JEE Main 2025 · 22 January · morning · Single correct · +4 / −1
A liquid when kept inside a thermally insulated closed vessel at 25° C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?
- (a) Δ U=0, q<0, w>0
- (b) Δ U>0, q=0, w>0
- (c) Δ U=0, q=0, w=0
- (d) Δ U<0, q=0, w>0
Answer and solution
Answer (b) Δ U>0, q=0, w>0
Solution
Chemistry ma’am · 4 steps
Insulated means no heat crosses, so the stirring work done on the liquid all goes into its internal energy.
- Heat term. The vessel is thermally insulated, so no heat can enter or leave the liquid. That makes the process adiabatic and q = 0.
- Work term. The stirrer is turned from outside, so the surroundings do work on the liquid. In the IUPAC convention, work done on the system is positive, so w > 0.
Δ U = q + w = 0 + w = w > 0
- Internal energy. All the stirring work stays inside the liquid, so its internal energy rises. You would see this as a small rise in temperature above 25°C.
- So. Δ U > 0, q = 0 and w > 0, which is option (b).
The idea to keep. Read the walls first: insulated means q = 0. Then ask who pushes whom. Work done on the system is positive, so in an adiabatic process Δ U = w.
Answer from the NTA final answer key. Solution by PrepWiser.