View
218
Download
1
Tags:
Embed Size (px)
Citation preview
1PHYS1001Physics 1 REGULARModule 2 Thermal Physics
ap06/p1/thermal/ptE_gases.ppt
PRESSURE
IDEAL GAS
EQUATION OF STATE
KINETIC THEORY MODEL
THERMAL PROCESSES
2
Overview of Thermal Physics Module:1. Thermodynamic Systems:
Work, Heat, Internal Energy0th, 1st and 2nd Law of Thermodynamics
2. Thermal Expansion
3. Heat Capacity, Latent Heat
4. Methods of Heat Transfer:Conduction, Convection, Radiation
5. Ideal Gases, Kinetic Theory Model
6. Second Law of ThermodynamicsEntropy and Disorder
7. Heat Engines, Refrigerators
3
Kinetic-Molecular Model of an Ideal Gas Thermal Processes
* Ideal gas, Equation of state (§18.1 p611) * Kinetic-molecular model of an ideal (results only – not the mathematical derivations) (§18.3 p619)* Heating a gas: heat capacities, molar heat capacity (§17.5 p582 §18.4 p626 §19.6 p658 §19.7 p659)* First Law of Thermodynamics: Internal Energy, Work, Heat, Paths between thermodynamic states (§19.1 p 723-725, §19.2 p725-728, § 19.3 p728-729, §19.4 p729-735)* Thermal Processes and pV diagrams: Isothermal, Isobaric Isochoric (constant volume gas thermometer), Adiabatic Cyclic (§19.5 p735-737, §19.8 p741-744, §17.3 p644-645
References: University Physics 12th ed Young & Freedman
4
HEAT ENGINES & GASES
5
Phases of matter
Gas - very weak intermolecular forces, rapid random motion
Liquid - intermolecular forces bind closest neighbours
Solid - strong intermolecular forces
low temphigh pressure
high templow pressure
6
Ideal Gas
* Molecules do not exert a force oneach other zero potential energy
* Large number of molecules
* Molecules are point-like
* Molecules are in constant random motion
* Collisions of molecules with walls ofa container and other molecules obey Newton's laws and are elastic
7
Quantity of a gasnumber of particles N
mass of particle m
molar mass M (kg.mol-1) mass of 1 mole of a substance
number of moles n ( mol) 1 mole contains NA particles
Avogadro's constant NA = 6.023x1023 mol-1
1 mole is the number of atoms in a 12 g sample of carbon-12
1 mole of tennis balls would fill a volume equal to 7 Moons
The mass of a carbon-12 atom is defined to be exactly 12 u
u atomic mass units, 1 u = 1.66x10-27 kg
(1 u)(NA) = (1.66x10-27)(6.023x1023) = 10-3 kg = 1 g
mtot = N m
If N = NA mtot = NA m = M M = NA m
n = N / NA = mtot / M
81.00 kg of water vapour H2O
M(H2O) = M(H2) + M(O) = (1 + 1 + 16) g = 18 g = 1810-3 kg
n(H2O) = mtot / M(H2O) = 1 / 1810-3 = 55.6 mol
N(H2O) = n NA = (55.6)(6.0231023) = 3.351025
m(H2O) = M / NA = (1810-3) / (6.0231023) kg = 2.9910-26 kg
1 amu = 1 u = 1.6610-27 kg
m(H2O) = 18 u = (18)(1.6610-27) kg = 2.9910-26 kg
9
Pressure P
pressure !!!
Is this pressure?
What pressure is applied to the ground if a person stood on one heel?
10Pressure P (Pa) Impact of a molecule on the wall of the
container exerts a force on the wall and the wall exerts a force on the molecule. Many impacts occur each second and the total average force per unit area is called the pressure.
P = F / A force F (N)area A (m2)pressure P (Pa)
Patmosphere = 1.013105 Pa
~1032 molecules strike our skin every day with an avg speed ~ 1700 km.s-1
11
Rough estimate of atmospheric pressure
air ~ 1 kg.m-3 g ~ 10 m.s-2 h ~ 10 km = 104 m
p = F / A = mg / A = V g / A = A h / A = g h
Patm ~ (1)(10)(104) Pa
Patm ~ 105 Pa
12
… and all the king's horses …
What force is required to separate the hemispheres? Is this force significant?
?
Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg
13
Famous demonstration of air pressure (17thC) by Otto Van Guerickle of Magdeburg
p = 1x105 PaR = 0.30 mA = 4R2
F = p AF = (105)(4)(0.3)2 NF = 105 N
14
Gauge and absolute pressures
Pressure gauges measure the pressure above and below atmospheric (or barometric) pressure.
Patm = P0 = 1 atm = 101.3 kPa = 1013 hPa = 1013 millibars = 760 torr = 760 mmHg
Gauge pressure Pg
Absolute pressure P
P = Pg + Patm0
100
200
300
400
Pg = 200 kPa Patm= 100 kPaP = 300 kPa
15
Ideal Gases – equation of state (experimental law)
p V = n R T = N k T
R, Universal gas constant (same value for all gases)
R = 8.314 J.mol-1.K-1
Boltzmann constant k = 1.38x10-23 J.K-1
k = R / NA R = k NA
must be in kelvin (K)
16
Ideal gas, constant mass (fixed quantity of gas)
All gases contain the same number of molecules when they occupy the same volume under the same conditions of temperature and pressure (Avogadro 1776 - 1856)
p V = n R T n = N / NA= p V / R T
1 1 2 2
1 2
p V p V
T T
17
Boyle's Law (constant temperature)
p = constant / V
Charles Law (constant pressure)
V = constant T
Gay-Lussac’s Law (constant volume)
p = constant T
18Isothermals pV = constant
0
20
40
60
80
100
120
140
160
180
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
volume V (m3)
pre
ssu
re p
(kP
a)
100 K
200 K
300 K
400 K
n RTp
V
19Thermodynamic system (ideal gas) work
p V T U S
mtot N n
p V = n R T
p V = N k T
k = R / NA
mtot = n M
N = n NA
heat
Q = n C T
CV or Cp
W p dV
dQS
T
internal energy
U = Q – W = n CV T
Q = 0p V = constantT V-1 = constant
20Ideal gas - equipartition of energy classical picture - not valid at low or high temperatures
Degrees of freedom - there is kinetic energy associated with each type of random motion
Translation f = 3
x
y
z
Rotationdiatomic molecule f = 2
Vibrationonly at high T
Provided the temperature is not too high (< 3000 K), a diatomic molecule has 5 degrees of freedom
21
Kinetic–Molecular model for an ideal gas (p619)
Large number of molecules N with mass m randomly bouncing around in a closed container with Volume V.
trKEpV3
2
TkNTRnpV
Kinetic-Molecular Model(Theory)
Experimental Law
For the two equations to agree, we must have:
TkNTRnKEtr 2
3
2
3
x
y
z
22 Total kinetic energy for random translational motion of all molecules, Ktr
Average translational KE of a molecule
2
2
1
2
3
2
3avgtr mvNTkNTRnKE
For an ideal gas, temperature is a direct measure of the average kinetic energy of its molecules.
Tkvvvmmv avgZavgYavgXavg 2
3
2
1
2
1 2,
2,
2,
2
221
avgmv is the average translational kinetic energy of a single molecule
23
At a given temperature T, all ideal gas molecules have the same average translational kinetic energy, no matter what the mass of the molecule
energy stored in each degree of freedom = ½ k T
Theorem of equipartition of energy (James Clerk Maxwell):
The thermal energy kT is an important factor in the natural sciences. By knowing the temperature we have a direct measure of the energy available for initiating chemical reactions, physical and biological processes.
24
Internal energy U of an ideal gas
1
2random random
U KE PE KE N f k T
2
fU N k T
Degrees of freedom (T not too high)
monatomic gas, f = 3 diatomic gas, f = 5, polyatomic gas, f = 6
2
fU N k T
Only translation possible at very low temp, T rotation begins, T oscillatory motion starts
PE = 0
25Heating a gas
tot
totA
Q m c T
N Mm N m n M
N
Q n M c T
Q
C M c
n C T
Molar heat capacity
26
Heating a gas at constant volume
Q
VQ n C T 2
fU N k T
1st Law Thermodynamics
U Q W
Constant volume process V = 0 W = 02 V
fN k T nC T
1 A An N N N k R
2V
fC R
Larger f larger CV smaller T for a given Q
All the heat Q goes into changing the internal energy U hence temperature T
U = n CV T
27
Heating a gas at constant pressure
Q
pQ n C T 2 Vf
U N k T nC T
1st Law Thermodynamics
U Q W
Constant pressure process W = p V
V pn C T n C T n R T
VCp C R
W
pV n RT p V n R T
It requires a greater heat input to raise the temp of a gas a given amount at constant pressure c.f. constant volume Q = U + W W > 0
28
22
22
2
2
1
monatomic 3 1.67
diatomic 5 1.4
21
21
32
501
V
p V
p V
V V
V V
V
monatomic
diatomic
fC R
fC C R R
fRC C R
fC CR
C C R
f
f
f
RC
29
n N mtot
Thermal processes
Q
W
T1
p1
V1
U1
S1
T2
p2
V2
U2
S2
VU WQ nC T p VC C R
21
dQS
T
pV n RT N k T
Reversible processesp VQ nC T Q nC T
30Isothermal change T = 0
U = 0 pV = n R T
22 21 1 1
2 1 11 1 2 2
1 2 2
2 2 2
1 1 1
ln
ln
ln
V VV V
W WnRT nRT
nRT VQ W pdV dV nRT
V V
V p pp V p V Q W n RT
V p p
V W V pe e
V nRT V p
Boyle’s Law (1627 -1691)T1= T2 p1V1 = p2 V2
31
Isothermals pV = constant
0
20
40
60
80
100
120
140
160
180
200
0.00 0.05 0.10 0.15 0.20 0.25
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 K
1
2W
Isothermal process
W is the area under an isothermal curve isotherm
32Isochoric (V = 0) W = 0 U = Q = n CV T
Isothermals pV = constant
0
20
40
60
80
100
120
140
160
180
200
0.00 0.05 0.10 0.15 0.20 0.25
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 K
1
2
W = 0
Isochoric process
1 to 2: Q > 0 T 1 < T 2 T > 0 U > 0
isochor
33
Isothermals pV = constant
0
20
40
60
80
100
120
140
160
180
200
0.00 0.05 0.10 0.15 0.20 0.25
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 K
1
2
W
Isobaric process
Isobaric (p = 0) W = p V Q = n Cp T
U = Q – W = n CV T
T2 > T1 U > 0 W > 0 Q > 0 W < Q
V1/T1=V2/T2
isobar
34
Adiabatic (Q = 0)
U = - W CV = (f / 2) R Cp = CV + R = (f / 2 +1)R
= Cp / CV = (f + 2) / f
p V = constant diatomic gas f = 5
T V -1 = constant = 7 / 5 = 1.4
V1 1 2 2 1 1 2 2
1
1
CW p V p V p V p V
R
35
Isothermals pV = constant
0
20
40
60
80
100
120
140
160
180
200
0.00 0.05 0.10 0.15 0.20 0.25
volume V (m3)
pre
ssu
re
p
(kP
a)
100 K
400 K
800 K
1
2
W
Adiabatic process
W
An adiabat steeper on a pV diagram than the nearby isotherms since > 1
1 to 2: Q = 0 T1 > T2 W > 0 U < 0
adiabat
36
Adiabatic processes can occur when the system is well
insulated or a very rapid process occurs so that there is not
enough time for a significant heat to be transferred eg rapid
expansion of a gas; a series of compressions and
expansions as a sound wave propagates through air.
Atmospheric processes which lead to changes in
atmospheric pressure often adiabatic: HIGH pressure cell,
falling air is compressed and warmed. LOW pressure cell,
rising air expands and cooled condensation and rain.
37
Q = 0U = - WT V -1 = constant
convergence divergence
divergence convergenceHIGH - more uniform conditions - inhibits cloud formation
LOW - less uniform conditions - encourages cloud formation
sunshine
Atmospheric adiabatic processes
Burma Cyclone5 May 2008 +50 000 killed ?
38
V
2 2 V2 1 2 1 1 21 1V V
( 1)p 2 1
p VV V 1 2
( 1)1 1 2 2 2 2 2 1
1 2 1 1 1 2
2 1
1
( 1)
2
ln( / ) ln( / ) ln (
con
/
stan
)
1 1
t
RT V CT V
nRTU W nC dT p dV dV
V
dT R dV RT T V V V V
T C V C
CR T VR C C
C C T V
p V p V T p V V
T T T p V V
p V
p V
TV
constantpV
39
Cyclic Processes:
U = 0 reversible cyclic process
40
Problem E.1
Oxygen enclosed in a cylinder with a movable piston
(assume the gas is ideal) is taken from an initial state A
to another state B then to state C and back to state A.
How many moles of oxygen are in the cylinder? Find
the values of Q, W and U for the paths A to B; B to C;
C to A and the complete cycle A to B to C to A and
clearly indicate the sign + or – for each process.
Does this cycle represent a heat engine?
41
0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
B
42Thermodynamic system (ideal gas) work
p V T U S
mtot N n
p V = n R T
p V = N k T
k = R / NA
mtot = n M
N = n NA
heat
Q = n C T
CV or Cp
W p dV
internal energy
U = Q – W = n CV T
Q = 0p V = constantT V-1 = constant
43
SolutionIdentify / Setup
oxygen diatomic f = 5
CV = (f / 2) R Cp = CV + R = (f / 2 +1) R
CV = 5/2 R Cp = 7/2 R
R = 8.315 J.mol-1.K-1
CV = 20.8 J.mol-1.K-1 Cp = 29.1 J.mol-1.K-1
p V = n R T = N k T
area under graph
V
V
p
U nC T
U Q W
W p dV pV
Q nC T
Q nC T
44
Execute
At A
4 2
2
4 10 2 10mol 0.96 mol 1 mol
8.31 10A A
A
p Vn
RT
0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
B
451 A to B is isobaric
T1= TB – TA = (400 – 100) K = 300 K
pA = pB = p1 = 40 kPa = 4.00104 Pa
V1 = (0.080 – 0.020) m3 = 0.060 m3
W1 > 0 since gas expands
W1 = p1 V1 = (4.0104)(0.06) = 2.4103 J
U1 > 0 since the temperature increases
U1 = n CV T1 = (1)(20.8)(300) J = 6.2103 J
Q1 > 0 since U1 > 0 and U1 = Q1 – W1 > 0 Q1 > W1 > 0
Q1 = n Cp T1 = (1)(29.1)(300) J = 8.7103 J
Check: First law
U1 = Q1 - W1 = (8.73103 – 2.4103) J = 6.3103 J
0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
BQ1
46
0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
B
472 B to C is isochoric
T2 = TC – TB = (800 – 400) K
= 400 K
V2 = 0 m3
W2 = 0 since no change in volume
U2 > 0 since the temperature increases
U2 = n CV T2 = (1)(20.8)(400) J = 8.3103 J
Q2 = U2 since W2 = 0
Q2 = 8.3103 J
0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
B
Q2
48
0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
B
493 C to AT3 = TA – TC = (100 – 800) K = -700 K
pA = 40 kPa = 4.0104 Pa pC = 80 kPa = 8.0104 Pa pCA = 4.0104 Pa
VA = 0.02 m3 VC = 0.08 m3 V3 = 0.06 m3
W3 < 0 since gas is compressed
W3 = area under curve = area of rectangle + area of triangle
W3 = - { (0.06)(4.0104) + (½)(0.06)(8.0104- 4.0104)} J = - 3.6103 J
U3 < 0 since the temperature decreases
U3 = n CV T3 = (1)(20.8)(-700) J
= - 14.6103 J
Q3 = U3 + W3
Q3 = (- 14.5103 - 3.6103) J = - 18.2103 J
0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
B
Q3
50
1A to B
2B to C
3C to A
cycle(total values)
W (kJ) + 2.4 0 - 3.6 - 1.2
Q (kJ) + 8.7 + 8.3 - 18.2 +1.2
U (kJ) +6.3 +8.3 - 14.6 0
Q – W (kJ) + 6.3 + 8.3 - 14.6 0
Complete cycle U = 0 J
Refrigerator cycle: |QH| = |QC| +|W|
THTC
|QH| |QC|
|W|
51Wcycle < 0 work is done on the system the system is not a heat engine because a heat engine needs to do a net amount of work on the surroundings each cycle.
The net work corresponds to the area unenclosed i.e. the area of the triangle:
Wcycle = - (1/2)(0.06)(4.0104) J = - 1.2103 J (value agrees with table)
Evaluate0
10
20
30
40
50
60
70
80
90
100
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20
volume V (m3)
pre
ss
ure
p (
kP
a)
100 K
400 K
800 KA
C
B
52Problem E.2 Typical 5 mark exam question
An ideal gas is enclosed in a cylinder which has a movable piston. The gas is heated resulting in an increase in temperature of the gas and work is done by the gas on the piston so that the pressure remains constant.
(a) Is the work done by the gas positive, negative or zero. Explain (b) From a microscopic view, how are the gas molecules effected? Explain.
(c) From a microscopic view, how is the internal energy of the gas molecules effected?
(d) Is the heat less than, greater than or equal to the work? Explain.
53SolutionIdentify / Setup
2
12 1 0
V
VW p dV p V V
U Q W First Law of Thermodynamics
V
p
T1
T2
T2 > T1
V1 V2
VU nC T
54
(a)Since work is done by the gas on the piston, the system expands as the volume increases (pressure remains constant)
The work done by the gas on the piston is positive.
(b)Since the temperature increases, the average translational kinetic energy of the gas molecules increases.
2
12 1 0
V
VW p dV p V V
55
(c)
The change in internal energy, U of an ideal gas is given by
where n is the number of moles of the gas and CV is the molar heat capacity of the gas at constant volume. Since the temperature increases, the internal energy must increase. Therefore, the total kinetic energy due to random, chaotic motion of the gas molecules increases.
VU nC T
56
(d)
Heat Q refers to the amount of energy transferred to the gas due to a temperature difference between the system and surroundings.
First law of thermodynamics
0 0 0U Q W U Q W Q W
The heat Q is greater the work done by the gas W.