For an isolated system, ∆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 ∆S > 0 and ∆U = 0 for this process.

Morever, ∆S = ${ q }_{ rev }$ /T = ∆H / T

= ∆U + p∆V /T = p∆V /T

i.e. T ∆S or ∆S > 0

# For an isolated system, ∆U = 0, what will be ∆5?

**prasanna**#1