Question

Everything in this section so far has ignored the boundary between two phases, as if each molecule were unequivocally part of one phase or the other. In fact, the boundary is a kind of transition zone where molecules are in an environment that differs from both phases. Since the boundary zone is only a few molecules thick, its contribution to the total free energy of a system is very often negligible. One important exception, however, is the first tiny droplets or bubbles or grains that form as a material begins to undergo a phase transformation. The formation of these initial specks of a new phase is called nucleation. In this problem we will consider the nucleation of water droplets in a cloud. The surface forming the boundary between any two given phases generally has a fixed thickness, regardless of its area. The additional Gibbs free energy of this surface is therefore directly proportional to its area; the constant of proportionality is called the surface tension, $\alpha$

$\sigma \equiv \frac{G_{\text{boundary}}}{A}$

ff you have a blob of liquid in equilibrium with its vapor and you wish to stretch it into a shape that has the same volume but more surface area, then u is the minimum work that you must perform, per unit of additional area, at fixed temperature and pressure. For water at $20°\mathrm{C},\sigma =0.073\mathrm{J}/{\mathrm{m}}^2$

(a) Consider a spherical droplet of water containing N1 molecules, surrounded by $N-N_1$molecules of water vapor. Neglecting surface tension for the moment, write down a formula for the total Gibbs free energy of this system in terms of $N,N_1$, and the chemical potentials of the liquid and vapor. Rewrite $N_1$in terms of $V_1$, the volume per molecule in the liquid, and $T$, the radius of the droplet.

(b) Now add to your expression for $G$a term to represent the surface tension, written in terms of $T$and u.

(c) Sketch a qualitative graph of G vs. T for both signs of µg - µ1, and discuss the implications. For which sign of ${\mu }_{g-{\mu }_1}$does there exist a nonzero equilibrium radius? Is this equilibrium stable?

(d) Let $T_C$represent the critical equilibrium radius that you discussed qualitatively in part (c). Find an expression for $T_C$in terms of ${\mu }_{g-}\mu$. Then rewrite the difference of chemical potentials in terms of the relative humidity (see Problem 5.42), assuming that the vapor behaves as an ideal gas. (The relative humidity is defined in terms of equilibrium of a vapor with a flat surface, or with an infinitely large droplet.) Sketch a graph of the critical radius as a function of the relative humidity, including numbers. Discuss the implications. In particular, explain why it is unlikely that the clouds in our atmosphere would form by spontaneous aggregation of water molecules into droplets. (In fact, cloud droplets form around nuclei of dust particles and other foreign material, when the relative humidity is close to $100\%$.)

Short answer

Result is:

(a). The formula for total Gibbs free energy of this system is$G=\frac{4\pi r^3}{3v_l}\left({\mu }_l-{\mu }_g\right)+N{\mu }_g$.

(b). The improved formula is $G=\frac{4\pi r^3}{3v_l}\left({\mu }_l-{\mu }_g\right)+N{\mu }_g+4\pi \sigma r^2$.

(c). The qualitative graph between$G$vs.$T$is

(c)$\frac{\partial G}{\partial r}=\frac{4\pi r_c^2}{v_l}\left({\mu }_l-{\mu }_g\right)+8\pi \sigma r_c=0$

$\text{(d)}h=e^{2\sigma v_l/kT{r}_c}$

Step-by-step solution

part(a) Step 1: Given information

we have been given that$G_{extra}=\sigma A$

part(a) Step 2:Simplify

The gibbs energy is given by:

$G=N_l{\mu }_l+\left(N-N_l\right){\mu }_g$

the volume of droplet

$V_r=\frac{4\pi r^3}{3}$

$v_l=\frac{4\pi r^3}{3N_l}\to N_l=\frac{4\pi r^3}{3v_l}$

part(b) Step 1: Given information

we have been given that replace$A$with$2\pi r^2$

part(b) Step 2:Simplify

After solving we get,

$G=\frac{4\pi r^3}{3v_l}\left({\mu }_l-{\mu }_g\right)+N{\mu }_g+4\pi \sigma r^2$

part(c) Step 1: Given information

we have been given that$G\propto \left({\mu }_l-{\mu }_g\right)r^3+r^2$

part(c) Step 2:Simplify

it is clear that

${\mu }_l>{\mu }_g$,$G\propto r^3+r^2+C$

${\mu }_l<{\mu }_g$,$G\propto -r^3+r^2+C$

part(d) Step 1: Given information

we have been given that$\frac{\partial G}{\partial r}=\frac{4\pi r_c^2}{v_l}\left({\mu }_l-{\mu }_g\right)+8\pi \sigma r_c=0$

part(d) Step 2:Simplify

The change in gibbs energy is:

$dG=-SdT+VdP+\mu dN$

at constant temperatur and number

$dG=VdP$

hence,$d\left({\mu }_g-{\mu }_l\right)=\left(v_g-v_l\right)dP$

solve for h to get:

$h=e^{2\sigma v_l/kT{r}_c}$