## delta S

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

katherinemurk 2B
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Joined: Wed Nov 15, 2017 3:02 am

### delta S

under what conditions does delta S = 0

Vy Lu 2B
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Joined: Fri Sep 28, 2018 12:24 am

### Re: delta S

Delta S (surroundings) is zero under free expansion; delta S (universe) is zero under reversible conditions and equilibrium; delta S (system) should be zero when there is no transfer of heat from and to the system and when the temperature and volume are constant in the system.

Sarah Fatkin 4I
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Joined: Fri Sep 28, 2018 12:27 am

### Re: delta S

Delta S of the system equals zero when there is no heat entering the system, because detla S = q(reversible)/T.

JiangJC Dis2K
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Joined: Fri Oct 12, 2018 12:16 am

### Re: delta S

For a system at equillibrium, the total delta s equals zero. For n irreversible expansion, given that there is no effect on the surroundings, delta s system equals delta s total.

Ian Marquez 2K
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### Re: delta S

Like the replies above, delta S of the universe, or when negative delta S of the system is equal to delta S of the surroundings, is 0 when the reaction is at equilibrium and is reversible. Also, delta U would be 0.

Ashe Chen 2C
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Joined: Mon Jan 07, 2019 8:23 am

### Re: delta S

$\Delta S = 0$ when q=0, or in free expansion.

Niveda_B_3I
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### Re: delta S

The change in entropy would be zero when the heat transferred into or out of the system is 0.

Jonny Schindler 1A
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Joined: Fri Sep 28, 2018 12:20 am

### Re: delta S

When volume is constant, changes in pressure/temperature result in a change in entropy