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(thermodynamics) chemical potential

  1. Sep 29, 2006 #1
    Why for systems containing several components and phases does it require that the chemical potential of ecah component must be idential in every phase?
  2. jcsd
  3. Sep 30, 2006 #2


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    You can mathematically derive this result using the differential gibbs free energy of each phase. I'll give an outline and you can derive the rest.

    If you have n moles of chemical species (ie, [itex]n_i[/itex] moles of the [itex]i^{th}[/itex] species), the differential of the gibbs free energy, for this single phase, open system is

    [tex] d(nG) = \left( \frac{\partial nG}{\partial P} \right)_{T,n} dP + \left( \frac{\partial nG}{\partial T} \right)_{P,n} dT + \sum_i \left( \frac{\partial nG}{\partial n_i} \right)_{P,T,n_j} dn_i [/tex]

    Since we define chemical potential as

    [tex] \mu_i = \left( \frac{\partial nG}{\partial n_i} \right)_{P,T,n_j} [/tex]

    and using maxwell's relation, the equation becomes

    [tex] d(nG) = nVdP - nSdT + \sum_i \mu_i dn_i [/tex]

    If you now have k phases with each phase having n species, you can think of each individual phase as an open system (cause it's free to transfer mass). So, you can apply the above equation to each phase individually.

    Now here's the important part. At equilibirium P & T are same throughout the whole system. Can you see that if you look at the entire system (with all the species and phases), it's closed? This means you can say that for the entire system,

    [tex]d(nG) = nVdP - nSdT [/tex]

    For the whole mass of the entire closed system.

    Now, the total gibbs free energy is given by the sum over each phase. But, this value must be equal to the the gibbs free energy of the entire closed system. So, compare the two equations and consider pairs of phases.

    Can you take it from here and get the result?
    Last edited: Sep 30, 2006
  4. Sep 30, 2006 #3
    Yes! Thank you very much for explaining!
  5. Sep 30, 2006 #4


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    Well, I didn't explain it fully, and there's quite a bit you still have to do to get the result. Are you sure you understood fully? Can you fill the gap and post the rest of the derivation here?
  6. Sep 30, 2006 #5
    Because my mathematics is not all that good, can I think of this question in another way?
    From looking at the word, chemical potential, it has to do with a component's ability to react. So the componets must have the same chemical potential, otherwise chemical reactions will occur, and the phase won't be stable?
  7. Sep 30, 2006 #6


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    That's a circular argument, so I don't think it will work. If your prof asked this in a test, I think he would expect a mathematical derivation.

    asdf, If your math is causing you a problem, you should work on improving it because it's really essential in both science and engineering. Have you thought about summer classes, or some external coaching?
  8. Oct 2, 2006 #7
    good suggestion... i agree with you. i'm trying to work on my math...
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