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Gibbs energy=chem potential (not convinced) |
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| Mar3-07, 09:35 PM | #1 |
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Gibbs energy=chem potential (not convinced)
my thermal book gives a hand-waving argument saying the followings:
firstly, Gibbs energy, defined by: [tex]G\equiv U+PV-TS[/tex] is an extensive quantity (proportional to N), and also [tex]\left (\frac{\partial G}{\partial N}\right ) _{T,P}=\mu[/tex] so far so good, but then it says: if P and T are held constant then [itex]\mu[/itex] is also constant, which implies whenever a particle is added to the system, G is increased by [itex]\mu[/itex]. Thus, [tex]G=N\mu[/tex] But why must [itex]\mu[/itex] be solely dependent on T and V??? why can't [itex]\mu[/itex] depend on.. let's say N? is there any algebraic prove for that? edit: oh yeah I see... the book skips a very Very important reason of why it works!!! since V, S and U are also extensive, [tex]V\sim N[/tex] [tex]S\sim N[/tex] [tex]U\sim N[/tex] Thus, [tex]\left (\frac{\partial G}{\partial N}\right ) _{T,P}=\mu= \frac{\partial U}{\partial N}+P\frac{\partial V}{\partial N}-T\frac{\partial S}{\partial N}[/tex] and each of the three partial derivatives is independent of N since V, S and U are directly related to N... don't you just hate it when books make some non-rigorous arguments, left out the important details and act as if the things are obvious and trivial!?!! |
| Mar4-07, 02:40 PM | #2 |
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But what they did is entirely correct. I can always rewrite the chemical potential as a function of other intensive/extensive variables because of the existence of equations of state.
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| Mar4-07, 07:18 PM | #3 |
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you can prove it rigorously, without reference to the macroscopic thermodynamics, by finding [tex]<N>\mu[/tex] in the grand canonical ensemble.
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| Mar5-07, 01:16 PM | #4 |
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Gibbs energy=chem potential (not convinced)
really...? I'm interested... can you provide more details please? I would really love a rigorous argument on this problem.
so, how would you go from the definition of G and mu?? |
| Mar5-07, 02:01 PM | #5 |
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