JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Heat & Thermodynamics

Over the full cycle $ABCA$ the internal energy returns to its initial value, so the net heat equals the net work done by the gas. The gas absorbs $40$J along $AB$, exchanges no heat along $BC$, and rejects $60$J along $CA$, giving net heat $40+0-60=-20$J and hence net work $-20$J. Along $BC$ the $50$J done on the gas means the gas does $W_{BC}=-50$J, and the part $AB$ is at constant volume, so $W_{AB}=0$. Writing $W_{AB}+W_{BC}+W_{CA}=-20$ then gives $0-50+W_{CA}=-20$, so the work done by the gas along $CA$ is $30$J.