Molar heat capacity of an ideal gas varies as , $C=C_v + \alpha T , C=C_v + \beta T and C=C_v + aP$ , where $\alpha , ß, and a$ are constants. Find the equations of the process for an ideal gas in terms of the variables $T and V.$

$\left(i\right)$ Form first law of thermodynamics, $dQ = dU + dW dQ = C dT = C_v dT + PdV$

$C=\frac{dQ}{dT}=C_v + P\frac{dV}{dT}=C_v + \frac{RT}{V}\frac{dV}{dT} ......\left(1\right)$

since it varies as, $C=C_v + \alpha T......\left(2\right)$

on comparing expression $\left(1\right)$ and $\left(2\right)$,

we have $\alpha T=\left(\frac{RT}{V}\right)\left(\frac{dV}{dT}\right) \Rightarrow \left(\frac{\alpha }{R}\right) dT = \frac{dV}{V}.......\left(3\right)$

on integrating eqn. $\left(3\right)$, we have $\ln V - \frac{aT}{R}=\ln k or V =k e^{\alpha T / R} or V{e}^{-(\alpha T / R)}= cons\tan t ....\left(4\right)$

Eqn. $\left(4\right)$ is the equation of process.

$\left(ii\right)$ We have $C=C_v + \frac{RT}{V}\frac{dV}{dT}$

Comparing it with the given molar heat capacity, $C=C_v + \beta V.$Hence , we have $\frac{RT}{V}\frac{dV}{dT}=\beta V \Rightarrow \frac{dV}{V^2}=\frac{\beta }{R}\frac{dT}{T}......\left(5\right)$

Above equation integration yields $-\frac{1}{V}=\frac{\beta }{R}\log T + cons\tan t$

$\log T + \frac{R}{\beta V}= another cons\tan t...\left(6\right)$

or $T{e}^{(R / \beta V) }= another cons\tan t ......\left(7\right)$

$\left(iii\right)$ On comparing $C=C_v + \frac{RT}{V}\frac{dV}{dT}$ 

or $C=C_v + aP$we have $\frac{RT}{V}\frac{dV}{dT}=aP$

since $PV=RT$for one mole , hence

$\frac{dV}{dT}=a or V=aT or V=anT for n$ moles