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Reversible and entropy (thermo)

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    In my book it says, "only if the transformation is reversible can we use [itex]dS=\frac{\delta Q}{T}[/itex]", but I don't know why? I think even if it's not reversible we also use [itex]\delta Q=TdS[/itex] to solve the thermo problems, don't we?

    I'm looking forward to your ideas, thanks.
     
  2. jcsd
  3. Sep 6, 2011 #2

    rude man

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    Entropy is a state variable. That means that, no matter how we get from state A to state B, the change in S is the same. Doesn't matter if the change is reversible or not.

    The point is that in order to CALCULATE the change in S, you must integrate along a reversible path.

    In fact, the integral TdS from state A to B is always less for an irreversible change than for a reversible one. E. Fermi proves this in his Dover book "Thermodynamics' but I never followed the proof. Since the states are undefined for an irreversible change I don't lose too much sleep over it. But Fermi actually invokes that fact later in the book on a different topic.

    If you want to get totally snowed, that is the book for you!
     
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