Transcript Slide 1

14.5 Distribution of molecular speeds
• For a continuum of energy levels,
N ( )
N  E j kT
 e
g ( )
Z
N  E j kT
N ( )d  e
g ( )d
Z
where
and
 2m kT 
Z V 

2
 h

3/ 2
4 2V 3 2 12
g ( )   s
m 
3
h
• Combining the above equations, one has

2N
1/ 2
N ( )d 

e
3/ 2
kT 
kT
d
• ε is a kinetics energy calculated through (1/2)mv2,
thus dε = mvdv
• The above equation can be transformed into (in class
demonstration)
 mv 2
m3 / 2
2
2 kT
N (v)dv  4N
v
e
dv
3/ 2
2kT 
14.6 Equipartition of energy
• From Kinetics theory of gases

 
f
kT
2
showing that the average energy of a molecule is the number of degrees
of freedom (f) of its motion.
• For a monatomic gas, there are three degrees of freedom, one for each
direction of the molecule’s translational motion.
• The average energy for a single monatomic gas molecule is (3/2)kT (in
class derivation).
• The principle of the equipartition of energy states that for every degree
of freedom for which the energy is a quadratic function, the mean
energy per particle of a system in equilibrium at temperature T is
(1/2)kT.
14.7 Entropy change of mixing
revisited
• From classical thermodynamics
Δs = - nR (x1lnx1 + x2lnx2)
where x1 = N1/N and x2 = N2/N
• Now consider mixing two different gases with the
same T and P, the increase in the total number of
configurations available to the system can be
calculated with
N!
N!
W 

N1! N 2! x1 N !x2 N !
• From Boltzmann relationship
S  k ln(W )  k ln N! ln(x1 N )! ln(x2 N )!
• Using Stirling’s approximation (see white board for details)
we get Δs = - nR (x1lnx1 + x2lnx2)
• From statistical point of view, when mixing two of the same
type of gases under the same T and V (i.e. nondistinguishable particles with the same Ej), there is no
change in the total number of available microstates, thus Δs
equals 0
14.8 Maxwell’s Demon
Demon (II)
Figure 14.4 Maxwell’s demon
in action. In this version the
demon operates a valve,
allowing one species of a twocomponent gas (hot or cold)
through a partition separating
the gas from an initially
evacuated chamber. Only fast
molecules are allowed
through, resulting in a cold
gas in one chamber and a hot
gas in the other.
• Problem 14.2: Show that for an assembly of N particles that
obeys Maxwell-Boltzmann statistics, the occupation numbers for the most
probable distribution are given by:


 ln Z 
N J   NkT 
  
j T

• Solution
re  organizedthe aboveequation
 1 Z 

N J   NkT 
 Z  
j 

Z   gJ  e
 Z

  J
 J
kT
 J

1 
  g j  e kT   

 kT 
T

J
NkT
 1 
NJ  
 g j  e kT   

Z
kT



J
N
  g j  e kT
Z
this is the MB distribution.The m ost probableone
• 14.3a) Show that for an ideal gas of N
molecules, g
Z 
Z kT   2m 
 e where

5
J
j
Nj
• Solution:
2
kT
N
N
P
3
2
 2 
 h 
PV  nRT  NkT
V kT

N
P
The partition function 14.12
3
Z V  2m kT 
 2m kT 
Z V
 


2
2
N N h
 h


From MB distribution
2

3
2

J
N J N kTJ
gJ
Z
kT
 e 
 e
gJ
Z
NJ N
gJ
kT  2m kT 



2
NJ
P  h

3
2
e
J
kT

kT 

P
5
2
 2m kT 


2
 h

3
2
• 14-3(b) For
calculate g
 
 J  3 2 kT , T  300k , P  10 3 Pa, and m  10 26 kg,
J
NJ
• Solution:
kT 
P
5
2

2

mkT


kT


e

2
 h

3
2
J

1.38  300  10 

2
3
2


10


2


e


3
2
 68
10
 6.62  10 
54
63
 1.1048  10  0.0542  10  4.538
 23
 2.67  10
8
5
2
 26
3