Question

An ideal gas has a molar heat capacity $C_v$ at constant volume. It undergoes a process given by $T=T_0{e}^{\frac{\alpha }{V}}$, where $T_0$ and $\alpha$ are constants. Find the molar heat capacity of the gas as a function of $V$.

• A. $C_v-\frac{VR}{\alpha }$
• B. $C_v+VR$
• C. $C_v+\frac{R\alpha }{V}$
• D. $C_v-\frac{V^{2}R}{\alpha }$

Answer 1

The correct option is A $C_v-\frac{VR}{\alpha }$

Given, the process equation is -
$T=T_0{e}^{\frac{\alpha }{V}}$
Differentiating w.r.t $T$ on both sides,
$\frac{dT}{dT}=T_0{e}^{\frac{\alpha }{V}}\times -\frac{\alpha }{V^2}\left(\frac{dV}{dT}\right)$
$=\frac{-\alpha }{V^2}T\left(\frac{dV}{dT}\right)$
$=\frac{\alpha }{V^2}\left(\frac{PV}{nR}\right)\left(\frac{dV}{dT}\right)\{\because PV=nRT\}$
$\Rightarrow 1=\frac{-\alpha }{VR}\left(\frac{P}{n}\frac{dV}{dT}\right)$
$\Rightarrow \frac{P}{n}\left(\frac{dV}{dT}\right)=\left(\frac{-VR}{\alpha }\right)\dots \dots \left(i\right)$

Also, we know that
$dQ=dU+dW$
$\Rightarrow nCdT=n{C}_{v}dT+PdV$
$\Rightarrow C=C_v+\frac{1}{n}\left(\frac{PdV}{dT}\right)\dots \dots \left(ii\right)$
Putting (i) in (ii),
$C=C_v-\frac{VR}{\alpha }$