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Q.
For an isolated system, $ \Delta U = 0 $ , then
AMUAMU 2012Thermodynamics
Solution:
For an isolated system, $\Delta U = 0$ and for a spontaneous process, total entropy change must be positive.
For example, consider the diffusion of two gases $A$ and $B$ into each other in a closed container which is isolated from the surroundings.
The two gases $A$ and $B$ are separated by a movable partition. When partition is removed, the gases begin to diffuse into each other and the system becomes more disordered. It shows that $\Delta S > 0$ and $\Delta U =0$ for this process.
Morever, $\Delta S = q _{ rev } / T =\Delta H / T$
$ =\Delta U + p \Delta V / T = p \Delta V / T $
i.e. $T \Delta S$ or $\Delta S > 0$