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JEE Main 2026 · 2 Apr · Shift 1, Physics question 47

A vessel contains 0.15 m³ of a gas at pressure 8 bar and temperature 140 °C with cp=3Rc_p=3R and cv=2Rc_v=2R. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is ________ kJ. (RR is gas constant)

Answer and worked solution

Official answer: 120

Adiabatic index. γ=cpcv=32\gamma=\dfrac{c_p}{c_v}=\dfrac32.

Final volume. V2=V1(P1P2)1/γ=0.15×82/3=0.6V_2=V_1\left(\dfrac{P_1}{P_2}\right)^{1/\gamma}=0.15\times8^{2/3}=0.6 m³.

Work.
W=P1V1−P2V2γ−1=8×105(0.15)−105(0.6)0.5=1.2×105 J=120 kJ.W=\frac{P_1V_1-P_2V_2}{\gamma-1}=\frac{8\times10^5(0.15)-10^5(0.6)}{0.5}=1.2\times10^5\text{ J}=120\text{ kJ}.

Answer: 120

Topics:Work done by a gasThermodynamic processes
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