Q 11-12-083JEE MainJEE Main 2022 (24 Jun, Shift 1)Medium
$0.056\ \text{kg}$ of nitrogen is enclosed in a vessel at a temperature of $127^\circ\text{C}$. The amount of heat required to double the speed of its molecules is ______ kcal. (Take $R = 2\ \text{cal mol}^{-1}\text{K}^{-1}$)
Numerical value type. Enter your answer.
Answer: 12
$n = \dfrac{56}{28} = 2$ mol, $T_1 = 400\ \text{K}$. Molecular speed $\propto\sqrt T$, so doubling the speed needs $T_2 = 4T_1 = 1600\ \text{K}$.
The gas is in a closed vessel (constant volume), and for diatomic $\text{N}_2$, $C_V = \dfrac52 R = 5\ \text{cal mol}^{-1}\text{K}^{-1}$:
$$Q = nC_V\Delta T = 2\times5\times1200 = 12000\ \text{cal} = 12\ \text{kcal}$$
Solution by Sreeraj P, M.Sc Physics