Q 11-11-042JEE MainJEE Main 2026 (2 Apr, Shift 1)Medium
A vessel contains $0.15\ \text{m}^3$ of a gas at pressure $8$ bar and temperature $140\ ^\circ$C with $c_p=3R$ and $c_v=2R$. It is expanded adiabatically till pressure falls to $1$ bar. The work done during this process is ______ kJ. ($R$ is gas constant)
Numerical value type. Enter your answer.
Answer: 120
$\gamma=\dfrac{c_p}{c_v}=\dfrac32$.
Final volume: $V_2=V_1\left(\dfrac{P_1}{P_2}\right)^{1/\gamma}=0.15\times8^{2/3}=0.15\times4=0.6\ \text{m}^3$
$$W=\frac{P_1V_1-P_2V_2}{\gamma-1}=\frac{8\times10^{5}\times0.15-1\times10^{5}\times0.6}{0.5}=\frac{1.2\times10^{5}-0.6\times10^{5}}{0.5}=1.2\times10^{5}\ \text{J}$$
$W=120$ kJ.
Solution by Sreeraj P, M.Sc Physics