JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
Match the thermodynamic processes taking place in a system with the correct conditions. In the table : ΔQ is the heat supplied, ΔW is the work done and ΔU is change in internal energy of the system.
| Process | Condition |
|---|---|
| (I) Adiabatic | (1) ΔW = 0 |
| (II) Isothermal | (2) ΔQ = 0 |
| (III) Isochoric | (3) ΔU ≠ 0, ΔW ≠ 0, ΔQ ≠ 0 |
| (IV) Isobaric | (4) ΔU = 0 |
- (a) (I) - (1), (II) - (1), (III) - (2), (IV) - (3)
- (b) (I) - (2), (II) - (4), (III) - (1), (IV) - (3)
- (c) (I) - (1), (II) - (2), (III) - (4), (IV) - (4)
- (d) (I) - (2), (II) - (1), (III) - (4), (IV) - (3)
Answer and solution
Answer (b) (I) - (2), (II) - (4), (III) - (1), (IV) - (3)
Solution
Physics sir · 3 steps
Ask one question per process: which quantity does this process force to zero?
- First law. Every row is a version of Δ Q = Δ U + Δ W. Each named process fixes one quantity, and that fixes which term vanishes.
Process What is held fixed What becomes zero Condition (I) Adiabatic no heat exchange Δ Q = 0 (2) (II) Isothermal temperature Δ U = 0, since U of an ideal gas depends only on T (4) (III) Isochoric volume Δ W = P Δ V = 0 (1) (IV) Isobaric pressure nothing: V and T both change, so Δ W ≠ 0, Δ U ≠ 0, Δ Q ≠ 0 (3) - Isobaric check. At constant pressure, heating the gas makes it expand (Δ W = P Δ V ≠ 0) and get hotter (Δ U ≠ 0). So Δ Q = Δ U + Δ W is also non-zero. It is the only process where all three survive.
- So. The match is (I) - (2), (II) - (4), (III) - (1), (IV) - (3), which is option (b).
The idea to keep. Adiabatic means no heat, isothermal means no change in U for an ideal gas, and isochoric means no work. Isobaric is the one process where none of the three is zero.
Answer from the NTA final answer key. Solution by PrepWiser.