JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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',

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

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.

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Solution

Using the first law of thermodynamics, $\Delta U = q + w$ .

For the process from state X to state Y:

Heat absorbed by the gas, $q_{X \to Y} = +10$ J

Work done by the gas, $w_{X \to Y} = -18$ J

Change in internal energy, $\Delta U_{X \to Y} = q_{X \to Y} + w_{X \to Y} = 10 - 18 = -8$ J

For the reverse process from state Y to state X:

Change in internal energy, $\Delta U_{Y \to X} = -\Delta U_{X \to Y} = +8$ J

Heat evolved by the gas, $q_{Y \to X} = -6$ J

Let the work done on the gas be $w_{Y \to X}$ .

$\Delta U_{Y \to X} = q_{Y \to X} + w_{Y \to X}$

$8 = -6 + w_{Y \to X}$

$w_{Y \to X} = +14$ J

Since the work done is positive, $14$ J of work is done on the gas by the surrounding.

Answer: $14$ J of the work is done on the gas 'A' by the surrounding.

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Solution

Using the first law of thermodynamics, $\Delta U = q + w$ .

For the process from state X to state Y:

Heat absorbed by the gas, $q_{X \to Y} = +10$ J

Work done by the gas, $w_{X \to Y} = -18$ J

Change in internal energy, $\Delta U_{X \to Y} = q_{X \to Y} + w_{X \to Y} = 10 - 18 = -8$ J

For the reverse process from state Y to state X:

Change in internal energy, $\Delta U_{Y \to X} = -\Delta U_{X \to Y} = +8$ J

Heat evolved by the gas, $q_{Y \to X} = -6$ J

Let the work done on the gas be $w_{Y \to X}$ .

$\Delta U_{Y \to X} = q_{Y \to X} + w_{Y \to X}$

$8 = -6 + w_{Y \to X}$

$w_{Y \to X} = +14$ J

Since the work done is positive, $14$ J of work is done on the gas by the surrounding.

Answer: $14$ J of the work is done on the gas 'A' by the surrounding.

JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry