and the sink respectivelyon the ideal gas scale. Let Q1 be the heat taken in from thesource during the isothermal expansion AB, and Q2 the heatgiven out to the sink during the isothermal compression CD.
Let us now calculate Q1 and Q2 in terms of T1 and T2 . Sincethe working substance is an ideal gas, its internal energy,depending on the temperature only, will remain unchangedduring the isothermal expansion AB. Therefore, the heat Q1absorbed by the gas will be equivalent only to the externalwork done by the gas in expanding from A to B at temperatureT1 i.e.
[ ]bb b
a a a
VV V bV V e eV
a
VdVQ p dV RT RT log V RT log
V V= ò = ò = =1 1 1 1
126 Thermal Physics
Similarly, Q2 will be equivalent to the work done on thegas during compression from C to D at temperature T2 i.e.
[ ]ccd
c d d
VV V cV V e eV
d
VdVQ p dV RT RT log V RT log
V V= ò = ò = =2 2 2 2
minus sign indicating that the work is done on the gas. Fromthe last two expressions, we get
1 1
2 2
be
a
ce
d
Vlog
VQ TVQ T logV
=...(vii)
The points B and C, and similarly the points D and A, lieon the same adiabatic. Therefore, we have, from Poisson’s law
cbT V T Vg g- - 1=11 2
and a dT V T Vg g- - 1=11 2
Dividing we get,
b c
a d
V VV V
=
Using this result in eq. (vii), we have
Q TQ V
=1 1
2 2
Now, if q1 and q2 be the temperatures of the source and
the sink measured on the Kelvin’s absolute scale, they wouldbe defined by
=1 1
2 2
Comparing the last two expressions , we get
TT
=1 1
2 2
127The Reversibility
i.e. the ratio of temperatures on the absolute scale is thesame as the ratio on the ideal gas scale. If q and T be thetemperatures of a given body on the absolute and ideal gasscales respectively, and t rq and Ttr the respective temperaturesof the triple point of water, we have
t r t r
TT
= 1
Since t rq = t rT = 273-16 K , we have
Tq =
Hence numerically q and T are also the same. Thus thetwo scales are identical in all respects.
Practical Realization of Absolute Scale
The Kelvin’s absolute scale cannot be directly realised inpractice. It is, however, exactly identical to an ideal gas scale.Hence the temperatures measured by a constant-volume (orconstant pressure) gas thermometer filled with an ‘ideal’ gaswould exactly be the same as the temperatures measured onan absolute scale. But, in practice, no gas is ideal, hence wecannot have an ideal gas thermometer.
However, the constant-volume hydrogen thermometer isthe closest approach to an ideal gas thermometer. Any readingtaken on the hydrogen thermometer can be corrected to obtainthe corresponding reading on the ideal gas thermometer. Thus,Kelvin’s absolute scale which coincides with the ideal gasscale, is realised in practice.
No Negative Temperatures on the Absolute Scale : Theefficiency of a Carnot’s engine is
Te
T= - 2
1
1 ,
128 Thermal Physics
where T1 is the temperature of the source and T2 that of
the sink. Since TT
=2 2
1 1, we may write
eqq
= - 2
1
1
To obtain 100% efficiency (i.e. e= 1), q2 must be zero. That
is, if a sink at absolute zero were available, all the heat takenin from the source would have been converted into work.Clearly, a negative temperature on the absolute scale wouldmean a temperature of the sink at which the efficiency of theengine is greater than unity. This would be a violation of thesecond law of thermodynamics. Hence, a negative temperatureon the absolute scale is impossible .
Steam Engine
A steam engine is one in which water is used as the workingsubstance. It consists essentially of a boiler (source), a heat-insulating cylinder closed with a lubricated piston, a condenser(sink) and a feed-pump (Fig.). Its working differs from that ofa Carnot engine in which the working substance remainsconfined in the cylinder throughout all the processes. In thesteam engine the working substance takes in heat in the boiler,performs work in the cylinder, gives out heat in the condenserand is transferred back to the boiler by the feed-pump. Thusits cycle, called the “Rankine’s cycle” consists of six processes:
129The Reversibility
1. The working substance (water) is heated in the boilerto its boiling temperature T1 at constant pressure p1
(the vapour pressure of water at T1). The process isrepresented by A A‘ in the p-V diagram (Fig. b).
2. The boiling water is converted into steam at the constanttemperature T1 and constant pressure p1. The processis represented by A’B’ in (Fig. b).
3. The steam so formed is passed through gas-heatedpipes (not shown) where it is superheated to atemperature T3 (say). It is then admitted into the cylinderthrough the valve V1. This process is represented byB’B in (Fig. b).
4. The valve V1 is automatically closed and the steamreached in the cylinder expands adiabatically, thusdoing external work against the piston until itstemperature and pressure fall to T2 and p2 respectively(say). During this process some of the steam condensesto form water droplets. This adiabatic expansion isrepresented by BC in (Fig. b).
5. The piston now returns, opening the valve V2 . Thesteam in the cylinder completely condenses into thecondenser at constant temperature T2 and constantpressure p2 . The complete condensation is representedby CD in the p-V diagram. The latter stage of thisprocess differs from the Carnot’s corresponding processin which the condensation stops at a point which lieson the adiabatic through A .
6. The water obtained by the condensation of steam istransferred from the condenser to the boiler by meansof the feed-pump and the pressure is again increasedfrom p2 to p1 . During this process the temperaturechanges slightly. This process also is different from thecorresponding Carnot process. It is represented by DAin (Fig. b) . The cycle is thus complete.
130 Thermal Physics
To obtain the net work done during the cycle, the workdone on the feed-pump must be allowed for. If the volume ofthe condensed water returned in each cycle by the feed-pumpfrom the condenser at pressure p2 to the boiler at pressure p1is Vw , the amount of work involved is (p1 - p2) Vw . This is givenby the shaded area ADp2 p1; the net work done by the workingsubstance being given by the area ABCD . If reversibility isassumed for all the processes, the Rankine cycle represents themaximum possible efficiency for the steam engine.
Efficiency : Let Q1 denote the total amount of heat takenin by the working substance during the three processes A® A’,A’ ® B’ and B’® B. Since all these processes take place atconstant pressure, we have
Q1 = Hb - Ha
where Hb and Ha are the total heat functions at the pointsB and A respectively.
Similarly, the heat Q2 given out during the process C ® D is
Qa = Hc - Hd .
The efficiency is given by
a cb d
ab
(H H ) (H H )Q Qe
Q H H- - --= =
-1 2
1...(i)
The values Hb, Hc and Hd are obtainable in the steam tables.The value of Ha is, however, not obtainable. Hence let useliminate it.
The equation defining the total heat function is
H = U + pV.
Differentiating:
dH = dU + pdV+Vdp.
But dU + p dV=dQ (by the first law of thermodynamics)
\ dH = dQ+Vdp.
131The Reversibility
For the adiabatic process DA , we have dQ = 0 .
\ dH = Vdp.
Integrating between the limits D and A , we get
A
a dD
H H V dp.- = òNow, V is practically constant during the process D® A,
and is equal to Vw, the volume of the water in the condenser.
\ 1 2
A
a w wdD
H H V dp V (P P ).- = = -ò\ a wdH H V (P P ).= + -1 2
Putting the value of Ha in eq. (i), we get
c wb
wb d
H H V (P P )e
H H V (P P )- - -
=- - -
1 2
1 2,
where the term Vw (p1 - p2) is the ‘feed-pump term.’
Actual steam engines have an efficiency only about60% of the efficiency so deduced because the actual processesare not strictly reversible. The adiabatic expansion of thesteam is p generally not complete and the valves take time intheir opening and closing. Hence in actual curve the angles atA, B , C and D are rounded (Fig.), instead of being sharp.Clearly, the actual curve has an area smaller than the Rankine’scurve.
132 Thermal Physics
Internal Combustion Engine
There are two general types of heat engines, the “steamengines” and the “internal combustion engines”. In both theengines, a working substance contained in a cylinder performsa cycle of processes in which it takes in heat from some source,converts a part of it into external work and gives out the restto a sink.
In the steam engine the heat is supplied to the workingsubstance by burning the fuel outside the cylinder (in a furnaceunder the boiler). In the internal combustion engine, the heatis developed by combustion taking place inside the cylinder.
One type of the internal combustion engine is theOtto engine, realised by Otto in 1876, whose plan is shownin Fig. It consists of a cylinder fitted with a piston and providedwith two valves, one inlet valve I and the other out-letvalve O.
These valves are automatically opened and closed at theright moments by a suitable mechanism. The working substanceis air.
133The Reversibility
The engine operates on a cycle, called ‘Otto cycle’, consistingof six processes. Out of these six processes, four involve themotion of the piston and are called ‘strokes’. Hence the cycleis a four-stroke cycle.
1. Suction Stroke : The inlet valve I is opened and theoutlet valve O is closed. The piston moves down, anda mixture of 98% air and 2% petrol vapour is suckedinto the cylinder at atmospheric pressure through thevalve I (Fig. a). The process is indicated by EC in theindicator (p-V) diagram (Fig.).
2. Compression Stroke : Both the valves are closed. Thepiston moves up and the mixture is compressed
adiabatically to about 15 of its original volume, its
temperature rising to 600 °C (Fig. b). The process isrepresented by CD in the p-V diagram.
3. Combustion : Just when the compression stroke ends,an electric spark ignites the mixture. Hence a largeamount of heat of combustion is produced. This createsa high pressure and raises the temperature of the airto 2000 °C at constant volume .
The process is indicated by DA in the p-V diagram.(The piston remains stationary during the process).
4. Expansion Stroke : The valves still remain closed. Thehot combustion products now expand adiabatically, sothat the piston moves down (Fig. c). As a result, thepressure as well as the temperature of the combustionproducts drops.
It is this stroke in which the engine does external workand hence it is also called as ‘working stroke’. It isrepresented by AB in the p-V diagram.
134 Thermal Physics
5. Valve Exhaust: Just when the working stroke ends, theoutlet Valve O is opened. The combustion products(still at a pressure and temperature higher than outside)escape through O until the pressure falls to theatmospheric pressure. In this process heat is given up.It is represented by BC in the p-V diagram. (The pistonremains stationary during this process also).
6. Exhaust Stroke : The piston moves up so that the burntgases escape out through the valve O (Fig. d). This isrepresented by CE in the p-V diagram.
The next cycle then occurs with a fresh charge of air andpetrol vapour.
The first and last processes EC and CE cancel each other.Therefore, we can imagine as if a fixed mass of air say, 1 mole,is always in the cylinder and is going round the cycle CDABC,taking in heat at constant volume along DA and giving up heatat another constant volume along BC .
Efficiency: Let Ta ,Tb , Tc and Td be the absolute temperaturescorresponding to the points A , B , C and D respectively. Thenif Cv is the molar specific heat of air at constant volume, theheat Q1 taken in during the process D® A is
135The Reversibility
1 v a dQ C (T T ),= -
and the heat Q2 given out during the process B ® C is
Q2 = Cv(Tb-Tc).
Hence the thermal efficiency is
cb
a d
T TQe .
Q T T-
= - = --
2
1
1 1 ...(i)
Let V1 and V2 be the volumes of the mixture at C and Drespectively. Since the points B and A , and similarly the points
C and D, lie on the same adiabatic ( TVg- 1 = constant), we have
abT V T Vg g- - 1=11 2 ...(ii)
and c dT V T Vg g- - 1=11 2 ...(iii)
Subtracting, we get
c ab d(T T )V (T T )Vg g- - 1- = -11 2
orcb
a d
(T T ) V(T T ) V
g g
r
- -æ ö æ ö- ÷ ÷ç ç÷ ÷= =ç ç÷ ÷ç ç ÷÷ ç÷ç- è øè ø
1 1
2
1
1
where r = V1/V2 , and is known as the compression ratio’.Putting this value in eq. (i), we get
eg
r
-æ ö÷ç ÷= - ç ÷ç ÷çè ø
11
1
This expression shows that the efficiency increases withincreasing compression ratio r . In the actual engine, r cannotbe made greater than about 10, otherwise the rise in temperatureduring the adiabatic compression of the air-fuel mixture willbe large enough to cause an explosion before the spark strikes.
For a typical compression ratio of 8 and g = 1 -4 , theefficiency would be
136 Thermal Physics
.
e . %.æ ö÷ç= - = =÷ç ÷çè ø
0 411 0 56 56
8
This is the theoretical efficiency predicted for an engineoperating in the idealised Otto Cycle. In real engines theefficiency is much lower, probably less than half the ideal valuebecause of effects such as acceleration, turbulence, friction,heat loss to the cylinder walls, and incomplete combustion ofthe air-fuel mixture.
Otto Engine and Carnot Engine: The efficiency of Ottoengine can also be expressed in terms of a ratio of temperatures.From eq. (ii) and (iii), we have
a d
cb
TTVV T T
g -æ ö÷ç ÷ = =ç ÷ç ÷÷çè ø
1
1
2
∴ V
eV
gg
r
-- æ öæ ö ÷÷ çç ÷÷= - = - çç ÷÷ çç ÷ç ÷÷çè ø è ø
11
2
1
11 1
b c
a d
T T.
T T- = -1 1
During the Otto cycle, the lowest temperature is Tc and thehighest temperature is Ta. The efficiency of a Carnot engineoperating between these two extreme temperatures would be
c(Carnot)
a
Te .
T= -1
which is greater than the efficiency of the Otto cycle.
Comparison of Steam Engines and Petrol Engines : Ascompared to steam engine, the internal combustion (petrol)engine is (i) cleaner, (ii) lighter, (iii) more compact, (iv) has noseparate boiler, (v) has easy fuel supply, and (vi) possesses ahigher thermal efficiency. Therefore these engines are used inmost forms of modern transport such as aeroplanes.
The steam engines are, however, cheaper than petrolengines because the fuel (coal) of the steam engines is cheaper
137The Reversibility
than the fuel (petrol) of the petrol engines. Hence they are usedin railways in which a very large amount of work is to beobtained.
Internal Combustion Engine : An internal combustionengine is one in which heat supplied to the working substanceis produced by combustion taking place inside the cylinder.
There are two types of internal combustion engines ; theOtto engine in which heat is supplied at constant volume , andthe Diesel engine in which heat is supplied at constant pressure.
Diesel Engine: The plan of this engine, proposed by Dieselin 1900, is shown in Fig. It consists of a cylinder closed by apiston and provided with three valves, air inlet valve I, oil inletvalve I’ and outlet valve O . These valves are automaticallyopened and closed at the right moments by a suitablemechanism. The working substance is air.
The engine operates on a four-stroke cycle, known as “Dieselcycle.”
(i) Suction Stroke : The air inlet valve I is opened, the oilinlet valve I’ and the outlet valve O are closed. The
138 Thermal Physics
piston moves down, and pure air is sucked into thecylinder at atmospheric pressure (Fig. a). This isindicated by EC in the p-V diagram (Fig.).
(ii) Compression Stroke : All the valves are closed. The pistonmoves up and compresses the air adiabatically to about
117
of its original volume, its temperature rising to
1000 °C. The is indicated by CD in the p-V diagram.
Just at the end of this stroke i.e. at the point D the oil inletvalve I’ is opened.
(iii) Working Stroke : A heavy oil (fuel) is now injected intothe cylinder through I’. As the temperature inside thecylinder is very high, this oil immediately burns. Itsrate of supply is so adjusted that as the piston movesdown, the burning oil supplies heat to the air at constantpressure. This portion of the working stroke isrepresented by DA in the p-V diagram. At A , whenthe temperature is about 2000 °C the supply of oil iscut off and for the rest of the working stroke theexpansion is adiabatic and so the temperature falls.This portion of the working stroke is shown by AB inthe p-V diagram.
139The Reversibility
At B the outlet valve O is opened so that the pressureimmediately falls to atmospheric pressure. The heat is givenup to the outside during the process which is indicated by BCin the p-V diagram.
(iv) Exhaust Stroke : The piston moves up so that the burntgases escape out through the valve O (Fig.). This isrepresented by CE in the p-V diagram.
The next cycle then occurs with a fresh supply of air.
The processes EC and CE cancel each other. Therefore, wecan imagine as if a fixed mass of air, say 1 mole, is always inthe cylinder and is going round the cycle CD ABC, taking inheat at constant pressure along DA and giving up heat at constantvolume along BC.
Efficiency : Let Ta ,Tb ,Tc ,Td be the absolute temperaturesat the points A, B, C, D respectively. Then if Cp be the molarspecific heat of air at constant pressure, the heat Q1 taken inat constant pressure during the process D® A is
Q1 = Cp(Ta - Td).
Similarly, if Cv be the molar specific heat of air at constantvolume, the heat given out during the process B® C is
Q2 = Cv(Tb - Tc).
Hence the thermal efficiency is
Qe
Q= - 2
1
1
v cb
p a d
C (T T )c (T T )
-= -
-1
cb
a d
(T T ),
(T T )g
æ ö- ÷ç ÷= - ç ÷ç ÷÷ç -è ø1
1 ...(i)
where γ = Cp/Cv.
140 Thermal Physics
Let V1 , V2 and V3 be the volume of air at C, D and Arespectively. Since the points B and A , and similarly the pointsC and D, lie on the same adiabatic, we have
abT V T Vg g- -=1 11 3
and c dT V T Vg g- -=1 11 2
Subtracting, we get
c ab d(T T )V T V T Vg g g- - -- = -1 1 11 3 2
or
1 1
3 2
1 1c ab d
V VT T T T
V V
g g- -æ ö æ ö÷ ÷ç ç÷ ÷- = -ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø
1 11 1
a de c
T Tg g
r r
- -æ ö æ ö÷ ÷ç ç÷ ÷= -ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø
where eρ (= V1/V3) is the ‘expansion ratio’ and cρ (= V1/V2 isthe ‘compression ratio’. Substituting this value of Tb - Tc in (i),we get
11 1
11
a de c
a d
T T
e .T T
g
r rg
-é ùæ ö æ ö÷ ÷ê úç ç÷ ÷-ç çê ú÷ ÷ç ç÷ ÷ç ç÷ ÷è ø è øê ú= - ê ú-ê úê úê úê úë û
...(ii)
Since the pressure along DA is constant ; we have
3 3 1
2 1 2
a c
ed
T V V VT V V V
rr
= = ´ =
ca d
e
T Tρ
=ρ
Putting this in eq. (ii), we get
141The Reversibility
1 11 1
11
cd d
e e c
cd d
e
T T
eT T
g grr r r
rgr
- -é ùæ ö æ ö÷ ÷ê úç ç÷ ÷-ç çê ú÷ ÷ç ç÷ ÷ç ç÷ ÷è ø è øê ú= - ê úê ú-ê úê úê úë û
or
1 11
11 1
e c
e c
e
g g
r rg
r r
é ùæ ö æ ö÷ ÷ê úç ç÷ ÷-ç çê ú÷ ÷ç ç÷ ÷ç ç÷ ÷è ø è øê ú= - ê úê ú-ê úê úê úë û
.
In Diesel engine, the air is simply compressed and thereis no danger of explosion. Hence its compression ratio canbe made much higher than that of an Otto engine. Thus,the efficiency of the Diesel engine can be made greaterthan that of an Otto engine. But, on account of the higherworking pressures, it must be more robust in construction.However, it runs cheaper as it requires lesser fuel and runs oncrude oil.
PROBLEMS
1. In a Carnot cycle the isothermal expansion of an idealgas takes place at 400 K and the isothermal compression at300 K. During then this case expansion 500 calories of heatenergy are transferred to the gas. Let’s determine (i) the workperformed by the gas during the isothermal expansion, (ii) theheat rejected from the gas during the isothermal compression,(iii) the work done on the gas during the isothermalcompression. (J = 4 18 joule/cal).
Solution : (i) The work performed by the gas during theisothermal expansion is equivalent to the heat Q1 absorbed bythe gas. Here
Q1 = 500 cal.
\ work = 500 × 4.18 = 2090 joule.
142 Thermal Physics
(ii) Let Q2 be the heat rejected. We know that
1
2
Q TQ T
= 1
2
or 2
TQ Q
T= ´ 2
11
300 K= 500 cal × = 375
400 Kcal.
(iii) The work done on the gas during the isothermalcompression is equivalent to the heat Q2 rejected andis thus
375 × 4.18 = 1567.5 joule.
2. A reversible engine works between two temperatureswhose difference is 110°. If it absorbs 746 joule of heat fromthe source and gives 546 joule to the sink, let’s calculate thetemperatures of the source and the sink.
Solution : For a reversible (Carnot) cycle, we have
1
2
Q TQ T
= 1
2
where Q1 is the heat taken in from the source at temperatureT1 and Q2 is the heat rejected to the sink at temperature T2 inone cycle.
Here , Q1 = 746 joule and Q2 = 546 joule .
∴1
2
746546
TT
= .
Also, T1-T2= 110 (given).
∴1
1
746546 110
TT
=-
Solving, we get
T1 = 410 K
and T2 = 300 K.
143The Reversibility
3. An ideal (Carnot’s) heat engine takes 1000 calorie ofheat at 627 ºC from a source and rejects a part at 27 °C to thesink. The work done by the engine per cycle, the heat rejectedto the sink per cycle and the efficiency of the engine may befound out.
Solution : For a Carnot cycle, we have
1
2
Q TQ T
= 1
2
where Q1 is the heat taken in from the source at temperatureT1 and Q2 is the heat rejected to the sink at temperature T2 inone cycle.
Here , Q1 = 1000 cal, T1 = 627 + 273 = 900 K and T2 = 27+ 273 = 300K.
∴9003002
1000Q
=
or1000 300
9002Q 333.3cal.´
= =
The balance of heat is converted into work W.
W = Q1 - Q2 = 1000 - 333.3 = 666.7 cal
= 666.7 × 4.18 = 2.79 × 103 joule. [Q 1 cal = 418 joule]The efficiency of the Carnot engine is
Te
T= - 2
2
1
300= 1 0.67= 67%.
900- =
4. A Carnot engine takes in 100 cal of heat from the sourceat temperature 400 K and gives up 80 cal to the sink. Thetemperature of the sink and Thermal efficiency of the engineis as follows.
144 Thermal Physics
Solution : For a Carnot engine, we have
Here Q{ = 100 cal, Q2 = 80 cal, T1 = 400 K, T2 = ?
∴2
11
2
QT T
Q= ´
80 cal
= (400 K) × = 100 cal
320 K.
The efficiency is2 2
1 1
Q Te = 1 or e = 1 -
Q T
æ ö÷ç ÷ç ÷ç ÷÷çè ø
80
= 1 - = 0.2 = 100
20%
5. A reversible (Carnot) engine takes 200 k-cal of heat froma source at 527 ºC , converts a part of it into work and rejectsthe remaining to a sink at 27 °C. Its efficiency work done bythe engine per cycle in joule in kilo-watt-hour in electron-volt,Given : J = 4.2 x 103 joule/kilo-calorie and leV = 1.60 × 10-l9Jare calculated as follows.
Solution : If T1 and T2 be the Kelvin temperatures of sourceand sink, the efficiency of a reversible engine is
2
1
Te=1- .
T
Here T1 = 527 + 273 = 800 K and T2 = 27 + 273 = 300 K.
∴ 300 5
e = 1 - = = 0.625 = 62.5%.800 8
We know, that the efficiency of an engine is defined as
1
We = ,
Q
where W is the work done by t he engine per cycle andQ1 is the heat taken from the source per cycle.
145The Reversibility
Here Q1 = 200 kilocalorie. Therefore,
W= eQ1 = 0.625 × 200 = 125 kilocalorie
= 125 × (4.2 × 103) [Q1 kilocalorie
= 4.2 × 103 J]
= 5.25 × l05 joule.
Now, 1 kilowatt-hour = 1000 x 3600 = 3.60 x 106 joule.
\ 5
6
5.25 × 10W =
3.60 × 10 = 0.146 kilowatt-hour.
Again , 1 electron-volt = 1.6 × 10-19 joule.
\5
24-19
5.25 × 10W = 3.28×10 eV.
1.60 ×10=
6. An ideal engine operating between 227 °C and 27 °Cdevelops 74600 watt. The efficiency of the engine, the heattaken per second from the hot reservoir and the heat rejectedper second to the cold reservoir may be found out as follows.
2
1
Te=1- .
T
Solution : The efficiency of an ideal (Carnot) engineoperating between temperatures of T1 and T2 Kelvin is givenby
Here T1 = 227+ 273 = 500 K and T2 = 27 + 273 = 300 K.
\300
0 4500
e = 1 . 40%- = =
Now, if Q1 be the heat taken from the hot reservoir (source)and Q2 the heat rejected to the cold reservoir (sink), then theefficiency of the engine is given by
1 2
1 1
Q - Q We - = ,
Q Q ...(i)
146 Thermal Physics
where W is the useful work produced by the engine. Here therate of useful work produced (power) is 74600 watt. Thus
W = 74600 joule/sec.[Q 1 watt = 1 joule/sec]
Hence, by eq. (i), the heat taken from the source is
WQ .
e .= = = ´ 5
1
746001 865 10
0 4 joule/sec
51.865 × 10=
.4 18 = 4.46 × 104 cal/sec.
The heat rejected to the sink is
Q2 = Q1 - W
= 1.865 x 105 - 0.746 x 105 = 1.119 × 105 joule/sec
51.119 × 10=
.4 18= 2.68 × 104 cal/sec.
7. A reversible engine takes in heat from a reservoir of heatat 527°C and gives out heat to a sink at 127°C. Let’s find asto how many calories per second it must take from the reservoirin order to produce useful mechanical work at the rate of 750watts.
Solution : Let T1 and T2 be the kelvin temperatures ofthe reservoir and the sink respectively. If the working substancetakes Q1 heat from the reservoir, performs work W andrejects the rest to the sink, then the efficiency of the engine isgiven by
1
We = .
Q
Since the engine is reversible, we have
2
1
( )Te = 1 - .
T (+
= - =+
127 2731 0 5
527 273
147The Reversibility
\ eq. (i) gives
wattWQ
e .= =1
7500 5 = 1500 joule/sec.
Now 1 cal= 4.18 joule.
∴ Q.
=1
15004 18 = 359 cal/sec.
8. A Carnot engine works between 100° C and 0°C. If thework done per cycle is 1200 kg-m2/s2, then the heat (in calories)taken in from the source may be calculated. (J = 4 18 joule/cal)
Hint: Q1 = W/e.A heat engine operating between temperatures 137 °C and
- 43 °C receives 250 kcal of heat per minute. The inventor ofthe engine claims to develop 12 HP with this engine. Let’s findwhether his claim is agreeable or not. 1 HP = 746 watt.
Solution : The maximum work obtainable from the engineis the work obtainable from a ‘Carnot’ engine operating betweenthe same temperatures 137 °C and -43°C.
Thus, the maximum possible efficiency of the engine is
2
1
Te = 1 -
T
Here T1, = 137 + 273 = 410 K and T2 = - 43 + 273 = 230 K.
∴230
0 44410
e = 1 - .=
Now, if Q1 be the heat taken from the source and W theuseful work produced by the engine, then we have
1
We = .
Q
Here Q1 = 250 kcal/min .
\ W = e Q1 = 0.44 × 250 = 110 kcal/min.
148 Thermal Physics
Now , 1 kcal = 4180 joule and 1 min = 60 sec.
∴ 60(110 × 4180) joule
W = sec
= 7663 joule/sec = 7663 watt
7663=
746 = 10.3 H.P. [Q1 H.P. - 746 watt]
This is the maximum possible rate of work (power) thatcan be produced by the engine. Hence the claim of developing12 HP is wrong .
9. For a Carnot engine using an ideal gas, the adiabaticexpansion ratio is 5 and the value of y= 1.40. The efficiencyof the engine may be calculated here.
Solution : The efficiency of Carnot engine in terms ofadiabatic expansion ratio p , is given by
11
1eg
r
-æ ö÷ç ÷= - ç ÷ç ÷çè ø
Here p = 5 and g =1.40.
∴∴∴∴∴0.40
0.401 = 1 - = 1 - (0.2)
5e
æ ö÷ç ÷ç ÷çè ø ...(i)
Let us put x = (0.2)0.40
so thatlog x = 0.40 log 0.2
= 0.40 × (-0.6990)
= -0.2796 = 1 .7204.
∴∴∴∴∴ x = 0.52.
Making this substitution in eq. (i), we get
e = 1 - 0.52 = 0.48 = 48%
149The Reversibility
10. The thermal efficiency of a reversible engine workingbetween temperatures 0 °C and 100 °C may be calculated asfollows.
Solution : If Q1 be the heat taken in by the working substancefrom the source at Kelvin temperature T1, and Q2 be the heatgiven up to the sink at Kelvin temperature T2 , then the efficiencyof any engine is given by
2
1
Qe = 1 -
Q
For a Carnot (reversible) engine, 2 2
1 1
Q T
Q T=
∴∴∴∴∴2
1
1T
eT
= -
Here , T2 = 0 + 273 = 273 K and T1 = 100 + 273 = 373 K.
∴∴∴∴∴ 0 27 27273
e = 1 - . %373
= =
11. The efficiency of a reversible heat engine workingbetween the temperatures 72 ‘C and 187 °C may be calculatedas follows.
Solution : The efficiency of a “reversible” (Carnot) engineis maximum and given by
2
1
Te = 1 -
T
where T2 and T1 are absolute temperatures of the sink (lowertemp.) and the source (higher temp.) respectively.
Here, T2 = 72 + 273 = 345 K and T1 = 187 + 273 = 460 K.
∴∴∴∴∴345
1 0 25460
Q = . 25%= - =
12. One of the most efficient engines ever developed operatesbetween 2100 K and 700 K. Its actual efficiency is 40%.
150 Thermal Physics
Percentage of its maximum possible efficiency is this is asfollows :
Solution : The maximum possible efficiency of a heat engineoperating between Kelvin temperatures T1 and T2 is given by
2
1
Te = 1 -
T
Here T1 = 2100 K and T2 = 700 K.
∴700
e = 1 - = 0.67 = 67%2100
The actual efficiency is only 40%.
∴actual efficiency
percentage fraction = × 100maximum efficiency
0 40100
0 67.
60%.
= ´ =
13. Let two Carnot engines working between thetemperatures (a) 1000 K and 500 K, (b) x K and 1000 K areequally efficient compute x .
Solution. For 1000 K and 500 K , we have
2
1
T 500e=1- 1 - = 0.5
T 1000=
For x K and 1000 K, we have
1000e =1- 0.5
x= (given)
\ x = 2000K.
14. A, B, C, D are four baths. An engine AC (workingbetween A and C) has efficiency halfway between the efficienciesof AB and AD. Show that the absolute temperature of C ishalf-way of those of B and D. All engines are reversible.
151The Reversibility
Solution : The efficiencies of AC , AB and AD are C B
A A
T T1 - ,
T T,
and D
A
T1 -
T respectively. We are given that
C B D
A A A
T T T11 - = +
T 2 T T
ì üæ ö æ öï ï÷ ÷ï ïç ç÷ ÷- -ç çí ý÷ ÷ç ç÷ ÷÷ ÷ï ç ç ïè ø è øï ïî þ1 1
or 1 12
C B D
A A
T T TT T
+- = -
orC B D
A A
T T + T =
T 2T
or B DC
T + TT
2
15. The temperature of the sink of a Carnot engine is 27°C. If the efficiency of the engine is 40%, the temperature ofthe source may be found in following way.
Solution : If T1 and T2 be the Kelvin temperatures of sourceand sink respectively, the efficiency of the Carnot engine is
2
1
Te=1-
T
Here e = 40% = 0.40, T2 = 27 + 273 = 300 K.
∴ 21
T 300T =
1 - e 1 - 0.40=
= 500 K = 227 ºC.
16. The efficiency of a Carnot cycle is 1/6. On reducing thetemperature of the sink by 60°C , the efficiency increases to1/3. Let us find the initial and final temperatures betweenwhich the cycle is working.
Solution : Let T1 and T2 be the initial Kelvin temperaturesof the source and the sink respectively. Then , the efficiencyis given by
152 Thermal Physics
2
1
T 1e=1-
T 6= ...(i)
When T2 is decreased to T2 - 60 (1 ºC = 1 K in size), thenew efficiency is
2
1
T - 60 1e'=1-
T 3=
From eq. (i) and (ii), we have respectively
2
1
T 1 51 - =
T 6 6=
and2
1
T 1 2 = 1- =
T 3 3- 60
Dividing, we get
2
2
T 5 =
T - 60 4
Solving, we get
T2 = 300 K = 27 °C.
Putting the value of T2 (in Kelvin) in eq. (i), we get
1
300 1 - =
T 61
This gives
T1= 360 K = 87 °C.
The cycle is initially working between 87 °C and 27 °C.
Finally, the temperature of the sink is reduced by 60 °C,so that the cycle works between 87 ºC and - 33 °C.
17. A reversible engine converts one-sixth of heat absorbedat the source into work. When the temperature of the sink isreduced by 82 °C, the efficiency is doubled. The temperaturesof the source and the sink would be as follow.
153The Reversibility
Solution : Suppose the engine absorbs heat Q1 from thesource at Kelvin temperature T1 and rejects heat Q2 to the sinkat Kelvin temperature T2 . Then the heat converted into workis Ql — Q2 . It is given that
1 2
1
Q - Q 1 =
Q 6
2
1
Q 11 - =
Q 6
For a reversible engine, 2 2
1 1
Q T =
Q T
\2
1
T 11 - =
T 6
which is the efficiency of the engine.
When T2 is reduced to T2 - 82 (a change of 82 °C is sameas a change of 82 K), the efficiency is doubled, that is, it becomes1/3. Thus
2
1
T - 82 11 - =
T 3
On solving eq. (i) and (ii), we get
T1 = 492 K = 219 ºC
and T2 = 410 K = 137 ºC.
18. A Carnot engine has an efficiency of 50% when its sinktemperature is 17 °C. Let’s the change in its source temperaturefor making efficiency 60%.
Solution : Let T1 be the initial temperature of the source.The temperature of the sink is T2 = 17 °C = 290 K and theefficiency e = 50% = 0.5 . Now
2
1 1
T 290e = 1 - = 1 - = 0.5
T T
\ T1 = 580 K .
154 Thermal Physics
Now, let the temperature of the source be raised to(580 + D T) K , when the efficiency becomes 60% i.e. 0.6 . Thus
2901- = 0.6
580 + TD
\ D T = 145 K
The temperature of the source should be raised by 145 Kor 145 °C.
19. A Carnot engine whose low-temperature reservoir is at7 °C has an efficiency of 50%. It is desired to increase theefficiency to 70%. We may find as to by how many degreeshould the temperature of the high-temperature reservoir beraised .
Solution : Let T1 be the initial absolute temperature of thehigh-temperature reservoir (source). The temperature of thelow-temperature reservoir (sink) is T2 = 7 + 273 = 280 K, andthe efficiency e is 50% = 0.5 . Now
2
1 1
T 280e = 1 - = 1 - = 0.5
T T
\ T1 = 560 K.
Now, let the temperature T1 of the source be raised to
(560 + D T) K in order to increase the efficiency to 70% i.e. 0.7.Thus
2801 - = 0.7
560 + TD
Solving, we get D T = 373 K.
The temperature of the source should be raised by 373 Kor 373 °C.
20. Two Carnot engines A and B are operated in series. Thefirst one, A , receives heat at 900 K, and rejects to a reservoir
155The Reversibility
at temperature T K. The second engine , B , receives the heatrejected by the first engine and in turn rejects to a heat reservoirat 400 K. Let’s calculate the temperature T for the situationwhen (i) the work outputs of the two engines are equal , (ii)the efficiencies of the two engines are equal.
Solution. (i) Let the engine A take in heat Q1 at temperatureT1 and reject heat Q at temperature T ; and the engine B takein heat Q at temperature T and reject heat Q2 at temperatureT2 . Then , the works done by them are given by
WA = Q1 - Q
and WB = Q - Q2 .
But here WA = WB.
∴ Q1 - Q = Q - Q2
or1 2Q = Q
= 2Q ...(i)
As both the engines are Carnot’s we have1 1 2 2Q T Q T
= and = Q T Q T
or1 2Q Q T T
= Q T+ +2 2
...(ii)
From eq. (i) and (ii), we getT T
T=
=1 2 2
Here T1= 900 K and T2 = 400 K.
∴900 400
2T+ =
or T = 650 K.
(ii) If efficiencies are equal, then
2
1
TTe = 1 - = 1 -
T T
156 Thermal Physics
or2
1
TT =
T T
or T = (T1 × T2)½ = (900 × 400)½ = 600 K.
21. A Carnot’s engine working as a refrigerator between260 K and 400 K extracts 600 calories of heat from the reservoirat lower temperature. The amount of heat delivered to thereservoir at the higher temperature and the work that must besupplied to operate the refrigerator are calculated as follows.
Solution : For a Carnot’s refrigerator, we have
1Q TQ T
= 1
2 2
where Q2 is the heat taken in from a cold body at temperatureT2 and Q1 is the heat delivered to the hot body at temperatureT1.
Here, Q2 = 600 cal, T1, = 400 K and T2 = 260 K . Thus
1Q = 400600 260
or600 400
260Q 923 cal.
´= =
Thus the excess of heat delivered to the reservoir at highertemperature is
Q1 - Q2 = 923 - 600 = 323 cal.
This comes from the work W done by the motor runningthe refrigerator. Thus
W = 323 cal.
Now, 1 cal = 4.18 joule .
W = 323 x 4.18 = 1350 joule.
22. A refrigerator is to be maintained at —73 °C and theoutside air is at 27 ºC. The minimum amount of work that
157The Reversibility
must be supplied to remove 1000 joules of heat from inside therefrigerator may be calculated as follows.
Solution : The refrigerant extracts heat from refrigerator’sinside space (which act as the cold body), has a net amountof work done on it by the electric motor fitted in the refrigerator,and delivers a larger amount of heat to the outside (room) air(which acts as the hot body).
For computing minimum work to be supplied, we treat thegiven refrigerator as Carnot refrigerator. Suppose its refrigerantextracts heat Q2 from the cold body at absolute temperatureT2 , has a net supply of work W and delivers heat Q1 to thehot body at absolute temperature T1 . Then
Q1 = Q2 + W
or W = Q1- Q2 = Q2 1
2
Q - 1
Q
æ ö÷ç ÷ç ÷ç ÷÷çè ø
But 1
2
Q T =
Q T1
2 (for Carnot refrigerator).
Here Q2= 1000 joule , T2 = - 73 + 273 = 200 K and T1 = 27+ 273 = 300K.
\ W = 1000 joule æ ö- ÷ç ÷ç ÷çè ø300 200
200 = 500 joule.
23. A refrigerator is maintained at 0 °C and the outsidetemperature of the room is 27 °C. The minimum energy (work)to be supplied to freeze 1 kg of water already at 0 °C iscalculated below. The coefficient of performance may also bechecked heat of ice = 80 cal/gram and 1 cal = 418 joule.
Solution : For computing the “minimum” energy tobe supplied, the refrigerator must be regarded as “Carnot”refrigerator. The heat that must be removed from insideand discarded to the room in order to freeze one kilogram ofwater is
158 Thermal Physics
mL = 1000g × 80cal/g = 80 × 104cal.
For Carnot refrigerator, the energy (work) that must besupplied to remove an amount of heat Q2 from its freezingcompartment at temperature T2 K is given by
1 22
2
T - TW =Q
T
where T1 K is the temperature of the outside room air.
Here, Q2 = mL = 8.0 × 104 cal , T1 = 27 + 273 = 300 K andT2 = 0 + 273 = 273 K.
\ W = (8.0 × l04cal) -300 273
273
= 7-9xl03cal
= (7.9 x 103) × 4.18 = 33 x 103 joule.
The coefficient of performance is given by
42
3
8 0 107 9 10
Q . calK 10.
W .´= =
´
Alternatively;
2
1 2
273300 273
TK 10.
T T= =
- -
24. A refrigerator works between - 10 °C and + 27 °C, whilethe other between - 20 °C and + 17 °C, both removing 2000 jouleheat from the freezer. The calculation shows us which onenormally one chooses.
Solution : For a (Carnot) refrigerator, the energy W thatmust be supplied to remove heat Q2 from the freezer attemperature T2 K is
159The Reversibility
2
1 2
T 273K = = 10
T - T 300 - 273
where T1 is the temperature of the outer atmosphere.
For the first refrigerator:
T1 = 27 + 273 = 300 K, T2 = -10 + 273 = 263 K.
\ W = (2000 joule) × 300 - 263
263 = 281 joule . 263
For the second refrigerator :
T1 = 17 + 273 = 290 K, T2 = - 20 + 273 = 253 K .
\ W = (2000 joule) x 290 - 253
263 = 292 joule .
Since the first refrigerator runs at less energy input, henceit will be our choice.
25. A refrigerator is fitted with a motor of 200 watt. Thetemperature of its freezing compartment is - 3 °C and that ofthe room air is 27 °C. The maximum amount of heat which therefrigerator can remove from the freezing compartment in 10minutes is calculated as follows.
Solution : For an ideal (Carnot) refrigerator thework (energy) that must be supplied to remove an amount ofheat Q2 from the freezing compartment at temperature T2 K isgiven by
1 22
2
T - TW = Q
T
where T2 K is the temperature of outside air. Therefore, theheat removed is given by
160 Thermal Physics
22
1 2
TQ = W
T - T
Here T2 = - 3 + 273 = 270 K , T1 = 27 + 273 = 300 K and
W = 200 watt = 200 joule/sec
= 200 × 600 joule (in 10 minutes)
= 1.2 ×l05 joule.
Substituting these values in eq. (i), we get
Q2 = (1.2 × l05 joule) × 270
270-300
= 10.8 × 105 joule
= 510 8 10
4 18.
.´
= 2.58 × l05 cal.
26. A refrigerator is driven by a 1000-watt electric motor,which is operating at an efficiency of 60%. If the refrigeratorcan be treated as a reversible heat engine operating between0 °C and room temperature, which is 20 °C, the time requiredby it to freeze 100 kg of water, which is at 0 °C is calculated asbelow. Heat losses may be neglected. (Latent heat of fusion ofice is 80 cal/gm and j = 4.18 joule/cal.)
Solution : The work (energy) that must be supplied to therefrigerator to remove an amount of heat Q2 from the freezingcompartment is given by
1 22
2
T - TW = Q
T
where T1 and T2 are the Kelvin temperatures of the outer airand the freezing compartment respectively. Therefore, the heatremoved is
161The Reversibility
22
1 2
TQ = W .
T - T
Now, the refrigerator is being driven by a motor of 1000watt with 60% efficiency
∴ energy input W = 1000 x 60100
= 600 watt.
Now, T1 = 20 + 273 = 293 K and T2 = 0 + 273 = 273 K .
Therefore, the heat extracted from the freezing compart-ment is
2
273Q = 600 ×
293-273 = 8190 joule/sec
= 81904.18 = 1.96 × 103 joule/sec.
But the heat which is to be extracted from 100 kg of waterto freeze it is
mL = 100,000 × 80 = 8 × 106 cal.
Hence the time required to extract 8 × 106 cal of heat is6 10 cal
. cal / sec´=´ 3
81 96 10 = 4082 sec.
27. The ice in a cold storage melts at the rate of 2 kg perhour when the external temperature is 20 °C. The power of themotor driving the refrigerator which may just prevent the icefrom melting is found out in following way. (Latent heat offusion of ice = 80 cal/gm, J = 4 2 joule/cal).
Solution : The energy that must be supplied to therefrigerator to remove an amount of heat Q2 from the freezingcompartment is given by
21
22
T - TW = Q
T ...(i)
162 Thermal Physics
where T1 and T2 are the Kelvin temperatures of the outer airand the freezing compartment respectively.
Now, the ice melts at the rate of 2 kg/hour, or 2000/(60 x 60) = 5/9 gram/sec.
Therefore, the heat that must be removed to prevent the icefrom melting would be
2
gram5 cal 400 calQ = × 80 =
9 sec gram 9 sec
Also, T1 = 20 + 273 = 293 K and T2 = 0 + 273 = 273 K .
Substituting these values in eq. (i), we get
400 293 2739 273
W-= ´
= 3.256 cal/sec
= 3.256 × 4.2 = 13.7 joule/sec
= 13.7 watt.
163Physical Significance
8
Physical Significance
Entropy
If a substance takes in an amount of heat Q in a reversibleprocess at a constant temperature T, then Q/T is called the“increase in entropy” of the substance. Similarly, if the substancegives up an amount of heat Q at constant temperature T, thenQ/T is called the “decrease in entropy” of the substance. Thechange in entropy is denoted by Δ S. Thus
Qs
TΔ = .
If the temperature of the substance does not remain constantduring the process, we may consider the heat to be taken inor given up in successive small elements dQ such that thetemperature remains sensibly constant for each element. Thechange in entropy will then be
dQs
TΔ = ∫ .
164 Thermal Physics
Like energy U and temperature T, the entropy S of asystem is a physical property that can be measured in thelaboratory. It provides an alternative statement of the secondlaw of thermodynamics according to which only those processesare possible for a system in which the entropy of the system plussurroundings increases . Thus we can decide whether or not anevent will occur spontaneously. An event would occur providedit will cause the entropy of the universe (system + surroundings)to increase. Events causing decrease in entropy of the universeare impossible. For example, heat can flow from higher tolower temperature, but the reverse is not possible. Freeexpansion of a gas is possible, free compression is not. Therefore,on account of the processes occurring in nature, the entropyof the universe is continuously increasing. Entropy does notobey conservation law (energy does).
Change in Entropy in Reversible Cycle—Clausius Theorem :The Clausius theorem states that “in any reversible cycle thenet change in entropy is zero,” that is,
0dQT
=∫ .
Let us first prove it for a (reversible) Carnot cycle (Fig.).Let us start with working substance in the state A . During theisothermal expansion AB, the working substance takes in anamount of heat Q1 from the source at the constant temperatureT1 of the source. Its entropy therefore increases by Q1/T1 . Thisis also the decrease in the entropy of the source. During theadiabatic expansion BC no heat is taken in or given up, so theentropy remains unchanged. During the isothermalcompression CD , the working substance gives up an amountof heat Q2 to the sink at constant temperature T2. Its entropytherefore decreases by Q2/T2 . This is also the increase in theentropy of the sink. During the adiabatic compression DAthere is again no change in entropy. Thus, for the whole cycle,the net increase in the entropy of the working substance is
165Physical Significance
1 2
1 2
Q Qs
T TΔ = − .
This is also the net decrease in the entropy of the source-sink system.
But, by the definition of Kelvin’s scale of temperature,1 2
1 2
Q QT T
− . Therefore
1 2
1 2
0Q Q
sT T
Δ = − = .
Thus , the net change in the entropy of the workingsubstance, and also of the surroundings, during a completeCarnot cycle is zero.
If we regard the heat taken in by the substance as positiveand heat given up as negative, then Q1 will be positive and Q2
will be negative. The above equation then becomes
1 2
1 2
0Q Q
sT T
Δ = + = . ...(i)
This equation indicates that the sum of the quantities Q/T is zero for a Carnot cycle.
Let us now consider the general case of any reversiblecycle, indicated by the curve A → B → A (Fig.). We mayconsider this cycle to be made up of a large number of
166 Thermal Physics
elementary Carnot cycles abcd, efgh, ijkl, etc. Let us imagine thatthe substance, instead of tracing the smooth curve A → B → A, traces successively the cycles abcd, efgh, ijkl. .........In followingthese cycles the portions bh ,fl,jp, nt and so on , are traversedtwice in the reverse order, and their effects are thus can celledout. The net effect of the whole process is that the substancegoes along the closed zigzag path abcfi ... ghcda . Now, for eachelementary Carnot cycle the above relation (i) holds and maybe written in the form
1 2
1 2
0Q Q
sT Tδ δ
δ = + = .
Taking the sum of such results for all the cycles we concludethat for the closed zigzag path abefi, ......ghcda , we have
0Q
sTδ
Δ =Σ = .
If the adiabatics of the elementary Carnot cycles beinfinitesimally close together, they will be connected withinfinitesimally small isothermals. Then the zigzag path willcoincide with the smooth curve A → B → A , and we maywrite
0dQ
sT
Δ = =∫ .
167Physical Significance
The symbol ∫ denotes the integration over the whole
cycle. Thus , in any reversible cycle, the net change in entropy iszero . This is ‘Clausius theorem.’
Change in Entropy in Irreversible Cycle : The efficiency ofa Carnot reversible cycle working between absolutetemperatures T1 and T2 is given by
2
1
1Q
eQ
= − ,
where Q1 is the amount of heat taken in at temperature T1 andQ2 that given up at temperature T2. Since the cycle is reversible,
we have 2 2
1 1
Q TQ T
= . Thus
2 2
1 1
1 1Q T
eQ T
= − = −
This is the maximum possible efficiency of an engineworking between temperatures T1 and T2. If the cycle of theengine be irreversible, the efficiency will be lowered. Thus, inthis case
2 2
1 1
1 1Q T
eQ T
= − < −
or2 2
1 1
Q TQ T
>
or2 1
2 1
Q QT T
>
Now, during a complete irreversible cycle, the entropy ofthe source decreases by Q1/T1 , while that of the sink increasesby Q2/T2. The net change in the entropy of the working substancein this case also is zero because when the cycle is completed,the working substance recovers its initial state. Thus the total
168 Thermal Physics
increase in the entropy of the system (working substance) plusthe surroundings (source and sink) is
2 1
2 1
Q QS
T TΔ = − ,
which is positive since 2 1
2 1
Q QT T
> . Thus, in an irreversible cycle
the entropy of the system plus its surroundings always increases .
In general, we can write
SΔ (universe) > 0,
where the equality sign holds for reversible process andthe inequality sign for irreversible process. Here the worduniverse means system + surroundings.
Motion Involve a Change : In a purely mechanical motionthere is no exchange of heat and hence no change in entropy.
The mixing of two gases is an irreversible process. Hencethe entropy will increase i.e. the change in entropy will bepositive.
Entropy is a State Function : Let two points i and f (Fig.)represent any two states of a system, characterised by definitevalues of pressure, volume and temperature. Suppose thesystem can pass ifreversibly from the initial state i to thefinal state f, either by the path 1 or by the path 2. Now, if thesystem goes through the complete cycle i → f→ i, the netchange in entropy for the cycle as a whole is zero, that is
169Physical Significance
0dQT
=∫ (Clausius theorem)
Let us put it as a sum of entropy-changes in two parts, i→1 → f and f→ 2 → i. Then we write
210
f i
i f
dQ dQT T
+ =∫ ∫ .
Now, if the direction of traverse of the part f—¥ 2—ti bereversed, the entropy-change will be reversed in sign, sincethe heat-change will now take place in the reverse direction.That is
22
i f
f i
dQ dQT T
=−∫ ∫Substituting this in eq. (i), we get
21
f f
i i
dQ dQT T
=∫ ∫ .
The quantity f
i
dQT∫ is the change in entropy in passing
from state i to state f. Thus , the change in entropy betweenany two equilibrium states of a system is independent of thereversible path connecting these states. In other words, the entropyof a system is a definite function of its state and is independent ofthe way by which that state has been achieved. Thus entropy is a‘state variable’.
Formulation of the Second Law of Thermodynamics : LetSi and Sf be the entropies of a system in an initial state i anda final state f (measured from some arbitrary zero). Then , theentropy-change will be
f
if i
dQS S S
TΔ = − = ∫ .
This relation applies only for reversible path. If the twostates are infinitesimally close to each other, the above equationwill be written as
170 Thermal Physics
dQdS
T=
or dQ = TdS.
This equation is the mathematical representation of thesecond law of thermodynamics.
Entropy and the Second Law of Thermodynamics : Thesecond law of thermodynamics is concerned with the directionin which any physical or chemical process involving energy-change takes place. It has emerged out from experience, andcannot be proved theoretically.
The consideration of entropy-changes in reversible andirreversible processes enables us to express the second law interms of entropy in the following way :
All natural (or irreversible) processes that proceed fromone equilibrium state to another of a system takes place in adirection that causes the entropy of the system plussurroundings to increase in the limiting (ideal) case of areversible process the entropy of the system plus surroundingsremains constant , that is, SΔ (universe) ≥ 0. A process forwhich SΔ (universe) < 0 is impossible.
The above entropy statement of the second law is consistentwith both the Kelvin-Planck statement and the Clausiusstatement of the law.
According to Kelvin’s statement of second law, we cannothave a ‘perfect’ heat engine in which the working substancewould convert into work all the heat Q taken in from thesource at temperature T. If this were so, the entropy of thesource would decrease by Q/T, whereas that of the workingsubstance would remain unchanged (because it returns to itsinitial state after completing the cycle). Thus, there would bea net decrease in the total entropy of the system plussurroundings, which is against the entropy statement.
171Physical Significance
Similarly, according to Clausius statement of second law,we cannot have a ‘perfect’ refrigerator in which the workingsubstance would transfer heat Q from a cold body attemperature T2 to a hot body at temperature T1; without takingenergy from an external source. If this were so, the entropyof the cold body would decrease by Q/T2, that of the hot bodywould increase by Q/T2, and that of the working substancewould remain unchanged because it undergoes a cycle. Thus,
as T2< T1, there would be a net decrease of 2 1
Q QT T
⎛ ⎞−⎜ ⎟
⎝ ⎠ in the
total entropy of the system plus surroundings, which violatesthe entropy statement of the second law.
Thus, all three statements of the second law ofthermodynamics are equivalent.
Entropy Change of an ‘Isolated’ System
According to the entropy-statement of the second law ofthermodynamics, the total entropy of a thermodynamic systemplus surroundings, that is, the entropy of the universe eitherremains constant (in reversible processes ) or increases (inirreversible processes):
SΔ (universe) ≥ 0 .
Processes involving SΔ (universe) < 0 are impossible.
If, however, the system is completely isolated from itssurroundings {i.e. it can neither exchange heat nor work withits surroundings) then the entropy of the system alone is theentropy of the universe which should either remain constantor increase. Thus
SΔ (isolated system) ≥ 0 .
Processes involving SΔ (isolated system) < 0 are impossible,that is, the entropy of an isolated system can never decrease.
172 Thermal Physics
Cosmic Rays Falling on Earth
Some processes in nature appear to involve a decrease inentropy and thus violate the second law of thermodynamics.For example, it is said that the cosmic rays falling on earthdecrease the entropy of the earth. Actually, this is not a violationof the second law of thermodynamics. The reason is that theearth is not an isolated system.
Its entropy can decrease but then somewhere else in theuniverse there will be an increase in entropy. In the givenexample, the entropy of the earth decreases but at the sametime the entropy in some remote parts of the universe fromwhere the cosmic rays come increases.
Isentropic Processes: A reversible process during which theentropy of a system remains constant is called “isentropic process”.
A reversible adiabatic process is an isentropic process. Forsuch a process, we have
dQ = 0 (no exchange of heat)
so that dS =dQT
= 0
or S = constant.
Thus, the entropy of a system remains constant during a reversibleadiabatic process in the system , and the process is isentropic. Thisis why the adiabatics are also known as “isentropics”.
Isothermal and Isentropic Processes in T-S Diagram : Forthe heat transferred in a reversible process, we have the relation
dQ = TdS.
It therefore follows that the total heat transferred in areversible process carrying a system from an initial state i toa final state f is given by
Q =f
iTdS∫ .
173Physical Significance
This integral can be interpreted graphically as the areaunder a curve on T-S diagram, in which T is plotted along they-axis and S along the x-axis.
An isothermal process (T constant) would be a horizontalline (parallel to S-axis) on a T-S diagram.
Anisentropic (adiabatic) process (5 constant) would be avertical line (parallel to T-axis) on a T-S diagram.
Carnot Cycle on T-S Diagram
A Carnot cycle consists of two reversible isothermalprocesses and two reversible adiabatic processes (Fig. a). Henceit forms a rectangle on a T-S diagram, no matter what theworking substance is (Fig. b).
During the isothermal expansion AB at constanttemperature T1, the entropy of the working substance increasesfrom S1 to S2. During the adiabatic expansion BC , thetemperature falls to T2 , but the entropy remains constant.During the isothermal compression CD at constant temperatureT2 , the entropy decreases from S2 to S1. Finally, during theadiabatic compression DA , the temperature rises to T1, theentropy remaining constant.
174 Thermal Physics
Let Q1 be the amount of heat absorbed by the workingsubstance during isothermal expansion AB and Q2 that rejectedduring isothermal compression CD . By the relation dQ = TdS,we have
Q1 = T1 (S2 – S1) = AH × AB = area ABGH
and Q1 = T2 (S2 -S1) = DH × DC = area DCGH.
The difference Q1 – Q2 is converted into useful work andis thus the available energy per cycle . It is given by
Q1– Q2 = area ABGH – area DCGH = area ABCD .
Thus, the area of the cycle on T-S diagram represents availableenergy.
Practical Utility : The T-S diagrams are very useful for thestudy of engines. On account of simplicity of figure the areaof the cycle can be easily computed. Efficiency of Carnot Engine: The efficiency of the engine is defined as
1
We
Q= =
1 2
1
Q QQ−
area ABCDarea ABGH
=
= AB BCAB AH
××
= BCAH
= 1 2
1
T TT−
or 2
1
1T
eT
= −
This is the required expression.
175Physical Significance
In the above given figure the entropy of a system isincreasing at constant temperature . Hence the process isisothermal.
In Fig. the temperature of a system is increasing, the entropyremaining constant. Hence the process is isentropic (or adiabatic).
If Q1 be the heat absorbed during the process in whichentropy increases, and Q2 the heat given up during the processin which entropy decreases, then the efficiency is given by
1 2
1
Q Qe
Q−
= = 1
areaADCQ ,
where Q1 – Q2 is the available energy which is equal to thearea of the cycle on the T-S diagram.
In Fig.(a) and (b) the areas of the cycles are equal. However,in each cycle, Q1 is given by the area ABED , which is greaterfor the cycle (a). Hence the efficiency of cycle (b) is larger.
176 Thermal Physics
For the heat transferred in a reversible process, we havethe relation
dQ = TdS. ...(i)
If 1 mole of a gas is heated at constant volume (isochoricprocess) at temperature T, then
dQ = CvdT. ...(ii)
Equations (i) and (ii) yield
dT TdS Cv
= , ...(iii)
where dT/dS is the slope of the isochoric curve on T-S diagram.If the gas is heated at constant pressure (isobaric process),then
dQ = CpdT ...(iv)
which gives dT TdS Cp
= ...(v)
Thus the slopes of isochoric and isobaric curves on T-Sdiagram are T/Cv and T/Cp respectively. Since Cv < Cp , the slopeof the isochoric curve is greater. The ratio of the slopes is
( )( )
slope isochoricslope isobaric =
v
p
TpC
TC v
C
C= =ϒ
The Entropy of the Universe is increasing (Principle ofIncrease of Entropy) : Whenever the entropy of a systemchanges, the entropy of its surrounding bodies also changes.The sum of all these entropy-changes of all the bodies takingpart in the process is called the ‘entropy-change of the universe’.
According to the principle of increase of entropy, wheneverany natural, that is, irreversible process takes place, the entropy ofthe universe increases .
177Physical Significance
Let us consider the entropy-change of the universe whena system goes from an initial equilibrium state i to a finalequilibrium state f through a natural (irreversible) process.There is an entropy-difference Sf—Si between the twoequilibrium states, but we cannot calculate it from the relation
f
if i
dQS S
T− = ∫ , because this relation applies only to reversible
paths. Hence, instead of the actual irreversible path , we choosesome reversible path connecting the states i and f and calculatethe entropy-change for that path. We are justified in doing sobecause entropy is the characteristic of the state only, no matterhow the state has been achieved. Thus
f
if i
dQS S
T− = ∫ (over any reversible path).
Irreversible processes are of various types. As an example,let us consider heat conduction.
Heat Conduction (Equalisation of Temperature) : Let a hotbody A at temperature T1 be placed in contact with a cold bodyB at temperature T2 in a box which is thermally insulated fromthe surroundings. An amount of heat Q (say) flows irreversiblyfrom the hot to the cold body so that both reach a commonintermediate temperature Tm (say).
To calculate the entropy-change of the system, we have tothink of some reversible process by which heat Q could beconducted from the hot (T1) to the cold (T2) body so that bothreach the temperature Tm. This is possible if we have a heat-reservoir whose temperature can always be changed andadjusted to any value. We adjust its temperature to T1 and putthe hot body A in contact with it.
Then we lower the reservoir temperature infinitesimallyslowly to Tm , so that heat (Q) from the hot body (A) flowsreversibly to the reservoir until the temperature of the body Afalls to Tm. (The flow is reversible because at every stage the
178 Thermal Physics
temperature of the reservoir is only infinitesimally lowerthan that of the hot body.) The loss in the entropy of the bodyA is
11m
QS
TΔ =− ,
where T1m is the average of T1 and Tm .
Now, we adjust the reservoir temperature to T2 and putthe cold body B in contact with it. Then we raise the reservoirtemperature infinitesimally slowly to Tm , so that heat (Q) fromthe reservoir flows reversibly to the cold body B until thetemperature of the body B rises to Tm. The gain in the entropyof the body B is
22m
QS
TΔ =+
where T2m is the average of T2 and Tm .
Both bodies are now at the same temperature Tm . Thechange in entropy for the system is thus
1 2S S SΔ = Δ + Δ
= – 1 2m m
Q QT T
+ = – 2 1m m
Q QT T
− ,
which is positive (since T1m > T2m). This is the increase inentropy of the system due to the reversible process selected.The same would be the increase in entropy of the system dueto the actual irreversible heat conduction.
Now, in the actual heat conduction the entropy-change ofthe surroundings is zero because the system is kept thermallyinsulated. Hence the entropy-change of the universe is
SΔ (universe) = 2 1m m
Q QT T
⎛ ⎞−⎜ ⎟
⎝ ⎠ =
2 1
Q QT T
⎛ ⎞−⎜ ⎟
⎝ ⎠,
179Physical Significance
which is positive (since T1m > T2m). Thus the entropy of theuniverse increases in the actual irreversible conduction of heat,as required by the principle of increase of entropy.
All natural processes, like conduction and radiation, areirreversible. These processes tend to equalize temperaturesbetween various bodies. It, therefore, follows that the entropyof the universe is continually increasing.
The Available Energy of the Universe is Decreasing : Heatenergy is available for conversion into work only when it is let downfrom a higher to a lower temperature. For example, in a perfectlyreversible Carnot engine, a quantity of heat Q1 is taken in froma hot body at temperature T1, a quantity Q2 is given up to acold body at temperature T2 , the balance Q1– Q2 being convertedinto work and is called the ‘available energy’. Thus
available energy = Q1 – Q2
= 2
11
1Q
⎛ ⎞= −⎜ ⎟⎝ ⎠
= 2
11
1T
QT
⎛ ⎞= −⎜ ⎟⎝ ⎠
.2 2
1 1
Q TQ T
⎡ ⎤=⎢ ⎥
⎣ ⎦Q
Clearly, the lower the temperature T2 of the cold body, thegreater will be the available energy. If T0 is the temperatureof the coldest body available, then
maximum available energy = 0
11
1T
QT
⎛ ⎞= −⎜ ⎟⎝ ⎠
.
Let us now consider the (irreversible) conduction of heatQ from a hot body at temperature T1 to a cold body attemperature T2. Let T0 be the lowest available temperature.Before conduction, the heat Q was available in a body at ahigher temperature T1. Hence the maximum amount of energy
available from this was 0
1
1T
QT
⎛ ⎞= −⎜ ⎟⎝ ⎠
. But, after conduction, the
180 Thermal Physics
heat Q is available at the lower temperature T2. Hence the
maximum available energy has reduced to 0
2
1T
QT
⎛ ⎞= −⎜ ⎟⎝ ⎠
. Thus
the energy that has become unavailable for work
= 0
1
1T
QT
⎛ ⎞= −⎜ ⎟⎝ ⎠
– 0
2
1T
QT
⎛ ⎞= −⎜ ⎟⎝ ⎠
= 0 0
2 1
T TQ
T T⎛ ⎞
= −⎜ ⎟⎝ ⎠
= 02 1
Q QT
T T⎛ ⎞
= −⎜ ⎟⎝ ⎠
= T0 ( SΔ ).
where 2 1
Q QS
T T⎛ ⎞
Δ = −⎜ ⎟⎝ ⎠
is the increase in entropy of the universe
brought about by the (irreversible) conduction of heat. Hence,we conclude that the energy that becomes unavailable for workduring an irreversible process is To times the increase in entropy ofthe universe due to that irreversible process .
Since irreversible processes are continually going on innature, energy is continually becoming unavailable for worki.e. the available energy of the universe is continually decreasing.This is known as the ‘degradation of energy’.
Heat energy is available for work only when there is adifference of temperature, so that heat can flow. In A there isno “available energy”. Hence the entropy of A is maximum (Asentropy increases, the available energy decreases).
Change in Entropy during Free Expansion : Let μ moles
181Physical Significance
of an ideal gas be contained in a vessel which is connected toanother ‘evacuated’ vessel by means of a stop-cock, both beingthermally insulated from the surroundings. As the stop-cockis suddenly opened, the gas rushes into the vacuum of thesecond vessel. This is the irreversible free expansion of the gas.As the system is thermally insulated, there is no exchange ofheat with the surroundings (Q = 0). Also no work is doneagainst the vacuum (W = 0). Hence, from first law UΔ = Q– W, we have
UΔ = 0.
The internal energy of the gas remains unchanged. Becausethe gas is ideal, its temperature T remains constant.
Suppose we wish to calculate the entropy-change of thesystem due to the irreversible free expansion from an initialvolume Vi to a final volume Vf. We can do so by consideringsome reversible process between the same initial and finalstates of the system. Because in the free expansion of the gasthe temperature remains constant, a convenient reversibleprocess, which can be substituted for the irreversible freeexpansion, is the isothermal expansion of the gas from theinitial volume Vi to the final volume Vf. (This we can do bybringing the gas in contact of a heat-source at temperature Twhich supplies heat to the gas). The entropy-change is
SΔ = Sf – Si = f
i
V
V
dQT∫ . ...(i)
Now, by the first law of thermodynamics, we have
dQ = dU + dW.
But dU = 0 (as the gas is ideal, there is no change in internalenergy in isothermal expansion).
∴ dQ = dW = pdV
= μRTdVV
(∴pV = μRT)
182 Thermal Physics
dQT
= μRdVV
.
Substituting this value of dQT
in eq. (i), we have
orf
i
V
V
dVS R
VΔ = μ ∫ = μR [log ] f
i
Ve VV
log fe
i
VS R
VΔ = μ
(reversible isothermal expansion)
which is positive, since Vf > Vi. This is the increase in theentropy of the ‘system’ (ideal gas) undergoing reversibleisothermal expansion.
Note that this is the increase in the entropy of the system(gas) alone due to the reversible isothermal expansion. Theentropy of the universe , however, remains unchanged duringthe process because there is an equal decrease in the entropyof the heat-source.
The above will also be the increase in entropy of the gasin the actual irreversible free expansion from volume Vi to Vf.
Now, in free expansion, the entropy-change of thesurroundings is zero because there is no exchange of heat withthe surroundings. Hence the increase in entropy of the universedue to the free expansion is
( ) log fe
i
VS universe R
VΔ = μ (free expansion).
Thus we conclude that the entropy of the universe increasesduring free expansion (an irreversible process).
Reversible Isothermal Compression
When 1 mole of an ideal gas undergoes reversible isothermalcompression from volume Vj to Vf, the change in entropy ofthe gas is
183Physical Significance
log fe
i
VS R
VΔ = .
Since Vf< Vi (compression), SΔ is negative which meansthat the entropy of the gas decreases .
There is also an equal increase in the entropy of the heat-reservoir in whose contact the gas has been put to carry outthe isothermal compression. Thus, the total entropy of theuniverse (gas plus surroundings) remains unchanged, acharacteristic of the reversible process.
Reversible Adiabatic Expansion : In adiabatic expansion(or compression) the gas does not exchange heat with thesurroundings. Hence there is no change in entropy.
Free Expansion: When 1 mole of an ideal gas undergoesfree expansion from volume Vi to Vf, the change in the entropyof the universe is
SΔ (universe) = log fe
i
VR
V .
Since Vf > Vi (expansion), SΔ is positive which means thatthe entropy of the universe increases, a characteristic ofirreversible process.
In free expansion, the increase in the entropy of the universeis same as that of the gas alone because in this case the entropyof the surroundings does not change.
Entropy-change on Heating a Substance : Suppose a liquidof mass m and specific heat c is heated from T1 to T2. We canconnect the initial and final states by a reversible process ofheating, making use of an infinite number of heat-reservoirswith temperature ranging from T1 to T2 . If dQ be the amountof heat taken in by the liquid for an infinitesimally smalltemperature-rise dT, then we have
dQ = mc dT.
184 Thermal Physics
The corresponding entropy-change of the liquid is
dS =dQT
= mc dTT
.
The entropy- change in the entire process is therefore
SΔ (liquid) =2
1
T
TdS∫ = mc
2
1
T
T
dTT∫
= mc 2
1
Te T[log T] = mc loge
2
1
TT ,
which is positive (since T2 > T1). Thus the entropy of the liquidincreases . The same would be the increase in the actualirreversible heating.
The above is the increase in the entropy of the system(liquid) alone due to the reversible heating. The entropy of theuniverse (system + surroundings) remains, however,unchanged because for every dQ absorbed by the liquid (system)at T, there is an equal amount of heat given up by the reservoir(surroundings) at T, so that there is an equal decrease in theentropy of the surroundings.
In the actual irreversible heating the liquid is placed incontact with a single reservoir. In this case the decrease inentropy of the reservoir (surroundings) is less than the increasein entropy of the liquid (system). Thus in the actual irreversibleprocess the entropy of the universe increases.
Entropy-change on Cooling : When a substance is cooledfrom temperature T1 to T2 , its entropy decreases, because
SΔ = me loge 2
1
TT ,
and T2 < T1 (cooling), so that SΔ is negative.
The total entropy of the universe, however, can neverdecrease ; if always increases .
Entropy-change in Melting of Ice : Suppose ice, of mass m
185Physical Significance
and latent heat L, melts at a (constant) Kelvin temperature T.It takes an amount of heat Q = mL . We can connect the initialand final states by a reversible process of melting, by bringingice in contact of a heat-reservoir at a temperature onlyinfinitesimally higher than T. Since the temperatureremains constant, the entropy-change in the entire process ofmelting is
SΔ =dQT∫ =
QT
= mLT
,
which is positive. This is the increase in entropy of the system(ice) alone . There is an equal decrease in entropy of thesurroundings (heat-reservoir), so that in reversible melting theentropy-change of the universe is zero. In the actual irreversiblemelting, however, the entropy of the universe increases.
Entropy-change when Ice is converted into Steam : Let T1
be the Kelvin temperature at which ice melts into water, andT2 the Kelvin temperature at which water is boiled to steam.Then, the entropy-change when ice is converted into water is
1SΔ =1
1
mLT ,
where L1 is the latent heat of fusion of ice. The entropy-changewhen water is heated from T1 to T2 is
2SΔ = mc loge 2
1
TT ,
where c is the specific heat of water. The entropy-change whenwater is converted into steam is
3SΔ =2
2
mLT ,
where L2 is the latent heat of vaporisation of water. Thus, thetotal entropy-change is
SΔ = 1SΔ + 2SΔ + 3SΔ
186 Thermal Physics
=1
1
mLT + mc loge
2
1
TT , +
2
2
mLT .
This is the entropy-change of the system (ice) alone.
Entropy-change on Mixing Liquids at DifferentTemperatures: Let c be the specific heat of the liquid. On mixingequal masses m of the same liquid at temperatures T1 and T2(T1
> T2), heat flows from the hotter to the colder liquid. Let T bethe equilibrium temperature of the mixture. By the principleof calorimetry, we have
mc (T1 - T) = mc (T - T2)
∴ T = 1 2
2T T+
The hotter liquid gives up heat in cooling from T1 to T.Replacing the actual irreversible process by a reversible processinvolving an infinite number of heat-reservoirs withtemperatures ranging from T1 to T, the entropy-change is givenby
1SΔ =1
T
T
dQT∫ = mc
1
T
T
dTT∫ = mc loge
1
TT .
Similarly, the entropy-change in the colder liquid whichtakes in heat in rising from T2 to T is given by
2SΔ =2
T
T
dQT∫ = mc
2
T
T
dTT∫ = mc loge
2
TT .
The entropy-change of the system is
SΔ = 1SΔ + 2SΔ
= mc e e1 2
log + logT TT T
⎡ ⎤⎢ ⎥⎣ ⎦
= mc 2
e1 2
logT
T T [∴logA + logB = logAB]
187Physical Significance
= mc loge
2
1 2
TT T
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= 2mc loge 1 2
TT T
= 2mc loge
1 2
1 2
( ) 2T T
T T
+
This is the required expression.
We know that the arithmetic mean of two numbers is
always greater than their geometric mean. Thus 1 2
2T T+
> 1 2T T
. Hence SΔ is necessarily positive which means that the entropyof the system increases in the irreversible process of mixing.
Since the system is thermally insulated, there is no entropy-change in the surroundings, and SΔ , as calculated above ,represents the entropy-change of the universe. We concludethat the entropy of the universe increases .
Entropy-change in Irreversible Heat Transfer
Let T1 > T2. Then heat will flow from the body m1 to thebody m2 until both reach at an equilibrium temperature T.
By the principle of calorimetry, we have
m1c1(T1 - T) = m2c2(T - T2).
∴ T =1 1 1 2 2 2
1 1 2 2
m c T m c Tm c m c
++ . (T1 > T > T2)
Replacing the actual irreversible process by two reversibleprocesses involving infinite succession of reservoirs, we have
1SΔ = m1c1 1
T
T
dTT∫
= m1c1 loge 1
TT < 0 (cooling from T1 to T)
188 Thermal Physics
and 2SΔ = m2c2 2
T
T
dTT∫ = m2c2loge
2
TT > 0
(heating from T2 to T).
:. total change in the entropy of the system is
SΔ = 1SΔ + 2SΔ
=1 1 2 2
1 2
log logm c m c
e e
T TT T
⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
=1 1 2 2
1 2
logm c m c
e
T TT T
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
> 0.
The entropy of the system increases.
Entropy of a Perfect Gas
Let us consider one mole of a perfect gas at pressure p,Kelvin temperature T and volume V. When an infinitesimalamount of heat dQ is added to it, the increase in entropy is
dS =dQT
.
If dU be the increase in the internal energy of the gas anddW the external work done, then by the first law ofthermodynamics, we have
dQ = dU + dW.
If Cv be the molar specific heat of the gas at constantvolume, dT the rise in temperature and dV the change involume, then
dU = CvdT
and dW = pdV = RT dVV
. [∴ pV = RT]
:. dQ = Cv dT + RTdVV
189Physical Significance
and so dS =dQT
= Cv dTT
+ RdVV
...(i)
The entropy of the gas, measured from an arbitrary zero,is then given by
S =dQT∫ =
dTCv
T∫ + dV
RV∫ . ...(ii)
From gas eq. pV = RT, we have dT = pdV Vdp
R+
and from
Mayer’s relation, R = Cp – Cv. Thus, from eq. (ii),
S = ( )v p v
pdV Vdp dVC C C
RT V+
+ −∫ ∫Again, RT = pV, so that
S = ( )v p v
pdV Vdp dVC C C
pV V+
+ −∫ ∫
= ( )v v p v
dpdV dVC C C C
V p V+ + −∫ ∫ ∫
= v p
dp dVC C
p V+∫ ∫
or S = CV logep + Cp loge V + constant....(iii)
This is the required expression.
If a gas passes from an initial state pi, Vi, Tf, to a final statePf,Vf, Tf then, from eq. (iii), the change in entropy of the gasis
SΔ =f f
i i
P V
v pP V
dp dVC C
p V+∫ ∫
or SΔ = Cv loge f
i
P
P + Cp loge f
i
V
V .
This is the entropy-change in terms of pressures, volumesand molar specific heats.
190 Thermal Physics
Entropy-change in a Perfect Gas : Proceeding as in lastquestion, we reach the following equation (i):
so that ds =dQT
= vC dT dVR
T V+ .
If the gas passes from an initial state i having Pi, Vi, andT1, as its pressure, volume and temperature respectively, to afinal state f with Pf,Vf, Tf as the corresponding quantities, thenthe change in entropy is
SΔ =f
idS∫
fv
i
C dT dVR
T V⎛ ⎞+⎜ ⎟⎝ ⎠∫
= Cv
f f
i i
dT dVR
T V+∫ ∫
or SΔ = Cv loge f
i
T
T + R loge
f
i
V
V . ...(i)
This is the expression for the entropy-change in terms oftemperatures, volumes and specific heat at constant volume.
Now,i i
i
p VT =
f f
f
p V
T or f
i
T
T =f f
i i
p V
p V and R = Cp – Cv.
∴ SΔ = Cv log log ( )logf f f
e e p v ei i i
p V VC C
p V V
⎛ ⎞+ + −⎜ ⎟
⎝ ⎠
or SΔ = Cv log logf fe p e
i i
p VC
p V+ ...(ii)
This is the expression for the entropy- change in terms ofpressures, volumes and specific heats.
Again, f
i
V
V = f f
i i
p T
p T , and so eq. (i) can be written as
SΔ = log log logf fiv e e e
i if
T TpC R
T p T
⎛ ⎞+ +⎜ ⎟⎜ ⎟
⎝ ⎠
191Physical Significance
log log logf f fv e e e
i i i
T p TC R R
T p T− +
= (Cp – R) log log logf f fe e e
i i i
T p TR R
T p T− +
SΔ = log logf fp e e
i i
T pC R
T p− . ...(iii)
This is the expression for the entropy-change in terms oftemperatures, pressures and specific heat at constant pressure.
Entropy- change in a Van der waals’ Gas : The equationof state for one mole of a Van der waals’ gas is
2
ap
V⎛ ⎞+⎜ ⎟⎝ ⎠
(V-b) = RT,
where a and b are Van der waals’ constants. When aninfinitesimal amount of heat dQ is added to the gas at Kelvintemperature T, the increase in entropy is
dS =dQT
.
By the first law of thermodynamics, we have
dQ = dU + dW.
If Cv be the molar specific heat at constant volume, thenthe increase in internal energy is
dU = CvdT
and the work done including internal work against themolecular attraction, is
dW = 2
ap
V⎛ ⎞+⎜ ⎟⎝ ⎠
dV = RT
dVV b− .
∴ dS =dU dW
T+
= v
dT dVC R
T V b+
−
192 Thermal Physics
The change in entropy when the gas passes from the state(Vi, Ti) to (Vf, Tf) is
sf – si =f
idS∫ =
f f
v i i
dT dVC R
T V b+
−∫ ∫
= log logf fv e e
i i
T V bC R
T V b
−+
− .
Let a small amount of heat dQ be given to 1 gm of waterat absolute temperature T. The change in entropy is
dS =dQT
.
If the temperature of the water rises by dT, then dQ = dT,the specific heat being unity.
∴ dS =dTT
.
Integrating, we get
S =dTT∫ = loge T + A,
where A is the constant of integration. If we take S = 0 at0 °C (T= 273 K), then
A = – loge 273.
Hence, for 1 gm of water, we have
Swater = loge T - loge 273 = loge 273T
.
Now, the amount of heat needed to convert 1 gm of waterinto steam at a temperature T is L, where L is the latent heat.Hence the increase in entropy will be L/T. Thus the total entropyof 1 gm of dry saturated steam at a temperature T will be
Ssteam = Swater + LT
= loge 273T
+ LT
.
193Physical Significance
Nernst Heat Theorem : Third Law of Thermodynamics :The second law of thermodynamics defines only difference inentropy between two equilibrium states connected by areversible path. Such a path , however, exists if both the stateslie on the same sheet of a p - V - T surface. If we consider twodifferent substances, or metastable phases of the same substance,such a reversible path may not exist. Therefore, the second lawdoes not uniquely determine even the difference in entropy.Nernst, in 1905, supplied a rule for unique determination ofentropy. It states :
The entropy of any system at absolute zero is a universal constant,which may be taken to be zero.
This statement is known as ‘Nernst heat theorem’ or ‘thirdlaw of thermodynamics’ ; it means that 5 = 0 at T = 0, whateverbe the values of any other parameters on which entropy S maydepend. Thus, the third law renders the entropy of any stateof any system unique.
Two important consequences of the third law are : (i) Heatcapacities of a system vanish at absolute zero, (ii) Coefficientof volume expansion of any substance vanishes at absolutezero.
An alternative statement of the third law is theunattainability of absolute zero. A fundamental feature of allcooling processes is that the lower the temperature achieved,the harder it is to go still lower. It can be generalised in thefollowing statement:
It is impossible by any procedure, no matter how idealised, toreduce any system to the absolute zero of temperature in a finitenumber of operations.
This statement implies that the entropy of a system cannotbe reduced to zero.
194 Thermal Physics
PROBLEMS
1. One end of a copper rod is in contact with a heatreservoir at 127 °C and the other with a reservoir at 27°C.During a certain time-interval, 1200 cal of heat is conductedthrough the rod. We can look into the total entropy-changeand whether the entropy of the rod change.
Solution : The hot reservoir at a constant temperature of127°C (= 400 K) gives up 1200 cal of heat to the cooler reservoirat a constant temperature of 27°C (= 300 K). The entropy-changes of the reservoirs are
SΔ (hot) =QT
= 1200400−
= – 3.0 cal/K.
(The heat given up is considered as negative.)
and SΔ (cold) =QT
= 1200300 = + 4.0 cal/K.
∴ SΔ total = – 3.0 + 4.0 = 1.0 cal/K.
The copper rod is in the steady state. It is neither absorbingnor giving up any heat. Hence there is no change in the entropyof the rod.
2. One mole of an ideal gas expands isothermally andreversibly to twice its original volume. Calculate the changein entropy of the gas, of the universe. We may see as whathappens if the gas had undergone free expansion.
Solution : The change in entropy of the gas in expandingfrom Vi to Vf (=2 Vi) is given by
SΔ = μR loge
f
i
V
V
= 1 mole × 8.31 joule/(mole-K) × loge 2
= 1 × 8.31 × 0.693 = + 5.76 joule/K.
195Physical Significance
This is an entropy increase. There is also an equal decreasein the entropy of the heat source which supplies heat to thegas for isothermal expansion. Thus, the entropy of the universe(gas-plus surroundings) remains unchanged.
If the gas undergoes free expansion, then the entropy ofthe surroundings does not change because the expanding gasis thermally isolated from its surroundings. Now the entropyof the universe increases by 5.76 joule/K .
3. Three grams of nitrogen expand isothermally to twicethe original volume. The change in entropy many be calculated.The molecular weight of nitrogen is 28. Given : R = 8.31joule/(mole-K), / = 4.18 joule/calorie and loge 2 = 0.693.
Solution : 3 gram nitrogen contains 3/28 mole (μ = 3/28).Now
SΔ = μR loge (Vf/Vi)
=328 mole × 8.31 joule/(mole-K) × loge 2
=328 × 8.31 × 0.693 = 0.617 joule/K
=0.6174.18 = 0.148 cal/K.
4. A 2-litre box is divided into two equal parts by a centralbarrier. One part is filled with hydrogen and the other withnitrogen at NTP. The barrier is removed and the gases mix. Letus find into volume of entropy increase due to mixing.
Solution : At NTP 1 mole of gas occupies 22.4 litre. Hence1 litre (each part of the box) contains (1/22.4) mole of gas.
Now, the entropy-increase of each gas is
SΔ = μR loge
f
i
V
V = 22.4
Rloge2.
196 Thermal Physics
The total entropy-increase is twice this , i.e.,
SΔ =2
22.4R
loge2 = 0.0619 R. [ Q loge2 = 0.693]
Now R = 8.31 joule/(mole -K) = 1.99 cal/(mole-K).
[Q 1 cal = 4.18 joule]
∴ SΔ = 0.0619 × 1.99 = 0.123cal/K.
5. An electric current of 10 amp is maintained for 1 sec ina resistor of 25 ohm while the temperature of the resistor iskept constant at 27°C. The entropy-change of the resistor andthe entropy-change of the universe are calculated as follows.
Solution : The heat dissipated in the resistor is
Q = i2 Rt = (10)2 × 25 × 1 = 2500 joule .
Since the resistor is maintained at 27°C (= 300 K), the heatQ is not absorbed by the resistor, but taken away by thesurroundings. Hence the entropy-change of the resistor is zero.
The entropy-change of the surroundings is
O 2500 SΔ (surroundings) = QT
= 2500300 = 8.33 joule/K.
This is also the entropy-change of the universe.
6. The same current is maintained for the same time in thesame resistor, but now thermally insulated. The initialtemperature of the resistor is 27°C. If the resistor has a massof 10 g and specific heat 0 84 joule/g-K, then entropy-changeof the resistor, and also of the universe many be calculated asfollows.
Solution : The thermally-insulated resistor now absorbsthe heat Q (= 2500 joule) dissipated in it and its temperaturerises. If Tf is its final temperature, then
Q = mc TΔ
197Physical Significance
or 2500 joule = 10 g × 0.84 joule/(g-K) × (Tf – 300)
∴ Tf – 300 = 298 K
or Tf = 598 K.
We know that when the temperature of a substance ofmass m and constant specific heat c changes from Ti to Tf, theentropy-change is
SΔ =f
i
T
T
dQT∫ =
f
i
T
T
mcdTT∫
= [ ]log f
i
T
e Tmc T = log f
ei
Tmc
T
= 2.3026 mc log10 f
i
T
T .
Hence the entropy-change of the resistor is
SΔ (resistor) = 2.3026 × 10 g × 0.84 Joule/(g-K) × log10 598300
= 2.3026 × 10 × 0.84 × log10 2
= 2.3026 × 10 × 0.84 × 0.3010
= 5.8 joule/K.
Since the resistor is thermally insulated, there is nocommunication of heat between it and the surroundings, sothat the entropy-change of the surroundings is zero. Hence
SΔ (universe) = 5.8 joule/K.
7. 1 kg of water at 273 K is brought into contact with aheat-reservoir at 373 K. When the water has reached 373 K,let’s see what would be the entropy-change of the water of theheat reservoir and of the universe.
Solution : The water (mass m = 1000 gm, specific heatc = 1 cal/gm-K) takes in heat from the reservoir and itstemperature rises from an initial value Ti = 273 K to a finalvalue Tf = 373 K. The change in its entropy is
198 Thermal Physics
SΔ (water) = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 1000 × 1 × log10 (373/273)
= 2.3026 × 1000 × 1 × 0.1354 = + 312 cal/K.
The water has been placed in contact with a single reservoirat 373 K which forms the surroundings. In this case heat flowsfrom the reservoir to water irreversibly. The quantity of heatflow is the heat required to raise the temperature of water from273 K to 373 K. Thus.
Q = mc (Tf – Ti)
= 1000 × 1 × (373 - 273) = 105 cal.
The reservoir gives up this heat at a temperature of 373 K.Hence the change in its entropy is
SΔ (reservoir) =QT
= 510
373−
= – 268 cal/K .
The entropy of the reservoir thus decreases.
The entropy-change of the universe is
SΔ (universe) = SΔ (water) + SΔ (reservoir)
= 312 cal/K – 268 cal/K
= + 44 cal/K.
This is the increase in the entropy of the universe.
8. If the water had been heated from 273 to 373 K by firstbringing it in contact with a reservoir at 323 K and then witha reservoir at 373 K, let us see the entropy-change of theuniverse.
Solution : The water undergoes the same rise of temperature(although in two steps). Hence its net entropy-change will besame as before i.e.
SΔ (water) = + 312 cal/K.
199Physical Significance
Now, the first reservoir heats the water from 273 to 323K so that the heat given up by this reservoir is
Q1 = mc (Tf – Ti)
= 1000 × 1 × (323 - 273) = 50,000 cal.
This heat is given up at a temperature of 323 K.
The second reservoir heats the water form 323 to 373 K sothat the heat given up by this reservoir is
Q2 = 1000 × 1 × (373 - 323) = 50,000 cal,
but this heat is given up at a temperature of 373 K. Hencethe total entropy-change of both the reservoirs is
SΔ (reservoirs) = – 50, 000 50,000
323 373−
= – 155 – 134 = – 289 cal/K.
The entropy-change of the universe in this case is therefore
SΔ (universe) = 312 – 289 = + 23 cal/K.
Thus, in this case the change (increase) in the entropy ofthe universe is less than before.
9. The water being heated from 273 to 373 K with almostno change in entropy of the universe is explained as below.
Solution : It is clear from above that larger is the numberof reservoirs ranging from 273 to 373 K used for heating thewater, lesser will be the change in the entropy of the universe.
10. When a body of mass 5 g is heated from 100 K to 300K the change in entropy is calculated as below. The specificheat of the body is 0.1 cal g–1 deg–1.
Solution : The change in entropy of a body (mass m, specificheat c) heated from Kelvin temperature Ti to Tf is given by
SΔ = 2.3026 mc log10 (Tf/Ti),
200 Thermal Physics
= 2.3026 × 5 g × 0.1 cal g–1 deg–1 × log10 (300/100)
= 2.3026 × 5 × 0.1 × 0.4771
= 0.55 cal/°C .
The entropy increases.
11. 10 gram of oxygen is heated from 50 °C to 150 °C atconstant volume. The change in entropy is found as follows.Given : Cv = 5 cal/(mole-K).
Solution : In usual notations :
SΔ = 2.3026 mc log10 (Tf/Ti).
Here m = 10g = 1032 mole (molecular weight of oxygen is 32),
c = Cv= 5 cal/(mole-K), Ti = 50 + 273 = 323 K and Tf= 150 + 273= 423 K .
∴ SΔ = 2.3026 × 1032 mole × 5
calmole k− × log10
423323
= 2.3026 × 1032 × 5 × (log 423 - log 323)
= 2.3026 × 1032 × 5 × (2.6263 – 2.5092)
= 0.42 cal/K.
The entropy increases.
12. For silver the molar specific heat at constant pressurein the range 100 K to 200 K is given by Cp = 0.076 T - 0.00026T2 - 0.15 cal/(mole-K), where T is Kelvin temperature. If 2 molesof silver are heated from 100 K to 200 K, calculate the changein entropy.
Solution : Let dQ be the heat taken in by μmoles of silverat temperature T for an infinitesimally small temperature-risedT. If Cp be the molar specific heat at temperature T, then wehave
201Physical Significance
dQ = μCpdT.
The corresponding entropy-change is
dS =dQT
= μCp dTT
.
The entropy-change for a temperature-rise from 100 K to200 K is therefore
SΔ = 200
100dS∫ =
200
100 p
dTC
Tμ∫
= 2200
100
0.076 0.00026 0.152
T TdT
T⎛ ⎞− −⎜ ⎟⎝ ⎠
∫ [∴ μ = 2]
= 200
100
0.152 0.076 0.00026T dT
T⎛ ⎞− −⎜ ⎟⎝ ⎠∫
= 2002
100
0.000262 0.076 0.15 log
2 e
TT T
⎡ ⎤− −⎢ ⎥
⎣ ⎦
= 2[0.076 (200 – 100) - 0.00013 (2002 – 1002) – 0.15 loge2]
= 2 [(0.076 × 100) – (0.00013 × 30000) – (0.15 × 0.693)]
= 2 [7.6 – 3.9 – 0.104]
= 7.192 cal/K .
13. Two blocks of copper, each of mass 850 g, are put intothermal contact in an insulated box. Their initial temperaturesare 52 °C and 12 ºC and the specific heat of copper is 0.1 cal(g-°C). Let us find the change in the entropy of the system andof the universe?
Solution : Let the final equilibrium temperature of theblocks be t °C. The heat lost by the hotter block must beabsorbed by the cooler one, that is,
mc (52 – t) = mc (t – 12)
∴ t = 32°C.
202 Thermal Physics
Now, change in entropy of the hotter block in cooling from52 °C (= 325 K) to 32 °C (= 305 K)
= 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 850 × 0.1 × (log 305 – log 325)
= 2.3026 × 850 × 0.1 × (2.4843 - 2.5119)
= – 5.40 cal/°C.
Again, change in entropy of the cooler block in heatingfrom 12 °C (= 285 K) to 32 °C (= 305 K)
= 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 850 × 0.1 × (log 305 – log 285)
= 2.3026 × 850 × 0.1 × (2.4843 – 2.4548)
= + 5.77cal/°C.
∴ net increase in the entropy of the system is
5.77 – 5.40 = 0.37 cal/°C .
Because the process is carried out in an insulated box, theentropy-change of the surroundings is zero so that the entropy-change of the universe is + 0.37 cal/°C, an increase.
14. The change in entropy when 40 g of water at 50 °C ismixed with 80 g of water at 20 °C is calculated as below.
Solution : Let the temperature of the mixture be t °C. Thespecific heat of water c = 1 cal/(g-deg). By the principle ofcalorimetry, we have
40 × 1 × (50 – t) = 80 × 1 × (t – 20)
∴ t = 30°C.
Now, change, in entropy of 40 g of water in cooling from50 °C (= 323 K) to 30 °C (= 303 K)
= 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 40 × 1 × (log 303 – log 323)
203Physical Significance
= 2.3026 × 40 × 1 × (2.4814 – 2.5092)
= – 2.56 cal/K .
The – sign indicates decrease in entropy.
Again, change in entropy of 80 g of water in heating from20 °C (= 293 K) to 30 °C (303 K)
= 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 80 × 1 x (log 303 – log 293)
= 2.3026 × 80 × 1 × (2.4814 – 2.4669)
= + 2.68 cal/K.
The + sign indicates increase in entropy.
Therefore, the net increase in the entropy of the system is
2.68 – 2.56 = 0.12cal/K.
15. 10 gm of ice at 0 °C melts into water at the sametemperature. The change in entropy is calculated below. Thelatent heat is 80 cal/g.
Solution : If the ice is to be melted reversibly then it mustbe put in contact with a heat-reservoir whose temperature ishigher than 0 °C by only an infinitesimally small amount. Thenwe can write the entropy-change of the ice-water system as
SΔ =dQT∫ .
But the temperature remains constant at 0 °C (= 273 K).
∴ SΔ =1 Q
dQT T
=∫ .
The heat which must be supplied to 10 g of ice to melt is10 × 80 = 800 cal; and T = 273 K.
SΔ =800273 = 2.93cal/K,
which is positive. Thus the entropy of the system increases.
204 Thermal Physics
In the reversible melting there will be an exactly equaldecrease in the entropy of the heat- reservoir so that the entropy-change of the system plus surroundings is zero, a characteristicof the reversible process.
16. If, in the above problem, the ice is melted by droppingit into a large quantity of water at 20 °C , then what wouldbe the change in the entropy of the water-reservoir and whatwould be the total entropy-change of the original ice and thewater-reservoir.
Solution : When the ice at 0 °C is melted by putting it ina large quantity of water at 20 °C, the melting of ice will beirreversible (heat will always flow from water to ice). In thiscase we shall have the following entropy-changes :
Entropy-change of ice will be
SΔ ice = + 2.93 cal/K . (Calculated in part a)
Entropy-change of water-reservoir which gives out 800 calof heat to the ice (since it is ‘large’, its temperature may beconsidered to remain constant at 20 °C i.e. 293 K) will be
SΔ reservoir =QT
= 800
293−
= – 2.73 cal/K.
Therefore, the total entropy-change of the system plussurroundings will be
SΔ total = SΔ ice + SΔ reservoir
= + 2.93 – 2.73 = + 0.20 cal/K .
Thus the total entropy increases, as is the characteristic ofan irreversible process.
17. The change in entropy when 10 g of ice at 0 °C isconverted into water at 50 °C by heating is calculated asflows. The latent heat of ice is 80 cal/g and specific heat ofwater is 1 cal/(g-°C.)
205Physical Significance
Solution : The change in entropy when a substance of massm and latent heat L melts at a constant Kelvin temperature Tis given by
SΔ =QT
= mLT
∴ the change in entropy when 10 g of ice melts at 0 C(= 273 K) is
10 80273×
= 2.93 cal/K.
Again, the change in entropy when 10 g of water is heatedfrom 0 °C (= 273 K) to 50 °C (= 323 K) is
2.3026 mc log10 (Tf/Ti)
= 2.3026 × 10 × 1 × (log 323 – log 273)
= 2.3026 × 10 × (2.5092 – 2.4362)
= 1.68 cal/K.
∴ total change in entropy = 2.93 + 1.68 = 4.61 cal/K.
This is entropy-increase.
18. The change in entropy when 1 g of water at 0 °C isconverted into steam at 100 °C is found out as follows. Takespecific heat of water to be constant at 1 cal/(g-°C) and latentheat of steam at 100 °C as 540 cal /g.
Solution : The change in entropy when the temperature ofa substance of mass m and specific heat c changes from Ti toTf is
SΔ = 2.3026 mc log10
f
i
T
T .
Hence the increase in entropy when the temperature of 1g of water rises from 0 °C (= 273 K) to 100 °C (= 373 K) is
2.3026 × 1 × 1 × log10
373273
206 Thermal Physics
= 2.3026 × (2.5717 – 2.4362)
= 2.3026 × 0.1355 = 0.312 cal/K .
Now, the heat absorbed by a substance of mass m andlatent heat L at temperature T is mL, so that the entropy changeis mL/T. Hence, the increase in entropy when 1 gm of waterat 100 °C (= 373 K) evaporates
=1 540
373×
= 1.448 cal/K .
∴ total increase in entropy = 0.312 + 1.448 = 1.76 cal/K.
19. The increase in entropy when 1 g of ice at 0 °C changesinto 1 gm of steam at 100 °C, assuming the specific heat ofwater as 1 cal/g-deg , latent heat of ice 80 cal/g and latent heatof steam 540 cal/g may be found.
Solution : The amount of heat Q given to 1 g of ice at0 °C (i.e. 273 K) to convert into water at the same tempe-rature is L cal where L is the latent heat. Hence the increasein entropy is
QT
=LT
= 80273 = 0.293 cal/K.
Now, the increase in entropy when the temperature of1 g of water (sp. heat c = 1 cal/g-deg) rises from Ti (= 0 °C= 273 K) to Tf (= 100 °C = 373 K)
= 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 1 × 1 × (log 373 – log 273)
= 2.3026 × (2.5717 – 2.4362)
= 2.3026 × 0.1355 = 0.312 cal/K .
Now, the heat required to convert 1 g of water at 100 °C(= 373 K) into steam is L cal where L is the latent heatof vaporisation. Therefore , the increase in entropy invaporisation is
207Physical Significance
LT
= 540373 = 1.448 cal/K.
∴ total increase in entropy = 0.293 + 0.312 + 1.448 = 2.053cal/K.
20. The increase in entropy when 1 kg of ice at 0 °C isconverted into steam at 100 °C is calculated below. Specificheat of water =10 kcal/(kg °C), latent heat of ice = 3 4 x 10s
joule/kg, latent heat of steam = 22 68 x 10s joule/kg ,7 = 42joule/cal.
Solution : The increase in entropy when 1 kg of ice ismelted into water at 0 °C (= 273 K) is
QT
=mLT
= 51 (3.4 10 /
273kg joule kg
K× ×
= 1.245 × 103 joule/K
=31.245 10
4.2×
= 0.296 × 103 cal/K.
The increase in entropy when 1 kg of water is heated from0 °C (= 273 K) to 100 °C (= 373 K) is
2.3026 mc log10 (Tf/Ti)
= 2.3026 × 1 kg × 1 kcal/(kg-K) × (log 373 – log 273)
= 2.3026 × 1 × 1 × (2.5717 × 2.4362)
= 0.312 kcal/K = 0.312 × 103 cal/K .
Finally, the increase in entropy when 1 kg of water isvaporised to steam at 100 °C (= 373 K) is
QT
=mLT
= 51 (22.68 10 /
273kg joule kg
K× ×
= 6.080 × 103 joule/K
208 Thermal Physics
=36.080 10
4.2×
= 1.448 × 103 cal/K.
∴ total increase in entropy = (0.296 + 0.312 + 1.448) × 103
= 2.056 × 103 cal/K.
21. The change in entropy upon the conversion of 10 g ofice at —20 °C into steam at 100 °C may be found. Given : sp.heat of ice =0.5 cal/g-K, latent heat of ice = 80 cal/g, latent heatof steam = 539 cal/g.
Solution : Ice (mass = 10 gm, sp. heat c = 0.5 cal/g-K) isfirst heated from –20 °C (= 253 K) to 0 °C (= 273 K). The increasein its entropy is
1SΔ = 2.3026 mc log10 (Tf/Tt)
= 2.3026 × 10 × 0.5 × (log 273 – log 253)
= 2.3026 × 10 × 0.5 × (2.4362 – 2.4031)
= 0.381 cal/K.
Ice now melts to water at 0 °C. The increase in entropy is
2SΔ =mLT
= 10 80
273×
= 2.930 cal/K
The water (m = 10 g , c = 1 cal/g-deg) is now heated from0 °C (= 273 K) to 100 °C (= 373 K). The increase in entropy is
3SΔ = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 10 × 1 × (log 373 – log 273)
= 2.3026 × 10 × 1 × (2.5717 – 24362)
= 3.120 cal/K.
Finally, the water is converted into steam at 100 °C(= 373 K). The increase in entropy is
4SΔ = mLT
= 10 539
373×
= 14.450 cal/K.
209Physical Significance
∴ total increase in entropy
= 1SΔ + 2SΔ + 3SΔ + 4SΔ
= 0.381 + 2.930 + 3.120 + 14.450
= 20.88 cal/K.
22. 10 g of steam at 100 °C is converted into water at thesame temperature. The change in entropy is computed. Thelatent heat of steam is 540 cal/g.
Solution : The steam (mass m = 10 g, latent heat L =540cal/g) condenses into water at 100 °C (= 373K), giving out heatmL. The change in entropy is
SΔ = QT
= mLT
− = –
10 540373×
= –14.48 cal/K.
The entropy of the steam decreases . The entropy of theuniverse, however, increases because the entropy of the cold-reservoir which absorbs the heat given out by the steamincreases by an amount greater than 14.48 cal/K.
23. 10 g of water at 20 °C are converted into ice at -10 °Cat constant pressure. Assuming specific heat of ice to be 0.5cal/(g-°C), and latent heat 80 cal/g, let us find the change inentropy.
Solution : The water (mass m = 10 g, specific heat c = 1cal/g-deg) first cools from 20 °C (= 293 K) to 0 °C (273 K). Thechange in its entropy is
1SΔ = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 10 × 1 × log10 (273/293)
= 2.3026 × 10 × 1 × (log 273 – log 293)
= 2.3026 × 10 × 1 × (2.4362 – 2.4669)
= – 0.707 cal/K .
The water now freezes into ice at constant temperature of273 K , giving out heat mL . The change in entropy is
210 Thermal Physics
2SΔ = mLT
− = –
10 80273×
= – 2.930 cal/K.
Finally, the ice so formed (mass m = 10 g, specific heatc = 0.5 cal/g-deg) cools from 0°C ( = 273 K) to –10 °C (= 263 K).The change in entropy is
3SΔ = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 10 × 0.5 × log10 (263/273)
= 2.3026 × 10 × 0.5 × (log 263 – log 273)
= 2.3026 × 10 × 0.5 × (2.4200 – 2.4362)
= –0186 cal/K.
∴ total change in entropy = 1SΔ + 2SΔ + 3SΔ
= – 0.707 – 2.930 – 0186 = – 3.823 cal/K.
The entropy of water decreases. In the actual (irreversible)process, the entropy of the universe (water + surroundings)increases.
24. 0.1 kg of steam at 100 °C is converted into ice at–10 °C. The change in entropy is calculated. Given : specificheat of ice = 0.5 cal/(g-°C), specific heat of water = 1.0 cal/(g-°C), latent heat of steam = 540 cal/g, latent heat of ice= 80 cal/g.
Solution : The change in entropy of the steam (m = 100 g,L = 540 cal/g) in condensing into water at 100 °C (= 373 K) is
1SΔ = QT
= mLT
− = –
100 540373×
= –144.8 cal/K.
The change in entropy of water (m = 100 g , c = 1.0 cal/g-°C)in cooling from 100 °C (= 373 K) to 0 °C (= 273 K) is
2SΔ = 2.3026 mc log10 (Tf /Ti)
= 2.3026 × 100 × 1.0 × (log 273 – log 373)
= 2.3026 × 100 × 1.0 × (2.4362 – 2.5717)
= – 31.2 cal/K.
211Physical Significance
The change in entropy of water (m = 100 g , L = 80 cal/g)in freezing into ice at constant temperature of 0 °C (= 273 K) is
3SΔ = QT
= mLT
− = –
100 80273×
= – 29.3 cal/K.
The change in entropy of ice (m = 100 g , c = 0.5 cal/g-°C)in cooling from 0 °C (= 273 K) to - 10 °C (=263 K) is
4SΔ = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 100 × 0.5 × (log 263 – log 273)
= 2.3026 × 100 × 0.5 × (2.4200 – 2.4362)
= –1.86 cal/K.
∴ total change in entropy = 1SΔ + 2SΔ + 3SΔ + 4SΔ
= – 144.8 – 31.2 – 29.3 – 1.86 = – 207 cal/K.
The entropy of the system decreases . In the actual processthe entropy of the universe would increase.
25. The change in entropy when 1 g of tin is heated from127 °C to 313 °C is calculated as follows. Given : melting pointof tin = 232 °C, latent heat of fusion of tin = 14 cal / g, sp. heatof solid tin = 0 055 cal/(g-°C), sp. heat of molten tin = 0 064cal/(g-°C).
Solution : On heating the tin, its temperature first risesfrom 127 °C to 232 °C, then it melts at constant temperature,and then the temperature of the molten tin rises from 232 °Cto 313 °C.
Now, the change in entropy when the temperature of 1 gof tin (specific heat c = 0.055) rises from Ti = 127 °C (= 400 K)to Tf= 232 °C (= 505 K)
= 2.3026 mc log10
f
i
T
T
= 2.3026 × 1 × 0.055 × log10
505400
212 Thermal Physics
= 2.3026 × 0.055 × (log10 505 – log10 400)
= 2.3026 × 0.055 × (2.7033 – 2.6021)
= 2.3026 × 0.055 × 0.1012
= 0.0128 cal/K.
The increase in entropy when 1 g of tin melts at 505 K is
QT
= mLT
= 1 14505×
= 0.0277 cal/K.
Finally, the increase in entropy when the temperature ofthe molten tin rises from 232 °C (= 505 K) to 313 °C (= 586 K)
= 2.3026 mc log10
f
i
T
T
= 2.3026 × 1 × 0.064 × log10
586505
= 2.3026 × 0.064 × (log10 586 – log10505)
= 2.3026 × 0.064 × (2.7679 – 2.7033)
= 2.3026 × 0.064 × 0.0646
= 00095 cal/K.
∴ total increase in entropy
= 0.0128 + 0.0277 + 0.0095 = 0.050 cal/K.
26. The change in entropy of 1 g of nitrogen when itstemperature rises from 40 °C to 80 °C while its volume increasesfourfold is calculated. For nitrogen cv = 0.18 cal/(g-°C) andmolecular weight M = 28g/mol. Take R = 2 .0 cal/mole-°C.
Solution : When (i moles of a gas pass from an initial statepi Vi, Ti to a final state pf, Vf, Tf, the change in entropy in termsof temperatures and volumes is given by
SΔ = log logf fv e e
i i
T VC R
T Vμ + μ ,
where R is universal gas constant.
213Physical Significance
1 g of gas has 1M mol, where M is molecular weight.
Therefore, for 1 g of gas, we have
SΔ = log logf fve e
i i
T VC RM T M V
+ .
Here vCM = Cv = 0.18
calg C−° ,
R = 2.0 = cal
mol C− ° and
soRM =
2.0 /2.8 /cal mol C
g mol−°
= 0.0714 cal
g C− ° ;
f
i
T
T =80 27340 273
++ =
353313 and
f
i
V
V = 4.
∴ SΔ = 0.18 loge 353313 + 0.0714 loge 4
= 2.3026 [0.18 × (log 353 – log 313)
+ 00714 log 4]
= 2.3026 [0.18 × (2.5478 – 2.4955)+ 0.0714 × 0.6020]
= 2.3026 [0.00941 + 0.04298]
= 0.12cal/”C.
27. When 10 g of hydrogen is heated from 27 °C to 327 °C,its volume increases to four times the initial volume.The change in entropy is calculated. For hydrogen : Cv = 5cal/mol-°C and molecular weight M = 2. Take R = 8 31 joule/mole-°C.
214 Thermal Physics
Solution : For μ moles of a gas ,the change in entropy is
SΔ = log logf fv e e
i i
T VC R
T Vμ + μ .
The molecular weight of hydrogen is 2. It means that1 mole of hydrogen has 2 g gas. Thus 10 g of hydrogen contains5 moles (μ = 5).
Now, Cv = 5 cal/mole–°C, R = 8.31 joule/mole-°C = 8.314.18
= 2.0 cal/mole-°C, f
i
T
T = 327 27327 273
++ =
600300 = and
f
i
V
V = 4.
∴ SΔ = 5 × 5 × loge2 + 5 × 20 × loge4
= 2.3026 (25 log10 2 + 10 log 10 4)
= 2.3026 [(25 × 0.3010) + (10 × 0.6020)]
= 2.3026 [7.525 + 6.020]
= 31 cal/°C.
28. The change in entropy per g of oxygen gas during aprocess in which its temperature changes from 10 °C to 50 °Cand the pressure changes from 76 cm to 100 cm of mercury iscalculated Cv = 70 cal/mole –°C, R = 2.0 cal/mole –°C.
Solution : For 1 g of a gas the change in entropy in termsof temperature and pressure is given by
SΔ = log logp f fe e
i i
T PC RM T M P
−
The molecular weight M of oxygen is 32. Thus
SΔ =7.0 273 50 2.0 100
log log32 273 10 32 76e e
+−
+
=1
32 [7 loge 1.141 – 2 loge 1.316]
=2.3026
32 [7 × 0.05728 – 2 × 0.1192]
215Physical Significance
=2.3026
32 [0.4010 – 0.2384]
=2.3026 0.1626
32×
= 0.0117cal/°C.
29. Two moles of an ideal gas occupy a volume of 10 litresat a pressure of 4 atmospheres. It is first heated at constantvolume until its pressure increases to 8 atmospheres, and thenat constant pressure until its volume becomes 40 litres. Thechange in entropy is calculated as follows. Given : Cv = 3 cal/(mole-deg) and R = 2 cal/(mole-deg).
Solution : The change in entropy of μ moles of an idealgas in terms of its pressure and volume is given by
SΔ = log logf fv e p e
i i
p VC C
p Vμ +μ .
In the first heating; Vf/ Vi = 1 and pf/pi = 8/4 = 2. Also μ= 2,Cv = 3 cal/(mole-deg) and Cp = Cv + R = 5 cal/(mole-deg).
∴ 1( )SΔ = 2 × 3 × loge2
= 6 × 0.6931 [∴loge 2 = 0.6931]
= 4.1586 cal/deg.
In the second heating ; pf/pi = 1, Vf/Vi = 40/10 = 4 .
∴ 2( )SΔ = 2 × 5 × loge 4
= 10 × (2 × 0.6931)
= 13.862 cal/deg.
The total entropy-change is therefore
SΔ = 1( )SΔ + 2( )SΔ
= 4.1586 + 13.862 = 18.020 cal/deg.
30. We can defend or refute the claim that with temperatureof the surroundings 27 °C, 12000 kilocalories of heat availablefrom a reservoir at 627 °C is more useful than 14000 kilocaloriesof heat available at 327 °C.
216 Thermal Physics
Heat energy is available for conversion into work onlywhen it is let down from a higher to a lower temperature. Ifa working substance in Carnot engine takes in heat Q1 athigher temperature T1 and gives up heat Q2 at lowertemperature T2, then the heat available for useful work
= Q1 – Q2 = Q1
2
1
1QQ
⎛ ⎞−⎜ ⎟
⎝ ⎠
=2
11
1T
QT
⎛ ⎞−⎜ ⎟
⎝ ⎠2 2
1 1
Q TQ T
⎡ ⎤∴ =⎢ ⎥⎣ ⎦
In the first case, Q1 = 12000 kcal, T1 = 627 + 273 = 900 Kand T2 = 27 + 273 = 300 K. Thus
useful energy = 12000 300
1900
⎛ ⎞−⎜ ⎟⎝ ⎠
= 8000 kcal.
In the second case, Q1 = 14000 kcal, T1 = 327 + 273 = 600K and T2 = 27 + 273 = 300 K. Thus
useful energy = 14000 300
1600
⎛ ⎞−⎜ ⎟⎝ ⎠ = 7000 kcal.
Hence the claim is correct.
217Changing State
9
Changing State
Clausius–Clapeyron’s Equation
A substance can exist in three states–solid, liquid and gas.Out of these three, only two states can generally coexist inequilibrium. Whenever there is a change of state, either a solidchanges into liquid or liquid into vapour, the temperatureremains constant as far as the change takes place. Thistemperature, although depends upon pressure, is characteristicof each substance. When the change is from solid to liquidstate, the characteristic temperature is called the melting pointof the solid and when it is from liquid to vapour state, thetemperature is called the boiling point of the liquid. The meltingpoint or boiling point have got a specific value at a specificpressure and vice versa.
It is possible to obtain a relation showing how the meltingand boiling points vary with pressure, by applying thesecond law of thermodynamics. The relation thus obtained isknown as Clausius Clapeyron equation or the first latent heatequation.
218 Thermal Physics
Let ABCD and EFGH represent the two isothermals atinfinitely close temperatures T and (T + dT) respectively.Referring to the Fig. below, the parts AB and EF correspondto the liquid state of the substance. At B and F, substance ispurely in the liquid state. Along BC and FG, the change of stateis in progress and the liquid and vapour states coexist inequilibrium. At C and G the substance is purely in vapourstate.
From C to D and G to H, the substance is in the vapourstate. Let P and (P + dP) be the saturated vapour pressures ofthe liquid at temperatures T and (T + dT) respectively. Let V1
and V2 be the volumes of the substance at F and G respectively.
Let us draw two adiabatics from F and G meeting the lowerisothermal at M and N respectively. Let us suppose that 1 gm.of the substance is taken round a reversible Carnot cycleFGMNF, allowing it to expand isothermally along FG,adiabatically along GN, compressing it isothermally along NMand adiabatically along MF.
The amount of heat Q1 absorbed along FG is equal to thelatent heat of vaporization (L + dL) at temperature (T + dT),as substance changes completely from liquid state at F to thevapour state at G. Also the quantity of heat Q2 rejected alongthe isothermal compression NM, is L, the latent heat attemperature T. Here latent heat is supposed to vary withtemperature.
219Changing State
Applying the principle of Carnot’s reversible cycle
1
1
QT =
2 1 1
2 2 2
orQ Q TT Q T
=
or1 2
1
Q QQ−
=1 2
2
T TT−
We have, here Q1 = L + dL, Q2 = L, T1 = T + dT and T2 = T
∴L dL L
L+ −
=T dT T
T+ −
ordLL
=dTT
or dL =L
dTT
... (1)
The amount of heat converted into work during the cycleFGMNF,
Q1 – Q2 = L + dL – L = dL ... (2)
But the work done during the Carnot cycle is given by thearea FGMNF, which may be treated as a parallelogram.
Hence,
dL (in work units)= Area FGMNF = FG × Perpendicular distance between
FG and NM
= (V2 – V1) × dP ... (3)
where V2 and V1 are the specific volumes (i.e. masses per unitvolume) in vapour and liquid states respectively, dP expressesthe difference of pressure between FG and NM.
Substituting this value of dL in (1),
dP(V2 – V1) =LT
dT
or ( )2 1
dP LdT T V V
=− ... (4)
220 Thermal Physics
This is called the Clapeyron’s latent heat equation andholds for both the changes of state, i.e., from liquid to vapourand solid to liquid. In the latter case L will represent the latentheat of fusion, V1 and V2 the volume occupied by 1 gm. ofsubstance in solid and liquid states respectively. It may benoted that L in eq. (4) is to be expressed in work units i.e. inergs/gm. or Joules/kg.
Applications
Effect of Pressure on Boiling Points of Liquids: When aliquid boils i.e. changes from liquid state to gaseous state, thereis an increase in its volume so that V2 > V1 or the quantity(V2 – V1). is positive. Hence from eq. (4) above, dP/dT is apositive quantity which means the boiling point of a liquidrises with increase in pressure or vice versa. Thus a liquid willboil at lower temperature under reduced pressure. Hence waterwill boil at a temperature less than 100°C, when the atmosphericpressure is less than 76cm. of mercury–the normal pressure.
Effect of Pressure on Melting Points of Solids: When asolid melts, there may be an increase in volume as in the caseof certain substances like wax and sulphur or there may be adecrease in volume as in the case of certain substances like ice,gallium and bismuth.
(a) When V2 > V1 (As for wax and sulphur) dPdT is a – ve
quantity. This means that melting point of such substancesrises with increase in pressure.
(b) When V2 < V1 (As for ice, gallium and bismuth), (V2 – V1)is a negative quantity. Hence dP/dT is also negativewhich means that melting point of such substancesdecreases with increase in pressure. Thus, ice will meltat a temperature lower than 0°C when the pressure ishigher than the normal pressure, 76 cm. of mercury.
221Changing State
Second Latent Heat Equation
The second latent heat equation, or the equation of Clausius,gives the variation of latent heat of a substance with temperatureand connects it with the specific heat of the substance in thetwo states.
Let C1 denote the specific heat of a liquid in contact withits vapours and C2 the specific heat of saturated vapours incontact with its liquid. Referring to the Fig. above, let us considerthat 1 gm. of the substance is taken round the cycle BFGCB.The quantity of heat absorbed by the substance (liquid) inpassing from B to F, when its temperature rises by dT is C1 dT.In passing along FG, when the substance changes from liquidto vapour at constant temperature T + dT, it absorbs a quantityof heat L + dL.
In passing from G to C, the temperature of the substance(vapours) falls by dT and hence it gives out a quantity of heatC2 dT, while in passing along CB, when it condenses fromvapour to liquid at constant temperature T, gives out a quantityof heat L.
Hence the net amount of heat absorbed during thecycle is
C1 dT + L + dL – C2 dT – L = (C1 – C2) dT + dL.
222 Thermal Physics
This must be equal to the work done which is equal to thearea of the cycle or the area FGNMF in the limiting case andhence
= ( )2 1
LdP V V dT
T− = (Proved in § above)
∴ ( )1 2C C dT dL− + =L
dTT
or ( )1 2C C dT− =L
dT dLT
−
or 2 1C C− = .dL LdT T
−
This is the latent heat equation of Clausius.
PROBLEMS
1. Calculate the change in the boiling point of water whenthe pressure is increased from 1 to 2 atmospheres.
Given: Specific volume of water and steam at 100°C are1 and 1601 cm3 /gm. Latent heat of steam = 540 cal /gm.
Boiling point of water at 1 atmospheric pressure = 100°C.
Solution: From Clausius–Clapeyron equation
( )( )
2 1
2 1or
dP LdT T V V
T V V dPdT
L
=−
−=
Given that dP = 2 – 1 = 1 atmosphere
= 106 dynes/cm2
T = Boiling point of water = 100°C = 373K
V2 – V1 = 1601– 1 = 1600 cm3/gm
L = 540 cal/gm = 540 x 4.2 x 107 ergs/gm
∴ Change in boiling point of water
223Changing State
6
7
373 1600 10540 4.2 10
dT× ×
=× ×
= 26.6K = 26.6°C
2. Calculate the latent heat of ice given that change ofpressure by one atmosphere changes the melting point of iceby 0.0074°C and when one gm. of ice melts its volume changesby 0.0907 c.c.
Solution: From Clausius Clapeyron equation
( )( )
2 1
2 1.or
dP LT
dT T V V
dP T V VL
dT
=−
−=
Here, dP = 1 atmosphere = 1.1013 × 106 dynes/cm2
T = 0°C = 273K, V2 – V1 = 0.0907 c.c.
dT = 0.0074°C = 0.0074 K61.013 10 273 0.09070.0074
L× × ×
=
99
7
3.39 103.39 10 ergs/gm. cal/gm
4.18 10×
= × =×
= 81.1 cal/gm.
3. Calculate the change in the boiling point of water whenthe pressure is increased from 1.0 to 10 atmospheres. Givenspecific volume of steam = 1677 c.c. /gm., latent heat of steam= 540 cal./gm., boiling point of water at one atmosphere pressure= 100°C = 373K and pressure of one atmosphere = 1.0 × 106
dynes/cm2.
Solution: From Clausius Clapeyron equation
( )( )
2 1
2 1We have
dP LdT T V V
T V V dPdT
L
=−
−=
224 Thermal Physics
Given that
dP = 10 – 1 = 9 atoms.
= 9 × 1 × 106 dynes/cm2
T = Boiling point of water = 100°C = 373K
V2 = 1677 c.c./gm., V2 = 1 c.c./gm.
V2 – V2 = 1676 c.c./gm.
and L = 540 cal./gm. = 540 × 4.2 × 107 ergs/gm.
Therefore, change in boiling point of water6
7
373 1676 9 10540 4.2 10
dT× × ×
=× ×
= 248.07°C = 248.07 K
4. Calculate the change in the melting point of wax for apressure of 50 atmospheres from the following data Meltingpoint = 64°C, volume at 0°C = 1 cc; volume of the solid at themelting point = 1.161 cc; volume of the liquid at the meltingpoint = 1.166 c.c, density of the solid at 64°C = 0.96gm. /c.c.latent heat = 97 cal. gm.
Solution: Here the melting point (64°C) is given to be atone atmosphere pressure and we are required to find thechange in the melting point for a pressure of 50 atmosphere,i.e. for a pressure change of (50 – 1) = 49 atmosphere.
∴ dP = 49 × 76 × 13.6 × 981 dynes/cm2
T = 64°C = 64 + 273 = 337 K,
L = 97 cal. = 97 × 4.2 × 107 ergs.
Now mass of the solid at melting point = Volume x density
= 1.161 × 0.96 gm.
∴ Specific volume of the solid at melting point
1
Volume 1.1611.0417 c.c./gm
mass 1.161 0.96V = = =
×
225Changing State
and specific volume of the liquid at melting point
V =Volume 1.166
1.0461c .c ./gmmass 1.161 0.96
= =×
∴ V2 – V1 = 1.0461 –1.0417 = 0.0044 c.c./gm.
Now Clausius Clapeyron equation is
dTdP =
2 1( )L
T V V−
So that dT = 2 1( )dP T V VL
× −
= 7
49 76 13.6 981 337 0.004497 4.2 10
× × × × ×× ×
= 0.018K = 0.018°C.
5. Calculate the depression in the melting point of ice byone atmospheric increase of pressure, given that the latentheat of fusion of ice L = 3.4 × 105 joule/kg. Specific volume of1 kg. of ice and water at 0°C are 1.091 × 10–3 m2 and 1.000× 10–3 m3 respectively. 1 atmospheric pressure = 105 N/m2.
Solution: Given
dP = l atmosphere = 105 N/m2
T = 0°C = 27311.
L = 3.4 x 105 joule/kg.
and specific volume of 1 kg. of ice
V1 = 1.091 × 10–3 m3
Specific volume of 1 kg. of water V2 = 1.000 × 10–3 m3
∴ V2 – V1 = (1.000 × 10–3 – 1.091 × 10–3) m3
= – 0.091 × 10–3
Now we have from Clapeyron equation
226 Thermal Physics
dPdT =
2 1 ( )L
T V V−
dT = 2 1 ( )dP T V VL
× −
=5 3
5
10 273 ( 0.091 10 )3.4 10
−× − ××
= – 0.0073 k = – 0.0073°C.
Hence the melting point of ice will be depressed by 0.0073°Cper atmosphere increase of pressure.
6. Calculate the depression of the melting point of ice(L = 80 Cals.) per atmosphere increase of pressure, if ratio ofthe densities of ice and water at 0°C is 10/11.
Solution: From Clausius–Clapeyron eqn.
2 1( )T V V dPdT
L−
=
Here T = 27311, L = 80 cal. = 80 × 4.2 × 107 ergs.
dP = 1 atmosphere = 1.03 x 106 dynes/cm2 and
2
1
Specific volume of waterand
Specific volume of iceVV
=
Den ity of ice 10 =
Den ity of water 11s
s=
But V2 = 1.0 cm3/4gm.
31
111.0 1.1 / .
10V cm gm∴ = × =
∴ V2 – V1 = 1.0–1.1 = –0.1cm3/gm.
( ) 6
7
273 0.1 1.013 10Hence
80 4.2 10dT
× − × ×=
× ×
= – 0.0082°C.
Thus, melting point of ice will decrease by 0.0082°C peratmosphere increase of pressure.
227Changing State
7. Calculate the pressure required to make water freeze at–1°C. Change of specific volume when 1 gm of water freezesinto ice = .091 cc., J = 4.2 x 107 ergs/cal., 1 atmosphere 106 dynes/cm2 and the latent heat of ice = 80 cal. /gm.
Solution: Given dT = – 1°C, V2 – V1 = –0.091 c.c.
L = 80 cal. = 80 x 4.2 x 107 ergs,
T = 0 + 273 = 273 K
Applying Clapeyron’s latent heat equation
2 1( )
dP LdT T V V
=−
( )( )
72
2 1
80 4.2 10 1or dynes/cm .
( ) 273 0.091L
dP dTT V V
× × × −=
− × −
7
6
80 4.2 10 atoms.
273 0.091 10× ×
=×− ×
= 135.2 Atoms.
Hence to lower the melting of ice by 1°C, the pressure mustbe raised by 135.2 atmospheres.
Now melting point of ice at atmospheric pressure is 0°C.Hence pressure required to make ice freeze at – 1°C
= 135.2 + 1 = 136.2 Atoms.
8. Calculate the change in vapour pressure of water as theboiling temperature changes from 100°C to 103°C. Given latentheat of steam = 540 cal. /gm. and specific volume of steam= 1670 cc. /gm.
Solution: The Clapeyrons’ equation describing the rate ofchange of vapour pressure with temperature is
2 1
2 1
( )
or( )
dP LdT T V VdP L dTdT T V V
=−
×=
−
228 Thermal Physics
Here L = 540 cal. = 540 × 4.2 × 107 ergs.,
T = 100 + 273 = 373K
dT = (103 – 100) = 3°, V2 = 1670 c.c./gm.,
V1 = 1 c.c./gm.
∴ Change in vapour pressure7
2
7
540 4.2 10 3dynes/cm
373 (1670 1)
540 4.2 10 3 atmosphere
373 1669 76 13.6 981
dP× × ×
=× −
× × ×=
× × × ×
= 0.17091 atmospheres.
9. Water boils at 100.5°C and 99.5°C when the atmosphericpressures are respectively 77.371 and 74.650 cm. of mercury.Calculate the volume of 1 gm. steam at 100°C the latent heatbeing 537 cal. /gm.
Solution: Given that
dT = 100.5–99.5 = 1°
L = 537 cal. = 537 × 4.2 × 107 ergs.
T = 100°C = 373 K
dP = (77.371 – 74.650) = 2.721 cm. of Hg
= 2.721 × 13.6 × 981 dynes/cm2.
Now Clapeyron’s latent heat equation is
2 1
2 1
7
( )
or
537 4.2 10 1 1658c .c .
373 2.721 13.6 981
dP LdT T V V
L dTV V
T dP
=−
− =
× × ×= =
× × ×
Since V1 = 1 c.c.
Hence the volume of 1 gm. of steam at 100°C
V2 = 1658 + 1 = 1659 c.c.
229Changing State
10. A Carnot’s engine having 100 gm. water–steam asworking substance has, at the beginning of the stroke, volume104 c.c. and pressure 788 mm. (B.P. = 101°C). After a completeisothermal change from water into steam, the volume is 167,404 c.c. and the pressure is then lowered adiabatically to 733.7mm. (B.P.° = 90°C). If the engine is working between 99°C and101°C, calculate the latent heat of steam.
Solution:Volume of 100 gm of water at the beginning ofstroke
= 104 c.c
∴ Specific volume 1104
1.04 . ./ .100
V c c gm= =
Similarly specific volume of steam after isothermal change
2167404
1674 . ./100
V c c gm= =
∴ V2 – V1 = 1674.04 – 1.04 = 1673 c.c.
Change in pressure
dP = 781 – 733.7 = 54.3 mm. = 5.43 cm. of Hg
= 5.43 × 13.6 × 981 dynes cm –2
Mean temperature 101 99
100 3732
T C K+
= =o
and dT = 101 – 99 = 2°.
Now according to Clapeyron’s latent heat equation
2 1
2 1
( )( )
or
dP LdT T V V
T V V dPL
dT
=−− ×
=
373 1673 5.43 13.6 981 ergs.
2× × × ×
=
7
373 1673 5.43 13.6 981 cal ./gm.
2 4.2 10× × × ×
=× ×
= 540.0 cal./gm.
230 Thermal Physics
11. Calculate the specific heat of saturated steam at 100°Cfrom the following data:
L at 90°C = 545.25 cals.L at 100°C = 539.30 cals.L at 110°C = 533.17 cals.
Specific heat of water at 100°C = 1.013 cals/gm.Solution: Given C1 = 1.013, C2 = ?
at110 at 90 533.17 545.2520 20
dL L C L CdT
− −= =
o o
12.08 0.604 cal ./gm.
20= − = −
Hence from 2 1
dL LC C
dT T− = −
2 1dL L
C CdT T
= + −
2
539.301.013 0.604
373C = − −
= – 1.037 cals./gm.12. If L = 800 –.705 T, show that the specific heat of steam
is negative.Solution: Given that L = 800 – 0.705 T
∴dLdT = – 0.705.
Now at normal B.P. of water, i.e. at 100°C = 373 K, we shallhave L = 800 – .795 × 373 = 537.
Also specific heat of water in liquid state,C1 = 1
Hence from second latent heat equation,
C2 = C1 + dL LdT T
−
5371 ( .705) 1.14.
337= + − − = −
Thus, the specific heat of saturated water vapour at 100°Cis negative.
231Changing State
13. The latent heat of water diminishes by 0.695 cal. /gm.for each degree centigrade rise in temperature at temperature100°C and latent of water vapour at 100°C, is 540 cal. /gm.calculate the specific heat for saturated steam.
Solution: By second latent heat eqn., the specific heat C2
of saturated steam is given by
2 1dL L
C CdT T
= + −
oHere 0.695cal ./gm. CdLdT
= −
T = 100°C = 373 K
L = 540 cal./gm.
and C1 = specific heat in liquid state (water)
= 1.0 cal./gm.
2
540 1 ( 0.695)
373C = + − −
= 1 – .695 – 1.448
= – 1.143 cal/gm.°C.
14. Generally the specific volume of a liquid is much lessas compared to specific volume of its vapour. Assuming thatthe vapour obeys ideal gas equation, prove that P = constante–(L/RT).
Solution : Clausius Clapeyron equation gives
2 1( )dP LdT T V V
=− ... (i)
Given specific volume of liquid V1 << specific volume ofvapour V2
or V2– V1 = V2 = V (say)
232 Thermal Physics
Hence eq. (i) becomes
dP LdT TV
= ... (ii)
But vapour obeys perfect gas equation
PV = RT
RT
VP
∴ =
Substituting value of V in eq. (ii),
2
2
or .
dP L LPRTdT RTTP
dP LdT
dT RT
= =⎛ ⎞⎜ ⎟⎝ ⎠
=
Integrating
[ ]( / ) Constant
Constant /
( / )
log Constant
or
Constant .
e
L RT
L RT
L RT
LP
RTP e
e e
e
− +
−
−
= − +
=
= ×
= ×
Triple Point
(i) The boiling point of water increases with increase inpressure and vice versa. The curve AB represents therelation between pressure and temperature and is calledthe steam line or vaporization line Fig. The liquid andvapour are in stable equilibrium together only alongthe line AB. At all points above AB the substance is allliquid and below it there exists vapour only. If at apoint M, pressure is raised keeping temperatureconstant, boiling point will consequently increase andall vapour will condense into liquid. Similarly if at M,pressure is decreased, all the liquid will vaporize andonly vapour will remain.
233Changing State
(ii) The melting point of ice decreases with increase inpressure i.e. ice melts below 0°C at a pressure higherthan the atmospheric pressure. The pressuretemperature relationship can be represented by a curveCD or C’D’ which is called the ice line or fusion line. Thecurve CD, which slopes to the left, is for ice typesubstances whose melting point is raised with increasein pressure; while C’D’, which slopes to the right, isfor wax type substances whole melting point is raisedwith increase in pressure. The substance is entirelysolid on the left of the curve while entirely liquid onthe right. The curve represents the equilibrium betweenthe solid and liquid state.
(iii) When the pressure on ice is raised, evaporation fromits surface slows down. The equilibrium between thesolid and vapour states of a substance can berepresented by a curve EF called the Hoar Forst line orsublimation line. Above the curve EF, the substance isall solid and below it all vapour.
These three curves, when plotted on the same diagram, arefound to meet in a single point O as shown in Fig. This pointis called the triple point. At the triple point, the pressure and
234 Thermal Physics
temperature are such that the solid, liquid and vapour states of thesubstance can exist simultaneously in equilibrium.
To Show that there is Only one Triple Point : Suppose thatthe three curves do not meet at a point but intersect enclosingan area ACF shown shaded in Fig. Then according to ice lineCD, the substance must be entirely solid in the shaded area asit is to the left of CD. According to the steam line AB thesubstance must be entirely liquid as it is above AB and accordingto hoar frost line EF the substance in the shaded portion mustbe entirely vapour as it is below EF. But these three conclusionscontradict one another and hence the shaded triangle ACFcannot exist. Thus, the three curves should meet in a singlepoint O called the triple point.
It should be noted that the triple point of water is not fixedbut is different for different allotropic forms of ice.
PROBLEMS
1. The coordinates of the triple point of water aret = 0.0075°C and p = 0.0060 atmosphere. Calculate the slopeof the ice line in atoms. /°C. What is the physical significanceof the negative sign, of the slope ?
Solution: Referring to Figures, let O be the triple point ofwater. Then the coordinates of point O are
235Changing State
P = 0.0060 atoms, t = 0.0075°C.
Now we know that the freezing point of water into ice at1 atmosphere pressure is 0°C. Hence the coordinates of pointD on the ice line are
P’ = 1 atoms., t’ = 0°C
Hence for ice line OD
dP = P’ – P = 1– 0.006 = 0.994 atoms.
and dT = dt = t’ – t = 0 – 0.0075 = 0.0075°C
Therefore, the slope of ice line is
dPdT =
o0.994atoms/ C
0.0075dPdt
= −
= – 132.5 atoms./°C.
Negative sign with the slope dP/dT indicates that the meltingpoint of ice decreases with an increase in pressure.
2. Calculate the pressure and temperature of the triplepoint of water. Given that the lowering of melting point of iceper atmosphere increase of pressure is 0.0072°C and thesaturated vapour pressure at 0°C = 460 m. m. while at 1 °C= 4.94 m. m.
Solution: Let the triple point correspond to a temperaturetoC and pressure p mm. of mercury.
Saturated vapour pressure at 0°C = 4.60 mm.
Saturated vapour pressure at 1°C = 4.94 mm.
∴ Increase in saturated vapour pressure for 1°C rise intemperature
= 4.94 – 4.60 = 0.34 mm.
Hence increase in vapour pressure for t°C = 0.34t
Therefore, saturated vapour pressure at triple point t°C isgiven by
p = (4.60 + 0.34 t) mm.
236 Thermal Physics
Now the melting point of ice at 760 m.m. (1 atmosphere)pressure is 0°C. But at the triple point the pressure is(4.60 + 0.34t).
Hence
Decrease in pressure = 760 – (4.60 + 0.34t)
= (755.4 – 0.34t) mm.
It is given that a decrease in pressure of 760 mm. (1 atoms.)will increase the melting point of ice by 0.0072°C. Thereforeby a decrease of (755.4 – 0.34t) mm. in pressure.
Increase in melting point = 0.0072 (755.4 0.34 )
760t× −
or Melting point = 0.0072 (755.4 0.34 )
0760
t× −⎡ ⎤+⎢ ⎥⎣ ⎦
But this is the temperature t at a triple point. Thus
0.0072 (755.4 0.34 )760
tt
× −=
or 760t = 5.4388 – 0.002448t
or 760.002448t = 5.4388
5.4388760.002448
t =
= 0.007156°C.
Substituting this value of t in the expression for p, we get
p = (4.60 – 0.34 × 0.007156)
= 4.60243 mm. of Hg.
3. The vapour pressure P (in mm. of mercury) of solidammonia is given by
3754log 23.03e P
T= −
while that of liquid ammonia is given by
237Changing State
3063log 19.49e P
T= −
where T is in K.Calculate the triple point of ammonia.
Solution: At triple point, the vapour pressure of thesubstance in each of the three states is identical. Hence equatingthe vapour pressure of solid ammonia
3754log 23.03e P
T= − ... (i)
with that of liquid ammonia
3063log 19.49e P
T= − ... (ii)
We have
3754 306323.03 19.49
T T− = −
( )1or 3754 3063 23.03 19.49
T− = −
691or 3.54
T=
∴ T = 195.2K
Substituting the value of T in eq. (i), we get
3754
log 23.03195.2e P = −
or loge P = 3.8
or 2.303 log10 P = 3.8
10
3.8or log 1.652
2.303P = =
∴ P = 44.87 mm Hg.
Thus the coordinates of triples point of ammonia areT = 195.2K and P = 44.87 m.m. of Hg.
238 Thermal Physics
Thermodynamical Potentials and Their Relations withThermodynamical Variables : The thermodynamic state of ahomogeneous system may be represented by means of certainselected variables, such as pressure P, volume V, temperatureT and entropy S. Out of these four variables, any two may varyindependently and when known enable the others to bedetermined. Thus there are only two independent variablesand the others may be considered as their functions.
There exists certain relations between thesethermodynamical variables. The first and second law ofthermodynamics provide two relations which may respectivelybe stated as,
dQ = dU + PdV
and dQ = TdS
Combining the two laws, we have
TdS = dU + PdV
or dU = TdS – PdV.
This expresses the change in internal energy of the systemin terms of four thermodynamical variables. However for acomplete knowledge of the system, certain other relations arerequired and for this purpose we introduce some functions ofthe variables P, V, T and S, known as thermodynamical potentialsor’ the thermodynamic functions. There are four principalthermodynamic potentials and we shall discuss them one byone.
Internal or Intrinsic Energy: We have seen already thataccording to first law of thermodynamics, there is a certainfunction, U, of the variables which characterises the systemand this function may be called the intrinsic energy or the internalenergy. When the system passes from one state to another, thechange in the internal energy is independent of the routefollowed between the two states. The internal energy of asystem is defined by the equation,
239Changing State
dU = dQ – dW
where dW is the external work done and may be replaced byPdV while dQ may be put equal to TdS by second law ofthermodynamics. Thus
dU = TdS – PdV
Taking partial differentials of the intrinsic energy withrespect to variables S, and V, we get
V
US
∂⎛ ⎞⎜ ⎟∂⎝ ⎠
= andS
UT P
V∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠
These are the relations connecting the internal energy Uwith the thermodynamical variables S, V, T and P.
Now since dU is a perfect differential*, we must have
V
UV S∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
=S
US V∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
∴S
TV∂⎛ ⎞
⎜ ⎟∂⎝ ⎠=
V
PS∂⎛ ⎞−⎜ ⎟∂⎝ ⎠
…(i)
This result is the first thermodynamical relation of Maxwell.
Helmholtz’s Function F: The first and second law ofthermodynamics combined together provide
dU = TdS – dW
If we consider processes in which the system exchangesheat with the surrounding and is thereby maintained at aconstant temperature T, then TdS = d (TS) and we can write
dU = d (TS)–dW
or d(U – TS)= – dW
or dF = – dW
where F = U – TS is known as the Helmholtz free energy ormore appropriately the work function (because eq. dF = – dWshows that for reversible changes the work done by the system
240 Thermal Physics
is equal to the decrease in this function F). Thus the Helmholtzfree energy is defined by the equation,
F = U – TS,
where the value of F depends only on the state of the substanceand dF is a perfect differential. Now
dF = dU – d (TS)
= dU – TdS – SdT.
But dU = TdS – PdV
∴ dF = TdS – PdV – TdS – SdT
= – PdV – SdT.
Here T and V are independent variables. Taking partialdifferentials of F
T
FV∂⎛ ⎞
⎜ ⎟∂⎝ ⎠= and
V
FP S
T∂⎛ ⎞− = −⎜ ⎟∂⎝ ⎠
and since dF is perfect differential
V
FV T∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
=T
FT V∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
T
SV∂⎛ ⎞
⎜ ⎟∂⎝ ⎠=
V
PT∂⎛ ⎞
⎜ ⎟∂⎝ ⎠…(ii)
This is the second thermodynamical relation of Maxwell.
Enthalpy or Total Heat H: Enthalpy is an extensivethermodynamical property and is of particular significance. Itis mathematically defined as
H = U + PV.
The differentiation of which yields
dH = dU + d (PV)
= (TdS – PdV) + PdV + VdP
= TdS + VdP.
241Changing State
At a constant pressure (dP = 0), dH = TdS = dQ, the quantityof heat given to the system from an external source. Thisexplains the name of heat function given to it. It also shows thatfor an isobaric process, change in enthalpy is equal to the heat givento the system.
Now taking partial differentials of H treating S and P asindependent variables.
P
HS
∂⎛ ⎞⎜ ⎟∂⎝ ⎠
= and .S
HT V
P∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
And since dH is a perfect differential
P
HP S∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
=S
HS P∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
orS
TP∂⎛ ⎞
⎜ ⎟∂⎝ ⎠=
P
VS
∂⎛ ⎞⎜ ⎟∂⎝ ⎠
…(iii)
This is the third thermodynamical relation of Maxwell.
Gibb’s Potential G : From the definition of enthalpy, wehave
H = U + PV.
or dH = dU + PdV + VdP
which on putting dU = TdS – PdV becomes
dH = TdS – PdV + PdV + VdP = TdS + VdP
So that if the process be isothermal [TdS = d (TS)] as wellas isobaric (dP = O), we get
d (H – TS) = O or dG = O
i.e. G = H – TS = constant.
The function G = H – TS = U + PV – TS
= U – TS + PV
= F + PV
242 Thermal Physics
is known as the Thermodynamic potential at constant pressure orGibb’s function. It is obvious that the function G remains constantif a thermodynamic process remains isothermal as well asisobaric. Thus the fundamental equation defining Gibb’sfunction G is
G = U – TS + PV
or dG = dU – d (TS) + d (PV)
= dU – TdS – SdT + PdV + VdP
= – PdV – SdT + PdV + VdP
(QdU = TdS – PdV)
= VdP – SdT.
Here the independent variables are P and T. Taking partialdifferentials
T
GP∂⎛ ⎞
⎜ ⎟∂⎝ ⎠= and
P
GV S
T∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠
and since dG is a perfect differential
T
GT P∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
=P
GP T∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
orP
VT∂⎛ ⎞
⎜ ⎟∂⎝ ⎠=
T
SP∂⎛ ⎞−⎜ ⎟∂⎝ ⎠
…(iv)
This is the fourth thermodynamical relation of Maxwell.
Thus we see that the four thermodynamical potentialU (S, V) ; F (T, V) ; H (S, P) and G (T, P) lead us to fourthermodynamical relations – known as Maxwell’s relations.
The four functions can be expressed in terms of measurablequantities P, V, T and S as follows :
dU = TdS – PdV,
dF = – PdV – SdT
dH = TdS + VdP
and dG = VdP – SdT.
243Changing State
The four quantities U (S, V), H (S, P), F (T, V) andG (T, P) are called thermodynamic potentials because thethermodynamic variables S, T, P and V can be derived fromthem by their differentiations with respect to the independentvariables associated with them in the same manner as thecomponents of a force are deduced from a force potential.These have already been derived in the equations above butare once again listed below :
S =P V
G FT T∂ ∂⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
T =V P
U HS S
∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
P =S T
U FV V∂ ∂⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
V =S T
H GP P
∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
The above equations give the values of thermodynamicalvariables in terms of thermodynamical potentials.
