Drops and bubbles: why the small one pushes harder.

Excess pressure 2S/r for a drop or air bubble and 4S/r for a soap bubble; two soap bubbles joined by a tube—the smaller one empties into the larger.

Predict which of two connected soap bubbles grows, then set the radius where the soap bubble's excess pressure is 29 Pa (S = 0.073, r ≈ 1 cm).

Using 2S/r for a soap bubble; thinking the bigger bubble has higher pressure.

• A liquid sticks to a solid if the solid-liquid surface energy is less than the sum of the solid-air and liquid-air surface energies.
• Angle of contact θ: where liquid meets solid, draw the tangent to the liquid surface; θ is the angle from the solid surface to that tangent, measured through the liquid. It depends on the particular liquid-solid pair.
• θ decides whether a liquid spreads or forms droplets: water beads up on a lotus leaf but spreads over a clean plastic plate.
• Balancing the three interfacial tensions at the line of contact: S_la cos θ + S_sl = S_sa.
• θ is obtuse when S_sl > S_la (water on a leaf, water on wax or oil, mercury on any surface): the liquid's molecules attract one another strongly and the solid's weakly, so the liquid does not wet the solid.
• θ is acute when S_sl < S_la (water on glass or plastic, kerosene on almost anything): strong liquid-solid attraction lowers S_sl, cos θ grows and the liquid spreads.
• Soaps, detergents and dyeing substances are wetting agents that make θ small so they penetrate well. Waterproofing agents make θ large between water and fibres.
• For a given volume, the sphere has the least surface area, so free drops and bubbles are spherical when gravity and air resistance can be ignored.
• Pressure inside a drop exceeds that outside. Growing a drop of radius r by Δr costs surface energy 8πrΔr S_la and gains work (P_i − P_o)4πr²Δr, so P_i − P_o = 2S_la/r.
• Across any curved liquid-gas surface the concave side is at the higher pressure. An air bubble (cavity) inside a liquid has one surface: excess pressure 2S/r.
• A soap bubble in air has two surfaces, inner and outer: P_i − P_o = 4S_la/r. That is why blowing a soap bubble needs a little extra pressure, but not too much.
• Capillary of diameter 2.00 mm dipped 8.00 cm into water (S = 7.30 × 10⁻² N m⁻¹, g = 9.80 m s⁻², 1 atm = 1.01 × 10⁵ Pa), hemispherical bubble of radius 1.00 mm at its end: P_o = 1.01 × 10⁵ + 0.08 × 1000 × 9.8 = 1.01784 × 10⁵ Pa; excess pressure 2S/r = 146 Pa; P_i = 1.02 × 10⁵ Pa to three significant figures.

Teaching this idea? Put the simulation on your own page or LMS with this code.