Question

A one mole of ideal gas goes through a process in which pressure $p$ varies with volume $V$ as $p=3-g{\left(\frac{V}{V_0}\right)}^2$, where, $V_0$ is a constant. The maximum attainable temperature by the ideal gas during this process is (all quantities are is SI units and $R$ is gas constant)

• A. $\frac{2V_0}{3R}$
• B. $\frac{2V_0}{R}$
• C. $\frac{3V_0}{2R}$
• D. None of these

Answer 1

Answer: (D)

( No option is matching)
Using $pV=RT$ and $p=3-g\left(\frac{V^2}{V_0^2}\right)$, we get
$T=\frac{3V}{R}-\frac{g{V}^3}{R{V}_0^2}$
Differentiating w.r.t. volume, we get
$\frac{dT}{dV}=\frac{3}{R}-\frac{g}{R{V}_0^2}\times 3V^2$
For maximum $T,\frac{dT}{dV}=0$
$\Rightarrow \frac{3}{R}-\frac{g}{R{V}_0^2}\times 3V^2=0\Rightarrow V=\frac{V_0}{\sqrt{g}}$
Substituting in Eq. (i), we get
$T_{\max }=\frac{3V_0}{R\sqrt{g}}-\frac{g{\left(V_0/\sqrt{g}\right)}^3}{R{V}_0^2}$
$=\frac{2V_0}{R\sqrt{g}}$