Chemistry Reference
In-Depth Information
where
h
6
{
(2
πmk
B
T
)
3
/
2
(2
πI
1
k
B
T
)(2
πI
2
k
B
T
)
1
/
2
(2
π
)
}
and is dependent on the molecular mass
m
, the inertial moments
I
1
and
I
2
and is a function of the temperature. Ω is the solid angle
A
=
d
Ω=sin
θdθdφ.
The remaining integrad is associated with the potential only. Two tricks
are adopted here:
(1) Substitute the integration over angles into the summation. Divide the
unit spherical surface into
K
solid angles of ∆
ω
, the number of molecules
whose axes point to each solid angle ∆
ω
being
N
α
(
N
α
is an integer). The
integral over dΩ is then equivalent to summing all the possible combinations
{
N
1
,N
2
,...,N
α
,...,N
k
}
, and
α
=1
N
α
=
N
. The partition function in
Equation 2.11 becomes
N
!
A
N
{
N
α
}
N
!(∆
ω
)
N
N
1
!
N
2
!
...N
α
!
...N
k
!
1
Z
(
N, V, T
)=
d
3
N
x
exp
V
ij
N
≡
{
N
α
}
×
···
−
β
Z
(
{
N
α
}
)
,
i<j
(2.12)
where
d
3
N
x
exp
V
ij
.
N
(∆
ω
)
N
A
N
α
=1
N
α
!
Z
(
{
N
}
)=
···
−
β
(2.13)
i<j
Denote the maximal term in Equation 2.12 by
Z
(
{
N
a
}
)
max
. Apparently the
following inequality exists
)
max
<Z
(
N, V, T
)
<N
k
Z
(
Z
(
{
N
a
}
{
N
a
}
)
max
.
Take the logarithm of each term in the inequality and divide by
N
.
The following inequality is obtained
1
N
ln
Z
(
N
ln
Z<
1
1
{
N
α
}
)
max
<
N
[ln
Z
(
{
N
α
}
)
max
+
k
ln
N
]
.
As
N
approaches infinity, ln
N/N
approaches zero. Therefore, ln
Z/N
can
be replaced by (1
/N
)ln(
Z
{
N
α
}
)
max
. As a result, one needs only to calculate
