Gibbs-Maxwell Relation: A Derivation Explained

[astropi]

So, this is not a homework problem. Merely for my own understanding.
This is (or should be) relatively simple I believe. Anyway, start of with Gibbs free energy and take the derivate, we arrive at:

dG = -SdT + VdP + \mu dN

take the partials we can see that with respect to dP we get V, with respect to dT we get -S, and with respect to dN we get mu. So far so good. Now, I know that if you take

\partial\mu / \partial P = V/N

but I'm just not seeing it? Is this not simply taking the derivative of G once more with respect to N? I think I'm missing something simple, so any help is appreciated. Thanks.

 

[Mapes]

You can get to

\left(\frac{\partial\mu }{ \partial P}\right)_T = \frac{V}{N}

from the http://en.wikipedia.org/wiki/Gibbs%E2%80%93Duhem_equation" . The constraint of constant temperature is required.

 

[astropi]

You can get to

\left(\frac{\partial\mu }{ \partial P}\right)_T = \frac{V}{N}

from the http://en.wikipedia.org/wiki/Gibbs%E2%80%93Duhem_equation" . The constraint of constant temperature is required.

Hi, could you please explain how the Gibbs-Duhem equation leads to V/N? I understand how to derive the Gibbs-Duhem equation, but fail to see how it helps us in this case. Sorry if it should be obvious and I just can't see it.

 

[Mapes]

Start with S\,dT-V\,dP+N\,d\mu=0, so that d\mu/dP=V/N-S/N(dT/dP). At constant temperature, the expression simplifies to V/N. Does this make sense?

 

[astropi]

Start with S\,dT-V\,dP+N\,d\mu=0, so that d\mu/dP=V/N-S/N(dT/dP). At constant temperature, the expression simplifies to V/N. Does this make sense?

It certainly does. In one of my stat-mech books it "showed" the derivation, but only said use -SdT + VdP +\mu dN which is clearly using the Gibbs free energy. I wonder if they just glossed over the part that you need to use the Gibbs-Duhem relation, or can you actually get V/N using Gibbs and not the Gibbs-Duhem relation? Thanks again!