Question

Two spherical soap bubble coalesce. If $V$ is the consequent change in volume of the contained air and $S$ the change in total surface area, show that
$3PV+4ST=0$
where $T$ is the surface tension of soap bubble and $P$ is atmospheric pressure.

Answer 1

Let $P_i$ and $R_i$ be the inside pressure and radius of the $ith$ soap bubble respectively.

$\therefore$ $P_1=P+\frac{4T}{R_1}$, $P_2=P+\frac{4T}{R_2}$ and $P_3=P+\frac{4T}{R_3}$

Also $P_1{V}_1+P_2{V}_2=P_3{V}_3$

$\therefore$ $(P+\frac{4T}{R_1})\frac{4\pi }{3}{R}_1^3$ + $(P+\frac{4T}{R_2})\frac{4\pi }{3}{R}_2^3$ = $(P+\frac{4T}{R_3})\frac{4\pi }{3}{R}_3^3$

OR $P(\frac{4\pi }{3}{R}_1^3+\frac{4\pi }{3}{R}_2^3-\frac{4\pi }{3}{R}_3^2)+\frac{4T}{3}(4\pi R_1^2+4\pi R_2^2-4\pi R_3^2)=0$

OR $P(V_1+V_2-V_3)+\frac{4T}{3}(S_1+S_2-S_3)=0$

OR $PV+\frac{4T}{3}S=0$ $⟹3PV+4ST=0$