JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

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.

Solution

For the forward process X $\to$ Y: $q = + 10$ J and work done by the gas $= 18$ J, so $w_{on gas} = - 18$ J.

$Δ U_{X \to Y} = q + w = 10 - 18 = - 8$ J.

For the reverse process Y $\to$ X, $Δ U_{Y \to X} = + 8$ J (state function reverses sign).

Heat is evolved: $q = - 6$ J. Applying the first law: $8 = - 6 + w$ , so $w = + 14$ J.

Positive $w$ indicates work done on the gas by the surroundings $= 14$ J.

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