PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 1
SPRING 2003
Friday, February 21, 2003
NAME: _________ SOLUTIONS_________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (20%) Consider the following game. You put up one dollar in order to play. You are asked to
choose a number from one to six, and then you are allowed to throw one die twice to see if
the number comes up. (If you prefer, you can imagine throwing a pair of dice once, to see if
the number appears.) If your number appears, you are paid three dollars; that is, the dollar
that you bet is returned along with two dollars more. If the number does not come up, you
lose the dollar that you bet.
a) (5%) What is the probability that your number will appear on the first roll of the die?
P(1) = p = 1/6 = 0.167
P(1) = 1/6 = 0.167______
b) (5%) What is the probability that your number will not turn up on the first roll, but will
appear on the second roll of the die?
P(2) = (1 – p)p = (5/6)(1/6) = 5/36 = 0.139
P(2) = 5/36 = 0.139_____
c) (5%) What is the combined probability that your number will turn up, and you will win?
P(1 or 2) = P(1) + P(2) = 11/36 = 0.306
P(1 or 2) = 11/36 = 0.306____
d) (5%) Suppose someone played the game, many times. Based on the probability that you
calculated, and the payoff offered for winning, how would this person do financially,
after playing for a long time, and making many bets? Circle the correct answer, and
support your choice with some numerical evidence.
WIN
LOSE
BREAK EVEN
If the player wins 0.305 of the time, and is paid $3 each time, then the average win is $3
× 0.306 = $0.917 for each $1 bet. After 100 plays, and $100 bet, the player would have
on average $91.70 left. In this game, that really means $91 or $92.
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2. (10%) Consider the expression, [ y 2 cos(ax)  axy2 sin( ax)]dx  2 xy cos(ax)dy . Is it an exact
differential?
 2
[ y cos( ax)  axy 2 sin( ax)]  2 y cos( ax)  2axy sin( ax)
y

2 xy cos( ax)  2 y cos( ax)  2axy sin( ax)
x
These are equal!
It IS an exact differential.
3. (30%) One mole of an ideal gas expands isothermally from point i to point f on the graph
shown. At point i the volume of the gas is V0, and the pressure is p0. At point f the volume is
3V0. (For one mole of an ideal gas, pV = RT.) Express all your answers in terms of p0, V0, and
R.
a) (10%) How much work is done by the gas as it expands isothermally from V0 to 3V0?
3V0
W   pdV . Since pV = RT, and T is constant, pV = p0V0, and p 
V0
3V0
Then, W  p0V0 
V0
p0V0
.
V
 3V 
dV
3V
 p0V0 ln V V  p0V0 ln  0 
V
 V0 
0
0
W  p0V0 ln 3
b) (10%) What is the change in internal energy as the gas expands isothermally from V0 to
3V0?
E 
kT
. Since T is constant, E is constant.
2
E = 0
2
c) (10%) How much heat is added to the gas as it expands isothermally from V0 to 3V0?
Q – W = E, so Q = W + E. Since E = 0, Q = W.
Q  p0V0 ln 3
4. (40%) Consider two systems, A and B, each containing the same number of molecules of gas,
NA = NB = 2.0 × 1022 molecules. Each system has the same internal energy, EA = EB = 160 J.
However, the two systems have different numbers of degrees of freedom. A = 6.0 × 1022 and
B = 10.0 × 1022. The volumes of the two systems, and the number of particles in each
system, remain constant throughout this problem.
a) (10%) Find the temperatures of the two systems.
2(160 J)
kT
2E
E 
, so T 
. Then, TA 
22
2
k
(6.0  10 )(1.38  10 23 J/K)
TA = 386 K
TB 
2(160 J)
(10.0  10 )(1.38  10 23 J/K)
22
TB = 232 K
b) (10%) Find the entropies of the two systems. (Your answers may be left in terms of k.)


6.0  10 22
S  k ln   k ln E 2  k ln E . Then, S A  k
ln( 160 J)  1.52  10 23 k
2
2
SA = 1.52 × 1023 k = 2.10 J/K
10.0  10 22
SB  k
ln( 160 J)  2.54  10 23 k
2
SB = 2.54 × 1023 k = 3.50 J/K
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The systems are placed in contact, and heat flows from one to the other. They come to
equilibrium when EA = 120 J and EB = 200 J.
c) (10%) Find the entropies of the two systems now. (Your answers may be left in terms of k.)
6.0  10 22
SA  k
ln( 120 J)  1.44  10 23 k
2
SA = 1.44 × 1023 k = 1.98 J/K
10.0 10 22
SB  k
ln( 200 J)  2.65 10 23 k
2
SB = 2.65 × 1023 k = 3.66 J/K
d) (5%)By how much did the entropy of the combined system (including both A and B)
change when they were placed in contact? (Your answer may be left in terms of k.)Did the
entropy increase? (Circle the correct answer.)
YES, ENTROPY INCREASED
NO, ENTROPY DID NOT INCREASE
Sbefore = 1.52 × 1023 k + 2.54 × 1023 k = 4.06 × 1023 k
Safter = 1.44 × 1023 k + 2.65 × 1023 k = 4.09 × 1023 k
S = Safter – Sbefore = 4.09 × 1023 k – 4.06 × 1023 k = 3 × 1021 k
S = 3 × 1021 k = 0.04 J/K
More accurate:




S  k A ln E A,after  k B ln E B ,after  k A ln E A,before  k B ln E B ,before
2
2
2
2
 E A,after 
 E

k
  B ln  B ,after 
S   A ln 
E

E

2 
 A,before 
 B ,before 
S 
k
 120 J 
 200 J 
22
21
(6.0  10 22 ) ln 
  (10.0  10 ) ln 
  2.53  10 k

2
 160 J 
 160 J 
S = 2.53 × 1021 k = 0.0349 J/K
e) (5%) Show that the total entropy of the two systems is a maximum (or at least an extreme
value) when EA = 120 J and EB = 200 J.


k
S  k A ln E A  k B ln EB  A ln E A  B ln EB  , and EA + EB = E, so EB = E – EA.
2
2
2
S
k
 0.
Then, S  A ln E A  B ln( E  E A ) . The maximum occurs when
E A
2
B 
A
S
k 
6
  A
 320 J
 120 J
  0, so E A  E
E A 2  E A E  E A 
 A  B
6  10
EB  E  E A  E  E
A
B
10
E
 320 J
 200 J
 A  B
 A  B
6  10
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Alternatively, one could point out that entropy is a maximized when the two systems come
to the same temperature, so show that they are at the same temperature.
2(120 J)
TA 
 290 K
22
(6.0  10 )(1.38  10 23 J/K)
2(200J)
TB 
 290 K
22
(10.0  10 )(1.38  10 23 J/K)
Entropy is a maximum.
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