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Change in entropy of an irreversible process.

Posted: Tue Feb 11, 2020 12:42 pm
by 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?

Re: Change in entropy of an irreversible process.

Posted: Tue Feb 11, 2020 1:08 pm
by 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

Re: Change in entropy of an irreversible process.

Posted: Tue Feb 11, 2020 8:33 pm
by 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.

Re: Change in entropy of an irreversible process.

Posted: Tue Feb 11, 2020 10:43 pm
by gabbymaraziti
Jack Riley 4f wrote: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?