JEE Main 2026 — Thermodynamics (C) Question with Solution

Let State $1$ have pressure $P_1$ and volume $V_1$, and State $2$ have pressure $P_2$ and volume $V_2$. For an isothermal expansion, $V_2>V_1$ and $P_1>P_2$.

For expansion from State $1$ to State $2$:
In a single-stage expansion, the external pressure is constant at the final pressure $P_2$. The magnitude of work is $|w_A|=P_2(V_2-V_1)$.
In a multi-stage expansion, the external pressure is reduced in steps, keeping it closer to the internal pressure of the gas. The work done by the gas increases with the number of stages, reaching a maximum for a reversible process ($|w_{rev}|$). Thus, $|w_{rev}|>|w_B|>|w_A|$.

For compression from State $2$ to State $1$:
In a single-stage compression, the external pressure is constant at the final pressure $P_1$. The magnitude of work is $|w_C|=P_1(V_2-V_1)$.
In a multi-stage compression, the external pressure is increased in steps. The work done on the gas decreases with the number of stages, reaching a minimum for a reversible process ($|w_{rev}|$). Thus, $|w_C|>|w_D|>|w_{rev}|$.

Since $P_1>P_2$, the single-stage compression work $|w_C|$ is the largest, and the single-stage expansion work $|w_A|$ is the smallest. The reversible work magnitude $|w_{rev}|$ is the same for both expansion and compression between the same states.

Combining these inequalities, we get:
$|w_C|>|w_D|>|w_{rev}|>|w_B|>|w_A|$

Therefore, the correct order of the magnitude of work is $|w_C|>|w_D|>|w_B|>|w_A|$.

Answer: $|w_C|>|w_D|>|w_B|>|w_A|$