JEE Advanced 2024 Paper 1 · Q14 · Kinetic Theory of Gases
One mole of a monatomic ideal gas undergoes the cyclic process $J\to K\to L\to M\to J$, as shown in the P-T diagram (vertices: $J = (T_0, P_0)$, $K = (3T_0, P_0)$, $L = (3T_0, 2P_0)$, $M = (T_0, 2P_0)$).
Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [$\mathcal{R}$ is the gas constant.]
List-I: (P) Work done in the complete cyclic process (Q) Change in the internal energy of the gas in the process JK (R) Heat given to the gas in the process KL (S) Change in the internal energy of the gas in the process MJ
List-II: (1) $\mathcal{R}T_0 - 4\mathcal{R}T_0\ln 2$ (2) $0$ (3) $3\mathcal{R}T_0$ (4) $-2\mathcal{R}T_0\ln 2$ (5) $-3\mathcal{R}T_0\ln 2$
[Figure: rectangular cycle on P-T axes with corners at $(T_0, P_0)$, $(3T_0, P_0)$, $(3T_0, 2P_0)$, $(T_0, 2P_0)$.]
Reveal answer + step-by-step solution
Correct answer:B
Solution
Identify each leg using $V = nRT/P$ ($n=1$): $J(T_0,P_0)$, $K(3T_0,P_0)$, $L(3T_0,2P_0)$, $M(T_0,2P_0)$.
JK isobaric ($P=P_0$, $T:T_0\to 3T_0$): $W_{JK} = R\Delta T = 2RT_0$. $\Delta U_{JK} = \tfrac{3}{2}R\Delta T = 3RT_0$. → Q $\to$ (3).
KL isothermal ($T=3T_0$, $P:P_0\to 2P_0$): $W_{KL} = RT\ln(V_L/V_K) = R(3T_0)\ln(P_0/2P_0) = -3RT_0\ln 2$. $\Delta U_{KL} = 0 \Rightarrow Q_{KL} = W_{KL} = -3RT_0\ln 2$. → R $\to$ (5).
LM isobaric ($P=2P_0$, $T:3T_0\to T_0$): $W_{LM} = -2RT_0$.
MJ isothermal ($T=T_0$, $P:2P_0\to P_0$): $W_{MJ} = RT_0\ln(2P_0/P_0) = RT_0\ln 2$. $\Delta U_{MJ} = 0$. → S $\to$ (2).
Total $W_{cycle} = 2RT_0 - 3RT_0\ln 2 - 2RT_0 + RT_0\ln 2 = -2RT_0\ln 2$. → P $\to$ (4). Mapping P→4, Q→3, R→5, S→2 = option (B).
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 →