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Thermodynamics, Entropy, Clausis Inequality, Reversible and Irreversib

  1. Nov 6, 2013 #1
    ∫dQ/T≤∫dQ(rev)/T * , where both integrals are evaluated between the same thermodynamic coordinates- A and B , say.

    - I am having trouble interpreting this inequality.

    -( I understand the derivation in my textbook via the Clausius diagram(considering a reversible and an ireversible process between the two points A and B, which form a closed loop, so together they from the closed loop in clausius inequality and * follows.)

    - I thought the change in entropy has to be greater for an irreversible process. So I also know that, we derive and introduce the state variable entropy by looking at when equality holds for the Clausius equality, and consequently we can not evaluate the entropy change of an irreversible process directly from dQ/T.

    - So I now interpret that the extra term required for the entropy of an irreversible process (let this be k) due to irreversibilities is large enough , such that, despite *, the entropy change in a irreversible process is greater than that of a reversible process. (as for a reversible process I have ds(rev)=dQ(rev)/T and for a irreversible process ds(irrev)=dQ/T+k)

    -My last question, I thought that entropy was a state function s.t dQ/T is the same regardless of the path ...oh wait, this is because entropy has only been defined as equivalent to dQ/T for a reversible process?

    So does this mean it is correct to say: entropy is path dependent for irreversible processes but is path independent for irreversible processes? One more question along these lines, ΔS system=0 seems to be used quite commonly in some text book examples in order to deduce ΔS surroundings, but surely ΔS system =0 is only correct if all processes which the working fluid undergoes are reversible - this doesn't hold for irreversible processes does it?

    Thanks alot for anyone who sheds some light on this !
    Last edited: Nov 6, 2013
  2. jcsd
  3. Nov 6, 2013 #2
    You've asked a lot of good questions here. The way I like to think of it is this: Consider two equilibrium states A and B for a closed system. There are an infinite number of process paths that you can take between these two equilibrium states, each of which features different histories for the heat added Q(t) and the work done W(t), where t is the elapsed time during the process. If you evaluate the integral of dQ/T over all these possible paths, you will find that these integrals will exhibit an upper limiting value. Since the integral has a maximum value over all possible paths between states A and B, the value must be a function only of the two end states A and B. We call this limit the change in entropy between A and B. It also turns out that the integral takes on its maximum value only for paths between A and B that are reversible paths. We use this fact to calculate the change in entropy between the end states. It isn't that "entropy has only been defined as equivalent to (the integral) of dQ/T for a reversible process." It's more like the entropy is the upper bound to the integral, and that upper bound is found to be obtained only for reversible paths. For all other paths, the integral is less.

  4. Nov 6, 2013 #3

    Andrew Mason

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    No. Entropy is a state function. A change in entropy of a system in moving from one state to another does not depend on the path between those two states. The change in entropy is defined as the integral of dQ/T along a reversible path between those two states. All reversible paths are equivalent. During a reversible process, the system is either in equilibrium with the surroundings with which it is in thermal contact or it is insulated from (ie. no thermal contact with) its surroundings. Irreversible paths between those two states of the system will result in a different end state for the surroundings.

    If the system returns to its original state (ie. a full cycle of an engine) there is no change in entropy of the system. We know this because entropy is a state function. If the cycle was reversible, there will be no entropy change in the surroundings. If it was irreversible, there will be a net increase in the entropy of the surroundings.

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