PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
FINAL EXAMINATION
SPRING 2004
Friday, May 7, 2004
NAME: __________ANSWERS__________________
There are seven pages to this examination. Check to see that you have them all.
CREDIT GRADE
Problem 1
20%
20%
Problem 2
50%
50%
Problem 3
30%
30%
TOTAL
100%
100%
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (20%) 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 atom 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 
 
1
N
b) (5%) 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
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) (5%) 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. (50%) The figure shows a p-V diagram, for n moles of a monatomic ideal gas, dreamed up by
an entrepreneur who wants to improve on the Carnot cycle. The cycle begins at a where the
pressure is p0, the volume is V0, and the temperature in T0. It proceeds along the isobaric path
ab to b, where the volume is 2V0. Then it goes along the isothermal bc to c where the volume
is 4V0. Next it goes along the isobaric cd to d where the volume is 2V0, and finally along the
isothermal da back to a. (Remember: pV = nRT)
a) (4%) Calculate the pressure and temperature at point b, in terms of p0, V0, and T0.
Since ab is isobaric, pb = p0 .
From the ideal gas law, at point a, p0V0 = NkT0. At b, pbVb = p02V0 = 2NkT0 = NkTb.
So, Tb = 2T0.
pb = p0
Tb = 2T0
b) (4%) Calculate the pressure and temperature at point c, in terms of p0, V0, and T0.
Since bc is isothermal, Tc = Tb = 2T0.
Then, pbVb = pcVc, or p02V0 = pc4V0, so pc = p0/2.
pc = p0/2
Tc = 2T0
3
c) (4%) Calculate the pressure and temperature at point d, in terms of p0, V0, and T0.
Since cd is isobaric, pd = pc = p0/2.
From the ideal gas law, at point d, pdVd = (p0/2)(2V0) = p0V0 = NkT0 = NkTd.
So, Td = T0.
pd = p0/2
Td = T0
d) (6%) Calculate the work done by the gas and the heat added to the gas as the system goes
from a to b. Express your answers in terms of p0 and V0.
b
Wab  a pdV . Since p is constant, this becomes, Wab = p0(Vb – Va) = p0(2V0 – V0) = p0V0
Qab = Wab +Eab, and
Eab = (3/2)Nk(Tb – Ta) = (3/2)Nk(2T0 – T0) = (3/2)NkT0 = (3/2) p0V0
Then, Qab = p0V0 + (3/2) p0V0 = (5/2) p0V0
Wab = p0V0
Qab = (5/2) p0V0
e) (6%) Calculate the work done by the gas and the heat added to the gas as the system goes
from b to c. Express your answers in terms of p0 and V0.
2p V
c
Wbc  b pdV . Since T is constant, pV  2 p0V0 , along the path, so p  0 0 .
V
V 
 4V 
c dV
Then, Wbc  2 p0V0 b
 2 p0V0 ln c   2 p0V0 ln 0   2 p0V0 ln 2
V
 Vb 
 2V0 
Since T is constant, E = 0, so Qbc = Wbc = 2p0V0 ln2
Wbc = 2p0V0 ln2
Qbc = 2p0V0 ln2
f) (6%) Calculate the work done by the gas and the heat added to the gas as the system goes
from c to d. Express your answers in terms of p0 and V0.
d
Wcd  c pdV . Since p is constant, Wcd = (p0/2)(Vd – Vc) = (p0/2)(2V0 – 4V0) = – p0V0
Qcd = Wcd +Ecd, and
Ecd = (3/2)Nk(Td – Tc) = (3/2)Nk(T0 – 2T0) = – (3/2)NkT0 = – (3/2) p0V0
Then, Qab = – p0V0 + [– (3/2) p0V0] = – (5/2) p0V0
Wcd = – p0V0
Qcd = – (5/2) p0V0
4
g) (6%) Calculate the work done by the gas and the heat added to the gas as the system goes
from d to a. Express your answers in terms of p0 and V0.
pV
a
Wda  d pdV . Since T is constant, pV  p0V0 , along the path, so p  0 0 .
V
V 
V 
a dV
Then, Wda  p0V0 d
 p0V0 ln a   p0V0 ln 0    p0V0 ln 2
V
 Vd 
 2V0 
Since T is constant, E = 0, so Qda = Wda = – p0V0 ln2
Wda = – p0V0 ln2
Qda = – p0V0 ln2
h) (3%) Calculate the net work done by the system as it goes around the complete cycle.
Express your answer in terms of p0 and V0.
W = Wab + Wbc + Wcd + Wda = p0V0 + 2p0V0 ln2 + [– p0V0] + [– p0V0 ln2]
W = p0V0 ln2
i) (3%) Calculate the net heat that goes into the system as it goes around the complete cycle.
Express your answer in terms of p0 and V0.
Q = Qab + Qbc + Qcd + Qda = (5/2)p0V0 + 2p0V0 ln2 + [– (5/2)p0V0] + [– p0V0 ln2]
Q = p0V0 ln2
j) (3%) Calculate the efficiency of this engine. (Hint: This should be a pure number. If you
can not get rid of constants, be sure you explain what you are trying to calculate.)
W

. W is the net work done, and Qin is what the textbook calls Q1. It is the heat from
Qin
the high temperature reservoir(s). In this case, Qin = Qab + Qbc = (5/2)p0V0 + 2p0V0 ln2
W
p0V0 ln 2
ln 2
 5
 5
Then,  
Qin 2 p0V  2 p0V0 ln 2 2  2 ln 2
 = 0.178 = 17.8%
k) (3%) What would be the efficiency of a Carnot engine that could be operated between the
highest and the lowest temperatures that this system reaches. (Hint: This should be a pure
number. If you can not get rid of constants, be sure you explain what you are trying to
calculate.)
T
T
1 1
 Carnot  1  2  1  0  1  
T1
2T0
2 2
Carnot = 0.50 = 50%
5
l) (2%) Based on your answers to (j) and (k), which would be the better engine? Which would
be the better heat pump? (Circle your choices.)
Better engine:
Carnot cycle
Cycle of this problem
Better heat pump:
Carnot cycle
Cycle of this problem
3. (30%) N molecules of an ideal diatomic gas are confined to a vessel of volume V that is
maintained at a constant temperature, T. At the temperature T the vibrational degrees of
freedom of the molecules are not active. (Each molecule has five degrees of freedom.)
a) (4%) What is the internal energy of this gas? Express your answer in terms of N, T, k, and
pure numbers.
E
5
NkT
2
b) (9%) For this gas, find the heat capacities, CV and Cp, and  = Cp/CV. Express your
answers in terms of N, k, and pure numbers. (For an ideal gas Cp – CV = Nk)
dE
d 5
 5

 NkT   Nk
dT dT  2
 2
5
7
C p  CV  Nk  Nk  Nk  Nk
2
2
C
2
2 7
7/2 7
 p 
 , or   1   1  

5 5
CV 5 / 2 5
CV 
c) (4%) The gas is irradiated with ultra-violet light, and one third (1/3) of the N molecules
dissociate, each into two atoms. What is the internal energy of this mixed gas?
Remember, T is held constant. (The vessel is in contact with a heat reservoir.) Express
your answer in terms of N, T, k, and pure numbers.
2
52 
5
N diatomic molecules with energy Emol   N kT  NkT
3
23 
3
2
32 
There are also N monatomic atoms with energy Eat   N kT  NkT
3
23 
5
8
E  NkT
The total energy is the sum of the two, E  NkT  NkT
3
3
There are now
6
d) (9%) For this mixed gas, find the heat capacities, CV and Cp, and  = Cp/CV. Express
your answers in terms of N, k, and pure numbers.
dE
d 8
 8

 NkT   Nk
dT dT  3
 3
4
4
8
4
12
Since there are now N particles, C p  CV  Nk  Nk  Nk  Nk  4 Nk
3
3
3
3
3
C p 12 / 3 12 3


  ,
CV
8/ 3
8 2
53
2
2
2
1 3
 4 , so   1 
or   1
, and  average 
1 1 
2
 average
 average
4
2 2
As before, CV 
e) (4%) Determine the ratio of the pressure of the gas after the irradiation to that before the
irradiation. Your answer should be a pure number.
pV = NkT. Since V and T do not change, pressure is proportional to N. Before the
4
irradiation there are N particles, and after, there are N particles. Therefore,
3
4
N
pafter
4
 3 
pbefore
N
3
7