Derive the ratio of $\frac{C_{P}}{C_{V}}$ for a diatomic gas.

(D) For a diatomic rigid rotator gas,the degree of freedom $(f)$ is $5$,and the energy associated with each degree of freedom is $\frac{1}{2} k_{B} T$.
From the law of equipartition of energy,the total internal energy $(U)$ of $1 \text{ mole}$ of gas is:
$U = 5 \times \frac{1}{2} k_{B} T \times N_{A}$
$U = \frac{5}{2} (k_{B} N_{A}) T$
Since $k_{B} N_{A} = R$,we have:
$U = \frac{5}{2} RT \quad \dots(1)$
The molar specific heat at constant volume $(C_{V})$ is given by:
$C_{V} = \frac{dU}{dT} \quad \dots(2)$
Substituting equation $(1)$ into $(2)$:
$C_{V} = \frac{d}{dT} \left[ \frac{5}{2} RT \right] = \frac{5}{2} R$
Using Mayer's relation for molar specific heat at constant pressure $(C_{P})$:
$C_{P} - C_{V} = R$
$C_{P} = C_{V} + R = \frac{5}{2} R + R = \frac{7}{2} R$
Therefore,the ratio $\gamma = \frac{C_{P}}{C_{V}}$ is:
$\frac{C_{P}}{C_{V}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} = \frac{7}{5} = 1.4$