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. $\frac{2 V_{0}}{3 R}$
• B. $\frac{2 V_{0}}{R}$
• C. $\frac{3 V_{0}}{2 R}$
• D. None of these

Answer 1

Answer: (D)

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