JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Heat & Thermodynamics

Over a complete cycle the internal energy returns to its initial value, so the heat absorbed equals the work done, which is the area enclosed by the loop on the pressure-volume diagram. The loop runs from $20$ kPa to $40$ kPa in pressure and from $20$ litre to $40$ litre in volume, so its semi-axes are

The enclosed area is $\pi \Delta P\Delta V=\pi \times 10^4\times 10^{-2}=100\pi$ J, so the heat absorbed in the cycle is $100\pi$ J.