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Gibbs energy=chem potential (not convinced)

by tim_lou
Tags: convinced, energychem, gibbs, potential
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tim_lou
#1
Mar3-07, 09:35 PM
P: 688
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!?!!
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StatMechGuy
#2
Mar4-07, 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.
quetzalcoatl9
#3
Mar4-07, 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.

tim_lou
#4
Mar5-07, 01:16 PM
P: 688
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??
StatMechGuy
#5
Mar5-07, 02:01 PM
P: 223
Quote Quote by quetzalcoatl9 View Post
you can prove it rigorously, without reference to the macroscopic thermodynamics, by finding [tex]<N>\mu[/tex] in the grand canonical ensemble.
I'm intrigued, since I've never seen this done before. I've always seen, starting from the microcanonical ensemble, a derivation that leads to something that we recognize as F or some such, and then that's the connection to thermodynamics.


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