# Maxwell Relations: when are they valid?

1. Aug 17, 2014

### John Perez

Hi, I have a question
If im not mistaken, the Maxwell relations of theromdynamics
-----for example: ∂G/∂T) = -S ; ∂G/∂P) = V -----
are valid only for reversible processes.

On the other hand,
dG = ∂G/∂T)*dT + ∂G/∂P)*dP + ∑ μi dni
is valid for any process.

This means that
dG = -SdT + VdP + ∑ μi dni
should only be applicable for reversible processes, and not necessarily for irreversible ones, Right??

2. Aug 17, 2014

### bolbteppa

The Maxwell relations are just the application of Poincare's lemma to the thermodynamic potentials.

http://johncarlosbaez.wordpress.com/2012/01/19/classical-mechanics-versus-thermodynamics-part-1/

The thermodynamic potentials are applicable to reversible and irreversible processes

http://www.quora.com/Thermodynamics...mics-only-applicable-for-reversible-processes

Thus the Maxwell relations apply to reversible and irreversible processes.

Here's a great way to derive/remember the massive amount of information this branch of the subject contains by drawing pictures:

http://web.mit.edu/course/8/8.593/spring10/bk/on_MRS_Thermodynamic_Mnemonic.pdf

3. Aug 17, 2014

### John Perez

Oh, I didnt know about that Poincare lemma. And the definition of the potentials does apply to any processes.
But I couldnt find anywhere on the first link a mention of the relations bieng valid in one or the other scenario.

dG = -SdT + VdP apllies to reversible processes, and while dG is the same no matter the "path", -SdT and VdP certainly depend on the path, so the equality should not hold for a irreversible process. In such case it should read dG < -SdT + VdP

4. Aug 17, 2014

### WannabeNewton

No they've valid for any quasi-static process.

This is also only valid for quasi-static processes.

The Maxwell relations are literally just a consequence of the commutativity of mixed partial derivatives.

5. Aug 17, 2014

### John Perez

Ok, thanks.
So all those things are only valid for quasi-static processes, and not "violent" ones. So, for example,
dG = -SdT + VdP + ∑ μi dni
is not valid for "violent" irreversible processes. Or I understand wrong?

After all, like you say it is just a consequence of mixed partial derivatives but the derivative they ultimately come from is U = TdS - PdV, which is a combination of the first and second Principles and (I think) only is valid for quasi-static processes (since reversible seems to be the wrong term here).

6. Aug 17, 2014

### WannabeNewton

No you understood it correctly. In the case of "violent" irreversible processes, i.e. non-quasi-static processes, we only have inequalities.

Last edited: Aug 17, 2014
7. Aug 17, 2014

### bolbteppa

The second link isn't wrong, the first comment seems to be talking about non-equilibrium thermodynamics, and the second comment makes the original point that dU = TdS - PdV is still valid for irreversible processes, I guess you're conflating this with quasi-static processes or something...

You seem to be saying something equivalent to: because ydx and xdy separately depend on the path, it implies that (ydx + xdy) is also dependent on the path, but you forget that putting them together implies ydx + xdy = d(xy)... I guess classical vector analysis is messing your intuition up so I recommend this article http://em.groups.et.byu.net/pdfs/publications/formsj.pdf

Once you have the differential form a thermodynamic potential established (i.e. dU or one of it's Legendre transforms) you take a second exterior derivative and get the Maxwell relations. This is Poincare's lemma (in direct form, i.e. d^2 w = 0) and is equivalent to commutativity of second partial derivatives (Clairaut's theorem).

8. Aug 18, 2014

### John Perez

Thanks, I think I got it, though I recieved kind of conflicting answers.
Im leaning for the non-quasi-static = inequalities.