Verified Solution • 2025–26 Session
hard
QJEEA-2024-14: JEE Advanced 2024 JEEA-2024-14 Solution: Q.14 One mole of a monatomic ideal gas undergoes the cyclic…
Q.14 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 List-II
(P) Work done in the complete cyclic process (1) ℛ𝑇0 − 4ℛ𝑇0 ln 2
(Q) Change in the internal energy of the gas in (2) 0
the process JK
(R) Heat given to the gas in the process KL (3) 3ℛ𝑇0
(S) Change in the internal energy of the gas in the (4) −2ℛ𝑇0 ln 2
(5) −3R𝑇0 ln 2
Multiple Choice Options
- A (A) P → 1; Q → 3; R → 5; S → 4
- B (B) P → 4; Q → 3; R → 5; S → 2
- C (C) P → 4; Q → 1; R → 2; S → 2
- D (D) P → 2; Q → 5; R → 3; S → 4
Step-by-Step Verified Solution
SOLUTION: (B) P → 4; Q → 3; R → 5; S → 2
- JK (Isobaric):
- W₁ = nR(3T₀ - T₀) = 2RT₀
- ΔU = n(3R/2)(3T₀ - T₀) = 3RT₀
- Q = ΔU + W = 5RT₀
- KL (Isothermal):
- W₂ = nR(3T₀) ln(P₀/2P₀) = -3RT₀ ln 2
- ΔU = 0
- Q = W = -3RT₀ ln 2
- LM (Isobaric):
- W₃ = nR(T₀ - 3T₀) = -2RT₀
- ΔU = n(3R/2)(T₀ - 3T₀) = -3RT₀
- Q = ΔU + W = -5RT₀
- MJ (Isothermal):
- W₄ = nRT₀ ln(2P₀/P₀) = RT₀ ln 2
- ΔU = 0
- Q = W = RT₀ ln 2
Matching List-I to List-II:
- (P) Work done in the complete cyclic process (W_net):
- W_net = W₁ + W₂ + W₃ + W₄
- W_net = 2RT₀ - 3RT₀ ln 2 - 2RT₀ + RT₀ ln 2
- W_net = -2RT₀ ln 2
- (P) → (4)
- (Q) Change in the internal energy of the gas in the process JK (ΔU_JK):
- ΔU_JK = 3RT₀
- (Q) → (3)
- (R) Heat given to the gas in the process KL (Q_KL):
- Q_KL = -3RT₀ ln 2
- (R) → (5)
- (S) Change in the internal energy of the gas in the process MJ (ΔU_MJ):
- ΔU_MJ = 0
- (S) → (2)
Final Answer:
- (B) P → 4; Q → 3; R → 5; S → 2