Molecular Partition Function
The Molecular Partition Function
The Boltzmann distribution can be written as
pi = exp(-i) / q
where pi is the probability of a molecule being found in a state i
with energy i. q is called the molecular partition function,
q = i exp(-i)
The summation is over all possible states (not the energy
levels).
Independent Molecules
Consider a system which is composed of N identical molecules.
We may generalize the molecular partition function q to the
partition function of the system Q
Q = i exp(-Ei)
where Ei is the energy of a state i of the system, and summation
is over all the states. Ei can be expressed as assuming there is
no interaction among molecules,
Ei = i(1) + i(2) +i(3) + … + i(N)
where i(j) is the energy of molecule j in a molecular state i
The partition function Q
Q = i exp[-i(1) - i(2) - i(3) - … -i(N)]
= {i exp[-i(1)]}{i exp[-i(2)]} … {i exp[-i(N)]}
= {i exp(-i)}N
= qN
where q  i exp(-i) is the molecular partition function. The
second equality is satisfied because the molecules are
independent of each other. The above equation applies only to
molecules that are distinguishable, for instance, localized
molecules. However, if the molecules are identical and free to
move through space, we cannot distinguish them, and the
above equation is to be modified!
Translational Partition Function of a molecule qT
Although usually a molecule moves in a three-dimensional
space, we consider first one-dimensional case. Imagine a
molecule of mass m. It is free to move along the x direction
between x = 0 and x = X, but confined in the y- and z-direction.
We are to calculate its partition function qx.
The energy levels are given by the following expression,
En = n2h2 / (8mX2) n = 1, 2, …
Setting the lowest energy to zero, the relative energies can then
be expressed as,
n = (n2-1)
 = h2 / (8mX2)
with
qx = n exp [ -(n2-1) ]
 is very small, then
qx = 1 dn exp [ -(n2-1) ] = 1 dn exp [ -(n2-1) ]
= 0 dn exp [ -n2 ] = (2m/h22)1/2 X
Now consider a molecule of mass m free to move in a container
of volume V=XYZ. Its partition function qT may be expressed
as
qT = qx qy qz
= (2m/h22)1/2 X (2m/h22)1/2 Y (2m/h22)1/2 Z
= (2m/h22)3/2 XYZ = (2m/h22)3/2 V
= V/3
where,  = h(/2m)1/2, the thermal wavelength. The thermal
wavelength is small compared with the linear dimension of the
container. Noted that qT  as T . qT  2 x 1028 for an O2
in a vessel of volume 100 cm3,  = 71 x 10-12 m @ T=300 K
Partition function of a perfect gas,
Q = (qT) N / N! = V N / [3N N!]
Energy
E = - (lnQ/)V = 3/2 nRT
where n is the number of moles, and R is the gas constant
Heat Capacity
Cv = (E/T)V = 3/2 nR
Diatomic Gas
Consider a diatomic gas with N identical molecules. A molecule is made
of two atoms A and B. A and B may be the same or different. When A
and B are he same, the molecule is a homonuclear diatomic molecule;
when A and B are different, the molecule is a heteronuclear diatomic
molecule. The mass of a diatomic molecule is M. These molecules are
indistinguishable. Thus, the partition function of the gas Q may be
expressed in terms of the molecular partition function q,
Q  q N / N!
The molecular partition q
q  i exp(  i)
where, i is the energy of a molecular state I, β=1/kT, and ì is the
summation over all the molecular states.
Factorization of Molecular Partition Function
The energy of a molecule j is the sum of contributions from its different
modes of motion:
 ( j)   T ( j)   R ( j)   V ( j)   E ( j)
where T denotes translation, R rotation, V vibration, and E the electronic
contribution. Translation is decoupled from other modes. The separation
of the electronic and vibrational motions is justified by different time
scales of electronic and atomic dynamics. The separation of the
vibrational and rotational modes is valid to the extent that the molecule
can be treated as a rigid rotor.
q  i exp(  i )  i exp[   ( iT   iR   iV   iE )]
 [i exp(  iT )][ i exp(  iR )][ i exp(  iV )][ i exp(  iE )]
 qT q R qV q E
The translational partition function of a molecule
qT  i exp(  iT )
ì sums over all the translational states of a molecule.
The rotational partition function of a molecule
q R  i exp(  iR )
 ì sums over all the rotational states of a molecule.
The vibrational partition function of a molecule
qV  i exp(  iV )
 ì sums over all the vibrational states of a molecule.
The electronic partition function of a molecule
q E  i exp(  iE )
 ì sums over all the electronic states of a molecule.
q  qT qV q R
qT  V / 3
w / qE  1
where
  h(  / 2M )1/ 2
  1 / kT
Vibrational Partition Function
Two atoms vibrate along an axis connecting the two atoms. The
vibrational energy levels:
 nV  (n  1 / 2)hv
n= 0, 1, 2, …….
If we set the ground state energy to zero or measure energy from the
ground state energy level, the relative energy levels can be expressed
as
 nV  nhv
5--------------5hv
4--------------4hv
3--------------3hv
2--------------2hv
1--------------hv
0--------------0
kT
  hv
Then the molecular partition function can be evaluated
q v  n exp(  n )  n exp( nhv)  1 /[1  exp(  hv)]
q v  1  e    e 2   e 3  ...
e   q v  e    e 2   e 3   ....  q v  1
Therefore,
1
q  1 /(1  e
)
1  e  hv
Consider the high temperature situation where kT >>hv, i.e.,
v
 
hv  1, q v  1 / hv  kT / hv
Vibrational temperature v
k v  hv
High temperature means that T>>v
 e  hv  1  h
v/K
v/cm-1
I2
309
215
F2
HCl
H2
1280 4300 6330
892 2990 4400
m
v

where
k
Rotational Partition Function
If we may treat a heteronulcear diatomic molecule as a rigid rod, besides
its vibration the two atoms rotates. The rotational energy
 JR  hcBJ ( J  1)
where B is the rotational constant. J =0, 1, 2, 3,…
q R  all rotationalstates exp[  JR ]
 all rotationalenergy levels g J exp[   JR ]
 J (2 J  1) exp[  hcBJ ( J  1)]
Bh/8cI2
where gJ is the degeneracy of rotational energy level εJ
hcB<<1
Usually hcB is much less than kT,

q R   (2 J  1) exp[  hcBJ ( J  1)]dJ
0
c: speed of light
R I: moment of Inertia
2
i
r im   I
i

 (1 / hcB) d{exp[  hcBJ ( J  1)]} / dJ dJ
0
 (1 / hcB){exp[  hcBJ ( J  1]}l0
=kT/hcB
Note: kT>>hcB
For a homonuclear diatomic molecule
q R  kT / 2hcB
Generally, the rotational contribution to the molecular partition function,
q R  kT / hcB
Where  is the symmetry number.
H 2O NH3 CH 4

2
3
12
Rotational temperature R
k R  hcB
Electronic Partition Function
q E  all electronicstates exp[   Ej ]  all electronicenergies g j exp[  Ej ]
 g 0 exp[  0E ]
=g0
=gE
where, gE = g0 is the degeneracy of the electronic ground state, and the
ground state energy 0E is set to zero.
If there is only one electronic ground state qE = 1, the partition function
of a diatomic gas,
Q  (1 / N!)(V / 3 ) N (kT / hcB) N (1  e hv )  N
At room temperature, the molecule is always in its ground state
Mean Energy and Heat Capacity
The internal energy of a diatomic gas (with N molecules)
U  U (0)  3N (1n /  )v  N ( ln  /  )v  N[1n(1  e  hv ) /  ]v
 (3 / 2) N1 /   N1 /   Nhv /( e hv  1)
 (5 / 2) NkT  Nhv /( e hv  1)
 (7 / 2) N kT (T>>1)
Contribution of a molecular to the total energy
Translational contribution
(1/2)kT x 3 = (3/2)kT
Rotational contribution
(1/2)kT x 2 = kT
Vibrational contribution
(1/2)kT + (1/2)kT = kT
kinetic
potential
the total contribution is (7/2)kT
qV = kT/hv
qR = kT/hcB
The rule: at high temperature, the
contribution of one degree of freedom
to the kinetic energy of a molecule
(1/2)kT
the constant-volume heat capacity
Cv  (U / T ) v
 (5 / 2) N k  N K ( hv) 2 e hv /( e hv  1) 2
 (7 / 2) N k (T>>1)
Contribution of a molecular to the heat capacity
Translational contribution
(1/2) k x 3 = (3/2) k
Rotational contribution
(1/2) k x 3 = k
Vibrational contribution
(1/2) k + (1/2) k = k
kinetic potential
Thus, the total contribution of a molecule to the heat capacity is (7/2) k
Partition Function
q = i exp(-i) = j gjexp(-j)
Q = i exp(-Ei)
Energy E= N i pi i = U - U(0) = - (lnQ/)V
Entropy S = k lnW = - Nk i pi ln pi = k lnQ + E / T
A= A(0) - kT lnQ
H = H(0) - (lnQ/)V + kTV (lnQ/V)T
Q = qN or (1/N!)qN
qq q q q
T
R
V
E