Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 22 January, Morning Shift — Question 33

A liquid when kept inside a thermally insulated closed vessel at $25^{\circ} \mathrm{C}$ was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters?

Option A:
$\Delta U>0, q=0, w>0$
Correct
Option B:
$\Delta U=0, q=0, w=0$
Option C:
$\Delta \mathrm{U}<0, \mathrm{q}=0, \mathrm{w}>0$
Option D:
$\Delta U=0, q<0, w>0$

Answer: A

Step-by-step solution

Thermally insulated $\Rightarrow q=0$

from I ${ }^{\text {st }}$ law

$\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$

$\Delta \mathrm{U}=\mathrm{w}$

$\mathrm{w}>0, \Delta \mathrm{U}>0$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Internal Energy and the First Law