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B Gibbs paradox

  1. Mar 26, 2016 #1
    I have heard that identical distinguishable classical particles having different ''statistics''.It is the limit of quantum case.Then we mix many parts(cells) of identical gases, the total entropy increases.I can not derive this limit from quantum particles to classical particles(please help me).I only know that by the thisness of classical particles we can ''paint'' each classical particle with one ''colour'',so that when we mix many cells of identical gases,the entropy will increase.
     
    Last edited: Mar 26, 2016
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  3. Mar 26, 2016 #2

    A. Neumaier

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    Please spend more time formulating proper questions (and edit your initial post!): The paradox is called after Gibbs, not Gibb. Distinguishable particles cannot be identical (which would make them indistinguishable). If you mix quantities of the same gas, nothing happens; the entropy doesn't increase.

    Thus your question is very unclear. However, the answer may well be already here: https://en.wikipedia.org/wiki/Gibbs_paradox; [Broken] so please study this first. If this is not sufficient then (with considerably more work to do for you) here: http://www.mdpi.org/lin/entropy/gibbs-paradox.htm
     
    Last edited by a moderator: May 7, 2017
  4. Mar 26, 2016 #3
    I will be more carefull!After reading Reichl carefully I think I might understand the resolution of Gibbs paradox.But I do not understand why there are identical classical particles because classical particles are distinguishable.Then if we mix many parts of same gas the entropy would increase.
     
    Last edited: Mar 26, 2016
  5. Mar 26, 2016 #4

    vanhees71

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    I'd say the Gibbs paradox has been already solved before the discovery of quantum theory by Boltzmann in introducing a ##1/N!## when counting the single-particle microstates in a classical phase-space volume. The idea behind it is indeed the indistinguishability of particles with the same intrinsic properties (where with intrinsic properties I mean those that characterize the particle's properties in its rest frame; classically that's mass and perhaps rotational degrees of freedom in the rigid-rotator model of molecules).

    The real problem with classical statistical physics is that there's no natural phase-space measure and no way to definitely count the single-particle microspace in a classical single-particle phase-space volume. This is only resolved by the introduction of Planck's constant, providing a natural scale of the dimension of action, and the quantum definition of states.

    For an elementary treatment, see

    http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf

    Sects. 1.5 and 1.8 for the (semi-)classical and quantum treatment of the entropy.
     
  6. Mar 26, 2016 #5
    Then depending on conceptional choice of considering classical particles as identical or distinguishable particles(having same intrinsic properties),we would have entropy increasing or not changing of the mixing of many cells of same gas.That resolves Gibbs paradox.Is this right?
     
  7. Mar 26, 2016 #6

    A. Neumaier

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    There is only one correct choice, and that resolves the problem: On must treat the classical particles as indistinguishable by weighting the contributions appropriately. This was Gibbs' solution (found at a time when only classical mechanics existed).
     
  8. Mar 26, 2016 #7
    Then where is the wrong of the application of reason for mixing different gases to the same gases that leads to the paradox.In Reichl page 66 the reason is general not depending on the types of gases.
     
  9. Mar 26, 2016 #8

    DrDu

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    I think that there are several solutions to the Gibbs paradox. One involves recognition of the indistinguishability of identical quantum particles. But the paradox is also resolvable for distinguishable classical particles: The point is that we don't know which particle is in which container even before mixing the two gasses. So entropy doesn't increase on mixing. If we would measure which particle is in which box, we would accumulate a lot of information which would be converted into entropy on deletion of the memory after mixing.
     
  10. Mar 26, 2016 #9
    I agree with many solutions to Gibbs paradox.But I mean I would like to find out ''the wrong point'' in the argue leading to the paradox,because the argue is general.
     
  11. Mar 26, 2016 #10

    vanhees71

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    Interesting, I thought it was Boltzmann who solved the problem, but I'm not sure. One should find some book/paper on the history of statistical mechanics.
     
  12. Mar 26, 2016 #11

    vanhees71

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    Of course, there's no paradox, if the gases in the two halves of the container are different originally and then mixed by taking (adiabatically) out the divider wall, because the state measurably changes from the two different particles separated ("ordered") to them being mixed and in complete thermal equilibrium ("less ordered"), i.e., the entropy must grow. If the two gases are made of the same particles, then nothing changes in the thermodynamical state when the divider wall is adiabatically taken out, and thus the entropy should not change.

    The solution is to put appropriate factors ##1/N_i!##, ##i## labeling the different species of the molecules making up the gases in the mixture.
     
  13. Mar 26, 2016 #12

    A. Neumaier

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    The wikipedia article on Gibbs' paradox mentions Boltzmann (''correct Boltzmann counting'') but gives no reference to his work. Jaynes refers to Gibbs 1875 for the resolution.
     
  14. Mar 26, 2016 #13
    In Reichl's book page 66,she write down the Gibbs free energy of mix gases,then calculate entropy through it.But Gibbs potential is additive,so entropy is also additive despite of the same type of gases.Then it leads to paradox.I would like find out the ''wrong'' in this argue,
     
  15. Mar 26, 2016 #14

    A. Neumaier

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    In the version I have (2nd ed. 1998), she discusses the Gibbs paradox in Section S2.B p.72ff and refers for its resolution to quantum mechanics. Jaynes article linked to in post #12 gives Gibbs' classical resolution.
     
  16. Mar 26, 2016 #15
    The Gibbs potential G(P,T.n1,...)=Sigma(njRT(Phi(T)+lnP+lnxj)),then what is wrong with partial pressure(xj) of identical particles?The increasing of entropy is result of xj.
     
  17. Mar 27, 2016 #16

    A. Neumaier

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    You can find yet another discussion of Gibbs' paradox in my online book, pp.237-239. But I am not going to do for you the work of understanding one of the explanations offered (by Reichl, Jaynes, or my book).

    By the way, you still haven't corrected the title of your thread. (Click on thread tools at the top of page 1 to make the correction.)
     
    Last edited: Mar 27, 2016
  18. Mar 27, 2016 #17
    The no answer still help me understand the Gibbs paradox,because it make me rethinking!
    I do not know how to correct the title of post 1,because having not the edit function.
    Thank you very much for your great helps.
     
    Last edited: Mar 27, 2016
  19. Mar 27, 2016 #18

    A. Neumaier

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    Don't you see a link to ''thread tools'' at the right top of the page? It is visible for me on the first page of all threads that I started. (But perhaps this feature is not available to anyone?)
     
  20. Mar 27, 2016 #19
    I do not see the tool!
     
  21. Mar 27, 2016 #20

    HPt

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    @fxdung: DrDu is right. In Eur. J. Phys. 35 (2014) 015023 http://dx.doi.org/10.1088/0143-0807/35/1/015023 (also available as arXiv preprint) you will find a detailed demonstration of the Gibbs paradox (in the form of a violation of the second law of thermodynamics) along with its resolution.

    @A. Neumaier: 'Identical' doesn't necessarily imply 'indistinguishable'. No inconsistency arises from regarding identical classical particles as distinguishable.
     
    Last edited: Mar 27, 2016
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