Q4(b). Vapour pressure of a curved water surface, radius 10 nm (Kelvin)

Reading: strictly, a bubble is a vapour cavity inside water (concave surface, lower pressure; point 5). The data are those of the standard droplet exercise, so the droplet (convex surface) is worked first; quote both values.

Given: r=1.0×10⁻⁸ m; T=293.15 K; p°=23.76 Torr; γ=72.76×10⁻³ N m⁻¹; ρ=998.2 kg m⁻³; M=18.015×10⁻³ kg mol⁻¹.

1. Origin: liquid in a droplet is under the Laplace pressure Δ P=2γ/r (1.46×10⁷ Pa here). Raising the pressure on a liquid raises its chemical potential by VₘΔ P, so its vapour pressure rises: RTln(p/p°)=Vₘ·2γ/r (Kelvin equation).
2. Molar volume: Vₘ=(M)/(ρ)=18.015×10⁻³998.2=1.805×10⁻⁵ m³ mol⁻¹
3. Exponent: (2γ Vₘ)/(rRT)=2(72.76×10⁻³)(1.805×10⁻⁵)(1.0×10⁻⁸)(8.314)(293.15)=0.1078
4. pp°=e⁰˙¹⁰⁷⁸=1.114; p=23.76×1.114, p=26.46 Torr
5. Cavity (bubble inside water): the surface is concave towards the vapour, so the sign changes: p=23.76 e⁻⁰˙¹⁰⁷⁸=21.33 Torr.
6. Size effect: the correction falls as 1/r: at r=1 μm the droplet's vapour pressure is only 0.1 % above p°; below a few nm it is large (figure).

Application. small droplets evaporate while large ones grow (Ostwald ripening); clouds need dust or salt nuclei because vapour must be supersaturated to form tiny droplets; concave menisci in pores cause capillary condensation below p°.