14.5 distribution of molecular speeds
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14.5 Distribution of molecular speeds
• For a continuum of energy levels,
where
and
dgeZ
NdN
eZ
N
g
N
kTE
kTE
j
j
)()(
)(
)(
2/3
2
2
h
mkTVZ
21
23
3
24)( m
h
Vg s
• Combining the above equations, one has
• ε is a kinetics energy calculated through (1/2)mv2,
thus dε = mvdv
• The above equation can be transformed into (in class demonstration)
dekT
NdN kT
2/12/3
2)(
dvev
kT
mNdvvN kT
mv22
2/3
2/3 2
24)(
14.6 Equipartition of energy• From Kinetics theory of gases
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.
kTf
2
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
!!
!
!!
!
2121 NxNx
N
NN
NW
• From Boltzmann relationship
• 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. non-distinguishable particles with the same Ej), there is no change in the total number of available microstates, thus Δs equals 0
)!ln()!ln(!ln)ln( 21 NxNxNkWkS
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 two-component 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:
• Solution
TjJ
ZNkTN
ln
oneprobablemostTheondistributiMBtheisthisegZ
N
kTeg
Z
NkTN
kTeg
Z
egZ
Z
ZNkTN
equationabovetheorganizedre
kTj
kTjJ
kTj
TJ
kTJ
jJ
J
J
J
J
.
1
1
1
• 14.3a) Show that for an ideal gas of N molecules,
• Solution:
23
2
25
2
h
m
P
kT
N
Zwheree
N
Z
N
gkT
j
jJ
23
2
25
23
2
23
2
23
2
22
22
12.14
h
mkT
P
kTe
h
mkT
P
kT
N
g
eN
Z
N
ge
Z
N
g
N
ondistributiMBFrom
h
mkT
N
V
N
Z
h
mkTVZ
functionpartitionTheP
kT
N
V
NkTnRTPV
kT
J
J
kT
J
JkT
J
J
J
JJ
• 14-3(b) For calculate
• Solution:
,10,10,300,23 263 kgmandPaPkTkT
J
J
J
N
g
8
6354
23
23
682
26
3
2523
23
2
25
1067.2
538.4100542.0101048.1
1062.6
102
10
1030038.1
2
e
eh
mkT
P
kTkTJ
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