Delta S for Reversible Processes

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

Bailey_Osborne_3A
Posts: 11
Joined: Wed Sep 21, 2016 2:55 pm

Delta S for Reversible Processes

I was wondering if someone could explain why delta S universe/total equals 0 for all reversible processes? Thanks!

Chem_Mod
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Re: Delta S for Reversible Processes

Because the process is carried out slowly so that at each moment, the system is in equilibrium with the surrounding. So qsys=-qsurr

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