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Entropy change in a reversible adiabatic process

  1. Sep 28, 2012 #1
    For a reversible process, I imagine it is correct to say that

    [itex] dS = \frac{dq}{T} [/itex] where all quantities refer to system quantities (not the surrounding).

    However, for an adiabatic process, [itex] dq = 0 [/itex].

    Thus, should it be the case that for an adiabatic reversible process,

    [itex] dS = \frac{dq}{T} = \frac{0}{T} = 0 [/itex]

    Please correct me if I am wrong, and point out the flaws in my reasoning.

    BiP
     
  2. jcsd
  3. Sep 28, 2012 #2

    DrDu

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    Science Advisor

    Entirely correct.
     
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