Derive the formula for excess of pressure (pressure difference) inside the drop and bubble.

In figure $(a)$ a gas bubble in liquid is shown and in figure $(b)$ this bubble is shown spherical and in figure $(c)$ a soap bubble in air is shown.

Suppose, bubble formed by this way, radius of bubble is $r$ and inside pressure is $\mathrm{P}_{i}$ and outside pressure is $\mathrm{P}_{0}$.

Let us obtain $\mathrm{P}_{i}-\mathrm{P}_{0}$ for bubble in liquid or drop of liquid.

The pressure inside the drop is $\mathrm{p}_{i}$ and outside is $\mathrm{P}_{\mathrm{o}} . \mathrm{P}_{i}$ is slightly higher than $\mathrm{P}_{\mathrm{o}}$.

When bubble has radius $r$, its area of surface $\mathrm{A}_{1}=4 \pi r^{2} \ldots .(1)$

If radius of bubble increase by $\Delta r$, new radius becomes $r+\Delta r$.

Area of this surface $\mathrm{A}_{2}=4 \pi(r+\Delta r)^{2}$

$\mathrm{A}_{2}=4 \pi\left(r^{2}+2 r \cdot \Delta r+\Delta r^{2}\right)$

$\Delta r^{2}$ is very small compared to others terms

$\therefore \mathrm{A}_{2}=4 \pi\left(r^{2}+2 r \Delta r\right) \quad \ldots .(2)$

The change in area of surface of drop

$\Delta \mathrm{A}=\mathrm{A}_{2}-\mathrm{A}_{1}$

$\Delta \mathrm{A}=4 \pi\left(r^{2}+2 r \Delta r\right)-4 \pi r^{2}$

$=4 \pi\left(r^{2}+2 r \Delta r-2 r^{2}\right)$

$=4 \pi(2 r \Delta r)$

$\Delta r=8 \pi r \quad \Delta r \quad \ldots .(3)$

The work done for increase area $\Delta \mathrm{A}$ of free surface of drop.

$\mathrm{W}=\mathrm{S}_{l a}(\Delta \mathrm{A})$

$\mathrm{W}=\mathrm{S}_{l a}(8 \pi r \cdot \Delta r)$