Statistical Thermodynamics
Partition Function
Sardar Vallabhbhai National Institute of Technology, Surat
Department of Chemistry
Faculty Coordinator– Dr. Ritambhara Jangir
Sawai Singh, Azhar Mahmood, Abhishek Mondal, Sahil, Harshit Mishra, Rathod Dhanraj,
Ripunjay Tanwar, Tanuj Singh, Thamminana Bhargava, Shubham Kumar
March 30, 2022
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
1 / 17
Introduction to Partition Function
Partition function indicates how the particles are distributed
in various energy levels.
From Boltzmann distribution law,
ni = gi e
N=
i
X
ni =
i=0
i
X
−ϵi
kT
gi e
where, k = 1.38 ∗ 10−23 J/K
(1)
−ϵi
kT
(2)
⇒ MolecularPartitionFunction(Q)
i=0
ϵT = ϵt + ϵr + ϵv + ϵe
(3)
gi = gt gr gv ge
(4)
From eqn: 2, 3, 4
Q = gt gr gv ge
X
e
−ϵT
kT
Q = Qt .Qv .Qr .Qe
(5)
(6)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
2 / 17
Translation Partition Function
Properties
It is unitless
Fraction of Particles in i th level
−ϵi
gi e kT
ni
=P
pi =
−ϵi
i
N
g e kT
(7)
i=0 i
pi =
e
−ϵi
kT
Q
For ground State, let g0 = 1 and ϵ0 = 0,
From eqn: 7,
g0 e 0
1
=
p0 =
Q
Q
1
Q=
p0
(8)
(9)
(10)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
3 / 17
Translation Partition Function
Properties
Q at low temperature (T → 0) Q = g0 . Therefore, at low
temperature only ground state is accessible.
Q at high temperature (T → ∞).
Q=
i
X
gi
(11)
i=0
Therefore, at high temperature all the levels are accessible.
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
4 / 17
Translation Partition Function
Qt(x) =
X
gt e
−ϵt
kT
Therefore,
Qt(x) =
X
e
(gt = 1)
(12)
−ϵt
kT
(13)
If the particle having mass m, velocity v, and wavelength in x-axis,
λx , then from de broglie wavelength equation,
h
= mv = Px
λx
(14)
1 2 2
P2
1
ϵt = mv 2 =
m v = x
2
2m
2m
(15)
ϵt =
h2
λ2x .2m
(16)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
5 / 17
Translation Partition Function
nλx
2
2lx
⇒ λx =
n
Now substituting λx in eqn: 15
lx =
⇒ ϵt =
n2 h2
8mlx2
(17)
(18)
(19)
Now substituting the value of ϵt in eqn:12
Qt(x) =
X
−n2 h2
e 8mlx2 kT
(20)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
6 / 17
Translation Partition Function
Z
∞
−n2 h2
e 8mlx2 kT dn
(21)
0
Let a =
h2
8mlx2 kT
Z
Qt(x) =
∞
2
e −an dn
(22)
0
⇒ Qt(x)
1
=
2
r
π
1 π8mlx2 kT
lx √
=
=
2πmkT
2
a
2
h
h
ly √
lx √
lz √
=
2πmkT
2πmkT
2πmkT
h
h
h
r
Qt = Qt(x) Qt(y ) Qt(z)
(23)
(24)
3
(2πmkT ) 2 lx ly lz
Qt =
h3
(25)
3
(2πmkT ) 2 V
Qt =
h3
(26)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
7 / 17
Rotational Partition Function
It is given by diatomic molecules only, e.g. H2 , N2 , CO and NO. As we
know from partition function
Qr =
X
ϵr
gr e − kT
(27)
The rotational energy level of a rigid diatomic rotor (that is a rotor whose
inter-nuclear distance remain fixed during rotation), obtained by solving
the Schrodinger wave equation are
ϵr =
J(J + 1)h2
8π 2 I
(28)
where the rotational quantum number J has the value 0, 1, 2, 3, ... and I
is moment of inertia.
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
8 / 17
Rotational Partition Function
The degeneracy gr of rotational energy level is equal to (2J + 1).
Qr =
X
(2J + 1)e −J(J+1)h
2 /8π 2 IkT
(29)
R
P
when all the energy levels are closely associated, so
is converted into .
Z ∞
2
2
Qr =
(2J + 1)e −J(J+1)h /8π IkT dJ
(30)
0
Let’s take β =
h2
,
8π 2 IkT
then
Z
Qr =
∞
(2J + 1)e −J(J+1)β dJ
(31)
0
Z = J(J + 1) = J 2 + J
(32)
dZ = (2J + 1) dJ
(33)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara Jangir
9 / 17
Rotational Partition Function
Z
∞
e −Z β dZ
(34)
8π 2 IkT
1
=
β
h2
(35)
Qr =
0
Qr =
This is valid only for hetero-nuclear molecules like NO, CO, HI.
For homonuclear molecules like O2 , N2 , ...
Qr =
8π 2 IkT
σh2
(36)
where σ is symmetry factor.
σ = 2 for homonuclear molecules.
σ = 1 for hetero-nuclear molecules.
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 10
Jangir
/ 17
Rotational Partition Function
Characteristic Rotational Temperature (θr )
Qr =
Let
8π 2 IkT
h2
(37)
8π 2 I
1
= 2
B
h
Then,
kT
T
=
B
B/k
(38)
Qr =
T
θr
(39)
θr =
T
Qr
(40)
Qr =
Therefore,
or
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 11
Jangir
/ 17
Vibrational Partition Function
For a diatomic molecule vibrating as a simple harmonic oscillator
(SHO), the vibrational energy levels, obtained by the solution of
Schrodinger wave equation are given by:
1
ϵvib = (v + )hν
2
(41)
where ν is the vibrational frequency and v is the vibrational quantum
number. The degenercy is one. Using equations
Q=
X
gi e
−ϵi
kT
i
and
1
ϵvib = (v + )hν
2
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 12
Jangir
/ 17
Vibrational Partition Function
The vibrational partition function of the diatomic molecule is given by
Qvib =
∞
X
v =0
e
−(v + 1
2 )hν
kT
−hν
= e 2kT
∞
X
e
−vhν
kT
(42)
v =0
In SHO, hν >> kT . Hence, summation cannot be replace by
integration. The summation in this can be as follows;
Let,
hν
x=
kT
∞
X
−vhν
1
1
e kT =
=
hν
−x
1−e
1 − e − kT
v =0
(43)
(44)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 13
Jangir
/ 17
Vibrational Partition Function
Using eqn
Qvib = e
−hν
2kT
∞
X
e
−vhν
kT
(45)
v =0
we get,
−hν
Qvib =
e 2kT
hν
1 − e − kT
(46)
If θvib is the ’characteristic vibrational temperature’, of the oscillator
as
hν
θvib =
(47)
k
then, vibrational partition function;
Qvib =
e
−θvib
2T
1 − e−
−θvib
T
(48)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 14
Jangir
/ 17
Vibrational Partition Function
According to Heisenberg’s uncertainty principle, the oscillator cannot
have a total zero energy. Thus, from
1
ϵvib = (v + )hν
2
even in the ground vibrational state (v=0), the oscillator posses an
energy ( hν
2 ), called the zero-point energy, above the classical energy
zero. Hence, energy level expression for the SHO is modified to,
ϵ′vib = vhν (v = 0, 1, 2, 3, .....)
(49)
Hence, the new vibrational partition function is given by,
′
qvib
=
∞
X
e
−vhν
kT
(50)
v =0
′
qvib
=
1
1−e
hν
− kT
=
1
1−e
−θvib
T
(51)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 15
Jangir
/ 17
Vibrational Partition Function
Case I:
When temperature is very low so that
θvib
T
>> 1. Using,
−hν
Qvib =
e 2kT
hν
1 − e − kT
we get;
Qvib = e
−θvib
2T
(52)
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 16
Jangir
/ 17
Vibrational Partition Function
Case II :
When temperature is very high so that
θvib
T
<< 1. Using,
−hν
Qvib =
e 2kT
hν
1 − e − kT
we get;
Qvib =
T −θvib
e 2T
θvib
(53)
The value of ν, and hence θvib , obtained from the infared spectrum of
a molecule.
Sardar Vallabhbhai National Institute of Technology, Statistical
Surat Department
Thermodynamics
of Chemistry Faculty Coordinator–
March 30,
Dr.2022
Ritambhara 17
Jangir
/ 17