0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $u_{∞}$

$\endgroup$
  • $\begingroup$ Thanks, but what I would like to understand is really that single passage: from $dU=...$ to $U=...$ $\endgroup$ – valerio Jul 14 '17 at 23:19
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.