PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 1
SPRING 2004
Friday, February 20, 2004
NAME: ________ANSWER KEY__________________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (25%) In the poker game seven card stud, seven cards are dealt to each player. The player
selects five of those cards to make the best possible hand.
a) (10%) How many different five card hands can be made from the seven cards dealt to a
player?
N5 
N!
7!
7!


n!( N  n)! 5!(7  5)! 5!2!
N5 = 21
b) (10%) How many different seven card hands can be made from a deck of 52 cards?
N7 
N!
52!
52!


 133784560
n!( N  n)! 7!(52  7)! 7!45!
N7 = 1.34 ×108
c) (5%) Based on your answer to part b), what is the probability that a player would be dealt
the exact same seven cards twice in a row? Think carefully about this one.
It is only necessary that the second hand match the first hand, whatever that happens to
be. Since there are 1.34 ×108 possible hands, the probability of being dealt that
particular hand is: Prob = 1/1.34 ×108
Prob = 7.47 ×10-9
1
2. (10%) Consider the expression, aeax cos( y 2 )dx  2 yeax sin( y 2 )dy . Is it an exact differential?
Circle the correct answer, and show why you made your selection.
It IS an exact differential
It IS NOT an exact differential


[ae ax cos( y 2 )] must equal
[2 ye ax sin( y 2 )]
y
x


[2 ye ax sin( y 2 )]  2 yae ax sin( y 2 )
[ae ax cos( y 2 )]  2 yae ax sin( y 2 )
x
y
For an exact differential,
These are not equal.
3. (35%) Electromagnetic radiation in an evacuated vessel of volume V, at thermal equilibrium
with the walls at temperature T, behaves like a gas of photons. The internal energy and the
pressure of the gas are given by the relations: E  aV T 4 , and p  13 aT 4 , where a is Stephan’s
constant. Initially the system is at pressure p0, volume V0 , and temperature T0. Express all
your answers in terms of p0, V0, T0, and a.
a) (10%) How much heat is added to the system if it expands isothermally from V0 to 2V0?
Q = W + E Since T is constant, p is constant, so W = p0V. Also, E = aT0 4V.
p  13 aT 4 , so aT 4 = 3p, and E = 3p0V. Then,
Q = p0V + 3p0V = 4p0 (2V0 – V0)
Q = 4p0V0 = 43 aT04V0
b) (5%) What is the change in entropy as the gas expands isothermally from V0 to 2V0?
S 
Q 4 p0V0

T
T0
S =
2
4 p0V0 4 3
 3 aT0 V0
T0
c) (10%) How much heat must be added to the gas if the pressure is to increase from p0 to 2 p0
at a constant volume of V0?
Q = W + E Since V is constant, W = 0, and Q = E.
E  aV T 4  3 pV , and E = 3V0 p = 3V0(2p0 – p0)
Q = 3p0V0 = aT04V0
d) (10%) How much heat must be added to the gas if the temperature is to increase from T0 to
2 T0 at a constant volume of V0?
Again, V is constant, so W = 0, and Q = E = a V0[(2T0) 4 – T04)] = a V0T04[16 – 1]
Q = 15 a V0T04 = 45 p0V0
3
4. (30%) Consider two systems, A and B. System A contains one mole of a monatomic ideal gas
at a temperature of 480 K. System B contains two moles of the same monatomic ideal gas at a
temperature of 240 K.
a) (10%) Find the internal energies of the two systems.
EA  32 N AkTA  32 RTA  32 (8.31 J/mole)( 480 K)
EB  32 2 N AkTB  3RTA  3(8.31 J/mole)( 240 K)
EA = 5980 J
EB = 5980 J
b) (10%) The two systems are placed in thermal contact, and heat flows between them. As a
result, one third (1/3) of the energy in system A is transferred to system B. Show that this
transfer results in an increase in the combined entropy of the two systems.
3NA
3NA
S  S A  S B  S A, f  S A,i  S B, f  S B,i  k ln( EA,2f )  k ln( EA,2i )  k ln( EB3 N, f )  k ln( EB3 N,i )
A
3NA
3NA
E  2
E 
3N Ak  E A, f
S  k ln  A, f   k ln  B , f  
ln 
2
 E A, i 
 EB , i 
 E A, i
3R  32 
S 
ln    0
(S = 0.255R = 2.12 J/K)
2  27 
 EB , f

 E
 B,i
A
2
2

  3N Ak ln  2  4 

2
 3  3 

c) (10%) Show that the total entropy of the two systems is a maximum after the transfer of
energy from system A to system B.
Mathematical solution:
S
0
For a maximum,
E A
3NA
2
A
3N Ak
[ln( E A )  2 ln( EB )]
2
3R
[ln( E A )  2 ln( E  E A )]
EA + EB = E, so EB = E – EA. Then, S 
2
  3R
2 
 3R  1

  0

 [ln( E A )  2 ln( E  E A )]  
E A  2
 2  EA E  EA 
S  S A  S B  k ln( E
2EA  E  EA , or E A 
)  k ln( EB3 N ) 
A
E
2E
, and E B 
; i.e. EB = 2EA which is the state of the system.
3
3
4
Or:
S
 ( S A  S B ) S A S B S A S B





0
E A
E A
E A E A E A EB
3N

2
S

ln(
E
)  ln( EB3 N
A
 k

 E
E A
EB
A

A
A

 3 N  ln( E A )
)
 ln( EB ) 
 k A
 3N A


2
E A
EB 


 1
1 
 3N Ak 

0
 2 E A EB 
EB = 2EA which is the state of the system.
Physical solution:
When the system comes to equilibrium, the entropy will be a maximum, and the
temperatures of A and B will be the same. Calculate the temperatures of A and B.
 2  5980 J 
 4  5980 J 
2



2EA
EA 
3
3


 = 320 K
= 320 K
TA 

TB 

3R
3(8.31 J/K)
3R
3(8.31 J/K)
Entropy is a maximum.
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