WAB=∫PdV (Assuming T to be constant) =∫V3RTdV=RT∫24 V−3dV=8×300×(−21[421−221])=225 J $\begin{aligned}& \mathrm{W}{\mathrm{BC}}=\mathrm{P} \int_4^2 \mathrm{dV}=10(2-4)=-20 \mathrm{J} \& \mathrm{W}{\mathrm{CA}}=0 \& \therefore \mathrm{W}_{\text {cycle }}=205 \mathrm{~J}
\end{aligned}$
JEE Main 2024 — Physics Thermodynamics
2024mcqhard
A real gas within a closed chamber at 27∘C undergoes the cyclic process as shown in figure. The gas obeys PV3=RT equation for the path A to B. The net work done in the complete cycle is (assuming R=8 J/molK ): 
Official previous-year question
Held on 9 Apr 2024 · Verified 6 Jul 2026.
Options
- A
20 J
- B
205 J
- C
−20 J
- D
225 J
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