Verified Solution • 2025–26 Session moderate

QJEEM25Ch4b23: JEE Main 2025 Chemistry - Class 11 JEEM25Ch4b23 Solution: A liquid when kept inside a thermally insulated closed vessel…

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 ?

Multiple Choice Options

  • A ΔU=0,q<0,w>0\Delta \mathrm{U}=0, \mathrm{q}<0, \mathrm{w}>0
  • B ΔU>0,q=0,w>0\Delta \mathrm{U}>0, \mathrm{q}=0, \mathrm{w}>0
  • C ΔU=0,q=0,w=0\Delta \mathrm{U}=0, \mathrm{q}=0, \mathrm{w}=0
  • D ΔU<0,q=0,w>0\Delta \mathrm{U}<0, \mathrm{q}=0, \mathrm{w}>0

Step-by-Step Verified Solution

Let’s analyze the situation step by step:

System: The liquid inside a thermally insulated (adiabatic) and closed vessel.

Process: The liquid is mechanically stirred from outside.

Heat Exchange ( qq )

Because the vessel is thermally insulated, there is no heat exchange between the system and surroundings. Hence,

q=0.q = 0.

Work ( ww )

The external mechanical stirring does work on the system by agitating the liquid. Work done on the system is conventionally taken as positive:

w &gt; 0.

Change in Internal Energy ( ΔU\Delta U )

From the First Law of Thermodynamics,

ΔU  =  q  +  w.\Delta U \;=\; q \;+\; w.

Since q=0q = 0 and w &gt; 0 , we get

\Delta U = w &gt; 0.

Hence:

\boxed{ \Delta U &gt; 0,\quad q = 0,\quad w &gt; 0. }

Answer: Option B is correct.

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