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Entropy for Reversible and Irreversible Processes

  1. Jan 18, 2014 #1

    Zag

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    Hello everyone,

    I've been reviewing some concepts on Thermodynamics and, even though I feel like I am gaining a level of comprehension about the subject that I could not have achieved before as an undergraduate, I am also running into some situations in which some thermodynamic concepts seem to contradict each other. Here is one of these situations involving Entropy of reversible and irreversible processes:

    From Clausius' Theorem, we know that: [itex]\oint\mathrm{\frac{dQ}{T}} \leq 0[/itex] (the equality holding for reversible processes)

    Now, following the text-book that I'm using (K. Huang - Statistical Mechanics), we can consider a transformation between two states [itex]A[/itex] and [itex]B[/itex] as being either reversible [itex](R)[/itex] or irreversible [itex](I)[/itex]:

    Untitled.jpg

    Now considering a closed path given by [itex]I + (-R)[/itex], we can rewrite Clausis' theorem in the following way:

    [itex]\int_{I}\mathrm{\frac{dQ}{T}} + \int_{-R}\mathrm{\frac{dQ}{T}} \leq 0[/itex]

    But since for a reversible path the integral yields the variation in the Entropy: [itex]\int_{R}\mathrm{\frac{dQ}{T}} = \Delta S[/itex], we get:

    [itex]\int_{I}\mathrm{\frac{dQ}{T}} \leq S(B) - S(A)[/itex]

    Now comes the tricky part that has been confusing me. If the system under consideration is isolated, not heat exchange occurs and [itex]\mathrm{dQ} = 0[/itex], from which we conclude that:

    [itex]S(B) - S(A) \geq 0[/itex]

    Again with the equality holding for a reversible process, and the inequality holding for irreversible processes. Even though I get the idea, for me this seems to somehow contradict other aspects of the Entropy (probably because I'm missing something). The point is: how come [itex]S(B) - S(A) \geq 0[/itex] for two fixed states [itex]A[/itex] and [itex]B[/itex] if the entropy is a function of state? Given two fixed states [itex]A[/itex] and [itex]B[/itex], I could in principle go from the first to the second either reversibly or irreversibly, and because the Entropy depends only on the initial and final state, [itex]S(B) - S(A)[/itex] should be the same for both cases!! How come this expression tells us it must be different? What am I missing?

    Thanks a lot guys!
     
  2. jcsd
  3. Jan 18, 2014 #2

    DrClaude

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    Staff: Mentor

    I think that Huang is not being very clear here. The point is that ##A## and ##B## are not the same in the different cases. You need to use the fact that [itex]S(B) - S(A)[/itex] is invariant for ##A## and ##B## fixed to derive the right formulas, and this is illustraed in Fig. 1.7. But when he writes
    $$
    S(B) - S(A) \ge 0
    $$
    he means that if you take a state ##A##, use a reversible process to reach a state ##B##, then ##S(B) - S(A) = 0##. If you take a state ##A'## and use an irreversible process to reach a state ##B'##, then ##S(B') - S(A') > 0##. He summarizes this in the equality above with ##A## and ##B##, but they are distinct sets of states.
     
  4. Jan 18, 2014 #3

    Zag

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    Thank you for your reply DrClaude, I really appreciate it.

    Alright, I understand your point. It would be equivalent then to say that it is not true that one can always find a reversible and an irreversible path connecting two different states for an isolated system, correct? Because if that was the case, the equation

    [itex]S(B) - S(A) \geq 0[/itex]

    would be contradicting itself for fixed [itex]A[/itex] and [itex]B[/itex], since [itex]S[/itex] is a function os the states only and not the path connecting them.

    But for and ideal gas contained in a volume with thermally insulating walls, isn't it true that one could let it expand freely through an irreversible process, but also let it expand adiabatically through a reversible process? Therefore, initial and final states would be the same and the equation [itex]\Delta S \geq 0[/itex] would be contradicting itself?

    EDIT: Oh wait, actually the final temperature would be different for these two states of an ideal gas, so indeed maybe there is no reversible process that could lead to the same state obtained from a free expansion. Alright! Makes sense! :D
     
    Last edited: Jan 18, 2014
  5. Jan 18, 2014 #4

    DrClaude

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    Staff: Mentor

    For isolated systems, yes. If you can't get the necessary extra entropy from heat, you can't have extra entropy!
     
  6. Jan 18, 2014 #5

    Zag

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    Great! Thank you very much DrClaude, I do understand Thermodynamics a bit better now. :D
     
  7. Jan 19, 2014 #6

    Andrew Mason

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    That is a good example. If states A and B were at the same temperature but with different volumes (and pressure = nRT/V). B can be reached from A by a free expansion. However, the only reversible paths from A to B involve heat flow (e.g. isothermal quasi-static expansion).

    AM
     
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