PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
FINAL EXAMINATION
SPRING 2003
Friday, May 9, 2003
Your grade will be sent to you by e-mail by 5:00 PM, Monday, May 12, 2003
NAME: ______ANSWERS_________________________
There are five 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. (10%) State the First, Second, and Third Laws of Thermodynamics in some reasonable form.
1st. In any process, the change in internal energy of a system is equal to the heat that goes into
the system minus the work that comes out of the system plus the change due to the
increase in number of particles in the systetm. (Conservation of energy.)
dE = dQ – dW + dN
2nd. For any two interacting systems, the entropy of the combined systems cannot decrease.
S  0
3rd. The entropy of a system goes to zero as the temperature of the system goes to zero.
2. (15%) When a single die (one half of a pair of dice) is tossed, the probability that the face
with only one dot on it will turn up is one in six. A naïve person might believe that tossing
the die a second time would double the probability of the “one” face turning up to two
chances in six. You can quickly see the fallacy of that reasoning if you extend it to six tosses,
since it predicts six chances in six; a sure thing. Calculate the actual probability of at least
one “one” turning up in six tosses of a die. (That means: the probability of a “one” showing
up once or more than once in six tosses.)
6
5
Easy way: Prob = 1 – Prob (0 “ones”) = 1    = 1 – 0.335
6
n
6n
N!
6!
1 5
p nq N n 
Harder way: Prob of n “ones” is P(n) 
   
n!( N  n)!
n!(6  n)!  6   6 
From this: P(1) = 0.402, P(2) = 0.201, P(3) = 0.054, P(4) = 0.008, P(5) = 0.00064, and
P(6) = 2.1 ×10-5.
Then, Probability = P(1) + P(2) + P(3) + P(4) + P(5) + P(6)
Either way, the result is the same.
Probability = 0.665 = 66.5%
1
NAME: ______ ANSWERS _________________________
3. (25%) Consider a collection of N identical, indistinguishable harmonic oscillators, all of
frequency . The energies that one these oscillators can take on (measured relative to the
ground state) are n = nh.

1
a) (5%) Find the partition function for one of these oscillators. (Hint:  x n 
for x < 1)
n 0
1 x


n0
n0
   e   nh    e   h  
n
1
1  e   h
 
1
1  e   h
b) (5%) Find the partition function for the collection of N oscillators.
Z
1 N 1 
1 
 


N!
N !  1  e   h 
N
Z
1 
1


  h 
N!  1  e

N
c) (5%) Find the average energy of the collection of oscillators.
E


h e  h
N h
ln Z  
 ln N ! N ln(1  e  h )  N

  h   h


1 e
e 1
E
N h
e 1
 h
For this system suppose the number of oscillators is N = 1000, and the temperature is such
h
that kT 
(i.e. h = 2).
2
d) (10%) How many oscillators are likely to be found in the n = 0 state, the n = 1 state and
the n = 2 state?
The probability of finding an oscillator in a particular state is given by:
e  nh
e  nh
p(n) 

 e  nh (1  e  h )  e 2 n (1  e 2 )
  nh

e
Then, N(n) = Ne-2n(1 – e-2) = 1000 e-2n(1 – e-2) = 864.7 e-2n
From this, we get:
N(0) = ______865___________
N(1) = ______117___________
N(2) = _______16___________
2
NAME: ______ ANSWERS _________________________
4. (40%) One mole of an ideal monatomic gas traverses the cycle shown in the figure. Process
12 is isothermal, process 23 takes place at constant pressure, and process 31 takes
place at constant volume. (Hint: for one mole of an ideal monatomic gas, the internal energy
is E  32 RT .)
The temperatures of the gas at points 1, 2, and 3 are:
T1 = 300 K,
T2 = 300 K,
T3 = 1500 K
a) (5%) Find V1/V2, the ratio of the volume of the gas at point 1 to the volume at point 2.
(Hint: V1 = V3)
V V
T 1500 K
23 takes place at constant pressure, so V  T. Then, 1  3  3 
V2 V2 T2 300 K
V1/V2 = _____5______
b) (10%) Find the magnitude of the heat added to the gas in the process 23. Express the
answer in terms of the gas constant R.
Constant pressure process, so:
Q23 = Cp(T3 – T2) = 52 R(1500 K  300 K)
Q23 = __(3000 K)R___
3
NAME: ______ ANSWERS _________________________
c) (5%) Find the magnitude of the heat removed from gas in the process 31. Express the
answer in terms of the gas constant R.
Constant volume process, so:
Q31 = CV(T1 – T3) = 32 R(300 K  1500 K) = – (1800 K)R
(Magnitude) Q31 = (1800 K)R
d) (10%) Find the magnitude of the heat removed from gas in the process 12. Express the
answer in terms of the gas constant R.
Process is isothermal, so E = 0, and Q12 = W12
V2
V2
V 
dV
1
W12   pdV  RT 
 RT ln  2   RT ln     RT ln 5   R(300 K) ln 5
V1
V1 V
5
 V1 
(Magnitude) Q12 = R(300 K) ln 5 = (483 K)R
e) (5%) How much work is done by the gas in one complete cycle? Express the answer in
terms of the gas constant R.
W = Qin – Qout = (3000 K)R – [(1800 K)R + (483 K)R]
W = (717 K)R
f) (5%) Find the efficiency of a heat engine operating with this cycle.

W
(717 K) R

Qin (3000 K) R
 = 0.239 = 23.9%
4
NAME: ______ ANSWERS _________________________
5. (10%) A Carnot refrigerator is to be used to lower the temperature of a container that has a
constant heat capacity of C. When an amount of heat dQ2 is removed from the container, its
dQ2
temperature decreases by an amount dT2 
. An amount of work dW is done to remove
C
dQ2 from the container, at T2, and deposit dQ1 into the high temperature reservoir at
temperature T1. The temperature of the high temperature reservoir does not change when heat
is added to it. Initially, the container is at the same temperature as the reservoir, T1. Show that
to lower the temperature of the container from T1 to some lower temperature Tf, the amount
of work required is:

T 
W  C T1  T f  T1 ln 1 
T f 

This is easy if you remember the coefficient of performance for a Carnot refrigerator.
T

Q
T2
T T
c 2 
. Then, W  Q2 1 2  Q2  1  1 . For an infinitessimal amount of work,
W T1  T2
T2
 T2 
T

T

dW  dQ2  1  1  CdT2  1  1 . Then, just integrate this from T1 to Tf.
 T2 
 T2 
T
T
T




dT2
  dT2  , and with a little work, W  C T1 ln f  (T f  T1 ) .
 dW  C T1 
T1


 T T2 T

f
f
1
1

T 
A few more manipulations yields: W  C T1  T f  T1 ln 1  .
T f 

If you did not recall the coefficient of performance, you could proceed:
dW  dQ1  dQ2  dQ1  CdT2 . To integrate this, dQ1 must be expressed in terms of dT2.
Q T
dQ1 T1
T
T
 , and dQ1  1 dQ2  1 CdT2 . Then,
For a Carnot cycle, 1  1 , so
Q2 T2
dQ2 T2
T2
T2
dW  dQ1  dQ2 
T

T1
CdT2  CdT2  C  1 dT2  dT2  . Then, proceed as shown above.
T2
 T2

5