One mole of a monatomic ideal gas undergoes the cyclic process J→K→L→M→J, as shown in the P-T diagram.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [ 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)
RT0−4RT0ln2
(2)
0
(3)
3RT0
(4)
−2RT0ln2
(5)
−3RT0ln2
(A)P→1;Q→3;R→5;S→4
(B)P→4;Q→3;R→5;S→2✓ Correct
(C)P→4;Q→1;R→2;S→2
(D)P→2;Q→5;R→3;S→4
Answer: (B)
Solution
From the P-T diagram: J(P0,V0,T0), K(P0,3V0,3T0), L(2P0,23V0,3T0), M(2P0,2V0,T0). JK is isobaric: ΔU=23nR(2T0)=3RT0 (Q→3). KL is isothermal: Q=W=nR(3T0)ln(21)=−3RT0ln2 (R→5). MJ is isothermal: ΔU=0 (S→2). Net work for the cycle: Wnet=−2RT0ln2 (P→4).
JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option