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

(N/A) Consider a liquid drop of radius $r$ with internal pressure $P_i$ and external pressure $P_o$. The excess pressure is $\Delta P = P_i - P_o$.
$1$. For a liquid drop:
The surface area is $A = 4\pi r^2$. If the radius increases by $\Delta r$,the change in area is $\Delta A = 8\pi r \Delta r$. The work done against surface tension $S$ is $W = S \Delta A = S(8\pi r \Delta r)$.
This work is done by the excess pressure: $W = (P_i - P_o) \Delta V = (P_i - P_o) (4\pi r^2 \Delta r)$.
Equating the two: $(P_i - P_o) (4\pi r^2 \Delta r) = S(8\pi r \Delta r) \implies P_i - P_o = \frac{2S}{r}$.
$2$. For a soap bubble:
$A$ soap bubble has two free surfaces (inner and outer). Thus,the work done is $W = 2 \times S \Delta A = 2S(8\pi r \Delta r)$.
Equating the work: $(P_i - P_o) (4\pi r^2 \Delta r) = 16\pi r S \Delta r \implies P_i - P_o = \frac{4S}{r}$.