Question

An ideal gas has a molar heat capacity at constant volume $C_V$. Find the molar heat capacity of this gas as a function of its volume $V$, if it undergoes a process $T=T_o{e}^{\alpha V^2}$, where $T_o$ and $\alpha$ are constants.

• A. $C_V+\frac{R}{\alpha V}$
• B. $C_V+\alpha RmV$
• C. $C_V+\frac{R}{\alpha V^2}$
• D. $C_V+\frac{R}{2\alpha V^2}$

Answer 1

The correct option is D $C_V+\frac{R}{2\alpha V^2}$

Given,
$T=T_o{e}^{\alpha V^2}$
Differentiating on both sides with respect to '$T$',
$\frac{d\left(T\right)}{dT}=T_o{e}^{\alpha V^2}\times 2\alpha V\frac{dV}{dT}$
$\Rightarrow 1=T\left(2\alpha V\frac{dV}{dT}\right).......\left(1\right)$
By using first law of thermodynamics, we can say that,
$\Delta Q=\Delta U+\Delta W$
or $nCdT=n{C}_{V}dT+PdV$
or $C=C_V+\frac{P}{n}\frac{dV}{dT}.....\left(2\right)$
Using ideal gas equation in $\left(1\right)$, we get,
$1=\left(\frac{PV}{nR}\right)2\alpha \frac{VdV}{dT}$
$\Rightarrow 1=\frac{2\alpha V^2}{R}\left(\frac{P}{n}\frac{dV}{dT}\right)$
$\Rightarrow \left(\frac{P}{n}\frac{dV}{dT}\right)=\frac{R}{2\alpha V^2}.......\left(3\right)$
Substituting $\left(3\right)$ in $\left(2\right)$,
$C=C_V+\frac{R}{2\alpha V^2}$
Thus, option (d) is the correct answer.