Getting the critical radius during nucleation from G(r) is straightforward - but in our lecture notes, a notation is used that I cannot quite wrap my head around:

The molar Gibbs Free Energy $G(r)$ is the sum of the surface term of a forming droplet and the volume term.

$$ G(r)=\bigg(\frac{4\pi r^3}{3}\cdot \frac{\rho_{l}}{N_A}\cdot(\mu_{l} -\mu_{g})\bigg)+4\pi r^2\sigma_{lg}$$

with $l\dots liquid$

and $g\dots gas$

The critical radius is then

$$ \frac{\partial G}{\partial r}=0=\frac{r\rho}{N_A}\cdot (\mu_l-\mu_g)+2\sigma_{lg} $$ $$ r\bigg(\frac{\partial G}{\partial r}=0\bigg)\rightarrow \underline{r_c=2\sigma_{lg}\cdot \frac{N_A}{\rho_l}\cdot \frac{1}{\mu_g-\mu_l}}$$

However, in the notation in our lecture notes, this is written as

$$\delta G=0=\delta(n_l\mu_l+n_g\mu_g+4\pi r^2 \sigma_{lg}) $$ $$=\delta n_l\bigg(\mu_l-\mu_g+\underbrace{\frac{2\sigma_{lg}}{r}\cdot \frac{N_A}{\rho_l}}_{\dots?}\bigg)$$

with $n_l=\frac{4 \pi r^3}{3}\frac{\rho_l}{N_A}=-n_g$

from which then follows the same $r_c$.

What is the motivation behind writing G not as a partial derivative, but as a variation - and what exactly happened to the surface term there?


1 Answer 1


At least I can see how the surface term has been transformed. The aim seems to be to express both the volume, and the surface area, of the droplet in terms of the number of moles of liquid $n_l$. From the definition $$ n_l = \frac{4\pi r^3}{3} \frac{\rho_l}{N_A} \equiv K r^3 $$ (where I have defined $K$ for convenience in a moment) the volume term in $G$ becomes, as you have noted, $n_l\mu_l+n_g\mu_g$, and if we differentiate with respect to $n_l$ and impose $\delta n_g = -\delta n_l$ we get the $\delta n_l(\mu_l-\mu_g)$ term in $\delta G$. (Formally $G$ represents the free energy change to form a droplet from the gas phase, I don't think we are meant to take literally the negative value of $n_g=-n_l$ in the OP).

For the surface area term, since $r^2=K^{-2/3}n_l^{2/3}$, $$ \frac{d\,(4\pi r^2)}{d n_l} = 4\pi K^{-2/3} \, \tfrac{2}{3}n_l^{-1/3} = 4\pi K^{-2/3} \, \tfrac{2}{3} \left(K^{-1/3} r^{-1}\right) = \frac{4\pi\times 2}{3Kr} = \frac{2N_A}{\rho_l\, r} $$ So finally $$ \frac{d G}{d n_l} = \mu_l-\mu_g + \frac{2\sigma_{lg}}{r} \frac{N_A}{\rho_l} $$ and this could be written in variational form, as you have it. Setting the right hand side to zero gives the expression for $r_c$ directly, again as you have pointed out.

As for the motivation for this, I am somewhat in the dark. It might be preparing the way for the description of nucleation in terms of density functional theory, in which the free energy (or more usually the grand potential) is expressed as a functional of the local density, including either square gradient terms or a nonlocal term, to represent the interface. The formalism would lead to a free energy minimum for a general heterogeneous system, but can be parameterized to take advantage of spherical symmetry in this case. In that case, writing a variational expression for the free energy is indeed the starting point (but in terms of a position dependent $\rho(r)$ which tends to $\rho_l$ or $\rho_g$ in the appropriate limits). This has been a very successful approach, dating from the work of Cahn and Hilliard in the late 1950s, and developed by David Oxtoby and Robert Evans in more recent times.

Or it might be intended to highlight some thermodynamic point, such as the appearance of the Laplace pressure $2\sigma_{lg}/r$ in the surface term. I'm afraid I'm just guessing here.


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