Derive Laplace’s law for spherical membrane of bubble due to surface tension.

Answer: For a soap bubble (two surfaces) the excess pressure inside is \(P_i - P_o = \dfrac{4T}{R}\); for a single spherical surface such as a liquid drop it is \(\dfrac{2T}{R}\).

Consider a spherical membrane (soap bubble) of radius \(R\) and surface tension \(T\). Let \(P_i\) be the pressure inside and \(P_o\) outside, \(P_i > P_o\). Because of the excess pressure the membrane tends to expand outwards, while surface tension tends to contract it.

Let the radius increase slightly from \(R\) to \(R + dR\) under the excess pressure.

Work done by the excess pressure: force \(= (P_i - P_o)\,4\pi R^2\); displacement \(= dR\):

Increase in surface energy: a bubble has two free surfaces (inner and outer), so the increase in area is

(neglecting \(dR^2\)). Increase in surface energy \(= T\,dA = 16\pi R T\,dR\).

The work done by the excess pressure is stored as surface energy:

This is Laplace’s law for a spherical membrane (bubble). For a liquid drop or an air bubble inside a liquid, there is only one surface, \(dA = 8\pi R\,dR\), and \(P_i - P_o = \dfrac{2T}{R}\).

The excess pressure is inversely proportional to the radius, so a smaller bubble has a larger excess pressure.