Q 11-11-002NEETNEET 2026Top questionEasy
In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ($\gamma = 5/3$) decreases from $60$ K to $50$ K. The work done by the gas in the process is :
(Take the universal gas constant as $R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}$)
Answer: (C) $124.5$ J
Concept. Adiabatic work by the gas: $W = \dfrac{nR(T_1 - T_2)}{\gamma - 1}$.
Step 1. Substitute.
$$W = \frac{8.3 \times 10}{2/3} = 124.5\ \text{J}$$
Answer. $124.5$ J (Option C).
Solution by Sreeraj P, M.Sc Physics