Question

Be able to explain how to find the allowed degrees of freedom for monatomic and diatomic molecules and the relationship between Cv and Cp

5) Derive 22.17 (22.7)

delta S= nR ln (Vf/Vi) (serway physics 9th)

6) Derive 22.18 (22.7)

delta S= Q(1/Tc - 1/Th) >0

Answer 1

Q4

Allowed degrees of freedom f = 3N - C

Where, N = no of atoms in a molecule

C = No of constraints

For mono atomic N=1 , C=0

f= 3×1 -0 = 3

For diatomic molecules N=2

C = 1

f= 3×2 - 1 = 5

Relation between Cp/Cv and f.....

Let us consider N molecules of a gas

and having degrees of freedom f for each molecules.

From law of equipartition of energy

Total energy associated with the molecules is,

E = N{(1/2 )fKT}

If we consider one mole of the gas, then

N is Na , Avagadro number and NK = R, universal gas constant.

So, E =( 1/2) fRT

Now, Cv = dE/dT

= 1/2 fCp R

Also Cp - Cv = R

Or, Cp = R + Cv

Or, Cp = R + 1/2 fR

Or, Cp = R[ 1 + f/2]

NOW, the ratio Cp/ Cv = 1 + 2/f

For, mono atomic f = 3

So, Cp/ Cv = 1 + 2/3 = 5/3 = 1.67

For diatomic f= 5

Cp/Cv = 1 +2/5 = 7/5 = 1.40