Gibbs energy=chem potential (not convinced)by tim_lou Tags: convinced, energychem, gibbs, potential 

#1
Mar307, 09:35 PM

P: 689

my thermal book gives a handwaving argument saying the followings:
firstly, Gibbs energy, defined by: [tex]G\equiv U+PVTS[/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 nonrigorous arguments, left out the important details and act as if the things are obvious and trivial!?!! 



#2
Mar407, 02:40 PM

P: 223

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.




#3
Mar407, 07:18 PM

P: 701

you can prove it rigorously, without reference to the macroscopic thermodynamics, by finding [tex]<N>\mu[/tex] in the grand canonical ensemble.




#4
Mar507, 01:16 PM

P: 689

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?? 



#5
Mar507, 02:01 PM

P: 223




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