Question

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

Answer 1

Consider a small, spherical, thin-filmed soap bubble with a radius R. Let the pressure outside the drop be p₀ and that inside be p. A soap bubble in the air is like a spherical shell and has two gas-liquid interfaces. Hence, the surface area of the bubble is:

A = 8 π R²    ...(1)

Hence, with a hypothetical increase in radius by an infinitesimal amount dR, the differential increase in surface area and surface energy would be:

dA = 16 π R . dR and

dW = T . dA 

= 16 π TR dR    ...(2)

We assume that dR is so small that the pressure inside remains the same, equal to the top. All parts of the surface of the bubble experience an outward force per unit area equal to p − p₀. Therefore, the work done by this outward pressure-developed force against the surface tension force during the increase in radius dR is:

dW = (excess pressure × surface area) . dR

= (p − p₀) × 4 π R² . dR    ...(3)

From Eqs. (2) and (3),

(p − p₀) × 4 π R² . dR = 16 π TR dR

∴ p − p₀ = `(4 T)/R`    ...(4)