JEE Main 2025 · 4 April · evening · Single correct · +4 / −1
Match List - I with List - II.
Δ Q= Heat supplied
Δ W= Work done by the system
Δ U= Change in internal energy
P= Pressure of the system
Δ V= Change in volume of the system
Choose the correct answer from the options given below :
| List - I | List - II | ||
|---|---|---|---|
| (A) | Isobaric | (I) | Δ Q=Δ W |
| (B) | Isochoric | (II) | Δ Q=Δ U |
| (C) | Adiabatic | (III) | Δ Q= zero |
| (D) | Isothermal | (IV) | Δ Q=Δ U+PΔ V |
- (a) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
- (b) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
- (c) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- (d) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
Answer and solution
Answer (b) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
Solution
Physics sir · 3 steps
Start from Δ Q = Δ U + Δ W and ask which term each process switches off.
- First law. For any process, Δ Q = Δ U + Δ W. Each special process makes one quantity fixed or zero, and that changes how the law looks.
Process What is fixed Effect First law becomes (A) Isobaric Pressure Δ W = PΔ V Δ Q = Δ U + PΔ V, (IV) (B) Isochoric Volume Δ V = 0, so Δ W = 0 Δ Q = Δ U, (II) (C) Adiabatic No heat exchange Δ Q = 0 Δ Q = zero, (III) (D) Isothermal Temperature Δ U = 0 for an ideal gas Δ Q = Δ W, (I) - Why U stays fixed. The internal energy of an ideal gas depends only on its temperature. Constant temperature means Δ U = 0, so all the heat supplied goes into work.
- So. The matching is (A)-(IV), (B)-(II), (C)-(III), (D)-(I), which is option (b).
The idea to keep. Isochoric kills work, isothermal kills Δ U, adiabatic kills heat. Find which term goes to zero and the first law tells you the rest.
Answer from the NTA final answer key. Solution by PrepWiser.