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Entropy of Reversible Process

  1. Oct 22, 2015 #1
    Even though entropy is a state function and thus independent of path why is it that the definition of entropy is restricted to a reversible process?

    Eq 1: dS=dqrev/T

    First of all, is this for an open system, closed system or isolated system? The Clausius inequality deals with the exchange between the surroundings and the system so the total entropy change is always greater than zero for a irreversible process and equal to zero for a reversible one.

    Eq 2: dS≥dqsystem/Tsurr.

    I thought that this is because a reversible process progresses so slowly that it is always in equilibrium and that the change in entropy at equilibrium is zero. But the definition above (eq. 1), although q is reversible, is not necessarily zero.

    One explanation I have gotten was that is that a reversible process minimizes heat and maximizes work. I understand this concept with engine performance and that reversible processes maximize work, but how this pertains to the definition of entropy I don't understand.
  2. jcsd
  3. Oct 23, 2015 #2
    The answers to all these questions are covered in the following short Physics Forums Insights article: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

    I wrote this article specifically for people like yourself who are struggling with these concepts (because they are presented to poorly in the available textbooks and online literature). Please don't hesitate to ask any followup questions that you may have.

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