Question

Derive an expression of excess pressure inside a liquid drop .

Answer 1

Consider a spherical drop of radius R and surface tension T

Let p_i be the pressure inside the drop and p₀ be the pressure outside it. As the drop is spherical in shape, the pressure, p_i, inside the drop is greater than p₀, the pressure outside.

Therefore, the excess pressure inside the drop is  p_i - p₀.

Let the radius of the drop increase from r to r + Δr, where Δr is very small,

so that the pressure inside the drop remains almost constant.

Let the initial surface area of the drop be

A₁ = 4πr²,

and the final surface area of the drop be

A₂ = 4π (r+Δr)²

∴ A₂ = 4π (r² + 2rΔr + Δr²)

∴ A₂ = 4π r² + 8πrΔr + 4πΔr²

As Δr is very small, Δr² can be neglected,

∴ A₂ = 4πr² + 8πrΔr

Thus, increase in the surface area of the drop is

dA = A₂ – A₁ = 4πr² + 8πrΔr - 4πr² = 8πrΔr --- (1)

Work done in increasing the surface area by dA is stored as excess surface energy.

∴ dW = TdA= T (8πrΔr) --- (2)

This work done is also equal to the product of the force F which causes increase in the area of the bubble and the displacement Δr which is the increase in the radius of the bubble.

∴ dW = FΔr --- (3)

The excess force is given by,

(Excess pressure) × (Surface area)

∴  F = (p_i – p₀) 4πr² --- (4)

Equating Eq. (2) and Eq. (3), we get,

T(8prΔr) = (p_i – p₀) 4πr² Δr

∴ (p_i – p₀) = 2T/r --- (5)

This equation gives the excess pressure inside a drop. This is called Laplace’s law of a spherical membrane or Young Laplace Equation in spherical form