JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Chemical Thermodynamics

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Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process during which 6 J of heat is evolved. In the reverse process of 'Y' to 'X',

(A)18 J of the work is done by the gas 'A'.
(B)2 J of the work is done by the gas 'A'.
(C)12 J of the work is done on the gas 'A' by the surrounding.
(D)14 J of the work is done on the gas 'A' by the surrounding.✓ Correct

Answer: (D)

Solution

For the forward change

q = + 10

J and the gas does

18

J of work, so

w = - 18

J and

Δ U_{X \to Y} = q + w = 10 - 18 = - 8

J. Internal energy is a state function, so the reverse change has

Δ U_{Y \to X} = + 8

J. With

6

J of heat evolved,

q = - 6

J, and the first law gives

8 = - 6 + w

, so

w = + 14

J. The positive sign means

14

J of work is done on the gas by the surroundings.

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Chemical Thermodynamics

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