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

Eileen Si 1G
Posts: 120
Joined: Fri Aug 30, 2019 12:17 am

What is the delta entropy of the surroundings and system during an irreversible expansion, and is it any different to an adiabatic system where q = 0?

Anthony Hatashita 4H
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am

### Re: Irreversible, Adiabatic

For the surroundings it's 0, the system is W=-PDeltaV I believe. I'm not sure about the second part of your question though, sorry about that.

Lauren Tanaka 1A
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am

### Re: Irreversible, Adiabatic

The delta S for an irreversible expansion means that the surroundings change in entropy is 0. In an adiabatic system the total heat would be 0. This doesn't necessarily mean the same for irreversible expansion.

Sydney Myers 4I
Posts: 100
Joined: Fri Aug 09, 2019 12:17 am

### Re: Irreversible, Adiabatic

When the delta S is 0, it doesn't necessarily mean that there's no heat transfer, just that whatever heat is being transferred is being converted to work or being transferred elsewhere. In an adiabatic system, there is no heat transferred.

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