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

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.

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.

• There’s much commonality with this answer, perhaps enough to consider the question a duplicate. Nov 29, 2022 at 17:10
• @Chemomechanics And indeed this from eight years ago. However I didn't think any existing questions were close enough to close this as a duplicate. Nov 29, 2022 at 17:15
• Your answer is the best I have read to date on the nucleation question, thanks for providing it. -NN Nov 29, 2022 at 18:58

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.