Question

Two soap bubbles of radii $a$ and $b$ combine under isothermal conditions to form a single bubble of radius $c$ without any leakage of air. If $P_0=$ atmospheric pressure and $T=$ surface tension of soap solution, show that $P_0=\frac{4T(a^2+b^2-c^2)}{c^3-a^3-b^3}$

Answer 1

As the process is isothermal

$P_1{V}_1+P_2{V}_2=PV(P_0-\frac{4T}{a})\frac{4}{3}\pi a^3+(P_0-\frac{4T}{b})\frac{4}{3}\pi b^3=(P_0-\frac{4T}{c})\frac{4}{3}\pi c^3{P}_0[c^3-(a^3+b^3)]-4T(a^2+b^2-c^2)=0P_0=\frac{4T(a^2+b^2-c^2)}{c^3-a^3-b^3}$