JEE Advanced 2023 Paper 1 · Q11 · Kinetic Theory of Gases
A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas $(\gamma = 5/3)$ and one mole of an ideal diatomic gas $(\gamma = 7/5)$. Here, $\gamma$ is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of $66$ Joule when heated at constant pressure. The change in its internal energy is ___ Joule.
Reveal answer + step-by-step solution
Correct answer:121
Solution
For mixture: $C_V^{\text{mix}} = (n_1 C_{V_1} + n_2 C_{V_2})/(n_1 + n_2) = (2\cdot \tfrac{3}{2}R + 1\cdot\tfrac{5}{2}R)/3 = \tfrac{11}{6}R$. $C_P^{\text{mix}} = C_V^{\text{mix}} + R = \tfrac{17}{6}R$. So $\gamma^{\text{mix}} = 17/11$. At constant pressure: $W = nR\Delta T$ and $\Delta U = nC_V \Delta T$, hence $\Delta U/W = C_V/R = 11/6$. Therefore $\Delta U = (11/6)\cdot 66 = 121$ J.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →