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delta S(system) vs delta S(surroundings)

Posted: Sun Feb 11, 2018 10:16 pm
by Megan Purl 1E
Under what conditions would delta S(sys) = -delta S(surr)?

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

Posted: Sun Feb 11, 2018 10:26 pm
by Niyanta Joshi 1F
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.

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

Posted: Mon Feb 12, 2018 8:33 pm
by Veritas Kim 2L
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.

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

Posted: Mon Feb 12, 2018 8:34 pm
by AtreyiMitra2L
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.

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

Posted: Tue Feb 13, 2018 1:36 am
by Rohan Chaudhari- 1K
Delta S of the universe must equal 0, which happens in reversible reactions at equilibriums.