Question

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

Answer 1

Consider a spherical drop as shown below. Let p_i be the pressure inside the drop and p_o be the pressure outside it. As the drop is spherical in shape, the pressure, p_i, inside the drop is greater than p_o, the pressure outside. Therefore, the excess pressure inside the drop is p_i − p_o.

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

The final surface area of the drop is: 

A₂ = 4π(r + Δr)²

= 4π(r² + 2rΔr + Δr²)

= 4πr² + 8πrΔr + 4πrΔr²

As Δr is very small, πrΔr² can be neglected.

∴ A₂ = 4πr² + 8πrΔr

Thus, the increase in surface area of the drop is:

dA = A₂ − A₁

= 8πrΔr

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

∴ dW = T . dA

= T(8πrΔr)    ...(i)

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

∴ dW = F . Δr

The excess force is given by, (Excess pressure) × (Surface area)

dW = (p_i − p_o) × 4πr² × Δr    ...(ii)

From equations (i) and (ii), we get,

(p_i − p_o) × 4πr² × Δr = T(8πrΔr)

∴ (p_i − p_o) = `(2 r)/r`

This equation gives the excess pressure inside a drop. This is called Laplace’s law of a spherical membrane.