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Free expansion and adiabatic process

  1. Sep 7, 2010 #1
    I hope someone in here can help me out of this problem.
    I know that for calculating the change in entropy of a system ([tex]\Delta S[/tex]) we can use this formula :

    [tex]\Delta S[/tex]=[tex]\int[/tex]δq/T
    Well the problem is here, when we calculating ΔS for an adiabatic process we know Q= 0 so ΔS=0 in this process, but why we just can't let Q be equal to zero in a free expansion (while we know that in free expansion W=Q=0) and get to the same result (ΔS=0) for the free expansion process?
    I should add that I know ΔS≠ 0 for free expansion process but, you know I just don't know where the problem is in my argument.
     
  2. jcsd
  3. Sep 7, 2010 #2
    The formula you wrote is only valid for a reversible process. And the free expansion is not a reversible process. To calculate the difference in entropy in this case, connect the initial and final state by any reversible path, for which generally you won't have dQ = dL = 0, and use your formula on this path.
     
  4. Sep 8, 2010 #3
    Thanks, that's really a convincing answer.
     
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