Consider a spherical drop of radius R and surface tension T
Let pi be the pressure inside the drop and p0 be the pressure outside it. As the drop is spherical in shape, the pressure, pi, inside the drop is greater than p0, the pressure outside.
Therefore, the excess pressure inside the drop is pi - 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 = (pi – p0) 4πr2 --- (4)
Equating Eq. (2) and Eq. (3), we get,
T(8prΔr) = (pi – p0) 4πr2 Δr
∴ (pi – 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