Derive an expression for excess pressure inside a drop of liquid.

#### Solution

Consider a spherical drop as shown in the figure. Let p_{i} be the pressure inside the drop and p_{0} be the pressure out side 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 + ∆r, where ∆r 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+∆r)^{2}.

∴ A_{2} = 4π(r^{2} + 2r∆r + ∆r^{2})

∴ A_{2} = 4πr^{2} + 8πr∆r + 4π∆r^{2} (As ∆r is very small, ∆r^{2} can be neglected)

∴ A_{2} = 4πr^{2} + 8πr∆r

Thus, increase in the surface area of the drop is

`dA = A_2 – A_1 = 8πrDeltar`

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

`therefore dW = TdA= T (8πrDeltar)` ... (1)

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

`therefore dW = F∆r` ... (2)

The excess force is given by,

(Excess pressure) ×(Surface area)

∴ F = (p_{i} – p_{0}) 4πr^{2} ... (3)

Equating Eq. (1), and Eq. (2), we get,

`T(8πrDeltar) = FDeltar`

`therefore T(8πrDeltar) = (p_i – p_0) 4πr^2Deltar` ... (using Eq. (3))

`therefore (p_i – p_0)=(2T)/r`

This equation gives the excess pressure inside a drop of liquid.