## Change in entropy of an irreversible process.

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

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gabbymaraziti
Posts: 111
Joined: Wed Sep 18, 2019 12:19 am

### Change in entropy of an irreversible process.

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
Posts: 100
Joined: Sat Aug 24, 2019 12:17 am

### Re: Change in entropy of an irreversible process.

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
Posts: 51
Joined: Wed Nov 14, 2018 12:21 am

### Re: Change in entropy of an irreversible process.

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
Posts: 111
Joined: Wed Sep 18, 2019 12:19 am

### Re: Change in entropy of an irreversible process.

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?

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