Question

If $\gamma$ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of the gas is

• A. $\frac{25}{2}(\gamma -1)$
• B. $\frac{3\gamma -1}{2\gamma -1}$
• C. $\frac{2}{\gamma -1}$
• D. $\frac{9}{2}(\gamma -1)$

Answer 1

Answer: (C)

$\frac{2}{\gamma -1}$

Since $C_v=n\frac{R}{2}$ and $C_p=\frac{nR}{2}+R=\left(\frac{n+2}{2}\right)R$

So,

$\gamma =\frac{C_p}{C_v}=\frac{n+2}{n}$

$⟹n=\frac{2}{\gamma -1}$