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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 22 January · Morning Shift · Chemistry · Thermodynamics (C)

A liquid when kept inside a thermally insulated closed vessel at...

A liquid when kept inside a thermally insulated closed vessel at 25∘C25^{\circ} \mathrm{C} was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?

Options

  1. AΔU<0,q=0,w>0\Delta \mathrm{U} \lt 0, \mathrm{q}=0, \mathrm{w}\gt0
  2. BΔU=0,q=0,w=0\Delta \mathrm{U}=0, \mathrm{q}=0, \mathrm{w}=0
  3. CΔU>0,q=0,w>0\Delta \mathrm{U}\gt0, \mathrm{q}=0, \mathrm{w}\gt0
  4. DΔU=0,q<0,w>0\Delta \mathrm{U}=0, \mathrm{q} \lt 0, \mathrm{w}\gt0

Correct answer

C. ΔU>0,q=0,w>0\Delta \mathrm{U}\gt0, \mathrm{q}=0, \mathrm{w}\gt0

Step-by-step solution

Thermally insulated ⇒q=0\Rightarrow q=0 from Ist \mathrm{I}^{\text{st }} law ΔU=q+wΔU=ww>0,ΔU>0\begin{aligned} & \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\ & \Delta \mathrm{U}=\mathrm{w} \\ & \mathrm{w}>0, \Delta \mathrm{U}>0 \end{aligned}
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