JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Chemical Thermodynamics

The cycle runs isobarically from state 1 to state 2, isochorically from 2 to 3, and isothermally back from 3 to 1, the last leg isothermal because $P_3{V}_3=P_1{V}_1=20\mathrm{bar}\mathrm{L}$. Along $1\to 2$ the gas expands at $1\mathrm{bar}$ through $20\mathrm{L}$, doing $20\mathrm{bar}\mathrm{L}$ of work; along $2\to 3$ the volume is fixed and no work is done; along $3\to 1$ the isothermal compression takes $P_1{V}_1\ln (V_3/V_1)=20\ln 2=20\times 2.3\times 0.3=13.8\mathrm{bar}\mathrm{L}$ of work on the gas. The net work over the cycle is $20-13.8=6.2\mathrm{bar}\mathrm{L}$, and with $1\mathrm{bar}\mathrm{L}=100\mathrm{J}$ its magnitude is $620\mathrm{J}$.