JEE Main 2025 · 7 April · evening · Single correct · +4 / −1
Match List - I with List - II .
Choose the correct answer from the options given below :
| List - I | List - II |
|---|---|
| (A) Isothermal | (I) ΔW (work done) = 0 |
| (B) Adiabatic | (II) ΔQ (supplied heat) = 0 |
| (C) Isobaric | (III) ΔU (change in internal energy) ≠ 0 |
| (D) Isochoric | (IV) ΔU = 0 |
- (a) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
- (b) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
- (c) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
- (d) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
Answer and solution
Answer (c) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
Solution
Physics sir · 2 steps
Ask what each process holds fixed, then see which quantity in Δ Q = Δ U + Δ W that kills.
- First law. Every match comes from Δ Q = Δ U + Δ W. For an ideal gas, Δ U depends only on temperature, and Δ W = P Δ V needs a volume change.
Process What stays fixed Consequence Match (A) Isothermal Temperature No temperature change, so Δ U = 0 (IV) (B) Adiabatic No heat exchange Δ Q = 0 by definition (II) (C) Isobaric Pressure V changes at fixed P, so T changes and Δ U ≠ 0 (III) (D) Isochoric Volume Δ V = 0, so Δ W = 0 (I) - So. The matching is (A)-(IV), (B)-(II), (C)-(III), (D)-(I), which is option (c).
The idea to keep. Isothermal fixes T, so Δ U = 0. Adiabatic fixes heat, so Δ Q = 0. Isochoric fixes V, so Δ W = 0. Isobaric fixes none of these three, so here Δ U ≠ 0.
Answer from the NTA final answer key. Solution by PrepWiser.