PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 1
SPRING 2002
Friday, February 22, 2002
NAME: _____________SOLUTION______________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (25%) A small time gambler is running a game on a street corner. He tosses five coins at
once, and bettors can bet on how many heads will appear. Anyone who bets on zero or five,
and wins, is paid $30 for each dollar bet. Those who win with one or four are paid $6 for
each dollar bet, and they get $3 for each dollar bet if they win with two or three heads.
a) (10%) Find the probabilities that 0, 1, 2, 3, 4, and 5 heads will appear when the coins are
tossed.
The probability of getting n heads out of N coins is given by: P(n) 
0
5!
1 1
Then, P(0) 
   
0!(5  0)!  2   2 
1
5!  1 
P(1) 
 
1!(5  1)!  2 
1
 
2
2
50
5 1
P(2) 
5!
1
 
2!(5  2)!  2 
P(3) 
5!  1   1 
  
3!(5  3)!  2   2 
3
1
 
2
5!  1 
P(0) 
 
5!(5  5)!  2 
5
5!  1 
1

 0.03125
  
0!(5)!  2  32
5
5!  1 
5

 0.15625
  
1!(4)!  2  32
5 2
5

5!  1  10
 0.3125
  
2!(3)!  2  32

5!  1  10
 0.3125
  
3!(2)!  2  32
53
5
4
5 4
5!  1 
5

 0.15625
  
4!(1)!  2  32
5
55
5!  1 
1

 0.03125
  
5!(0)!  2  32
5!
1 1
P(4) 
   
4!(5  4)!  2   2 
1
 
2
N!
p n q N n .
n!( N  n)!
5
5
b) (5%) Suppose someone played the game, and always bet that three heads would show up.
Based on the probability that you calculated, and the payoff offered for winning (3 to 1),
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 0.3125 of the time, and wins $3 each time, then the average win is $3 ×
0.3125 = $0.9375 for each $1 bet. After 100 plays, and $100 bet, the player would have
on average $93.75 left. In this game, that really means $93 or $94.
1
c) (5%) A street-smart character plans to improve his odds by switching the coins used in
the game. He has a set of five coins, each of which comes up heads 60% (0.60) of the
time. What is the probability that three heads will come up when these five loaded coins
are tossed?
3
P(3) 
5!  3   2 
  
3!(5  3)!  5   5 
53
3

2
5!  3   2 
216
 0.3456
   
3!(2)!  5   5 
625
d) (5%) Based on the probability that you just calculated, and the payoff offered for
winning, how would this character do after playing for a long time with the loaded coins,
and always betting on three heads? Circle the correct answer, and support your choice
with some numerical evidence.
WIN
LOSE
BREAK EVEN
If the player wins 0.3456 of the time, and wins $3 each time, then the average win is $3 ×
0.3456 = $1.0368 for each $1 bet. After 100 plays, and $100 bet, the player would have
on average $103.68 left. In this game, that really means $103 or $104.
Here are the overall odds:
For p = 0.5
n
prob
0
0.03125
1
0.15625
2
0.3125
3
0.3125
4
0.15625
5
0.03125
"payoff"
30
6
3
3
6
30
"gain"
0.9375
0.9375
0.9375
0.9375
0.9375
0.9375
For p = 0.6
n
prob
0
0.01024
1
0.0768
2
0.2304
3
0.3456
4
0.2592
5
0.07776
"payoff"
30
6
3
3
6
30
"gain"
0.3072
0.4608
0.6912
1.0368
1.5552
2.3328
2. (10%) Consider the expression, xy2 cos( 2 x 2 )dx  y sin( 2 x 2 )dy . Is it an exact differential?


[ y sin( 2 x 2 )]  4 xy cos( 2 x 2 )
[ xy2 cos( 2 x 2 )]  2 xy cos( 2 x 2 )
x
y
These are not equal, so the expression is not an exact differential.
2
3. (25%) Energy produced in the center of the sun has a hard time finding its way out. On the
average, in the sun, a photon will go a distance of s0 = 1.0 cm (0.010 m) between collisions
with hydrogen atoms, and on the average, it is held about 10-8 s by the hydrogen atom before
being reemitted in a completely random direction. Hence, the photon takes about 108 “steps”
per second.
a) (10%) What is the average distance traveled in any one dimension per step. (You do not
have to do any calculations if you know the answer and can give a short explanation.)
This problem is practically the same as problem 4-9 in the textbook, which was part of
the second homework assignment. There you actually did the integration to find that the
average distance traveled was equal to zero. Without calculation, one can deduce that
since the photon is equally likely to go in any direction, when you average over all the
possibilities, it goes nowhere. The average distance traveled is zero (0).
b) (10%) The radius of the sun is about 7.0 ×108 m. How many “steps” must a photon take
before having a 32% chance of being beyond the edge of the sun in any one dimension?
(Hint: In a homework problem you found that the square of the standard deviation about
s2
the value given in part a) for any one step is given by (sz ) 2  0 ,where s0 is the average
3
step size.)
This asks how many steps until the standard deviation of the distribution equals the radius
of the sun.
s 2 (0.010 m) 2
 3.33  10 5 m 2 . Then,
 2  N (sz )2 , and (sz ) 2  0 
3
3
2
(7.0  108 m) 2
N

 1.47  1022
5
2
2
3.33  10 m
(sz )
c)
(5%) At the rate of 108 steps per second, how many years are needed to take the number
of steps calculated in part b)? (1 year = 3.16 ×107 s)
t
1.47  1022 steps
1.47  1014 s

 4.65  106 yr
1 108 steps/s
3.16  107 s/yr
3
4. (40%) One mole of an ideal gas is taken through the process shown on the graph. It starts at
point 1, with pressure p0 and volume V0. It expands, at constant pressure, to point 2, where its
volume is 2V0. Then it is compressed isothermally to point 3, where its pressure is 2p0 and its
volume is again V0. (For one mole of an ideal gas, pV = RT.) Express all answers in terms of
p0, V0, and R.
a) (5%) The internal energy of the gas at point 1 is E  32 p0V0 . What is the internal energy of
the gas at point 3?
Since the path from 2 to 3 is isothermal, the internal energy at 3 is the same as at 2. Then,
E3  3 p0V0
E1  32 p0V0 , so E3  E2  32 p0 (2V0 )
b) (5%) How much work is done by the gas as it expands from 1 to 2?
2V0
2V0
V0
V0
W  p0V0
W   pdV  p0  dV  p0 (2V0  V0 )
c) (10%) How much work is done by the gas as it is compressed from 2 to 3?
Since T is constant for this process, pV = p0(2V0) so p 
V0
V0
2V0
2V0
W   pdV  2 p0V0 
V 
dV
1
 2 p0V0 ln  0   2 p0V0 ln  
V
2
 2V0 
4
2 p0V0
. Then,
V
W  2 p0V0 ln 2
d) (10%) How much heat is added to the gas as it goes from 1 to 3?
Q = W + E. From 1 to 3, W = p0V0 – 2p0V0 ln 2, and E  3 p0V0  32 p0V0  32 p0V
Then, Q  p0V0  2 p0V0 ln 2  32 p0V0
Q   52  2 ln 2 p0V0
e) (10%) Find the change in entropy of the gas as it goes from 1 to 3.
Method 1:
V N E 
 (2 E1 ) 
 
  k ln 2
S  S3  S1  k (ln 3  ln 1 )  k ln  3   k ln  N 3   k ln 
V E 
 E 
 1 
1 

 1 

S  k ln 2 . For this gas, there are three degrees of freedom per molecule, and there are
2
3N A
ln 2  32 kN A ln 2 . However,
NA molecules in a mole. Then,   3N A , so S  k
2
kNA = R, so
S  32 R ln 2
A

2
A

2

2

2

2
dQ
.
T
dQ = dE + dW, but from 1 to 3, V does not change, so dW = pdV = 0. Then dQ = dE.
For one mole of an ideal gas, E  N A 32 kT  32 RT , so dE  32 RdT , and
3
Method 2: Integrate by a direct path from 1 to 3, S  1
dS 
T 
 2T 
dQ dE 3 dT
dQ 3 3 dT 3

 2R
. Then S  13
 2 R 1
 2 R ln  3   32 R ln  1   32 R ln 2
T
T
T
T
T
 T1 
 T1 
2
Method 3: Integrate along the path in the diagram, S  1
dQ 3 dQ
 2
.
T
T
From 1 to 2, dQ = dE + dW, so
dQ 2 dE 2 dW 3 2 dT
pdV 3 2 dT
dT 5 2 dT 5
S1 2  12
 1
 1
 2 R 1
 12
 2 R 1
 R 12
 2 R 1
 2 R ln 2
T
T
T
T
T
T
T
T
From 2 to 3, dE = 0, so dQ = dW, and
dQ 3 dW
pdV
RT dV
dV
1
S 2  3  23
 2
 23
 23
 R 23
 R ln   R ln 2
T
T
T
V T
V
2
5
3
Then, S13  S1 2  S2 3  2 R ln 2  R ln 2  2 R ln 2
5