Gas Laws & Ideal Gas Equation question from JEE Main 2021 (31 Aug, Shift 1)

Answer: (a)

1. Integrate $\frac{dp}{p} = -a\,dv$ with $p = p_0$ at $v = 0$: $p = p_0 e^{-av}$.
2. For one mole $T = \frac{pv}{R} = \frac{p_0 v e^{-av}}{R}$.
3. $\frac{dT}{dv} = \frac{p_0}{R}e^{-av}(1 - av) = 0 \Rightarrow v = \frac{1}{a}$.
4. $T_{max} = \frac{p_0}{R}\cdot\frac{1}{a}\cdot e^{-1} = \frac{p_0}{aeR}$
5. So the answer is (a) $\frac{p_0}{aeR}$.