Previous Year NEET Question- Kinetic Theory L-6

Question: The molar specific heats of an ideal gas at constant pressure and volume are denoted by $C_p$ and $C_v$ respectively. If $\gamma=\frac{C_\rho}{C_v}$ and $R$ is the universal gas constant, then $C_V$ is equal to#

A) $\frac{1+\gamma}{1-\gamma}$

B) $\frac{R}{(\gamma-1)}$

C) $\frac{(\gamma-1)}{R}$

D) $\gamma R$

Answer: $\frac{R}{(\gamma-1)}$#

Solution:#

As we know that and $$ \begin{aligned} & C_p-C_V=R \ & C_p=R+C_V \end{aligned} $$ and $$ \frac{C_p}{C_V}=\gamma $$ (given)

So, $$ \frac{R+C_V}{C_V}=\gamma \Rightarrow \gamma C_v=R+C_V $$ $\Rightarrow \quad \gamma C_v-C_v=R$ $\Rightarrow \quad C_v=\frac{R}{\gamma-1}$