Q 11-11-163JEE MainJEE Main 2018 (16 Apr, Shift 1)Medium
One mole of an ideal monatomic gas is taken along the path $ABCA$ as shown in the $PV$ diagram. The maximum temperature attained by the gas along the path $BC$ is given by:
Answer: (C) $\dfrac{25}{8}\dfrac{P_0V_0}{R}$
$B = (V_0, 3P_0)$ and $C = (2V_0, P_0)$. The straight line $BC$ is
$$P = 5P_0 - \frac{2P_0}{V_0}V$$
For one mole, $RT = PV = 5P_0V - \dfrac{2P_0}{V_0}V^2$. This is maximum where
$$\frac{d(PV)}{dV} = 5P_0 - \frac{4P_0}{V_0}V = 0 \quad\Rightarrow\quad V = \frac54V_0$$
which lies between $B$ and $C$. There $P = 5P_0 - \frac52P_0 = \frac52P_0$, so
$$T_\text{max} = \frac{1}{R}\cdot\frac52P_0\cdot\frac54V_0 = \frac{25}{8}\frac{P_0V_0}{R}$$
Solution by Sreeraj P, M.Sc Physics