An ideal gas has a molar heat capacity  $C_v$ at constant volume. The molar heat capacity of this gas a function of its volume, $V$

According to first law of thermodynamics, $dT=\alpha T_0{e}^{\alpha V}dV$
$dQ=dU+PdV$ (For one mole) 
Now molar specific heat is given by
$C=\frac{dQ}{dT}=\frac{dU+PdV}{dT}=\frac{C_{v}dT+(RT/V)dV}{dT}$

$=C_v+\left[\frac{RT}{V}\right]\left[\frac{dV}{dT}\right] \dots .\left(i\right)$

Given,  $T=T_0{e}^{\alpha V}$

or  $dT=\alpha T_0{e}^{\alpha V}dV$

$\therefore \frac{dV}{dT}=\frac{1}{\alpha T_0{e}^{\alpha V}} \dots .\left(ii\right)$

Substituting the value of  $(dV/dT)$ from eq. (ii) in eq. (i),
We have  
$C=C_v+\left(\frac{RT}{V}\right)\left[\frac{1}{\alpha T_0{e}^{\alpha V}}\right]$

$C=C_v+\left(\frac{R{T}_0{e}^{\alpha V}}{\alpha V{T}_0{e}^{\alpha V}}\right)=C_v+\frac{R}{\alpha V} \dots ..\left(iii\right)$

(C) Given that  $P=P_0{e}^{\alpha V}$

$\frac{RT}{V}=P_0{e}^{\alpha V} or T=\frac{P_0}{R}V{e}^{\alpha V}$

Now,  $C=C_v+\left[\frac{RT}{V}\right]\left[\frac{dV}{dT}\right]$

$$\begin{gathered} =C_v+\left(P_0{e}^{\alpha V}\right)\left[\frac{R}{P_0{e}^{\alpha V}(1+\alpha V)}\right] \\ =C_v+\frac{R}{(1+\alpha V)} \end{gathered}$$