JEE Advanced 2024 — Paper 1 — PHY-IV solution

JEE Advanced2024Paper 1PHY-IV

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.

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	List-II
(P) Work done in the complete cyclic process	(1) $R T_{0} - 4 R T_{0} ln 2$
(Q) Change in the internal energy of the gas in the process $J K$	(2) $0$
(R) Heat given to the gas in the process $K L$	(3) $3 R T_{0}$
(S) Change in the internal energy of the gas in the process $M J$	(4) $- 2 R T_{0} ln 2$
	(5) $- 3 R T_{0} ln 2$

A
P $\to$ 1; Q $\to$ 3; R $\to$ 5; S $\to$ 4
B
P $\to$ 4; Q $\to$ 3; R $\to$ 5; S $\to$ 2
C
P $\to$ 4; Q $\to$ 1; R $\to$ 2; S $\to$ 2
D
P $\to$ 2; Q $\to$ 5; R $\to$ 3; S $\to$ 4

Solution

Identify each leg from the P-T diagram. Note that constant-P legs on a P-T plot are horizontal lines, and a constant-T leg is vertical. With $n = 1$ mole, $C_{V} = \frac{3 R}{2}$ .

J $\to$ K (isobaric at $P_{0}$ , $T_{0} \to 3 T_{0}$ ): $W_{J K} = n R Δ T = 2 R T_{0}$ , $Δ U_{J K} = n C_{V} Δ T = \frac{3 R}{2} \cdot 2 T_{0} = 3 R T_{0}$ .

K $\to$ L ( $3 T_{0}$ to $3 T_{0}$ in P-T means isothermal): In a P-T diagram, the segment KL goes from $(3 T_{0}, P_{0})$ to $(3 T_{0}, 2 P_{0})$ . Temperature constant means this is an isothermal process where pressure doubles, hence volume halves. For an isothermal: $W_{K L} = n R T ln (V_{f} / V_{i}) = R (3 T_{0}) ln (1/2) = - 3 R T_{0} ln 2$ , $Δ U_{K L} = 0$ , so $Q_{K L} = W_{K L} = - 3 R T_{0} ln 2$ .

L $\to$ M (isobaric at $2 P_{0}$ , $3 T_{0} \to T_{0}$ ): $W_{L M} = n R Δ T = - 2 R T_{0}$ , $Δ U_{L M} = \frac{3 R}{2} (- 2 T_{0}) = - 3 R T_{0}$ .

M $\to$ J (isothermal at $T_{0}$ , $2 P_{0} \to P_{0}$ , volume doubles): $W_{M J} = n R T_{0} ln 2 = R T_{0} ln 2$ , $Δ U_{M J} = 0$ .

Net cyclic work: $W_{net} = 2 R T_{0} - 3 R T_{0} ln 2 - 2 R T_{0} + R T_{0} ln 2 = - 2 R T_{0} ln 2$ . So P $\to$ 4.

$Δ U_{J K} = 3 R T_{0} \Rightarrow$ Q $\to$ 3.

$Q_{K L} = - 3 R T_{0} ln 2 \Rightarrow$ R $\to$ 5.

$Δ U_{M J} = 0 \Rightarrow$ S $\to$ 2.

$∴$ Option (B) is correct.

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