Why condensation nuclei is required for condensation and formation of clouds?

Question

Why condensation nuclei is required for condensation and formation of clouds?

What is the physics behind it?

Can you please explain it intuitively?

Thank you.

Answer 1

Suppose we form a sphere of ice with radius $r$ from water vapour in a cloud. Then the mass of the ice will be $\tfrac43\pi r^3 \rho$ and if the associated latent heat for the process is $L$ the enthalpy change of the water will be:

$$ \Delta H_1 = -\tfrac43\pi r^3 \rho L \tag{1} $$

where $\Delta H$ is negative because this is an exothermic process. However the formation of a drop creates a surface with an area $4\pi r^2$ and this area has an associated surface energy $\gamma$ joules per square metre. That means to form the sphere we have to supply an enthalpy:

$$ \Delta H_2 = +4 \pi r^2 \gamma \tag{2} $$

where this time $\Delta H$ is positive because we have to supply energy to create the surface. Incidentally the surface energy $\gamma$ is the same as the surface tension in liquids. All interfaces, air-liquid, air-solid and liquid-solid, have an interfacial energy that is the same physical property as the surface tension.

Anyhow the total enthalpy change is just the sum of these two enthalpies:

$$ \Delta H = 4 \pi r^2 \gamma - \tfrac43\pi r^3 \rho L \tag{3} $$

and as a rough guide the sphere of ice will form only if the total enthalpy change is negative. This isn't quite true as I'm ignoring the entropy change, but it is a useful first approximation. If we write equation (3) as:

$$ \Delta H = 4 \pi r^2 (\gamma - \tfrac13 r \rho L) $$

then $\Delta H$ will be negative only if the bracket is less than zero:

$$ \gamma - \tfrac13 r \rho L < 0 $$

or:

$$ r_c > \frac{3\gamma}{\rho L} \tag{4} $$

where $r_c$ is the critical radius for nucleation. For radii less than $r_c$ it is energetically unfavourable to form the sphere of ice and it would simply evaporate again.

The problem is that if we are growing our ice sphere from nothing the radius starts out at zero and since that is smaller than $r_c$ the sphere would just evaporate instead of growing. This is an example of a nucleation barrier. The sphere needs to reach a certain size before it can grow, but it cannot reach that size without growing.

And this is where the nuclei come in. A nucleus like a dust particle provides a surface that ice can grow on without having to start from nothing, and that evades the constraint in equation (4), though I confess I am unsure exactly how the nucleation works.

Answer 2

To address the only question remaining in the wake of John Rennie's excellent exposition, my understanding of the physics of "seeded nucleation" is as follows:

Let's take a dust grain in a water vapor-laden parcel of air as an example. The formation of condensed water droplets is inhibited by the Young-La Place energy barrier described above by John. Now, enter the dust particle as a nucleation seed.

If the dust particle is composed of something that is wetted by water, then there will be a tendency for water vapor molecules that impinge upon it to stick and over time the particle's surface will become fully populated with water molecules. At this point the dust particle stops looking like dust to other water molecules; instead it looks like a glob of water molecules that have already condensed, and to which more molecules can then get stuck- and the condensation process is thus "seeded" by the presence of dust in the air.

Note that this process relies again on surface-energy arguments, just as in John's description.

Answer 3

Feynman describes that liquid water exhibits a force in water vapour molecules that are closwby. The force accelarates the water vapour molecules. This is the heat of condensation.

So really the liquid water held together by strong cohesive forces, is puling in more water.

Point is that we do not learn a lot about these cohesive forces. That is what chemists do. In the liquid phase water molecules are held together by hydrogen bonds, that form because water molecules are dipôles with an uneven distribution of charge. So once you have a liquid you have these cohesive forces thst collectively suck out water vapour out of the air.

Air on the other hand, does nothing. The molecules are too far apart for repulsive or attractive forces to act.

So basically water is holding water and air is holding nothing as it is a gas (and almost ideal). But in the beginning of a droplet there is no starting point. But solid particles are made up of many molecules that are held together by strong forces. So they can pull on the water vapour molecules as well. Moreover these solid particles often have a huge ibterfacial area, so many places where water molecule can attach itself.

So I think the condensation nuclei are just a question of minimum critical mass to get the growing going. I used to believe in infinite surface tension and stuff, but surface tension seems even stranger than thinking air can hold water.

Answer 4

Boiling point of water droplet depends on its size, because of surface tension. This is only significant for very small droplets. Droplets containing several water molecules evaporate quickly at room temperature. When water vapor condenses it first need to form tiny droplets so unless you can somehow skip this step droplets will evaporate faster than they can grow big enough. Solid particles that can adsorb water on its surface function as droplets of sufficient size to avoid this problem. Shaking or sound can also do this.