PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 2
SPRING 2002
Friday, April 19, 2002
NAME: __________SOLUTIONS__________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (30%) Enthalpy is defined as H = E + pV, where E is internal energy, p is pressure, and V is
volume.
a) Show that: dH = TdS + Vdp + dN
dH = dE + pdV +Vdp, but the First Law states, dE = TdS – pdV + dN. Then dH becomes,
dH = TdS – pdV + dN + pdV +Vdp = TdS + Vdp + dN
b) Use the equation given in part a) to show that: T 
H is a function of S, p, and N. Therefore, dH 
H 

 S  p , N
H 
H 
H 
 dS 
 dp 
 dN
 S  p, N
 p S , N
 N  S , p
A term by term comparison with the equation given in part a) shows that T 
c) Use the equation given in part a) to show that:
T
V
 
 p S , N  S


 p, N
(This is one of the Maxwell relations.)
Using the method of part b), we also see that V 
H 
 . Then,
 p  S , N
V 
 2H
T 
 2H
 
 
, and
. Clearly, these are equal.
 S  p , N  S p
 p  S , N  p S
1
H 

 S  p , N
2. (25%) Consider a collection of 3N identical, distinguishable harmonic oscillators, all of
frequency . The energies that one these oscillators can take on (measured relative to the
ground state) are n = nh. Where n is an integer that can take on values from 0 to .
a) Find the partition function , for one of these oscillators.

1
(Hint:  x n 
for x < 1)
n 0
1 x


n0
n0
   e   nh    e   h  
n
1
1  e   h
b) Find the partition function Z, for the collection of oscillators. (Since the oscillators are
distinguishable, no N! is needed in the denominator.)
Z 
3N
1



  h 
1 e

3N
c) Find the Helmholtz function F, for this collection of oscillators.
(Hint: F = – kT ln Z)
1


F  kT ln Z  kT ln 
  h 
1 e

  h
F  3NkT ln 1  e


3N



 3NkT ln 1  ln 1  e   h  3NkT ln 1  e   h

d) Find the entropy S, for this collection of oscillators. Hint: S  
F 
 . This will not
T V , N
give a particularly neat expression.
S 

h


F 


  h 
kT



3
NkT
ln
1

e


3
NkT
ln
1

e



T V , N
T
T
V , N



h


h 
kT 

e
 2
h

 kT   ln 1  e  kT
S  3Nk T
h




kT
1

e


h



 h  1


S  3Nk 
 ln 1  e kT

 kT  hkT

e 1












I said it was not particularly neat.
2



 V , N
e) Show that the expression for the entropy that you obtained in part d) goes to zero when T
goes to zero, in agreement with the third law of thermodynamics.
h
gets very large. The exponential in the denominator of the left
kT
term gets very large, and that term goes to zero. The exponential in the ln gets very small,
and the ln 1 = 0, so S goes to zero.
As T goes to zero,
3. (15%) The number density of the gas, mainly hydrogen, that fills interstellar space is about
one molecule per cubic centimeter ( = 1 ×106 m-3). The diameter of the molecules is about
d = 1 ×10-10 m, and the temperature of interstellar space is about 10K.
a) Find the mean free path of the molecules in interstellar space.
l
1

2 
1
2 (1  10 m ) (1  1010 m) 2
6
3
l = 2.25 ×1013 m
b) Find the average speed of the molecules in interstellar space. (The mass of molecular
hydrogen is m = 2 amu, and 1 amu = 1.66 ×10-27 kg.)
v
8kT
8(1.38  1023 J/K)(10 K)

m
 (2  1.66  10- 27 kg)
v = 324 m/s
c) Find the average time between collisions for molecules in interstellar space. Express your
answer in centuries. (1 year = 3.16 ×107 s, and 1 century = 100 years.)
t
l 2.25  1013 m

 6.94  1010 s  2.20  103 years
v
324 m/s
t = 22 centuries
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4. (30%) One mole of an ideal monatomic gas traverses the cycle shown in the figure. Process
12 takes place at constant volume, process 23 is adiabatic, and process 31 takes place
at constant pressure. In the work that follows, express all answers in terms of the gas constant
R. (Hint: for one mole of an ideal monatomic gas, the internal energy is E  32 RT .)
a) Compute the heat added, and the work done for the process 12.
Since the process is at constant volume, W12 = 0. Then, Q12 = CV T12
For one mole of an ideal monatomic gas, CV  32 R , so Q12  32 R(600 K  300 K)
Q12 = 450 R
W12 = 0
b) Compute the heat added, and the work done for the process 23.
Since the process is adiabatic, Q23 = 0. Then, from the first law,
W23 = Q23 – E23 = – E23 =  32 R(T3  T2 )   32 R(455 K  600 K)  32 R(145 K)
Q23 = 0
W23 = 217.5 R
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c) Compute the heat added, and the work done for the process 31.
Since this is at constant pressure, Q31 = Cp T31.
For one mole of an ideal monatomic gas, C p  CV  R  32 R  R  52 R . Then,
Q31  52 R(300 K  455 K)   52 R(155 K)
W31  Q31  E31   52 R(155 K)  32 R(300 K  455 K)   52 R(155 K)  32 R(155 K)
Q31 = = – 387.5 R
W31 = – 155 R
d) Compute the net heat added, and the net work done for the entire cycle.
Qnet = Q12 + Q23 + Q31 = 450 R + 0 + (– 387.5 R)
Wnet = W12 + W23 + W31 = 0 + 217.5 R + (– 155 R)
Qnet = 62.5 R
Wnet = 62.5 R
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