Process equation is

$p V^\alpha=\text { constant }(k) \Rightarrow p=\frac{k}{V^\alpha}$

Work done by the gas in given process is

$\Delta W=\int \limits_{V_i}^{V_f} p d V$

$=\int \limits_{V_i}^{V_f} k d V$

$\left.=\left[\frac{p V}{1-\alpha}\right]_{V_i}^{V_f}=\frac{p\left(V^{1-\alpha}\right.}{1-\alpha}\right]_{V_i}^{V_f}$

$=\frac{p \Delta V}{1-\alpha}=\frac{n R \Delta T}{1-\alpha}$

The change of internal energy of gas in this process will be

$\Delta U=C_V \Delta T$

And if $\Delta Q$ is heat supplied to the gas then,

$\Delta Q=C \Delta T$

Now, by first law of thermodynamics, we have

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

$\Rightarrow \quad C \Delta T=C_V \Delta T+\frac{n R \Delta T}{1-\alpha}$

Heat capacity of the gas is

$\Rightarrow \quad C=C_V+\frac{n R}{1-\alpha}$