PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
FINAL EXAMINATION
SPRING 2002
Wednesday, May 8, 2002
Your grade will be sent to you by e-mail by midnight, Saturday, May 11, 2002
NAME: _______ANSWERS____________________
There are six pages to this examination. Check to see that you have them all.
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (25%) In class it was shown that the partition function for one atom of an ideal monatomic
3/ 2
 2 m 
1
gas was given by,  monatomic  V  2  , where  
, and V is the volume of the
kT
h  
container to which the molecule is confined. If the same procedure is followed for a diatomic
gas molecule, at a temperature where the rotational energy levels are well excited, but the
vibrational levels are not, the partition function becomes,
3/ 2
 2 m   2  2 I
 diatomic  V  2   
.
h    h  
In the additional term, which is due to the rotation, I is the moment of inertia (or rotational
inertia) of the diatomic molecule.
a) (5%) Consider that there are N identical, indistinguishable diatomic molecules in the
container. What is the partition function Z for the N molecules?
2
N
3/ 2
N
2
3/ 2
2
1
1   2 m   2  2 I 
1  2 m   2    V 
N
 

Z  ( diatomic) 
V 
 

 
 2I  

N!
N !   h 2    h   
N !  h 2   h     5 / 2 


N
N
N
3/ 2
2
 V 
1  2 m   2  
Z  C  5 / 2  , where C is a constant and C   2    2 I  .
N!  h   h  
 
 V 
Z  C  5 / 2 
 
N
b) (8%) Use the result of part a) to calculate the average energy of the N molecules as a
function of T.
E 

 
5N
 5N 1 5
(ln Z )  
ln   
 NkT
 ln C  N ln V 

 
2
2  2

E
1
5
NkT
2
c) (5%) Use the result of part a) to calculate the Helmholtz function F for the N molecules.
5N


F  kT ln Z  kT  ln C  N ln V 
ln  
2


5N


F  kT  ln C  N ln V 
ln  
2


d) (7%) Use the Helmholtz function to calculate the pressure of the gas as a function of
temperature and volume. This should give the same ideal gas formula as a monatomic
gas.
p
F
 
5N
1



 kT  ln C  N ln V 
ln    kTN ,

V
V 
2
V


or pV = NkT.
p
2
NkT
V
2. (30%) A quantity of gas is taken reversibly around the cycle a-b-c-d-a shown on the T-S
diagram shown in the figure below.
a) (4%) The system goes around the cycle in the direction a-b-c-d-a. Is it operating as a heat
engine or a refrigerator? (Circle the correct answer.)
HEAT ENGINE
REFRIGERATOR
b) (12%) Calculate the heat transferred in each step of the cycle. The sign of the heat
transferred in each step is important. You may leave your answers in terms of R.
Q = TS
Qa-b = T(Sb – S a) = (600 K)(2R – R/2) = (600 K)(1.5 R)
Qa-b = (900 K)R
For b-c, S , so
Qb-c = 0
Qc-d = T(Sd – S c) = (200 K)( R/2 – 2R) = (200 K)(– 1.5 R)
Qc-d = – (300 K)R
For d-a, S , so
Qd-a = 0
c) (5%) How much work is done by the system in the complete cycle? You may leave your
answer in terms of R.
Q = W + E. When this is applied to the entire cycle, E = 0, so Q = W.
W = Q = (900 K)R + 0 – (300 K)R + 0
W = (600 K)R
3
d) (5%) From your results in parts b) and c), determine the efficiency of the system when it
operates as a heat engine.

W
W
(600 K)R 2



Qin Qa  b (900 K)R 3
 = 67%
e) (4%) The cycle shown in the figure is actually a Carnot cycle. Calculate the efficiency of
the cycle with the formula for the efficiency of a Carnot engine to see if it gives the result
that you obtained in part d).
  1
T2
200 K
1 2
 1
 1 
T1
600 K
3 3
Carnot = 67%
4
3. (30%) A container is divided into two equal chambers, each of the same volume V. One
chamber contains N molecules of an ideal gas at temperature T, and the other chamber is
completely empty, a perfect vacuum. A small hole, of area A, is punched in the wall
separating the two chambers, and gas begins to leak into the empty chamber. The
temperature of the gas is kept constant.
a) (10%) Find an expression for the rate at which molecules leave the filled chamber at the
instant the hole is punched. Express your answer in terms of A, V, N, and v , the average
speed of an air molecule. (Hint: the flux of molecules moving in the + x direction can be
written, f  x  14  v , where  is the number of molecules per volume.)
dN
N
 f  x A  14  v A  14  v A
dt
V 
dN N vA

dt
4V
b) (10%) The number of molecules in the chamber that was initially filled will decrease
until the two chambers are equally populated (each with N/2 molecules). Then, the rate at
which molecules leave the chamber will be equal to the rate at which they enter it from
the other side. Derive an expression for n, the number of molecules in the initially filled
chamber, as a function of time after the hole is punched. Again, the result can be in terms
of A, V, N, and v .
If there are n molecules in the first chamber, there are N – n molecules in the other. Then
nv A
based on part a), the rate at which molecules are leaving is
, and the rate at which
4V
dn
nvA ( N  n)vA
vA
( N  n)v A



( 2n  N )
they are entering is
, Then
dt
4V
4V
4V
4V
dn
vA 
N

 n   . This differential equation can be separated and solved.
dt
2V 
2
dn
vA
N
vA


dt , and after integrating, ln  n    
t  const . Then,
N
2
V
2
2
V


n
2
vA
vA 
vA

t

t
N
N N  2V t N 
N
2V 
2V

1 e
. At t = 0, n = N, so C  . Then, n   e
n   Ce

2
2 2
2 
2

c) (10%) As the gas redistributes itself between the two chambers, the entropy of the system
increases. Calculate the difference between the entropy of the final state with both
chambers holding N/2 molecules, and the initial state when all the molecules were in one
chamber.
S  k ln   k ln V N E 3 N / 2 . The volume doubles, while E and N do not change.



S  k ln  f  k ln i  k ln  f
 i

 (2V ) N E 3 N / 2 
  k ln 
  k ln 2 N  kN ln 2
N 3N / 2 
 V E


S = kN ln 2
5
4. (15%) Assume that Earth and the Sun are perfect blackbodies, and that Earth’s only source of
heat is the Sun. What will be the temperature of the surface of Earth when it is radiating
energy at the same rate that it is absorbing it from the sun, i.e. when it reaches steady state.
The following information may be useful.
Temperature of the surface of the Sun: TS = 5800 K
Radius of the Sun:
RS = 6.96 ×108 m
Radius of Earth:
RE = 6.37 ×106 m
Distance from Earth to the Sun:
r = 1.50 ×1011 m
Stephan-Boltzmann constant:
 = 5.67 ×10-8 W/m²·K4
Be sure to make your work clear. In case you do not complete this correctly, that will make it
possible to assign partial credit.
The sun radiates energy at the rate: PS  TS4 4RS2 .
When this reaches the orbit of Earth, it is spread over a sphere of area 4 r 2 .
The fraction of the emitted energy that is absorbed by Earth is equal to the fraction of the
 RE2
RE2
area just calculated that Earth occupies. That fraction is
.

4 r 2 4r 2
2
RE2
4
2 RE
 TS  RS 2
The rate at which energy is absorbed by Earth is, T 4 R
4r 2
r
4
2
The rate at which Earth radiates energy is, TE 4 RE . This must equal the expression above.
4
S
2
S
2
2
R
RE2
4
4 RS
4
4 RS
T 4 R  T  R 2 , so 4TE  TS 2 , and TE  TS 2 . Then, TE  TS S .
r
r
4r
2r
4
E
2
E
4
S
2
S
6.96 108 m
= 279 K = 6° C = 43° F.
TE  (5800 K)
2(1.5 1011 m)
6