Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

  2. jcsd
  3. Sep 28, 2012 #2


    User Avatar
    Science Advisor

    Entirely correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Entropy change in a reversible adiabatic process