Past exam of the atmospheric physics course of the University of Cambridge / 2012 / Paper 1 / 2 / a

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Solution

A convex liquid surface has additional surface free energy. The resulting Kelvin equation for a pure spherical water droplet is

\frac{e _{s} ( r )}{e _{s} ( \infty )} = exp (\frac{2 γ M _{w}}{ρ _{w} RT r}) > 1.

Here gamma is surface tension and

M_{w}

is molar mass. Smaller droplets therefore need a higher environmental vapour pressure to avoid evaporating; forming pure-water embryos by homogeneous condensation has a large nucleation barrier.

An atmospheric droplet normally forms on a cloud condensation nucleus. Dissolved salt reduces water activity and hence equilibrium vapour pressure, opposing the curvature effect. Köhler theory combines them:

\frac{e _{eq}}{e _{s} ( \infty )} = a_{w} e^{A / r}, ln \frac{e _{eq}}{e _{s} ( \infty )} ≃ \frac{A}{r} - \frac{B}{r ^{3}}, A = \frac{2 γ M _{w}}{ρ _{w} RT}, B > 0.

Using the radius form, with radius and diameter conventions in Köhler activation kept distinct, the Köhler activation radius is

r_{c} = 3 B / A

, with

ln S_{c} = 2 A / (3 r_{c})

. Supersaturation above this maximum activates a wet aerosol into a growing cloud droplet. Solubility, particle size and wettability determine the threshold; not every particle activates. Cooling by ascent commonly supplies the necessary modest supersaturation. ^[ref]

Figure 1.
Köhler activation
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