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Explaining (ir)reversibility with S.M.

  1. May 16, 2010 #1

    Everyone knows: imagine the isolated system consisting of a reservoir and a box with a gas and one of its sides is a piston. If we (inside the system) compress the piston quasi-statically, the entropy increase of the isolated system is zero (or negligible if we do it slowly enough). If we on the other hand charge and bump into the piston and compress it all at once, the entropy change is considerable.

    Now I was wondering how I could imagine this quantity S in this situation and more specifically: try to explain it (intuitively).

    So I was thinking of doing it with classical thermodynamics with the use of Clausius' relation, but then your argument depends on you classifying it as a reversible or irreversible, which is kind of what you are trying to explain.

    If we look at Statistical Mechanics, I see no connection at all with (ir)reversibility. If we denote the entropy of the isolated system as S, then S = k log(omega). Now is there anything in the theory of S.M. that predicts (and maybe even explains...) that in the first scenario, omega is constant, and in the latter, omega grows?
  2. jcsd
  3. May 21, 2010 #2
    Irreversibility is a statistical statement of probabilities being sufficiently high. Where is the problem?
  4. May 23, 2010 #3
    It's worth thinking about what the actual question is.
    Both thermodynamics and statmech say that there are reversible and irreversible processes. You need to know the exact process to judge which one it is. So what is the question now?

    Maybe a better question is: Whatever you see as a quasi-static process (slow?), how does physics justify that it is reversible?
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