Change in entropy of an irreversible process.

[gabbymaraziti]

Why is deltaS of an irreversible process not equal to zero?

For example, 4I.9 states that for the irreversible process, deltaS is the same as deltaS for the reversible process. This does not make sense to me, because the process in this example is also free expansion. Therefore, w= 0, deltaU= 0, and q= 0. Therefore, shouldn't deltaS = 0 because deltaS = q/T?

[Jack Riley 4f]

delta S of the surroundings does equal 0 because it is free expansion, but delta S total must still increase so that only leaves delta S of the system to contribute to this. Thus, delta S total = delta S system =/= 0

[Caroline Beecher 2H]

When calculating delta S(total) of an irreversible system, it will equal delta S(system). Since no work is done in free expansion, w = 0. And because delta U = 0, and delta U = w + w, q also = 0. With this, we know that no heat is transferred into the surroundings, so delta S(surroundings) = 0.

[gabbymaraziti]

delta S of the surroundings does equal 0 because it is free expansion, but delta S total must still increase so that only leaves delta S of the system to contribute to this. Thus, delta S total = delta S system =/= 0

Why does deltaS still have to increase?