Thermodynamics Partha 2015 upto30th-Jan.pptp.hazra/Thermodynamics_Partha... · Thermodyyynamics:...

Classical Thermodynamics

CHM102CHM102

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Ch i l B diCh i l B di DD SS P lP lChemical BondingChemical Bonding: Dr. : Dr. SauravSaurav PalPal

SpectroscopySpectroscopy: Dr Pankaj Mandal: Dr Pankaj MandalMolecular propertiesMolecular properties

SpectroscopySpectroscopy: Dr. Pankaj Mandal: Dr. Pankaj Mandal

KineticsKineticsKineticsKinetics

Thermodynamics Thermodynamics Macroscopic propertiesMacroscopic properties

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Thermodynamics: “Heat” + “Study of motion”Thermodynamics: “Heat” + “Study of motion” Heat TransferHeat Transfery yy y

ThemodynamicsThemodynamics: Heat, work, energy: Heat, work, energyyy , , gy, , gy

Wide ApplicationsWide Applications: :

1. Energy change associated to all chemical and physical processes.1. Energy change associated to all chemical and physical processes.gy g p y pgy g p y p

2. Mutual transformation of different kinds of energy.2. Mutual transformation of different kinds of energy.

3. To predict the direction and extent of chemical reaction.3. To predict the direction and extent of chemical reaction.

Birth:Birth:3

Birth:Birth:

Industrial revolution (late 18th, early 19th century)Industrial revolution (late 18th, early 19th century)

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ThermodynamicsThermodynamics

Basically is based on four laws. Basically is based on four laws.

These laws are not formulated, rather these are generalization These laws are not formulated, rather these are generalization ggdeduced from our long experience in nature.deduced from our long experience in nature.

Cl l h d h l Classical thermodynamics… is the only physical theory of universal content which I am convinced that, within the applicability of its basic concepts, will pp y f p ,never be overthrown.”

Alb t Ei t i5

Albert Einstein

Classical ThermodynamicsClassical Thermodynamics::

Describes Describes macroscopic properties macroscopic properties of the systems.of the systems.

E ti l E i i lE ti l E i i lEntirely Empirical Entirely Empirical

Based on 4 Laws:Based on 4 Laws:

00th th Law Law Defines Temperature (T) Defines Temperature (T)

11st st Law Law Law of conservation of energy. It tells that system may Law of conservation of energy. It tells that system may exchange energy with its surroundings strictly by heat flow or exchange energy with its surroundings strictly by heat flow or

kk D fi I t l E (U)D fi I t l E (U)work. work. Defines Internal Energy (U)Defines Internal Energy (U)

22nd nd Law Law Defines Entropy (S) Defines Entropy (S) py ( )py ( )

33rd rd Law Law Gives Numerical Value to Entropy Gives Numerical Value to Entropy

Systems

• A system is part of the universe chosen for observation, separately from the rest of the universe.

• The system plus surroundings comprise a universe.

• The boundary between a system and its surroundings is the system wallsystem wall.

• If heat cannot pass through the system wall, it is termed an adiabatic wall, and the system is said to be thermally isolated, y yor thermally insulated.

• If heat can pass through the wall, it is termed a diathermal llwall.

• Two systems connected by a diathermal wall are said to be in thermal contact

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thermal contact.

Isolated, Closed and Open Systemsy• An isolated system cannot exchange mass or energy with its

surroundings.g

• The wall of an isolated system must be adiabatic.

• A closed system can exchange energy, but not mass, with its surroundings.

• The energy exchange may be mechanical (associated with a volume change) or thermal (associated with heat transfer th h di th l ll)through a diathermal wall).

• An open system can exchange both mass and energy with

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• An open system can exchange both mass and energy with its surroundings.

Open SystemOpen System

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Closed SystemClosed Systemyy

10Isolated SystemIsolated System Isolated SystemIsolated System

Isolated, Closed and Open Systems

Isolated Closed OpenIsolated System

Neither energy b

ClosedSystem

Energy, but not

OpenSystem

Both energy and

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nor mass can be exchanged.

gy,mass can be exchanged.

gymass can be exchanged.

Properties of a SystemProperties of a System

Thermodynamic variables which are experimentallyThermodynamic variables which are experimentallyThermodynamic variables which are experimentally Thermodynamic variables which are experimentally measurable. Such variables are macroscopic properties measurable. Such variables are macroscopic properties such as P, V, T, m, composition, viscosity etc.such as P, V, T, m, composition, viscosity etc.

Thermodynamic variables Thermodynamic variables

e tensi e ariablese tensi e ariablesIntensive variablesIntensive variables extensive variablesextensive variables

T,T, P,P, density,density, specificspecific gravity,gravity,surfacesurface tension,tension,

Volume, energy, enthalpy,Volume, energy, enthalpy,Entropy free energy massEntropy free energy mass,,

specificspecific heatheat capacity,capacity, dielectricdielectricconstantconstant..

DoesDoes notnot dependdepend uponupon massmass ofof

Entropy, free energy, massEntropy, free energy, mass

Depends on the mass Depends on the mass of the systemof the system

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DoesDoes notnot dependdepend uponupon massmass ofofthethe systemsystem

of the system.of the system.

States of a systemHow do we define a State of a SystemHow do we define a State of a System??yy

Macroscopic state of a system can be specified by Macroscopic state of a system can be specified by thermodynamic variablesthermodynamic variables which are experimentallywhich are experimentallythermodynamic variables thermodynamic variables which are experimentally which are experimentally measurablemeasurable

C itiC iti f h h i l i th t if h h i l i th t i• Composition • Composition –– mass of each chemical species that is mass of each chemical species that is present in the system.present in the system.

• pressure (p or P), volume (V), • pressure (p or P), volume (V), TemperatureTemperature (T), (T), densitydensity, , etc.etc.

• field strength, if magnetic/electrical field act on the • field strength, if magnetic/electrical field act on the systemsystem

13• gravitational field• gravitational field

Equilibrium StatesEquilibrium States

1. Composition remains fixed and definite.1. Composition remains fixed and definite.

2. Temp. at all parts of the system is the same.2. Temp. at all parts of the system is the same.

3. No unbalanced forces between different parts 3. No unbalanced forces between different parts of the system or between system and surroundings.of the system or between system and surroundings.

System at equilibrium must have definite P, T, and compositionSystem at equilibrium must have definite P, T, and compositionSystem at equilibrium must have definite P, T, and compositionSystem at equilibrium must have definite P, T, and composition

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ProcessesProcesses

AA processprocess refers to the change of a system from one equilibrium refers to the change of a system from one equilibrium state to another.state to another.

Isothermal (T)Isothermal (T)( )( )

Adiabatic (Adiabatic (no heat exchange between system and surroundingsno heat exchange between system and surroundings))..

Isobaric (p),Isobaric (p),

Isochoric (V)Isochoric (V)

CyclicCyclic

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yy

Reversible ProcessReversible Process::

• Change must occur in • Change must occur in successive stages of infinitesimal quantitiessuccessive stages of infinitesimal quantities• • Infinite durationInfinite duration• Changes of the thermodynamic quantities in the different stages will be the• Changes of the thermodynamic quantities in the different stages will be thesame as in the for ard direction b t opposite in signsame as in the for ard direction b t opposite in sign rtrt for ardfor ardsame as in the forward direction but opposite in sign same as in the forward direction but opposite in sign wrtwrt forward forward directiondirection

Irreversible ProcessIrreversible Process::

•• Real / SpontaneousReal / Spontaneous• • Real / SpontaneousReal / Spontaneous• Occurs suddenly or • Occurs suddenly or spontaneously spontaneously without the restriction of without the restriction of occurring in successive stages of infinitesimal quantities.occurring in successive stages of infinitesimal quantities.• • Not remain in the Not remain in the virtual equilibrium during the transition.virtual equilibrium during the transition.• The work (w) in the • The work (w) in the forward and backward processes would forward and backward processes would be unequal.be unequal.

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be u equabe u equa

StateState functionfunction:: ThoseThose thermodynamicthermodynamic propertiesproperties dependsdepends onon thethe statestate ofof thethesystem,system, notnot onon thethe pathpath throughthrough whichwhich itit hashas beenbeen broughtbrought ininthatthat statestate.. PotentialPotential energy,energy, InternalInternal energy,energy, entropy,entropy,gy,gy, gy,gy, py,py,enthalpyenthalpy..

Path function: Path function: Those thermodynamic Those thermodynamic propertiesproperties depends on the path depends on the path through which it has been brought in that statethrough which it has been brought in that state..

((H t kH t k))((Heat, workHeat, work).).

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Mathematical formulation of State function (Mathematical formulation of State function (TUTORIAL 1TUTORIAL 1):):

1 If any thermodynamic property or function z = f(1 If any thermodynamic property or function z = f(x yx y)) dependsdepends1. If any thermodynamic property or function, z = f(1. If any thermodynamic property or function, z = f(x,yx,y) ) depends depends on the initial and final values of thermodynamic variableson the initial and final values of thermodynamic variables, then the , then the change of z, i.e., change of z, i.e., dzdz is a perfect differentialis a perfect differential, ,

2. In case Z is state function, it will follow the following 2. In case Z is state function, it will follow the following mathematical relationship,mathematical relationship,

3. If Z = f(x,y) 3. If Z = f(x,y) depends on the initial and final values of depends on the initial and final values of thermodynamic variables, then also thermodynamic variables, then also ∮dz= 0∮dz= 0

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Concept of Heat and workConcept of Heat and workJoule's experiment

W = JQW = JQW = JQW = JQ

W = work expended in the W = work expended in the d ti f h td ti f h tproduction of heat orproduction of heat or

obtained from heatobtained from heat

Heat Heat WorkWork

1 calorie = 4.184 Joules1 calorie = 4.184 Joules

Both heat and work represents energy in transit.Both heat and work represents energy in transit.

Work involved in a process , heat change involved duringWork involved in a process , heat change involved duringa process depend on the path of transformation.a process depend on the path of transformation.

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Work and heat are the two methods by which energy is exchangedWork and heat are the two methods by which energy is exchangedbetween system and surroundingsbetween system and surroundings..

Sign Convention of heat and work:Sign Convention of heat and work:

Work:The work done by the system is defined to be nagative (-). The work done on the system – the external work of

mechanics is positive (+)mechanics – is positive (+).

Heat:The heat absorbed by the system is defined to be positive (+). The given out by the system is negative (-).

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21U = U = UUff –– UUii MeasurableMeasurable

(TUTORIAL(TUTORIAL--1)1)Concept of Internal energy:Concept of Internal energy:

Internal Energy, U. is the total energy within a system.Internal Energy, U. is the total energy within a system.U is the internal energy of the body (due to molecular motions U is the internal energy of the body (due to molecular motions gy y (gy y (and intermolecular interactions)and intermolecular interactions)

•• Extensive propertyExtensive property•• Extensive property.Extensive property.•• State function, independent of path.State function, independent of path.•• For cyclic process,For cyclic process,

22dUdU is a perfect or exact differential is a perfect or exact differential

Thermal Equilibrium and Thermal Equilibrium and Zero’thZero’th Law of ThermodynamicsLaw of Thermodynamics

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First Law of ThermodynamicsFirst Law of Thermodynamics::

Energy cannot be destroyed, it can be transformedEnergy cannot be destroyed, it can be transformed totoone form to another. (Law of conservation of energy).one form to another. (Law of conservation of energy).

K.E.= 0, P.E. = maximumK.E.= 0, P.E. = maximum

K E = maximum P E = 0K E = maximum P E = 0K.E.= maximum, P.E. = 0K.E.= maximum, P.E. = 0

Energy is conservedEnergy is conserved24

Energy is conservedEnergy is conserved

11stst law of Thermodynamicslaw of Thermodynamics

The first law for a closed system or a fixed mass may be expressedThe first law for a closed system or a fixed mass may be expressedThe first law for a closed system or a fixed mass may be expressed The first law for a closed system or a fixed mass may be expressed as:as:

net energy transfer to (or from)net energy transfer to (or from)net energy transfer to (or from)net energy transfer to (or from)the system as heat and work the system as heat and work = = net increase (or decrease) in thenet increase (or decrease) in the

total energy total energy of the systemof the system

q+ w = q+ w = UU

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Moving Boundary Work

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How the expansion takes place in a pistonHow the expansion takes place in a piston--cylinder device?cylinder device?

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Equation of state for Ideal and Real gasEquation of state for Ideal and Real gas

Ideal gas: PV =Ideal gas: PV = nRTnRTIdeal gas: PV Ideal gas: PV nRTnRT

Real gas: van Real gas: van derder Waal’s equationWaal’s equation

nRTnbVV

anP

2

2NN·b·b

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Work and heat change associated in Isothermal ReversibleWork and heat change associated in Isothermal ReversibleTUTORIAL 2TUTORIAL 2--33

Work and heat change associated in Isothermal ReversibleWork and heat change associated in Isothermal Reversibleand irreversible expansionand irreversible expansion

w

(b) Irreversible process:(b) Irreversible process:

WWirrirr = = PPextext (V(V22 –– VV11))

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Adiabatic process in an Ideal GasTUTORIAL 4TUTORIAL 4

Adiabatic process in an Ideal Gas • Since dQ = 0 for an adiabatic process,dU = – P dV and dU = CV dT, so that dT = – (P/CV) dV .V , ( V)

• For an ideal gas, PV = nRT, so that P dV +V dP = nR dT = – (nRP/CV) dV.so that P dV V dP nR dT (nRP/CV) dV.

Hence V dP + P (1 +nR/CV) dV = 0.

Th C dP/P (C R) dV/V 0Thus, CV dP/P + (CV + nR) dV/V = 0.

For an ideal gas, CP – CV = nR.

so that CV dP/P + CP dV/V = 0, or dP/P + γ dV/V = 0.• Integration gives ln P + γ ln V = constant, so that

30PVγ = constant.

Adiabatic process in an Ideal Gas W k d i ibl di b ti• Work done in a reversible adiabatic process

For a reversible adiabatic process, PVγ = K.p ,• Since the process is reversible, W = -CV∆T =Cv(T1-T2) ,

For 1 mole of gas , T=PV/R gso that Wadi = = Cv[P1V1/R-P2V2/R]

= Cv/R[P1V1-P2V2] Cv/R[P1V1 P2V2]

W = 1/(γ –1) [P2V2 – P1V1].• For an monatomic gas, γ = 5/3, so that

W = –(3/2)] [P2V2 – P1V1].

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Reversible Processes for an Ideal Gas

Adiabatic process

Isothermal process

Isobaric process

Isochoric processprocess process process process

PVγ = KC /C

T constant P constant V constantγ = CP/CV

W = – [1/(γ –1)] [P V P V ]

W = nRT ln(V2 /V1) W = P V W = 0.[P2V2 – P1V1]

ΔU = CV ΔT ΔU = 0 ΔU = CV ΔT ΔU = CV ΔT

PV = nRT, U = ncVT, cP – cV = R, γ = cP/cV.

32Monatomic ideal gas cV = (3/2)R, γ = 5/3.

Thermal insulationThermal insulationFFFFOOOO

CCOOCC

OO

550044

120120

100100

ValveValve

4400330022001100

00110022003300440055

8080

00

2020

2020

4040

6060

6060

4040

AA BB

ValveValve5500

6060

TT11, V, Vm,1m,1, P, P11BB

StirrerStirrer

1. 1. No change in temperature No change in temperature was detected,was detected,

dqdq = 0= 0

22 As expansion is taking place againstAs expansion is taking place against zero pressurezero pressure2. 2. As expansion is taking place against As expansion is taking place against zero pressurezero pressure,,

dwdw = 0,= 0,

As a result, As a result, dUdU = 0= 0

U = f(V,T)U = f(V,T)VUT T

UVT

VU

0

UVT ,

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TVUHence ,

0

U

((VALID for Ideal gas onlyVALID for Ideal gas only))0

TV

((VALID for Ideal gas onlyVALID for Ideal gas only))

U U 0

VU

T

((VALID for REAL gas onlyVALID for REAL gas only))TV

U

= a/v= a/v22

Actually the gas which had expanded into B was somewhat Actually the gas which had expanded into B was somewhat cooler and when thermal equilibrium was finally established the cooler and when thermal equilibrium was finally established the q yq ygas was at a slightly different temperature from that before the gas was at a slightly different temperature from that before the expansion. expansion.

Because the system used by Joule Because the system used by Joule had a very large heat had a very large heat capacitycapacity compared with the heat capacity of air, the small change compared with the heat capacity of air, the small change

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of temperature that took place was not observed. of temperature that took place was not observed.

Concept of Enthalpy and heat CapacitiesConcept of Enthalpy and heat Capacities

Prove that, Prove that, H = H = qqPP

Prove that, CProve that, CPP -- CCVV = R (for one mole of Ideal gas) = R (for one mole of Ideal gas)

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Joule-Thompson Effect

AA gasgas passespasses throughthrough aa POROUSPOROUS PLUGPLUG fromfromaa regionregion wherewhere itit isis atat highhigh pressurepressure toto aa regionregionwherewhere itit isis atat lowerlower pressurepressure.. TheThe gasgas expands,expands,andand thethe temperaturetemperature ofof thethe gasgas cancan bebe loweredlowered..ThisThis isis anan importantimportant tooltool inin lowlow temperaturetemperaturephysicsphysics..

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JouleJoule--Thompson ApparatusThompson Apparatus

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39

40

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HPH

T

1JT

TP

T

H PH

CHP

PT

1

PT

VT 1

Later we can showLater we can show

V

TVT

CPT

PPJT

H

1

PP

In an adiabatic throttle process the gas pressure is reduced (In an adiabatic throttle process the gas pressure is reduced (PP <P<P ))TT

In an adiabatic throttle process, the gas pressure is reduced (In an adiabatic throttle process, the gas pressure is reduced (PP22<P<P11), ), and thus and thus

If the temperature of the gas is reduced,If the temperature of the gas is reduced,T2<T1T2<T1 , which , which 0)(

HTJT

produces a cooling effect;produces a cooling effect;

If th t t f th i i dIf th t t f th i i d T2>T1T2>T1

0)( HTJ P

T If ,the temperature of the gas is raisedIf ,the temperature of the gas is raised, T2>T1, T2>T1, ,

which produces a heating effect; which produces a heating effect; 0)(

HTJ PT

43 If , the temperature of the gas has no change, i.e., If , the temperature of the gas has no change, i.e.,

T2=T1T2=T10)(

HTJ PT

Second law of ThermodynamicsSecond law of Thermodynamics

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Second Law of ThermodynamicsSecond Law of Thermodynamics

QQ WWIs it possible?Is it possible?

It is impossible for a system to undergo a cyclic process whose sole effectsIt is impossible for a system to undergo a cyclic process whose sole effectsIt is impossible for a system to undergo a cyclic process whose sole effects It is impossible for a system to undergo a cyclic process whose sole effects are are the flow of heat into the systemthe flow of heat into the system from a heat reservoir and the performance from a heat reservoir and the performance

of an of an equivalent amount of workequivalent amount of work by the system on the surroundings.by the system on the surroundings.--------------KelvinKelvin--plank statementplank statement

45The second types of perpetual motion machineThe second types of perpetual motion machine

--------------KelvinKelvin--plank statementplank statement

Kelvin statement (1851)Kelvin statement (1851)

No process can completely convert heat into work; i.e. it No process can completely convert heat into work; i.e. it is impossible to build a “perfect” heat engine.is impossible to build a “perfect” heat engine.

Efficiency = work done/heat absorbedEfficiency = work done/heat absorbed

Hot Reservoir, Hot Reservoir, TTHH

QQHH

= W/Q= W/QHH= Q= QHH-- QQCC/Q/QHH

EngineEngine

HH

QQ

WW

Later we will showLater we will show

Cold Reservoir, Cold Reservoir, TTCC

QQCC Later we will show,Later we will show,W = QW = QHH**T/TT/THH

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CC

Heat EngineHeat Engine

A heat engine is a cyclic process that absorbs heat fromA heat engine is a cyclic process that absorbs heat fromA heat engine is a cyclic process that absorbs heat from A heat engine is a cyclic process that absorbs heat from a heat bath and converts it into work. We shall see that a heat bath and converts it into work. We shall see that in the cyclic process the engine also dissipates some in the cyclic process the engine also dissipates some y p g py p g pheat to a heat bath at a lower temperature.heat to a heat bath at a lower temperature.

Hot Reservoir, Hot Reservoir, TTHH

QQHH

Hot Reservoir, Hot Reservoir, TTHH

QQHH

EngineEngine

QQHHWW

EngineEngine

QQHHWW

Cold ReservoirCold Reservoir TTCC

QQCC

Cold Reservoir,Cold Reservoir, TTCC

QQC C = 0= 0

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Cold Reservoir, Cold Reservoir, TTCC

Real engine. Real engine. QQHH = = QQCC + + WWCold Reservoir, Cold Reservoir, TTCC

Impossible engine. Impossible engine. QQHH = = WW

QQ QQ WWQQ11 = Q= Q22 + W+ W

• It is impossible for a system to undergo a cyclic process whose sole p y g y peffects are the flow of heat into the system from a cold reservoir and the flow of an equal amount of heat out of the system into a hot reservoir.

• -------Clausius statement

Refrigeration engine takes away heat from the colder reservoirRefrigeration engine takes away heat from the colder reservoirt h t i ith th h l f t l l t i l kt h t i ith th h l f t l l t i l k

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to hot reservoir with the help of external electrical workto hot reservoir with the help of external electrical work

Carnot engine or Carnot cycleCarnot engine or Carnot cycle

1. The engine must operate in 1. The engine must operate in complete cycles to exclude any complete cycles to exclude any k i l d i it hk i l d i it hwork involved in its own changework involved in its own change..

2. To obtain maximum work in a cycle of operations, every step 2. To obtain maximum work in a cycle of operations, every step should be carried out in a reversible fashion.should be carried out in a reversible fashion.

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Carnot CycleCarnot CyclePressurePressure

aa••aa

QQ11Step Step --II

••bb TT11Q=0Q=0

Step Step --IVIV

••Q=0Q=0

Q=0Q=0

Step Step --IIIIStepStep --IIIIIIdd

QQ22 TT22••ccStep Step --IIIIII

VolumeVolume

Step 1 aStep 1 a--b (Isothermal Reversible Expansion)b (Isothermal Reversible Expansion): :

The gas enclosed in a The gas enclosed in a cyclindercyclinder fitted with frictionless fitted with frictionless Piston. To start with cylinder containing the gas is kept in a Piston. To start with cylinder containing the gas is kept in a large thermostat at higher temp. Tlarge thermostat at higher temp. T11 (source), and suppose the (source), and suppose the a ge t e ostat at g e te pa ge t e ostat at g e te p 11 (sou ce), a d suppose t e(sou ce), a d suppose t evolume of the gas be Vvolume of the gas be V11. .

VV11 VV2211 22

As, temperature remains constant, hence the change inAs, temperature remains constant, hence the change inInternal energy (Internal energy (U = 0).U = 0).gy (gy ( ))

Hence, Hence, U = q + WU = q + W

Heat absorbed by the system = (Heat absorbed by the system = ( ) of work done by the system) of work done by the system

The heat absorbed by the gas QThe heat absorbed by the gas Q == RTlnRTln(V(V /V/V ))

Heat absorbed by the system = (Heat absorbed by the system = (--) of work done by the system) of work done by the system

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The heat absorbed by the gas, QThe heat absorbed by the gas, Q11 = = RTlnRTln(V(V22/V/V11))= work done by the gas= work done by the gas= w= w11

Step 2 (Adiabatic Reversible Expansion)Step 2 (Adiabatic Reversible Expansion): :

The The cyclindercyclinder is taken out from the thermostat and kept in a is taken out from the thermostat and kept in a thermally insulated enclosure. The gas is allowed to expand thermally insulated enclosure. The gas is allowed to expand further from volume Vfurther from volume V22 to Vto V33 adiabatically and adiabatically and reverasiblyreverasibly until until u t e o o u eu t e o o u e 22 toto 33 ad abat ca y a dad abat ca y a d e e as b ye e as b y u tu tthe temperature falls down to that of the sink Tthe temperature falls down to that of the sink T22. .

The heat absorbed by the gas = nilThe heat absorbed by the gas = nil

The work done by the gas wThe work done by the gas w == CC (T(T TT ) (T) (T > T> T ))The work done by the gas, wThe work done by the gas, w22 = = CCvv (T(T2 2 –– TT11) (T) (T1 1 > T> T22))

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Step 3 (Isothermal Reversible Compression)Step 3 (Isothermal Reversible Compression): :

The cylinder is then placed in a thermostat at lower temp. TThe cylinder is then placed in a thermostat at lower temp. T22(sink), and the gas is compressed isothermally and reversibly (sink), and the gas is compressed isothermally and reversibly from V3 to V4from V3 to V4from V3 to V4. from V3 to V4.

VV33 VV44

The work done on the gas, wThe work done on the gas, w33= (= (--)RT)RT22ln(Vln(V44/V/V33) ) ((as as VV4 4 < V< V33))

The heat given out by the gas, QThe heat given out by the gas, Q22 = = -- ww33= RT= RT22ln(Vln(V44/V/V33))

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Step 4 (Adiabatic Reversible Compression)Step 4 (Adiabatic Reversible Compression): :

The The cyclindercyclinder is taken out from the thermostat and kept in a is taken out from the thermostat and kept in a thermally insulated enclosure. The gas is allowed to compress thermally insulated enclosure. The gas is allowed to compress reversibly to its volume Vreversibly to its volume V11 and its original temperature Tand its original temperature T11 is is e e s b y to ts o u ee e s b y to ts o u e 1 1 a d ts o g a te pe atu ea d ts o g a te pe atu e 11 ssattained.attained.

The heat absorbed by the gas = nilThe heat absorbed by the gas = nil

The work done on the gas wThe work done on the gas w == CC (T(T TT ))The work done on the gas, wThe work done on the gas, w44 = = CCvv (T(T1 1 –– TT22))

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1.1. Calculate total work involved in all these steps:Calculate total work involved in all these steps:

WWnetnet = R = R lnln(V(V22/V/V11)[T)[T11--TT22] = Q] = Q11--QQ22

2 Net heat absorbed :2 Net heat absorbed :2. Net heat absorbed :2. Net heat absorbed :

Only step 1, heat absorption takes place, Only step 1, heat absorption takes place, QQ11 = RT= RT11ln(Vln(V22/V/V11))

3. Efficiency of the engine:3. Efficiency of the engine:

η= net work done/net heat absorbed= (Tη= net work done/net heat absorbed= (T11--TT22)/T)/T11= (= (QQ11--QQ22)/ Q)/ Q11(( 11 22)) 11

WWnetnet= Q= Q11××((T/T)T/T)

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netnet 11 (( ))

1)1) OnlyOnly ΔΔT/TT/T fractionfraction ofof thethe totaltotal quantityquantity ofof heatheat takentaken fromfromsourcesource cancan bebe transformedtransformed inin toto workwork..

2)2) TheThe efficiencyefficiency ofof engineengine ((ηη==11--TT22/T/T11)) dependsdepends onlyonly onon thethetemperaturestemperatures ofof thethe sourcesource andand thethe sinksink andand isis independentindependent ofoftemperaturestemperatures ofof thethe sourcesource andand thethe sink,sink, andand isis independentindependent ofofthethe workingworking substancesubstance ofof thethe engineengine,, providedprovided itit isisreversiblereversible..

3)3) TheThe wholewhole ofof heatheat suppliedsupplied QQ willwill bebe transformedtransformed toto work,work, ii..ee..,,W=Q,W=Q, onlyonly whenwhen TT22==00.. ThatThat is,is, thethe engineengine shouldshould workwork betweenbetweenQ,Q, yy 22 ,, ggabsoluteabsolute zerozero andand aa higherhigher temperaturetemperature.. SinceSince thisthis cannotcannot bebe realizedrealizedinin practice,practice, completecomplete conversionconversion ofof heatheat intointo workwork isisi ti bli ti blimpracticableimpracticable..

44)) WhenWhen TT11 =T=T22 ,, ii..ee..,, thethe temperaturestemperatures ofof thethe sourcesource andand thethe sinksink areare thethe)) 11 22 ,, ,, ppsame,same, WW isis zerozero.. ThatThat isis nono workwork willwill bebe obtainedobtained byby operatingoperating thetheengineengine underunder isothermalisothermal conditioncondition..

Carnot realized that in reality it is not possible to build a Carnot realized that in reality it is not possible to build a y py pthermodynamically reversiblethermodynamically reversible engine, so real heat engines are engine, so real heat engines are

less efficient than less efficient than idealized reversible engineidealized reversible engine. .

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ProblemProblem:: AA CarnotCarnot engineengine workingworking betweenbetween 0000CC andand 10010000CC takestakes upupProblemProblem:: AA CarnotCarnot engineengine workingworking betweenbetween 0000CC andand 10010000CC takestakes upup840840 joulesjoules fromfrom thethe highhigh temperaturetemperature reservoirreservoir.. CalculateCalculate thethe workworkdone,done, thethe heatheat rejectedrejected andand thethe efficiencyefficiency..

AnsAns:: 00..225225 kJ,kJ, 146146 calories,calories, 00..268268

Home AssignmentHome Assignment

Draw Carnot cycle in (TDraw Carnot cycle in (T--S) diagram and show that the net workS) diagram and show that the net workDraw Carnot cycle in (TDraw Carnot cycle in (T--S) diagram, and show that the net work S) diagram, and show that the net work done by the cycle is given by, done by the cycle is given by, WWnetnet = = S*S*(T(T22--TT11), where ), where S = net S = net entropy change in this cycle.entropy change in this cycle.

11 Efficiency of Carnot engine is maximum (Efficiency of Carnot engine is maximum (Proof is done in class)Proof is done in class)1.1. Efficiency of Carnot engine is maximum. (Efficiency of Carnot engine is maximum. (Proof is done in class)Proof is done in class)

2.2. Efficiency of any reversible engine is more than the efficiency ofEfficiency of any reversible engine is more than the efficiency ofI ibl i (I ibl i ( f i d i lf i d i l ))Irreversible engine. (Irreversible engine. (proof is done in clasproof is done in class).s).

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Second law of ThermodynamicsSecond law of Thermodynamics

1. Concept of entropy1. Concept of entropy

2. Concept of time arrow2. Concept of time arrow2. Concept of time arrow2. Concept of time arrow

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Limitations of first lawLimitations of first law

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62TheThe totaltotal energyenergy isis disperseddispersed intointo randomrandom thermalthermal motionmotion ofof thetheparticlesparticles inin thethe systemsystem..

• The thermodynamic property of a system that is related to itsdegree of randomness or disorder is called entropy (S).degree of randomness or disorder is called entropy (S).

• Entropy is a measure of the extent to which energy isdispersed.

Th t S d th t h ΔS S S t t• The entropy S and the entropy change ΔS=S2-S1 are statefunctions.

• The entropy S has a unique value, once the pressure P,temperature T and the composition n of the system arespecified, S = S(P,T,n).

Th t i t i t i i ith th• The entropy is an extensive property, i.e. increases with theamount of matter in the system. 63

Entropy is state Entropy is state functionfunction? Home Assignment? Home Assignment

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ClausiusClausius InequailityInequaility::

dsds ≥≥ dqdq/T/Tdsds ≥ ≥ dqdq/T/T

Entropy Criteria in different ProcessesEntropy Criteria in different Processes

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Change in the extent to which energy is dispersed depends on Change in the extent to which energy is dispersed depends on g gy p pg gy p phow much energy is transferred as heat.how much energy is transferred as heat.

In case of closed system:In case of closed system:

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Combination of first and 2Combination of first and 2ndnd law:law:

dSdS == nn

nn

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Entropy change due to Phase TransitionEntropy change due to Phase Transition

At the transition temperature, any transfer of energy as heat At the transition temperature, any transfer of energy as heat between system and its surroundings is reversible because thebetween system and its surroundings is reversible because thebetween system and its surroundings is reversible because thebetween system and its surroundings is reversible because the

two phases in the system are in equilibrium two phases in the system are in equilibrium

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06/02/201506/02/2015

How to calculate the entropy change in Surroundings?

2

Entropy change in case of Adiabatic  Cases

Entropy of Mixing

4

Understanding Entropy Considering Molecular Picture

If N b th t t l b f di ti i h bl l l iIf N be the total number of distinguishable molecules in asystem, and N1, N2, N3……. etc. be number of moleculesdistributed in different energy levels, then the number ofgy ,microstates corresponding to the given distribution is calledThermodyamic Probability,

S= klnW

6

Let us a take a sample containing 16 distinguishable molecules,sharing a total energy of 16 E and quantum states differing

E

by unit E.

3

4

1

2

W=1 W=8.9  10⁵ W=1.44  10⁷

0

Number of microstates increases with increase in the randomnessin the distribution of the molecules. Hence, entropy increases. 

S= klnW

Third LAW  of Thermodynamics

W =1S= klnW = 0

W >1S= klnW 0S= klnW = 0 S  klnW 0

Total: 191.05

Third LAW of Thermodynamics

Absolute zero ????

Vapour cycle Refrigeration 

Magnetic refrigeration

Energy levels of Paramagnetic Substance in Absence and Presence of Magnetic Field:g

Concept of Helmholtz work Function and Gibbs Free energy

Helmholtz work Function A

Helmholtz work Function A

Helmholtz work Function A

G = H – TSWhy it is called  free energy or available energy?

If a system undergoes reversible changedG = dH – d(TS)

= d(U + PV) – d(TS)= dU + d(PV) – TdS – SdT= dU + VdP + PdV – TdS – SdT

At constant external P & Isothermal conditionAt constant external P & Isothermal conditiondGP,T = dU + PextdV – TdS

= dU + PextdV – dq= -(dW) + P dV= -(dW) + PextdV= -(dWP-V + dWnonP-V ) + PextdV= -(PdV + dWnonP-V ) + PextdV

dW= - dWnonP-V

-dGP,T = dWnonP-VIt is the amount of work required for any externalIt is the amount of work required for any external

use exclusive of the expansion or mechanical work. This work may be electrical work for pushing electron through a circuit or work required for transmitting nerve impulse or work of contracting muscles.

Isothermal processes occur in living cells. 

Surroundings act as heat sink in most of the casescases.

Problem: How much energy is available for sustainingl d i i f h b i f 1muscular and nervous activity from the combustion of 1

mol of glucose molecules under standard conditions at37 °C (blood temperature)? The standard entropy ofreaction is +259.1 JK-1 mol-1. ΔrH = -2808 kJ mol-1

Method: The available energy from the reaction isgyequal to the change in standard Gibbs energy for thereaction (ΔrG). To calculate this quantity, it is legitimatereaction (ΔrG ). To calculate this quantity, it is legitimateto ignore the temperature dependence of the reactionenthalpy to obtain Δ H and to substitute the data intoenthalpy, to obtain ΔrH, and to substitute the data into

STHG rrr

Answer: Because the standard reaction enthalpy is-2808 kJ mol-1, it follows that the standard reactionGibbs energy isgy

1

111r )molJK1.259()K310(kJmol2808G

Therefore, W dd = -2888 kJ.mol-1 for the combustion

1kJmol2888

Therefore, Wadd,max 2888 kJ.mol for the combustionof 1 mol glucose molecules, and the reaction can use2888 kJ for external work2888 kJ for external work.

To place this results in perspective, consider that aperson of mass 70 kg needs to do 2 1 kJ of work to climbperson of mass 70 kg needs to do 2.1 kJ of work to climbvertically through 3 m; then at least how much amountf l i d d t l t th t k?of glucose is needed to complete the task?

Gibbs Free Energy

Gibbs Free Energygy

1. If G is negative, the g ,forward reaction is spontaneous.p

2. If G is 0, the system is at equilibrium.at equilibrium.

3. If G is positive, the reaction is spontaneousreaction is spontaneous in the reverse direction.

Free Energy and EquilibriumFree Energy and Equilibrium

Remember from above:If G is 0, the system is at equilibrium.

So Gmust be related to the equilibrium constant, q ,K. The standard free energy, G°, is directly linked to Keq by: 

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Masters Equations of Chemical Thermodynamics

A closed system (constantcomposition, and only p‐V work)

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α

κ

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The "cyclic relation" (sometimes called Euler's chain relation) is a calculus identity which is very useful in thermodynamics. This relation can appear in several different forms all of which are equivalent. The form that we will find most useful is,

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Maxwell’s Relation:

Imp

IImp

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Maxwell Relation for G The Gibbs function (or free energy) is defined as

G = U – TS + PVG   U  TS   PVdG = dU – TdS – SdT + PdV + Vdp .

dU = TdS – PdV,so that dG = – SdT + VdP ; i.e. G = G(T,P).

dG (G/T) dT + (G/P) dPdG =(G/T)PdT + (G/P)TdP,so that S = – (G/T)P and V = (G/P)T...

2G/TP = 2G/PT,so that (S/P)T =  –(V/T)P .

Note that Maxwell’s relation equates (S/P)T , a theoreticalq antit to (V/T) α V both of hi h ma be

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quantity, to (V/T)P = α V , both of which may be measured.

Maxwell’s Relations

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