Work done in the process is

$W =\int p d V$

$=\int \frac{C}{V^3} d V=\frac{C V^{-2}}{-2}$

$=\frac{p V^3 \cdot V^{-2}}{-2}=\frac{p V}{-2}$

$\Rightarrow \Delta W=-\frac{n R \Delta T}{2}$

Also, for ideal gas,

$\Delta U=n C_V \Delta T$

So, by first law of thermodynamics, we have

$\Delta Q=\Delta U+\Delta W$

$\Rightarrow \quad n C \Delta T=n C_V \Delta T-\frac{n R \Delta T}{2}$

$\Rightarrow \quad C=C_V-\frac{R}{2} \Rightarrow C=\frac{3}{2} R-\frac{R}{2}=R$

${\left[\therefore \text { for monoatomic ideal gas, } C_V=\frac{3}{2} R\right]}$

$\therefore$ Heat capacity for given process is

$C=R$