JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Heat & Thermodynamics

In a cyclic process the net work done equals the area enclosed by the loop on the pressure-volume diagram. The cycle $CAB$ encloses a triangle whose base is the volume change $\Delta V=5-1=4{\text{m}}^3$ and whose height is the pressure change $\Delta P=5\text{Pa}$. The enclosed area is $\frac{1}{2}\times 4\times 5=10\text{J}$, so the work done in the cycle is $10\text{J}$.