Q 11-11-072JEE MainJEE Main 2024 (9 Apr, Shift 1)Medium
A sample of 1 mole gas at temperature $T$ is adiabatically expanded to double its volume. If adiabatic constant for the gas is $\gamma = \dfrac32$, then the work done by the gas in the process is:
Answer: (C) $RT\left[2 - \sqrt2\right]$
$TV^{\gamma-1}$ is constant: $T_2 = T\left(\dfrac12\right)^{1/2} = \dfrac{T}{\sqrt2}$.
$$W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{RT\left(1 - \frac{1}{\sqrt2}\right)}{1/2} = RT\left(2 - \sqrt2\right)$$
Solution by Sreeraj P, M.Sc Physics