# Why Gibbs energy for nucleation theory?

In nucleation theory, the free energy is given by (https://en.wikipedia.org/wiki/Classical_nucleation_theory but in many other places also): $$\Delta G=\frac{4\pi }{3}r^3\Delta g +4\pi r^2\sigma,$$ with $$\Delta g$$ the difference in free energy between the two phases (liquid and gas for instance), and $$\sigma$$ the surface tension. $$r$$ is the radius of the "nucleus" of the new phase.

My question is why don't we take the pressure into account? Because we are not at constant pressure, since Laplace law says for the difference of pressure inside and out of the "nucleus": $$\Delta P= \frac{2\sigma}{r}$$ ? So why do we use a Gibbs energy with no pressure?

• To answer this you need a full derivation of the formula, to see what is going on and what the symbols mean. I don't have time to write this out now, but the pressure difference you mention does have to be considered. Sorry to ref my own work, but I did it v. fully in my book on thermodynamics (OUP). Dec 17, 2018 at 9:41
• What's the title of the book ?
– J.A
Dec 17, 2018 at 14:11
• "Thermodynamics, a complete undergraduate course", Oxford University Press. (see section 24.2.1); global.oup.com/ukhe/product/… Dec 17, 2018 at 17:45

The formula you quote gives the Gibbs energy of a system [single liquid droplet plus vapor of same chemical species].

You say "we are not at constant pressure". This is true in the sense that pressure of liquid inside the droplet is different from pressure of vapor just above the droplet.

The formula you quote actually takes into account these differences of pressure, through $$\Delta g$$.

What the formula gives is not the basic Gibbs energy as defined in textbooks ($$U-TS+PV$$), but something related but more technical: it is the ordinary Gibbs energy of the system minus the ordinary Gibbs energy the system would have if it was all pure vapor at the same $$T,P$$, minus some constant of no consequence.

The ordinary Gibbs energy of liquid can be expressed as

$$G_L = \mu_{liquid} N_{liquid} = g_{liquid}V_{liquid}$$ and similarly for the vapor. Thus, we can regard the Gibbs energy as having a density, either per molecule ($$\mu$$), or per volume ($$g$$).

There is also Gibbs energy per unit boundary surface, so we have a separate contribution

$$G_{s} = \sigma S .$$

Chemical potential $$\mu$$ is a function of $$T,P$$, but we have a different function for liquid ($$\mu_L(T,P)$$), and different function for vapor($$\mu_V(T,P)$$). And we have same temperature $$T$$ everywhere (by assumption), but different pressures for liquid $$P_L$$ and vapor $$P_V$$ (due to curved surface of the droplet).

Now, with this, we can understand how different pressures enter the expression for $$\Delta G$$.

$$\Delta G(T,P_V) = g_L(T,P_L) V_L + g_V(T,P_V)V_V + \sigma(T) S$$

Because $$P_V$$ is easy to measure and control directly (by putting in more vapor), it is taken as "the pressure" that $$\Delta G$$ is regarded as function of, but this is just arbitrary choice, we could use $$P_L$$ instead. The difference in pressures enters non-trivially in the right-hand side: the Gibbs energy densities $$g_L,g_V$$ are to be taken at different pressures.

We can assume that all evaporation/condensation happens inside a big container of constant volume $$V$$, so $$V_V = V - V_L$$ and rewrite this as

$$\Delta G(T,P_V) = g_L(T,P_L) V_L + g_V(T,P_V)V - g_V(T,P_V)V_L + \sigma(T) S$$ This is almost the original formula, but the second term is getting in the way. However, provided the condensation or evaporation happens while $$T$$ and $$P_V$$ do not change (big container so it acts as fixed reservoir), this term does not change and we can drop it and define $$\Delta G'$$:

$$\Delta G'(T,P_V) = \left( g_L(T,P_L) - g_V(T,P_V)\right) V_L + \sigma(T) S.$$

So we see that the original $$\Delta g$$ in the equation actually takes into account different pressures.

It may be convenient to express it as function of vapor pressure only (and $$T$$ and radius $$r$$):

$$\Delta g(T,P_V) = g_L(T,P_V+\frac{2\sigma}{r}) - g_V(T,P_V).$$