Question

Two spherical soap bubbles of radii $a$ and $b$ in vacuum coaleasce under isothermal conditions. The resulting bubbles has a radius given by:

• A. $\frac{(a+b)}{2}$
• B. $\frac{ab}{a+b}$
• C. $\sqrt{a^2+b^2}$
• D. $a+b$

Answer 1

Answer: (C)

$\sqrt{a^2+b^2}$

Two bubbles coaleasce to form a bigger bubble at isothermal conditions i.e $T=constant$

Excess pressure inside the soap bubble $P=\frac{4T}{r}$ where $T$ is the surface tension of soap bubble

From $PV=nRT$ we get $n=\frac{PV}{RT}$

As no mole of soap bubble is lost, thus $n=n_1+n_2$

$\therefore$ $\frac{PV}{RT}=\frac{P_1{V}_1}{RT}+\frac{P_2{V}_2}{RT}$

$⟹$ $PV=P_1{V}_1+P_2{V}_2$ $(\because T=constant)$

$\frac{4T}{r}\times \frac{4}{3}\pi r^3=$ $\frac{4T}{a}\times \frac{4}{3}\pi a^3+$ $\frac{4T}{b}\times \frac{4}{3}\pi b^3$

$⟹r^2=a^2+b^2$ OR $r=\sqrt{a^2+b^2}$