Chapter 14: The Classical Statistical Treatment of an

Ideal Gas

14.1 Thermodynamic properties from the Partition Function

• All the thermodynamic properties can be expressed in terms of the logarithm of the partition function and its derivatives.

• Thus, one only needs to evaluate the partition function to obtain its thermodynamic properties!

• From chapter 13, we know that for diluted gas system, M-B statistics is applicable, where

S= U/T + Nk (ln Z – ln N +1)

F= -NkT (ln Z – ln N +1)

μ= = – kT (lnZ – lnN)

• Now one can derive expressions for other thermodynamic properties based on the above relationship.

1. Internal Energy:

We have …

kTE

j

jj

n

jj

j

eZ

N

g

NandENU

1

j

kT

E

jjj

jjkT

E jj

eEgZ

NEge

Z

NU

j

kTE

j

j

egZ

differentiating the above equation with respect to T, we have …

= (keeping V constant means E(V) is constant)

Therefore,

or

))((2

j

jkT

E

jj

kT

E

jV kT

Eeg

dT

kT

Ed

egT

Z jj

j

kT

E

jj

j

eEgkT 21

VT

ZkT

Z

NU

)( 2

VT

ZNkTU

ln2

2. Gibbs Function since G = μ N

G = - NkT (ln Z – ln N)

3. Enthalpy G = H – TS → H = G + TS

H =-NkT(lnZ - ln N) + T(U/T + NklnZ – NklnN + Nk)

= -NkTlnZ + NkTlnN + U + NkTlnZ – NkTlnN + NkT

= U + NkT = NkT2 + NkT

= NkT [1 + T · ]

4. Pressure

TV

FP

TV

NZNkTP

))1ln(ln(

TV

NZNkTP

)1ln(ln

TV

ZNkTP

00)(ln

TV

ZNkTP

)(ln

14.2 Partition function for a gas

assuming g1 = g2= g3 = … gn = 1, E1 = 0

Z = 1 + e + e + …

j

kTE

j

j

egZ

kT

E

kT

E

kT

E

egegegZ321

321

• For a gas in a container of macroscopic size, the energy levels are very closely spaced, which can be regarded as a continuum.

• Recall

• Since the gas is composed of molecules rather than spin ½ particles, the spin factor does not apply, i.e. γs = 1.

• For continuum system

dmh

Vdg s

21

23

3

24)(

0

/)( degZ RT2/3

2

2

h

mkTVZ

14.3 Properties of a monatomic ideal gas

Previous sections illustrate that many of the thermodynamic properties is a function of lnZ, which Z itself is a function of T and V.

ln Z = ln V + 3/2 ln ( )

= lnV + (3/2) ln + 3/2 lnT

TT

Z

VV

Z

V

T

1

2

3ln

1ln

2/3

2

2

h

mkTVZ

• Recall

• Now we have

• Thus, PV = NkT (note that k = R/NA)

• Similarly, we have and

• Therefore, U = 3/2 NkT

TV

ZNkTP

)(ln

VV

Z

T

1ln

VT

ZNkTU

ln2

TT

Z

V

1

2

3ln

• From equation 14.1

S= U/T + Nk (ln Z – ln N +1)

and U = 3/2 NkT ;

We have

The above equation become invalid as T 0

2/3

2

2

h

mkTVZ

2/3

2

2ln

2

5

h

mkTVNkS

0lnln2

5S

N

VTNkS

2/3

20

2ln

2

5

h

mNkS

• Example (textbook 14.1)

(a) Calculate the entropy S and the Helmholtz function for an assembly of distinguishable particles.

(b) Show that the total energy U and the pressure P are the same for distinguishable particles as for molecules of an ideal gas while S is different. Explain why this makes sense.

• Solution: