Main 2016 | JEE 2016 | Discussion Page

n moles of an ideal gas undergoes a process A and B shown in the figure. The maximum temperature of the gas during the process will be

532020193_process 1.JPG

$\frac{9}{4}\frac{p_{0}V_{0}}{nR}$

$\frac{3}{2}\frac{p_{0}V_{0}}{nR}$

$\frac{9}{2}\frac{p_{0}V_{0}}{nR}$

$\frac{9p_{0}V_{0}}{nR}$

Option A

p-V equation for path AB

$p= -(\frac{p_{0}}{V_{0}})V+3p_{0}$

$\Rightarrow pV=3p_{0}V -\frac{p_{0}}{V_{0}}V^{2}$

OR $T=\frac{pV}{nR}=\frac{1}{nR}(3p_{0}V-\frac{p_{0}}{V_{0}}V^{2})$

For maximum temperature

$\frac{dT}{dV}=0$

$\Rightarrow 3p_{0}-\frac{2p_{0}V}{V_{0}}=0\Rightarrow V=\frac{3}{2}V_{0}$

and $p= 3p_{0}-\frac{p_{0}}{V_{0}}=\frac{3p_{0}}{2}$

Therefore, at these values,

$T_{max}=\frac{(\frac{3p_{0}}{2})(\frac{3V_{0}}{2})}{nR}=\frac{9p_{0}V_{0}}{4nR}$

Discussion Forum