Q2(b)(ii). Pressure difference across an ethanol droplet of radius 220 nm

Given: γ=22.39mN m⁻¹=22.39×10⁻³N m⁻¹; r=220 nm=2.20×10⁻⁷ m; T=293 K.

Which law, and why: surface tension tries to shrink the surface, so the liquid inside a curved surface is under extra pressure. The balance of forces on the droplet gives the Laplace equation.

1. Enlarging the sphere from r to r+dr adds area dσ=8π r dr, which costs work γ 8π r dr. The surface therefore pulls inwards with a force 8πγ r.
2. At equilibrium the outward force balances the inward forces: 4π r²pᵢₙ=4π r²pₒᵤₜ+8πγ r, so pᵢₙ-pₒᵤₜ=(2γ)/(r)
3. A droplet has one surface, so the factor is 2.
4. Δ p=2(22.39×10⁻³N m⁻¹)2.20×10⁻⁷ m=2.04×10⁵N m⁻²
5. In other units: 2.04×10⁵ Pa=204 kPa=2.04 bar=2.01 atm

Reading the answer: the pressure inside the droplet is about two atmospheres above the outside. Δ p grows as the radius shrinks: at 22 nm it would be ten times larger. A soap bubble, with two surfaces, would have 4γ/r (4.07×10⁵ Pa at this radius). The extra pressure raises the vapour pressure of the droplet (Kelvin equation, p=p^*e^(2γ Vₘ/rRT)): with Vₘ=58.4cm³ mol⁻¹ for ethanol, p/p^*=1.0049, a rise of 0.49 %.

Application. in a capillary the same pressure difference 2γ/r lifts the liquid to h=2γ/ρ gr. Measuring this capillary rise is a simple laboratory method for surface tension.