delta S(system) vs delta S(surroundings)


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Megan Purl 1E
Posts: 49
Joined: Fri Sep 29, 2017 7:07 am

delta S(system) vs delta S(surroundings)

Postby Megan Purl 1E » Sun Feb 11, 2018 10:16 pm

Under what conditions would delta S(sys) = -delta S(surr)?

Niyanta Joshi 1F
Posts: 30
Joined: Thu Jul 13, 2017 3:00 am

Re: delta S(system) vs delta S(surroundings)

Postby Niyanta Joshi 1F » Sun Feb 11, 2018 10:26 pm

delta S (universe)= delta S (sys) + delta S (surr)
When the reaction is at equilibrium and it is a reversible reaction, the total delta S of the universe is zero.
Hence, delta S (sys) = - delta S (surr) for these conditions.

Veritas Kim 2L
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

Re: delta S(system) vs delta S(surroundings)

Postby Veritas Kim 2L » Mon Feb 12, 2018 8:33 pm

You can always assume that delta S(system)= -delta S(surrounding) for a reversible expansion at equilibrium because the heat that flows to the surrounding is being released by the system.

AtreyiMitra2L
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Re: delta S(system) vs delta S(surroundings)

Postby AtreyiMitra2L » Mon Feb 12, 2018 8:34 pm

This can only occur in isothermal, reversible reactions. It would not be considered a violation of the second law of thermodynamics (law that says the entropy is always increasing) because these never actually occur in nature.

Rohan Chaudhari- 1K
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Joined: Fri Sep 29, 2017 7:06 am

Re: delta S(system) vs delta S(surroundings)

Postby Rohan Chaudhari- 1K » Tue Feb 13, 2018 1:36 am

Delta S of the universe must equal 0, which happens in reversible reactions at equilibriums.


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