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
2m kT
Z V
2
h
3/ 2
4 2V 3 2 12
g ( ) s
m
3
h
• Combining the above equations, one has
2N
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 4N
v
e
dv
3/ 2
2kT
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 2m
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 2m kT
2m 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 2m kT
2
NJ
P h
3
2
e
J
kT
kT
P
5
2
2m 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