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Is dn = dS = 0 when two phases are equilibrium?

  1. Dec 20, 2015 #1
    Hello, every one! I'm studying the Section 9.2 - Phase Equilibrium of "Classical and Statistical Thermodynamics" written by Carter.

    In equation (9.21) and (9.22)

    He said:

    [itex]dS_{A}=\frac{1}{T_{A}}(dU_{A}+P_{A}dV_{A}-\mu _{A}dn_{A})[/itex]-----(9.21)

    [itex]dS_{B}=\frac{1}{T_{B}}(dU_{B}+P_{B}dV_{B}-\mu _{B}dn_{B})[/itex]-----(9.22)

    I think dS and differentials of all other state variables should be zero! That is, dSA=dSB=dnA=dnB=dVA=dVB=...=0

    Is it correct? If it's correct, then I cannot get the following conclusions which needs to assume those increments dUA,dVA,dnA,... could be arbitrary.

    TA = TB (thermal equilibrium)
    PA = PB (mechanical equilibrium)
    μA = μB (diffusive equilibrium)

    Thanks in advance!!
     
  2. jcsd
  3. Dec 21, 2015 #2

    DrDu

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    No, why should these differential be constant?
     
  4. Dec 21, 2015 #3
    It's due to the definition of equilibrium. In other words, if those macroscopic state variables are not constant, then it's not at equilibrium.
    For example, if the color of the solution is changing, then it's not at equilibrium.

    I don't know why I was wrong..
     
  5. Dec 21, 2015 #4

    DrDu

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    But e.g. you can have a mixture of ice and water and even if some of the ice is melting (i.e. both the amount of ice and water is changing), the two are in equilibrium.
     
  6. Dec 21, 2015 #5
    Really? Even if some of the ice is melting, the two are still in equilibrium?
    So... What's the definition of equilibrium? It seems that the exact definition of equilibrium is "the system is at equilibrium if and only if its entropy is a maximum".

    Is that correct?

    Thank you very much :)
     
  7. Dec 21, 2015 #6

    BvU

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    I think DD is straying somewhat. A phase equilibrium is not enough for a thermodynamic equilibrium: matter and energy can still flow macroscopically.
    So I'm siding with PhE post #5.
    However, physics isn't a democratic thingy, so if someone knows better, I'm eager to be put right !
     
  8. Dec 21, 2015 #7

    DrDu

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    I'd rather say that phase equilibrium is a special case of thermodynamic equilibrium.
     
  9. Dec 21, 2015 #8

    DrDu

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    At equilibrium, the system is insensitive to changes in particle number of the two phases.


    This depends on the system. Is it thermally isolated or constant temperature, constant volume or constant pressure. E.g. at constant temperature and pressure, free enthalpy has to be maximal, not entropy. Which case is your book considering?
     
  10. Dec 21, 2015 #9

    BvU

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    We aren't making progress here if I rather say that thermodynamic equilibrium is a special case of phase equilibrium, namely phase equilibrium plus constant P, U, G, A, H, S etc... :smile:
     
  11. Dec 21, 2015 #10

    DrDu

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    I don't know what you have in mind, but in equilibrium thermodynamics (I suppose we are talking about this) we are considering only thermodynamic equilibrium states. E.g. considering the state diagram of water as a function of p and T, there is one line where both phases are in equilibrium. In this sense, phase equilibria are a subset of thermodynamic equilibria.
     
  12. Dec 21, 2015 #11

    BvU

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    Somewhat circular, to me that seems.

    The 'subset' issue is a matter of language, it seems to me. I think we agree: phase equilibrium isn't thermodynamic equilibrium, but thermodynamic equilibrium is phase equilibrium.

    (The "fruit isn't orange but orange is fruit" comes to mind. For me that makes oranges a subset of fruit. Not the other way around).

    Yet another approach: there are a lot more phase equilibria than thermodynamic equilibria. A TE is a PE with some extra restrictions. So TE is a subset of PE.

    If a system is moving back and forth over this line where both phases are in equilibrium, I can not consider that system to be in thermodynamic equilibrium: T changes. [edit] woops, V changes.
     
    Last edited: Dec 21, 2015
  13. Dec 21, 2015 #12

    DrDu

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    This is true if the system is isolated and there is also no work done on the overall system. Then it is clear that ##dU_B=-dU_A##, ##dV_B=-dV_A## and ##dn_B=-dn_A## as total internal energy, volume and particle number of the two phases taken together is constant. As total S is maximal, we can write ##0=\partial S/\partial U_A=1/T_A-1/T_B##, so that ##T_A=T_B##. Analogous equations hold for p and ##
    mu##.
     
  14. Dec 21, 2015 #13
    Yes. I think this is more like what the original question was getting at. The idea is to look at the combined system in a situation where the combination is isolated.
     
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