Question

For the case of an ideal gas, find the equation of the process in which the molar heat capacity varies as $C=C_v+\alpha T^2$, where $\alpha$ is constant. Take $V_0$ to be the initial volume of the gas.

• A. $V=V_0{e}^{\frac{\alpha T^2}{2R}}$
• B. $V=V_0{e}^{\frac{2\alpha T}{R}}$
• C. $V=V_0{e}^{\frac{\alpha T^2}{R}}$
• D. $V=V_0{e}^{\frac{T^2}{\alpha R}}$

Answer 1

The correct option is A $V=V_0{e}^{\frac{\alpha T^2}{2R}}$

$C=C_v+\alpha T^2\dots \dots \left(i\right)$
Also, we know that
$C=C_v+\frac{P}{n}\left(\frac{dV}{dT}\right)\dots \dots \left(ii\right)$
Comparing (i) and (ii)
$\alpha T^2=\frac{P}{n}\left(\frac{dV}{dT}\right)$
$\Rightarrow \alpha T^{\text{/}2}=\frac{\text{/}nR\text{/}T}{V\text{/}n}\left(\frac{dV}{dT}\right)\{\because PV=nRT\}$
$\Rightarrow \alpha T=\frac{R}{V}\left(\frac{dV}{dT}\right)$
Integrating on both sides,
$\Rightarrow \int \frac{dV}{V}=\int \frac{\alpha }{R}TdT$
$\Rightarrow \ln V=\left(\frac{\alpha }{R}\right)\frac{T^2}{2}+\text{constant}$
For $T=0,V=V_0$
$\Rightarrow \ln \left(\frac{V}{V_0}\right)=\left(\frac{\alpha }{R}\right)\frac{T^2}{2}$
$\Rightarrow V=V_0{e}^{\frac{\alpha T^2}{2R}}$