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Homework Help: Entropy change for non-thermally-isolated system

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data
    This isn't an actual problem, it's just a question based on my reading. Reif says "If the external parameters of a thermally isolated system are changed quasi-statically by any amount, ΔS = 0."

    I don't understand why it has to be thermally isolated. Looking at the entropy relation

    [tex] dS = dq/T [/tex]

    There isn't any work in there, and the derivation didn't set work = 0. So if a system undergoes a reversible process and dQ is non-zero, it seems that work also has no effect on the entropy change?
  2. jcsd
  3. Mar 6, 2012 #2

    rude man

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    Homework Helper
    Gold Member

    We have the system and we have the surroundings. The two together form the "universe".

    If the system is thermally isolated from the surroundings, Q = 0 by definition and so is ΔS = Q/T, no matter how T may change during the process. So ΔS = 0 for the system, surroundings and universe.

    If on the other hand the system is not thermally isolated, then heat can, for example, get into the system from the surroundings. If the process is reversible though, entropy gained by the system = entropy lost by the surroundings. Entropy change of the universe = 0.

    Example: isothermal expansion of an ideal gas. We supply heat to the gas, keeping T constant, allowing it to expand from V1 to V2 and p1 to p2. ΔU = 0 so 1st law reads Q = W = ∫pdV = nRT∫dp/p from p1 to p2 = nRT*ln(p2/p1). So the gas increases entropy by Q/T. This amount obviously is lost by the surroundings. The universe entropy change is Q/T - Q/T = 0.
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