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Why is entropy zero in an adiabatic process?

  1. May 21, 2014 #1
    No heat exchange is facilitated during an adiabatic process. Change is heat is zero.
    How does this relates to the entropy being zero?
    ∫dQ/T?
    But this could really just mean that the integral is of any constant.
     
  2. jcsd
  3. May 21, 2014 #2
    Heat exchange and change in entropy are related.. dS=dQ/T. If the process is adiabatic then dQ=0 and change in entropy dS=0/T=0 or ΔS=∫dQ/T=∫0/T=0. So entropy is constant (S=constant) in reversible adiabatic process. It doesn't mean that the entropy of the system is 0, it means that the change in the entropy of the system is 0. ie. entropy of the system is same before and after the process.
     
    Last edited: May 21, 2014
  4. May 21, 2014 #3

    D H

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    The correct expression is ##dS \ge \frac {dQ}T##, not ##dS = \frac {dQ}T##. The latter holds if and only equal if the process is reversible. Free expansion is the canonical example of a non-reversible adiabatic process. Entropy increases in free expansion even though there is no heat transfer to/from the system and no work done on/by the system.
     
    Last edited: May 21, 2014
  5. May 21, 2014 #4
    Hi DH. I think you meant "even though there is no heat transfer..."

    Chet
     
  6. May 21, 2014 #5
    Thanks! The first post cleared things up. The second refined the understanding.
     
  7. May 21, 2014 #6

    ZapperZ

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    Please note that it is the CHANGE in entropy, not the absolute value of the entropy, that is zero. This needs to be clarified in light of the title of the thread.

    Zz.
     
  8. May 21, 2014 #7
    You are right. But it was a legitimate mistake. I thought the entropy was zero and so asked why could it have been other arbitrary constant. Perhaps, the mod could do some editing.
     
  9. May 21, 2014 #8

    D H

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    Yes, I did. Thanks. Post edited to reflect the missing "no".
     
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