Q 11-11-160JEE MainJEE Main 2018 (15 Apr, Shift 1)Easy
One mole of an ideal monatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, $27^\circ\text{C}$. The magnitude of work done on the gas will be:
Answer: (B) $300R\ln 2$
For an isothermal process, $pV$ is constant, so doubling the pressure halves the volume.
Work done on the gas:
$$W = nRT\ln\frac{V_1}{V_2} = nRT\ln\frac{p_2}{p_1} = (1)(R)(300)\ln 2 = 300R\ln2$$
Solution by Sreeraj P, M.Sc Physics