JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Heat & Thermodynamics

For a complete cyclic process, the internal energy returns to its initial value, so $\Delta U=0$ over the cycle. By the first law of thermodynamics, $Q=\Delta U+W$, so for the full cycle

and the magnitude of the net heat exchanged equals the magnitude of the net work done, which in turn equals the area enclosed by the closed loop on the P–V diagram.

From the diagram, the loop is bounded between $P=200$ kPa and $400$ kPa, and between $V=200$ cc and $400$ cc, with the boundary between the extreme points $A$ and $C$ traced out as a semicircular arc through $B$ rather than a straight line. Taking the half-range in each variable as the radius,

the area enclosed by this semicircular loop is


Since $|Q|=|W|$, the magnitude of heat exchanged over the cycle is $5\pi$ J.