If a gas has ‘f’ degress of freedom, find the ratio of C_p and C_v
(or)
How specific heat capacity of mono atomic, diatomic and poly atomic gases can be explained on the basis of Law of equipartition of Energy ?

The law of equipartition of energy has successfully been applied to calculate the specific heats of monoatomic, diatomic and polyatomic gases. Consider one mole of an ideal gas at
temperature T, has f degrees of freedom. According to law of equipartition of energy. Energy associated with each molecule per degree of freedom $=\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}$
Total internal energy of 1 mole of the gas. i.e.,
$\mathrm{U}=\mathrm{f}\left(\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\right){\mathrm{N}}_{\mathrm{A}}=\frac{\mathrm{f}}{2}\mathrm{RT}\left(\text{ as }{\mathrm{k}}_{\mathrm{B}}=\frac{\mathrm{R}}{{\mathrm{N}}_{\mathrm{A}}}\right)$
Thus ${\mathrm{C}}_{\mathrm{V}}=\frac{\mathrm{dU}}{\mathrm{dT}}=\frac{\mathrm{d}}{\mathrm{dT}}\left(\frac{\mathrm{f}}{2}\mathrm{RT}\right)=\frac{\mathrm{f}}{2}\mathrm{R}$.................(1)
Since C_P – CV = R is true for any ideal gas, whether mono, di or polyatomic,
$\begin{array}{l}\to {\mathrm{C}}_{\mathrm{P}}={\mathrm{C}}_{\mathrm{V}}+\mathrm{R}=\frac{\mathrm{f}}{2}\mathrm{R}+\mathrm{R}\\ =\left(\frac{\mathrm{f}}{2}+1\right)\mathrm{R}=......\left(2\right)\end{array}$
The ratio of two molar specific heats (C_P and C_V), i.e.,
$\mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{\frac{\mathrm{f}}{2}\mathrm{R}+\mathrm{R}}{\frac{\mathrm{f}}{2}\mathrm{R}}=\left(1+\frac{2}{\mathrm{f}}\right)$