An ideal gas has molar heat capacity  $C_V$ at constant volume. The gas undergoes a process as  $T=T_0(1+\alpha V^2)$ where T and V are temperature and volume respectively, $T_0$ and $\alpha$ are positive constants. The molar heat capacity C of the gas is given as $C=C_V+Rf\left(V\right),$ where $f\left(V\right)$ is a function of volume. The expression for $f\left(V\right)$ is

Given,   

$$\begin{gathered} C=C_V+Rf\left(V\right)\mathrm{.......}\left(i\right) \\ \because T=T_0(1+\alpha V^2)=T_0+T_0\alpha V^2 \\ dT=0+T_0\alpha .2V.dV \\ dT=2T_0\alpha VdV\Rightarrow \frac{dT}{dV}=2T_{0}V\alpha \end{gathered}$$

According to the first law of thermodynamics 

$$\begin{gathered} dQ=dU+dW \\ nCdT=n{C}_{V}dT+PdV \\ C=C_V+\frac{P}{n}\frac{dV}{dT}\mathrm{...........}\left(ii\right) \\ \therefore PV=nRT \\ \frac{P}{n}=\frac{R}{V}.T=\frac{R}{V}\left[T_0(1+\alpha V^2)\right] \end{gathered}$$

Putting this value  $\left(\frac{dT}{dV}\right)\text{\hspace{0.17em}and\hspace{0.17em}}\frac{P}{n}\text{\hspace{0.17em}in\hspace{0.17em}eqn}\left(\text{ii}\right)\text{\hspace{0.17em}we\hspace{0.17em}get.}$
 $$\begin{gathered} C=C_V+\frac{R}{V}\left[T_0(1+\alpha V^2)\right]\times \frac{1}{2T_0\alpha V} \\ C=C_V+R\left[\frac{1+\alpha V^2}{2\alpha V^2}\right] \end{gathered}$$
 Compare with equation (i), we get,

$f\left(V\right)=\frac{1+\alpha V^2}{2\alpha V^2}$