Question

Derive an expression for the excess pressure inside a soap bubble. 

OR 

Derive Laplace’s law for spherical membrane of a 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 Po and that inside be p. A soap bubble in air is like a spherical shell and has two gas-liquid interfaces. Hence, the surface area of the bubble is

A = 8πR² ………. (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 = 16πR ∙ dR and 

dW = T ∙ dA = 16πTRdR ………….. (2)

We assume that dR is so small that the pressure inside remains the same, equal to p. All parts of the surface of the bubble experiences 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) × 4πR² ∙ dR ………. (3)

From Eqs. (2) and (3),

(p – p₀) × 4πR² ∙ dR = 16πTRdR

∴ p – p₀ = \(\frac{4T}R\)…………… (4)

which is the required expression.

[Note : The excess pressure inside a drop or bubble is inversely proportional to its radius : the smaller the bubble radius, the greater the pressure difference across its wall.]