Physical Chemistry Third Edition

1084 26 Equilibrium Statistical Mechanics. II. Statistical Thermodynamics

The expression forW for a given distribution in a dilute gas was given in

Eq. (25.2-13).

W({N})

∏∞

j 1

g

Nj

j

Nj!

Replacement of ln(Ω)byln(Wmp) and substitution into Eq. (26.1-1) gives

SkBln(Wmp)kBln

⎛

⎝

∏

j

g

Nj

j

Nj!

⎞

⎠kB

∑

j

[

Njln(gj)−ln(Nj!)

]

≈kB

∑

j

[

Njln

(

gj

Nj

)

+Nj

]

(26.1-4)

where we have used the fact that the logarithm of a product is the sum of the log-

arithms of the factors and have used Stirling’s approximation for ln(Nj!), shown in

Eq. (25.2-21). We now replace gi/Nj in Eq. (26.1-4) by its expression from

Eq. (25.3-22):

SkB

∑

j

[

Njln

(z

N

eεj/kBT

)

+Nj

]

kB

∑

j

[

Njln

(z

N

)

+

Njεj

kBT

+Nj

]

kB

[

Nln

(z

N

)

+

E

kBT

+N

]

S

U

T

+NkBln

(z

N

)

+NkB (26.1-5)

where the mechanical energyEis identified with the thermodynamic energyU.

Equation (26.1-5) is our desired result: a formula for the entropy of a dilute gas in

terms of the partition function. It contains the internal energyU, but we have already

obtained a formula forUin Eq. (25.3-26).

The Pressure of a Dilute Gas

Since the molecules of a dilute gas move independently, the pressure of the gas is equal

to the sum of the pressures exerted by the individual molecules. If the system is in

microstate numberk, its pressure is

PkPk,1+Pk,2+Pk,3+ ··· +Pk,N (26.1-6)

wherePk,1is the pressure exerted by the first molecule given that the system is in

microstatek, and so on. This sum can be written in another way by summing over

molecular states instead of over molecules. We number the molecular states 1, 2, 3,...:

PN 1 P 1 +N 2 P 2 +N 3 P 3 ··· 

∑

i

PiNi (26.1-7)

whereNiis the number of molecules in molecular statei, and wherePiis the pressure

exerted by a molecule in statei.