PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Homework Solutions
Assignment 7. Due Tuesday 3/21/06 : 11-1, 11-11, 11-12, 11-14, 11-18
N (v)dv A  v v
A  v v
 ve dv . Then,
ve dv  1
N
N
N 0
A   v v0
Av02   v   v v0  v  Av02   x
 e d   
ve
dv

xe dx  1 .

0
N 0
N 0  v0 
v
N
 0
11-1. a) f (v)dv 
0
0
N
A  v v
Av02
ve
dv

 1 . Then, A  2 , and

v0
N 0
N
f (v ) 
0
xe x dx  1 , so:
A v v  v  v v
ve
  2 e
N
 v0 
0
0
f (v )
0


0
1
2
3
v/v 0
4
5
6

  v 

1 
b) v   vf (v)dv   v 2 e v v0 dv  2  v 2e v v0 dv  v0  (v v0 )2 e v v0 d (v v0 )
0
0
0
v0 0
 v0 

v  v0  x 2e x dx

0

0
x 2e x dx  2
v  2v0



v
1 
v 2   v 2 f (v)dv   v 2  2 e v v dv  2  v3e v v dv  v02  (v v0 )3 e v v d (v v0 )
0
0
0
v0 0
 v0 
0


v 2  v02  x3e x dx

0
0
0
x3e x dx  6
v 2  6v02
f (v)   v   v v   1   v v  v   v v
ev v
  2 e
  3 e
 2
c)
   2 e
v
v  v0 
v0
 v0 
  v0 
0
0
0
0
vrms  v 2  6v0
0

v
1    0
 v0 
vm = v0
d)  2  (v  v )2  v2  2vv  v 2  v2  2v v  v 2  v2  v 2
  v 2  v 2  6v02  (2v0 ) 2
  2v0  1.41v0
1
Note: The integrals needed in this problem are all of the form


0
x ne x dx and could be
evaluated several ways. Maple, or a similar program, could be used, tables of integrals
could be used, they could be integrated by parts, or the following method could be used.
Call


0
xne x dx  I n . For the integrals in the problem,  = 1.

Then, consider I 0   e  x dx  
0
 x 
0

1

0  1  1 .

I 0
 1 1

.
 
0
0

     2

   x
I
  1  2
I 2   x 2e  x dx  
xe dx   1  
.
 

0
0


   2   3

  2  x
I
  2  (2)(3)
I 3   x 3e  x dx  
x e dx   2  
.
 

0
 0

   3 
4

Next, I1   xe x dx  


e 

1


e  x dx  
n
In general: I n  
11-11.  
n
I n 1   
n!
   1
 
 I0   
    n 1 .

  
      
N
P
1
n
. To get n , consider, PV = NkT. Then,
. Also,  = d2.
V
kT
8
n

1
kT
(1.38  1023 J/K)(300 K)


.
8 P
8 Pd 2
8 (1.01  105 Pa)(4.6  10-10 m) 2
2
d
 kT
 3.86  108 m

d 4.6  1010 m

 = 3.86 ×10-8 m

= 84
d
Book’s method for fc:
8kT
8(1.38 1023 J/K)(300 K)

 380 m/s
m
 (44 amu)(1.66 10 27 kg/amu)
fc 
v

fc 
380 m/s
3.86  108 m
v
fc = 9.8 ×109 s-1
More consistent method:
fc 
vm

fc 
337 m/s
3.86  108 m
vm 
2kT
2(1.38  1023 J/K)(300 K)

 337 m/s
m
(44 amu)(1.66  10 27 kg/amu)
fc = 8.7 ×109 s-1
2
1

8
n
11-12.  

fc 
1
8


(1  106 m-3 ) (1  1010 m) 2
1  1014
m
8
 = 2.0 ×1013 m
vm 1 2kT
1
2(1.38  1023 J/K)(10 K)


  m
2.0  1013 m (2 amu )(1.66  10 27 kg/amu)
f c  (1.44  1011 s -1 )(3.16  107 s/yr)(100 yr/century )
fc = 0.0456 century-1
That is about one collision every 22 centuries or one in 2200 years.
1
, and n can be found as in problem 118
n
11-14. a) The limit occurs for D < 0.4  .  

1
kT

where  = d2 has been used. Then,
2
8 P
8 Pd
d2
 kT
0.4kT
0.4(1.38  1023 J/K)(273 K)
0.4kT
and P 

D
D 8 d 2 (100  10 6 m) 8 (4.6  10-10 m) 2
8 Pd 2
11, so  
P = 14 Pa
b)
dN
1
1 N 
dN
 vA 
 vA 
 A   nv A    v A  
 
 N . Then,
dt , and
dt
4
4V 
N
 4V 
 4V 
 vA 
ln N  
t  const ,
 4V 
 vA 

t
 4V 
so N  N 0e
 vA 
 vA 

t
 4V 
. For V and T constant, P  N, so
 vA 
 vA 


t
t

t
P
1
. If P  0  P0e  4V  , then e  4V   , or e 4V   2 , so
P  P0e
2
2
4V ln 2 4V ln 2 4V ln 2  m
 vA 



t1 2  ln 2 , so t1 2 
vA
A
8kT
8kT
 4V 
A
m
t1 2 
12
4(103 m3 ) ln 2  (44 amu)(1.66  1027 kg/amu)
 (50  10 6 m) 2
8(1.38  10 23 J/K)(273 K)
1
1  mv 1  m 8kT
mkT
11-18.   nmv  
. Then,


2
3
6 2 
6 2  d  m 3 d 2
d
12
12
t1/2 = 976 s = 16.3 min
d
(39.94 amu)(1.66  10- 27 kg/amu) (1.38  10 23 J/K)(288 K)
3 (22  10 6 Pa  s)
3
mkT
.
3
d = 2.80 ×10-10 m