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: