PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Homework Solutions
Assignment 8. Due Tuesday 3/28/06 : 12-1, 12-4, 12-6, 12-11, 12-12
 = 1.13 ×1015
12-1. a)   250
b) 25 heads is most probable. Then,
c) p 
wmax 
50!
25!25!
wmax = 1.26 ×1014
wmax 1.26  1014


1.13  1015
p = 0.112
1000!
, so ln w(500)  ln( 1000!)  2 ln( 500!)
500!500!
ln w(500)  1000 ln 1000  1000  2(500 ln 500  500)  1000 ln 1000  1000 ln 500
 1000 
ln w(500)  1000(ln 1000  ln 500)  1000 ln 
  1000 ln 2
 500 
12-4. w(500) 
w(500)  e1000ln 2
log[ w(500)]  1000 ln 2 log e  301
w(500)  10301
1000!
, so ln w(600)  ln( 1000!)  ln( 500!)  ln( 400!)
600!400!
ln w(600)  1000 ln 1000  1000  600 ln 600  600  400 ln 400  400
ln w(600)  1000 ln 1000  600 ln 600  400 ln 400  673
w(600) 
w(600)  e673
log[ w(600)]  673 log e  292
w(600)  10292
w(600) 10292
 301  10 9
w(500) 10
Students who used the book’s value of w(500) = 10300 will get a ratio of 10-8.
That answer will get full credit.
1
12-6. a) The table lists how many particles are in each energy state for each configuration k.
k=
1
1
6
5
4
3
2
1
0
b) wk 
2
3
4
1
1
5
6
7
2
1
1
1
1
1
8
9
3
2
2
1
1
1
2
3
N!
 Ni !
2
w1  w7  w8 
2
1
2
4!
4
1!3!
3
w2  w3  w4 
1
4!
 12
1!1!2!
4!
4!
6
w6 
 24
2!2!
1!1!1!1!
   wk  3(4)  3(12)  2(6)  24  84
w5  w9 
c) N j 
12-11.   n 2j
N
k
jk
wk
1(4)
1(12)
 0.05 N5 
 0.14

84
84
1(12)  1(12)
2(6)  1(24)  1(4)
N 4 
 0.29 N 3 
 0.48
84
84
1(12)  1(24)  3(4)  2(6)
N 2 
 0.71
84
1(12)  2(12)  1(24)  3(4)  2(6)
N1 
 1.00
84
3(4)  2(12)  2(12)  1(12)  2(6)  1(24)  1(4)
N0 
 1.33
84
N 6 
2
2
( ) 2
( ) 2
mO L2O
2 ( )
2
2 ( )
2
2

n
Then,
,
so
n

n
n

n
He
O
O
He
j
2mHe L2He
2mO L2O
mHe L2He
2mV 2 3
2mL2
nO  nHe
LO
LHe
mO
1102 m 16 amu
 3
mHe
2 10-10 m 4 amu
2
nO = 1.73 ×108
V  2m 


12-12. g ( ) 
4 2   2 
3/ 2
V  2m 


g ( ) 
4 2   2 

3/ 2
For one kilomole at STP, V = 22.4 m³.
3
kT 
2
(22.4 m3 )  2(4 amu)( 1.66 10 27 kg/amu)


4 2 
(1.054 10 34 J  s) 2



3/ 2
3
(1.38 10 23 J/K)(273 K)
2
g() = 5.57 ×1052 J-1
The units on the answer indicate that g() is a density of states. To get a number of states,
a small range of energy should be designated, and dN = g()d would be used.
g ( ) 5.57 1052 J -1

 9.3 1025 J -1  1026 J -1
26
NA
6.02 10
3