PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS SPRING 2001
Quiz 1 Tuesday, February 13, 2001
NAME: ______________ SOLUTION______________________
To receive credit for a problem, you must show your work, or explain how you arrived at your answer.
1. (15%) Suppose you get involved in a game in which six coins are tossed simultaneously. The object of the game is not to guess how many heads, or tails will show up in a toss. Instead, you are to guess what the distribution will be; i.e. will all six be the same, or will there be one of one type and five of the other, or two of one type and four of the other, or three of each. In such a game, what distribution has the greatest probability of occuring, and what is that probability?
The probability of getting six of one kind and none of the other equals the probability of getting six heads plus the probability of getting no heads (six tails). The probability of getting n heads is given by:
Then, P ( 6 , 0 )
P ( 5 , 1 )

5 !
( 6
6 !


5
6 !
( 6
6 !

6 )!
)!
1
2

5
P ( n )
 n !
( N
N !
 n )!
p n q
N
 n
1
2
1
2

6
6

5
1
2
6

6


1 !
( 6
6 !

1 )!
0 !
( 6
6 !

0 )!
1
2

1
1
2
1
2

0
6

1
1
2
6

0

6 !
2
5 !
1 !
P ( 4 , 2 )

P ( 3 , 3 )

4 !
( 6
6 !

4 )!
3 !
( 6
6 !

3 )!
1
2
1
2

4
3

2
6 !
6 !
0 !
1
2

6

12
64
1
2

6


2
64
0 .
188

0 .
031
1
2

6

4
1
2

6

3


2
6
!
!
(
3 !
3 !
6
6 !

2 )!
1
2

6

1
2

2
20
64

1
2

6

2
0 .
313

6 !
2
4 !
2 !
1
2

6

30
64

0 .
469
. There is no second distribution.
Most probable distribution: ______4 and 2______
Probability = ___0.469__________
2. (10%) Consider the expression
 yx e x

3

 x e dy . Is it an exact differential?

 y
( yx
3 e x 
3 yx
2 e x
)
 x
3 e x 
3 x
2 e x

 x
( x
3 e x
)

3 x
2 e x  x
3 e x
These are equal, so it is an exact differential.
1

3. (30%) Consider a particle that is constrained to move in one dimension. Suppose this particle executes a random walk. Every 10
-6
s, it moves either exactly 1 nm to the right or exactly 1 nm to the left. a.) (10%) If the probability that the particle moves to the right is 0.6, what is the average distance the particle will have moved from its starting point after 10 s.
S

N
For p
(
 p

0 .
6 q )
, s
0 and
 s
N ( 2 p

1 ) s
0
{Book calls ( 2 p

1 ) s
0
0

1 nm, s

0.20
nm, S

N ( 0 .
 s
20 ) nm
}

N ( 2

10

10 m)
After 10 s, N

10 s
10
6 s

10
7
, so S

10
7
( 2

10

10 m)
S = _2 × 10
-3
m = 2 mm__ b.) (10%) If the probability that the particle moves to the right is 0.5 instead, what will the average displacement from the starting point be after 10 s.
For p = 0.5, s

0
S = _ 0 _______________ c.) (10%) How long must you wait if you want the probability that the particle is more than 1 cm from its starting point to be greater than 32%. In this part, you should again assume that the probability that the particle moves to the right is 0.5.
The distribution of molecules is Gaussian. Therefore, 68% of them are within one standard deviation of the mean, and 32% are farther away. This problem asks how long it will take for the standard deviation to become 1.0 cm.
 
2 s
0
Npq , so

2 
4 s
0
2
Npq , and N


2
4 s
0
2 pq

4 ( 1 .
0
( 1 .
0

10

2 m )
2

10

9 m )
2
( 0 .
5 )( 0 .
5 )

1 .
0

10
14 t = (10
14
steps)(10
-6
s/step) = 10
8
s t = _10
8
s = 3.16 y_____
2

4. (45%) 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 V
0
, and the pressure is p
0
. At point f the volume is
3 V
0
. (For one mole of an ideal gas, pV = RT .) Express all your answers in terms of p
0
, V
0
, and
R . a) (8%) How much work is done by the gas as it expands isothermally from V
0
to 3 V
0
?
W

3 V
0
 pdV
V
0
. Since pV = RT , and T is constant, pV = p
0
V
0
, and p

Then, W
 p
0
V
0
3 V
0

V
0 dV
V
 p
0
V
0
 
3 V ln V 0
V
0
 p
0
V
0 ln


3 V
0
V
0

 p
0
V
0
V
.
W
 p
0
V
0 ln 3 b) (5%) What is the change in internal energy as the gas expands isothermally from V
0
to 3 V
0
?
E
  kT
2
. Since T is constant, E is constant.

E = 0 c) (7%) How much heat is added to the gas as it expands isothermally from V
0
to 3 V
0
?
Q – W =

E , so Q = W +

E . Since

E = 0, Q = W .
Q
 p
0
V
0 ln 3
3

d) (5%) Based on your answer to c), calculate the change in entropy of the gas as it expands isothermally from V
0
to 3 V
0
(Hint: Q = T

S )

S

Q
T
 p
0
V
0 ln 3
T
 p
0
V
0 ln 3 p
0
V
0
/ R

S = R ln 3 e) (10%) The number of states available to the system increases as the gas expands isothermally from V
0 that available at V
0
to 3 V
0
? What is the ratio of the number of states available at 3 V
. (Hint:

= V
N
E

/2
)
0
to

3 V
0

V
0

( 3 V
0
)
N
( V
0
)
N
E

/
E

/ 2
2

3
N

3 V
0
/

V
0
= 3
N
¯¯¯¯¯¯¯¯¯¯¯ f) (10%) Based on your answer to e), calculate the change in entropy of the gas as it expands isothermally from V
0
to 3 V
0
?
S
3 V
0
– S
V
0
= k ln

3 V
0
– k ln

V
0
= k ln (

3 V
0
/

V
0
) = k ln 3
N
= kN ln 3 = R ln 3
Since this is one mole of gas, N = N
A
, so kN = kN
A
= R .

S = R ln 3
4