# According to Thermodynamics Everything should Pop into Existence?

1. Jun 10, 2010

### nonequilibrium

(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?

2. Jun 10, 2010

### Andy Resnick

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 $\Delta G$.

3. Jun 10, 2010

### nonequilibrium

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)

4. Jun 10, 2010

### Mapes

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

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.

5. Jun 10, 2010

### nonequilibrium

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

6. Jun 10, 2010

### Mapes

As with the first derivation, I'm not seeing how this hypothesized system obeys energy conservation.

7. Jun 10, 2010

### nonequilibrium

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)