JEE Main 2023 · 25 January · 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 Process | I. | Work done by the gas decreases internal energy |
| B. | Adiabatic Process | II. | No change in internal energy |
| C. | Isochoric Process | III. | The heat absorbed goes partly to increase internal energy and partly to do work |
| D. | Isobaric Process | IV. | No work is done on or by the gas |
- (a) A-I, B-II, C-IV, D-III
- (b) A-II, B-I, C-III, D-IV
- (c) A-II, B-I, C-IV, D-III
- (d) A-I, B-II, C-III, D-IV
Answer and solution
Answer (c) A-II, B-I, C-IV, D-III
Solution
Physics sir · 3 steps
Write Δ Q = Δ U + W, then ask which term each process forces to zero.
- First law. For any process on a gas, Δ Q = Δ U + W, where W is the work done by the gas. Each special process fixes one of these three terms, and that tells you its List II match.
Process What is fixed First law becomes Match A. Isothermal T constant, so Δ U = 0 for an ideal gas Δ Q = W II, no change in internal energy B. Adiabatic No heat exchange, Δ Q = 0 W = -Δ U I, work by the gas lowers U C. Isochoric V constant, so W = 0 Δ Q = Δ U IV, no work on or by the gas D. Isobaric P constant, nothing is zero Δ Q = Δ U + PΔ V III, heat splits between U and work - Why isobaric is III. At constant pressure the gas heats up, so Δ U > 0, and it expands, so W = PΔ V > 0. The heat absorbed pays for both.
- So. The matching is A-II, B-I, C-IV, D-III, which is option (c).
The idea to keep. Isothermal kills Δ U, adiabatic kills Δ Q, isochoric kills W. Isobaric kills nothing, so the heat is shared between Δ U and work.
Answer from the NTA final answer key. Solution by PrepWiser.