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