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mr. vodka
Jun10-10, 07:12 AM
(In the following discussion, when I use the word "always", I mean "as good as always" if you're willing to ignore exotic systems with negative temperature and such)

In the following discussion I will assume we're working in a heat bath with constant T and P:

So there are several ways to see the total Gibbs free energy of an object, defined G = U - TS + PV, is negative.

Two simple ways:
(*) Chemical potential is defined as \mu = -T \left( \frac{dS}{dN} \right)_{U,V} and thus is always negative. We also can prove G = \mu N.

(*) We know T \Delta S \geq Q = Q + P \Delta V - P \Delta V \geq Q + P \Delta V + W = \Delta U + P \Delta V so TS \geq U + PV or G \leq 0.

So now the problem is, when we 'make something' its G function goes from zero to something negative (as was just shown in two ways). This implies it should happen spontaneously, since in constant T and P the second law becomes "G goes to a minimum".

So does this say things should randomly pop into existence? Obviously there is a thinking error?

Andy Resnick
Jun10-10, 12:01 PM
Like potential energy, the only meaning free energy has is in terms of *changes* (dG vs. G). Changes can be positive or negative. I'm not sure the (absolute) free energy has any meaning.

Sometimes (especially in biochemistry), you will enounter notation like \Delta \Delta G , which corresponds to changes to \Delta G .

mr. vodka
Jun10-10, 12:12 PM
Well, that's what I used, didn't I? First G = 0 and then G is negative, so the net change is negative (well the basic principle is that G goes down when something is created, check my math above)

Mapes
Jun10-10, 12:32 PM
(*) Chemical potential is defined as \mu = -T \left( \frac{dS}{dN} \right)_{U,V} and thus is always negative. We also can prove G = \mu N.

Here you are assuming constant energy. How do you propose to change the amount of matter in a system without changing the total energy?

(*) We know T \Delta S \geq Q = Q + P \Delta V - P \Delta V \geq Q + P \Delta V + W = \Delta U + P \Delta V so TS \geq U + PV or G \leq 0.

How do you justify replacing \Delta S, \Delta V, and \Delta U with S, V, and U? I don't see how that's valid.

mr. vodka
Jun10-10, 12:45 PM
Okay, drop my first "derivation" then.

About the second: well, if I create the whole system, V_i = 0 and V_f = V, same for S and U, don't you agree?

EDIT: Btw thanks for the critique, I hope to discover my error before my exam in the morning, it's quite troubling I can't see where my reasoning goes astray

Mapes
Jun10-10, 12:52 PM
As with the first derivation, I'm not seeing how this hypothesized system obeys energy conservation.

mr. vodka
Jun10-10, 03:11 PM
Everything enters as heat from the environment (that's the meaning of -TS in the definition of G, right? And in this case: TS > U + PV (as shown in the 2nd derivation)

But I've come to the conclusion "dS_tot > 0 <=> dG < 0" under constant P and T is only an equivalence if the system has a constant amount of particles! That's probably where I made my error. (G is still < 0, but now it just doesn't matter, really)