JEE Main 2026 — Thermodynamics (C) Question with Solution

Using the first law of thermodynamics, $\Delta U=q+w$.

For the process from state X to state Y:
Heat absorbed by the gas, $q_{X\to Y}=+10$ J
Work done by the gas, $w_{X\to Y}=-18$ J
Change in internal energy, $\Delta U_{X\to Y}=q_{X\to Y}+w_{X\to Y}=10-18=-8$ J

For the reverse process from state Y to state X:
Change in internal energy, $\Delta U_{Y\to X}=-\Delta U_{X\to Y}=+8$ J
Heat evolved by the gas, $q_{Y\to X}=-6$ J
Let the work done on the gas be $w_{Y\to X}$.
$\Delta U_{Y\to X}=q_{Y\to X}+w_{Y\to X}$
$8=-6+w_{Y\to X}$
$w_{Y\to X}=+14$ J

Since the work done is positive, $14$ J of work is done on the gas by the surrounding.

Answer: $14$ J of the work is done on the gas 'A' by the surrounding.