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

Hint To find the value we should keep in mind the relative comparison of the radius of the two surfaces. Apart from that the basic mechanical approach to find work done is implemented.

Complete step-by-step solution:So the pressure inside the liquid drop be ${P_i}$ and the pressure outside the liquid drop be ${P_o}$
Therefore the excess pressure inside would be $ = {P_i} - {P_o}$
Let $T$ be the surface tension of the liquid and the increase in drop radius due to excess pressure $ = \Delta r$
The work done by the excess pressure is given by
$dW$= Force x Displacement
= (Excess pressure x Area) x (increase in radius)
$ = [({P_i} - {P_o}) \times 4\pi {r^2}] \times \Delta r$
Suppose the initial surface area of the liquid drop be ${A_1} = 4\pi {r^2}$
And the final surface area of the liquid drop be
$
  {A_2} = 4\pi {(r + \Delta r)^2} \\
 \Rightarrow {A_2} = 4\pi ({r^2} + 2r\Delta r + \Delta {r^2}) \\
 \Rightarrow {A_2} = 4\pi {r^2} + 8\pi r\Delta r + 4\pi \Delta {r^2} \\
 $
As $\Delta r$ is very small, which implies that$\Delta {r^2}$is almost negligible, i.e. $4\pi \Delta {r^2} \approx 0$
Therefore, ${A_2} = 4\pi {r^2} + 8\pi r\Delta r$
Therefore, increase in the surface area of the drop
 $
  dA = {A_2} - {A_1} \\
 \Rightarrow dA = 4\pi {r^2} + 8\pi r\Delta r - 4\pi {r^2} \\
 \Rightarrow dA = 8\pi r\Delta r \\
 $
Work done to increase the surface area:
$dW = T \times dA$
where $T$ is the surface energy
$ \Rightarrow dW = T \times 8\pi r\Delta r$
Comparing, we get
$ \Rightarrow {P_i} - {P_o} = 2T/r$

Note: While calculating the value the nature of the liquid plays a vital role in determining the nature of the surface. In case of soap or film multiple surfaces should be kept in mind.Apart from that the difference should only be neglected in case of extensive gap between values.