Consider a spherical drop of radius R and surface tension T
Let p_{i} be the pressure inside the drop and p_{0} be the pressure outside it. As the drop is spherical in shape, the pressure, p_{i}, inside the drop is greater than p_{0}, the pressure outside.
Therefore, the excess pressure inside the drop is p_{i }- p_{0}.
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
A_{1} = 4πr^{2},
and the final surface area of the drop be
A_{2} = 4π (r+Δr)^{2}
∴ A_{2} = 4π (r^{2} + 2rΔr + Δr^{2})
∴ A_{2} = 4π r^{2} + 8πrΔr + 4πΔr^{2}
As Δr is very small, Δr^{2} can be neglected,
∴ A_{2} = 4πr^{2} + 8πrΔr
Thus, increase in the surface area of the drop is
dA = A_{2} – A_{1} = 4πr^{2} + 8πrΔr - 4πr^{2 }= 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} – p_{0}) 4πr^{2} --- (4)
Equating Eq. (2) and Eq. (3), we get,
T(8prΔr) = (p_{i} – p_{0}) 4πr^{2 }Δr
∴ (p_{i} – p_{0}) = 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