JEE Chemistry 2025 Question with Solution

Given: A liquid is kept in a thermally insulated closed vessel and is mechanically stirred from outside.

Find: The correct signs of $\Delta U$, $q$, and $w$.

Since the vessel is thermally insulated, there is no heat exchange with the surroundings.

$$q = 0$$

Mechanical stirring means external work is done on the system, so work is positive.

$$w > 0$$

Using the first law of thermodynamics,

$$\Delta U = q + w$$

Substituting the known values,

$$\Delta U = 0 + w > 0$$

Therefore, the internal energy increases.

Hence,

$$\Delta U > 0, \; q = 0, \; w > 0$$

Therefore, the correct option is A.

Given: The vessel is thermally insulated and closed, and the liquid is stirred mechanically from outside.

Find: Which option correctly represents $\Delta U$, $q$, and $w$.

1. System description: A thermally insulated vessel does not allow heat exchange, so

$$q = 0$$

2. Work done: Mechanical stirring transfers energy to the liquid as work done on the system, therefore

$$w > 0$$

3. Change in internal energy: By the first law of thermodynamics,

$$\Delta U = q + w$$

Putting $q=0$,

$$\Delta U = w > 0$$

Thus the internal energy of the liquid increases.

So the correct thermodynamic parameters are $\Delta U>0, q=0, w>0$, which corresponds to option A.