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JEE Main 2022 · 28 Jun · Shift 2, Physics question 38

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the gas during the part BC. The internal energy of the gas at A is 1560 J. The workdone by the gas during the part CA is :

P against V: cycle A → B vertical (pressure rises at constant volume), B → C down a straight line to lower volume and pressure, C → A horizontal at constant pressure.
Choose one option
Answer and worked solution

Official answer: B

Cycle. ΔU=0\Delta U=0, so Wnet=Qnet=40+0−60=−20W_{net}=Q_{net}=40+0-60=-20 J.

Parts. AB is isochoric (W=0W=0), BC has W=−50W=-50 J, so WCA=−20+50=30W_{CA}=-20+50=30 J.

(B)

Topics:First law and internal energyThermodynamic processes
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