Nucleation - Internal energy of a droplet

Question

From K. Huang's Statistical Mechanics, par. 2.2:

Suppose a droplet of liquid is placed in an external medium that exerts a pressure $P$ on the droplet. Then the work done by the droplet on expansion is empirically given by

$$dW=P dV - \sigma da$$

where $da$ is the increase in the surface area of the droplet and $\sigma$ the coefficient of surface tension. The first law now takes the form

$$dU = dQ - P dV + \sigma da$$

Integrating this, we obtain for the internal energy of a droplet of radius $r$ the expression

$$U = \frac 4 3 \pi r^3 u_\infty + 4 \pi \sigma r^2$$

where $u_\infty$ is the internal energy per unit volume of an infinite droplet.

I don't understand this last passage.

I do understand the second term, since

$$\sigma \int da = 4 \pi \sigma r^2$$

but frankly I don't understand where the first term is coming from. Why should

$$\int dQ - \int P dV = \frac 4 3 \pi r^3 u_\infty$$

hold?

Answer 1

I do not have Huang so I cannot be sure of the full context of the question but this is what Pippard has. Imagine an experiment when a small spherically shaped liquid drop of radius $r$ is formed from a bulk liquid at the end of a pipette whose other end is the reservoir, and the drop is surrounded by at atmospheric pressure $p_0$. Assuming that the liquid is incompressible its volumetric free energy density $f_0$ is independent of pressure. Because of the surface tension over the drop there is pressure difference of $\Delta p = 2\sigma/r$ so that the reservoir pressure must be $p=p_0 + \Delta p$ to maintain equilibrium.

Now increase the radius of the drop reversible by $\delta r$, the resulting amount of differential volume change through the pipette is $4\pi r^2 \delta r$, and the work required to effect this change is equal to the change in the free energy of the system $\delta F = 4\pi r^2 \delta r \Delta p = 8 \pi \sigma r \delta r$. But the free energy change of the reservoir is $4 \pi r^2 f_0 \delta r$ where $f_0$ is the volumetric free energy density of the fluid. Therefore $$\left (\frac{\partial F_{drop}}{\partial r} \right )_T = 4\pi r^2 f_0 + 8\pi \sigma r .$$ Integrated by $r$ you get $$F_{drop}= \frac{4\pi}{3}\pi r^3 f_0 + 4\pi \sigma r^2 = Vf_0 + \mathcal{A} \sigma$$

So you get two terms: one term is $\mathcal{A} \sigma$ that is proportional to the surface area $\mathcal{A}$ and surface tension coefficient $\sigma$, and another term $Vf_0$ that is proportional to the volume of the drop $V$ and to the liquid's volumetric energy density $f_0$. This $f_0$ is probably what Huang writes as $u_{∞}$

Answer 2

Maybe we can look at the problem this way. We want to compute the energetic cost of creating a droplet of liquid of radius $r$ in the vapor phase.

This energetic cost will be

$$dU = dU_{vol} + dU_{sup}$$

where $dU_{vol}$ is the "usual" volumetric energy change while $dU_{sup}$ is the energetic cost of creating the interface.

The energetic cost of creating a droplet of radius $r$, neglecting the interface, is given by the usual form of the second law of thermodynamics for an hydrostatic system:

$$U_{vol} = \int dU_{vol} = \int (dQ - PdV)$$

We may be tempted to try to solve this integral "formally", but I think that it is not possible, because $dQ$ will in general be non-zero and $P$ will change, so that we cannot take it out of the integral. If we could, we would have

$$\int (dQ - PdV) = Q- \frac 4 3 P\pi r^3$$

but this cannot be the right expression because the final result cannot depend on $Q$ and $P$.

So we cannot formally solve the integral, but we know what the result must be. Indeed, since we are neglecting the cost of creating the interface, we must have

$$U_{vol} = \frac 4 3 \pi r^3 u_\infty$$

It is really important that we use $u_\infty$ and not $u_r$ because -I know, I have repeated it too many times times already- we are here neglecting the energy of the interface: an infinite droplet has no interface by definition, therefore we take the value $u_\infty$ for the internal energy density.

To this term, we finally add the cost of creating the interface:

$$U_{sup} = 4 \pi \sigma r^2$$

So the answer is that Huang is probably not formally solving the integral $\int (dQ-PdV)$, but he is just plugging in the result that he knows must hold.