Q 11-11-115JEE MainJEE Main 2021 (31 Aug, Shift 2)Medium
A sample of gas with $\gamma = 1.5$ is taken through an adiabatic process in which the volume is compressed from $1200\ \text{cm}^3$ to $300\ \text{cm}^3$. If the initial pressure is $200$ kPa, the absolute value of the work done by the gas in the process $=$ ______ J.
Numerical value type. Enter your answer.
Answer: 480
$P_2 = P_1\left(\dfrac{V_1}{V_2}\right)^\gamma = 200\times4^{1.5} = 1600$ kPa.
$P_1V_1 = 200\times10^3\times1200\times10^{-6} = 240$ J; $P_2V_2 = 1600\times10^3\times300\times10^{-6} = 480$ J.
$$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{240 - 480}{0.5} = -480\ \text{J}$$
Absolute value: $480$ J.
Solution by Sreeraj P, M.Sc Physics