PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
FINAL EXAMINATION
SPRING 2006
Wednesday, May 10, 2006
NAME: _________ANSWERS______________________
There are six pages to this examination. Check to see that you have them all.
CREDIT GRADE
Problem 1
20%
Problem 2
30%
Problem 3
25%
Problem 4
15%
Problem 5
10%
TOTAL
100%
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (20%) In winter, around Troy, a layer of ice usually forms on the surface of a pond or lake,
but the water remains liquid below this layer. One possible explanation for this phenomenon
is based on the variation of pressure with depth in the water. Deeper in the water, the
pressure is greater, so the freezing point is lower. Do you think that this is the entire
explanation? Support your answer by calculating the thickness of the ice layer that would
form on a lake at a uniform temperature of – 1.0ºC, if this explanation is correct.
Useful information: v = 1.09 ×10-3 m3/kg
v = 1.00 ×10-3 m3/kg
 12 =3.34 ×105 J/kg
P
 12
12
, so P 

T . P due to a layer of water of depth y is gy.
T T (v  v)
T (v  v)
 12
 12
 12
T
v T
T
 12


gy 
T , so y 
T (v  v)
T (v   v ) g T (v   v ) g
 v   g
T   1
 v 
y
(3.34  105 J/kg)
(1.0 K)
3
3
 1.00  10 m /kg  9.8 m/s 2
(273 K)
 1
3
3
 1.09  10 m /kg 
Thickness = 1510 m
units
Too thick! Not the whole story.
1
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.
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
2
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 + U. When this is applied to the entire cycle, U = 0, so Q = W.
W = Q = (900 K)R + 0 – (300 K)R + 0
W = (600 K)R
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
T1
200 K
1 2
 1
 1 
T2
600 K
3 3
Carnot = 67%
3
3. (25%) In class it was shown that the partition function for an ideal monatomic gas was given
3/ 2
 2 m 
1
 2 mkT 
by, Z  V 
, where V is the volume of the container
  V  2  , since  
2
kT
 h

h 
to which the gas is confined and T is its temperature. If the same procedure is followed for a
diatomic gas, at a temperature where the rotational energy levels are well excited, but the
vibrational levels are not, the partition function becomes,
3/ 2
3/ 2
 2 m   2  2 I
 2 mkT   2 
.
Z diatomic  V 
 
 2 IkT  V  2  

2
 h
  h 
h   h  
In the additional term, which is due to the rotation, I is the moment of inertia (or rotational
inertia) of a diatomic molecule.
a) (8%) Find an expression for the average energy per molecule of the diatomic gas.
3/ 2
2
2
3/ 2
2
3/ 2
2
U

   2 m   2  2 I 
   2 m   2 
1 

(ln Z diatomic)  
ln V  2  
ln V  2  


 2I 5 / 2 
N

   h    h   
   h   h 
 


3/ 2
2
3/ 2
2


U
    2 m   2  
    2 m   2   5
5/ 2

 2 I   ln    
 2 I   ln  
ln V  2  
ln V  2  
N
    h   h  
    h   h   2


U 51 5

 kT
N 2 2
U 5
 kT
N 2
b) (5%) Based on the result of part a) find the internal energy of the gas when N molecules
are present.
U 5
5
 kT , so U  N kT
N 2
2
U
4
5
NkT
2
c) (5%) Find the Helmholtz function F for the N molecules.
   2 mkT 3 / 2  2 2



F   NkT (ln Z  ln N  1)   NkT ln V 
2
IkT

ln
N

1
  

   h2   h 


 

   2 mkT 3 / 2  2  2



F   NkT  ln V 
2
IkT

ln
N

1




   h2   h 


 

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.

   2 mkT 3 / 2  2  2



2
IkT

ln
N

1




 NkT  ln V 
2

h
h






 



 2 mkT 3 / 2  2  2

 




P

NkT
ln
V

ln
2
IkT

ln
N

1







2


V 
  h 
 h




NkT
P
V
F

P

V
V
P
5
NkT nRT

V
V
4. (15%) An astronomer observes the light from a distant gas cloud in space. This is the same
astronomer that appeared in the second quiz, but it is a different gas cloud. A wavelength
analysis of the light reveals two sharp lines which are characteristic of two energy transitions
in a particular molecule. The lower energy line (2.1 × 10-3 eV) corresponds to the energy for
a transition between the first excited state and the ground state while the higher energy line
(3.2 × 10-3 eV) corresponds to the transition between the second excited state and the ground
state. (As before, these energies are determined in the rest frame of the cloud; all relativistic
corrections due to motion of the cloud with respect to earth have already been included.
Please, do not worry about this complication here.) Based on the measured intensities of the
two lines, the astronomer determines that there are 4 times more molecules in the first excited
state than there are in the second excited state, and neither state is degenerate. Compute the
temperature of the gas cloud based on this information.
fi  N
T
e

i

kT
Z
, so
1
kT
f1 e
  e

f2
e kT
2
 2  1
kT
 f   
  1
. Then, ln  1   2 1 , so T  2
kT
 f 
 f2 
k ln  1 
 f2 
3.2 103 eV  2.1103 eV
(8.62 105 eV/K) ln 4
T = 9.2 K
5. (10%) This question deals with the laws of thermodynamics. You need not write them out,
just answer the questions by circling the correct choices. Your instructor trusts you.
a) (2%) Are you able to state the Zeroth Law of Thermodynamics in some reasonable form?
YES
NO
b) (2%) Are you able to state the First Law of Thermodynamics in some reasonable form?
YES
NO
c) (2%) Are you able to state the Second Law of Thermodynamics in some reasonable form?
YES
NO
d) (2%) Are you able to state the Third Law of Thermodynamics in some reasonable form?
YES
NO
e) (2%) Are you able to state the Fourth Law of Thermodynamics in some reasonable form?
YES
NO
6