Entropy change in a reversible isothermal process

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In reversible processes, the change in entropy (∆S) for the system is zero, indicating that the system can return to its original state without any net change. However, for a reversible isothermal process, the change in entropy is quantified by the equation ∆S(system) = nRln(Vf/Vi), where Vf and Vi represent the final and initial volumes, respectively. This highlights that while the system undergoes a change, the total entropy change (∆S(total)) for both the system and surroundings remains zero, as expressed by the equation ∆S(total) = ∆S(surroundings) + ∆S(system). This balance is crucial in understanding thermodynamic processes and the nature of reversibility.
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Why does ∆S = 0 for a reversible process, but for a reversible isothermal process, ∆S is given by nRln(Vf/Vi) (or other variations of that equation)?
 
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You are confusing ΔS for the system with ΔS for the combination of system and surroundings.
 
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Chestermiller is correct.∆S(system) = nRln(Vf/Vi)

∆S(total)= ∆S(surroundings) + ∆S(system) = 0
 

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