Q 11-11-066JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
A total of $48\ \text{J}$ of heat is given to one mole of helium kept in a cylinder. The temperature of the helium increases by $2^\circ\text{C}$. The work done by the gas is (Given $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$)
Answer: (D) $23.1\ \text{J}$
Helium is monoatomic: $\Delta U = nC_V\Delta T = 1\times\dfrac32\times8.3\times2 = 24.9\ \text{J}$.
$$W = Q - \Delta U = 48 - 24.9 = 23.1\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics