Q 11-11-046JEE MainJEE Main 2026 (24 Jan, Shift 2)Medium
10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ is $\alpha$ Joule ( $P_1 = 21.7$ Pa and $P_2 = 30\ \text{Pa}, \text{C}_v = 21\ \text{J/K.mol}, R = 8.3\ \text{J/mol.K}$). The value of $\alpha$ is ______ .
Answer: (A) 21
From $P_1$ to $P_2$ the gas goes straight up at $V = 1\ \text{m}^3$: an isochoric process, so $W = 0$ and $Q = nC_v\Delta T$.
From $PV = nRT$ at constant $V$: $n\Delta T = \dfrac{V\Delta P}{R}$.
$$Q = C_v\frac{V\Delta P}{R} = 21\times\frac{1\times(30 - 21.7)}{8.3} = 21\times1 = 21\ \text{J}$$
So $\alpha = 21$.
Solution by Sreeraj P, M.Sc Physics