Q 11-11-062JEE MainJEE Main 2024 (4 Apr, Shift 2)Medium
A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The adiabatic constant for the gas is $\gamma = 3/2$. The work done by the gas in the process is ($\mu = 1$ mole)
Answer: (C) $RT[2 - \sqrt2]$
Final temperature: $T' = T\left(\dfrac{V}{2V}\right)^{\gamma - 1} = \dfrac{T}{\sqrt2}$.
$$W = \frac{\mu R(T - T')}{\gamma - 1} = \frac{RT(1 - 1/\sqrt2)}{1/2} = RT(2 - \sqrt2)$$
Solution by Sreeraj P, M.Sc Physics