Gibbs Free Energy of a Water Droplet and Bubble Nucleation [Thermal & Statistical Mechanics I Studied #65]

Example 28.6. Calculate the Gibbs function of a liquid droplet of radius $r$ that is in equilibrium with vapour. Assume the temperature is one where the liquid is the thermodynamically stable phase.

Since it says “in equilibrium”, let’s say everything is constant. Then $p$ and $T$ are constant too, and so $\mu$ is constant as well.

So, since $G = \mu N$, we can write $dG = \mu\,dN$, and since we’re dealing with both vapour and liquid,

let’s write it like this.

But we’re also going to take surface tension into account, so we add a surface-tension term onto the Gibbs function above. Since $dW = \gamma\,dA$,

And then there’s the conservation of particle number, Particle conservation: dN_vap = −dN_liq , so

It’s totally fiiine to do it like this~

Let’s also throw in the relation we used before, Sphere surface: dA = 8πr dr , too.

Putting it in: dG = −μ_vap dN_liq + μ_liq dN_liq + γ8πr dr = −(μ_vap − μ_liq) dN_liq + γ8πr dr

Now we’re going to write $dN$ in terms of $r$, so let’s think about it like this.

The droplet (blue) and the same droplet grown by dr (outer circle), with dr marked by a red double arrow

If the droplet’s radius grows by $dr$, the volume grows by Volume of the new shell: 4πr² dr this much, right?

How many liquid water molecules are in this extra volume???

The liquid in a volume of The shell volume 4πr² dr has a mass of Its mass: 4πr² dr · ρ_liq, in kg . And if we divide that by $m$, the mass of one liquid molecule, out comes the number!!! That is,

Number of molecules in the shell: dN_liq = 4πr² dr · ρ_liq / m

$Substituting dN_liq: dG = −(μ_vap − μ_liq)·4πr² dr ρ_liq/m + γ8πr dr = {−(μ_vap − μ_liq)4πr²ρ_liq/m + γ8πr} dr$

Now if we slap an integral on both sides,

Integrating: G(r) = G(0) + 4πγr² − Δμ 4πr³ρ_liq/(3m), after tidying up

We’ve derived the Gibbs function of a droplet of radius $r$.

Shall we plot it? Aaaaah, right!

Why Δμ = μ_vap − μ_liq is positive: the liquid is the stable phase, so μ_vap is bigger than μ_liq

Now shall we really plot it? I’ll just copy the one in the book.

(It’s coming up a little further down, with my circles drawn on it.)

So then. Let’s compute the extremum of that graph, the point where $dG = 0$.

$Setting dG = {−Δμ 4πr²ρ_liq/m + γ8πr} dr = 0$

Since that’s what we’re looking for,

Solving: Δμ 4πr²ρ_liq/m = γ8πr, so r = 2γm/(Δμ ρ_liq)

That means right at that $r$, the droplet is in equilibrium. Let’s call that $r$ “r star”.

Now, when droplets form, one that comes out smaller than this reacts in the direction that minimizes the Gibbs function, so

The book’s plot of (G(r) − G(0))/G_0 against r/r*, with a blue circle left of the peak and a blue arrow sliding back down to r = 0

$r \to 0$: it “disappears."

And a droplet that comes out bigger than $r^*$ also reacts in the direction that minimizes the Gibbs function, so

The same plot with a red circle right of the peak and a red arrow sliding down toward bigger r

$r \to \infty$: “the droplet grows without limit."

So the thing from the previous post, the bubbles that keep growing when water bubbles away at a boil, can be explained by the principle of Gibbs function minimization, I guess~ hehe hahaha

(It’s the same story with liquid and vapour swapped: a vapour bubble inside the liquid.)

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