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 |
Choose the correct answer from the options given below:
- A.A-II, B-I, C-IV, D-III
- B.A-I, B-II, C-IV, D-III
- C.A-II, B-I, C-III, D-IV
- D.A-I, B-II, C-III, D-IV
Show answer and solution
Correct option: A
Solution
In an isothermal process, temperature is constant, so internal energy doesn't change (ΔU = 0), matching II. In an adiabatic process, Q = 0, so by the first law ΔU = -W, meaning work done BY the gas comes entirely at the expense of internal energy, matching I. In an isochoric process, volume is constant so W = 0 — no work is done on or by the gas, matching IV. In an isobaric process, both heat absorption and work done are non-zero, so Q = ΔU + W — heat goes partly to internal energy and partly to do work, matching III. This gives A-II, B-I, C-IV, D-III.
[JEE Main 2023 · 25 Jan · Shift 2]