This includes the derivation of partition function and its relation with different thermodynamic parameters

2. The rotational energy for a single molecule is then
given by
𝐸𝑟 =
ℎ2
8𝜋2 𝐼
𝐽 𝐽 + 1 … … . . 1
Where I is moment of inertia of diatomic molecule
J= rotational quantum number (J=0,1,2,3…..)
𝑬 𝒓 = 𝑩𝒉𝑪 𝑱 𝑱 + 𝟏 ………..(2)
𝑤ℎ𝑒𝑟𝑒, 𝐵 =
ℎ
8𝜋2 𝐼𝐶

3. Rotational partition function for single molecule is given as
𝑞 𝑟 = 𝑔 𝑟 𝑒𝑥𝑝
−𝐸𝑟
𝐾𝑇∞
𝐽<0 …….(3)
Since, (2J+1) Eigen states ≅ (2𝐽 + 1)
∴ 𝑞 𝑟 = (2𝐽 + 1)𝑒𝑥𝑝
;𝐵ℎ𝐶 𝐽 𝐽:1
𝐾𝑇
∞
𝐽<0
… … … . 4
Ignoring the nuclear spin term
Rotational partition function is expressed as
𝑞 𝑟 = (2𝐽 + 1)𝑒𝑥𝑝;𝜌𝐽 𝐽:1 … … . . 5
∞
𝐽<0
Where, = Bhc/KT

4. Equation (5) can be solved by Euler-Maclaurin formula
𝑞 𝑟 =
1
𝜌
(1 +
𝜌
3
+
𝜌2
15
+
4𝜌3
315
+ ⋯ … … )
 is very small, i.e <0.05 then,
𝑞 𝑟 =
1
𝜌
=
8π2
IKT
ℎ2
For the molecule having some symmetry, symmetry number  is
introduced,
𝑞 𝑟 =
8π2IKT
 ℎ2
Hetero nuclear diatomic()= 1
Homo nuclear diatomic()=2
Simple symmetrical (CO2), S2()= 2
Asymmetric triatomic HOD()= 1
Benzene ()= 6
Chair form cyclohexane()=12

5. 𝐸𝑟 = 𝐾𝑇2
𝜕𝑙𝑛𝑄
𝜕𝑇 𝑉
Since, 𝑄 𝑟 = 𝑞 𝑟
𝑁
Or, 𝑙𝑛𝑄 𝑟 = 𝑁𝑙𝑛
8π2IKT
 ℎ2
Or,
𝜕𝑙𝑛𝑄
𝜕𝑇 𝑉
=
1
𝑇
× 𝑁
Substituting this rotational energy is found to be
𝐸𝑟 = 𝐾𝑇2
×
1
𝑇
× 𝑁
Er= NKT
Er= nRT

6. 𝐶 𝑉 =
𝜕
𝜕𝑇
𝐾𝑇2
𝜕𝑙𝑛𝑄
𝜕𝑇
𝑉
𝐶 𝑉 =
𝜕
𝜕𝑇
𝐾𝑇2
×
1
𝑇
× 𝑁
𝐶 𝑉 = 𝑁𝑇
= 𝑛𝑅
Similarly,
entropy, free energy and enthalpy is related as
∴ 𝑆𝑟 = 𝑛𝑅[1 + 𝑙𝑛
8π2
IKT
 ℎ2 ]
∴ 𝐺𝑟 = −𝑛𝑅𝑇𝑙𝑛
8π2
IKT
 ℎ2
Hr= nRT

7. Q. Calculate the rotational partition function for H2molecule
at 0°C given that K= 1.38X10-16erg deg-1, h= 6.624X10-
27ergsec, =2 and I= 0.459X10-40g cm2
Solution:
𝑞 𝑟 =
8π2
IKT
 ℎ2
=
8π2X4.59X10−48X1.38X10−23 𝑋273
2𝑋 (6.624𝑋10−34)2
𝑞 𝑟 =1.554

8. Thank You