Two soap bubbles of radii \(r_1\) and \(r_2\) in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to

In Iso-thermal process: \(p V=\mathrm{C}\). The pressure inside bubble is: \(p \propto 1 / r\) and volume is: \(V \propto r^3\).
\(\therefore p V \propto r^2\)
And \(n=\frac{p V}{R T}\), therefore number of moles of gas inside the bubble is proportional to the square of its radius
\(\therefore n \propto r^2\)
On combining bubbles, the total number of moles is conserved: Therefore, \(n_1+n_1=N\)
\(\therefore r_1^2+r_2^2=R^2\)
The above result can also be obtained by total surface energy conservation:
\(2 \times\left(4 \pi R^2\right) \sigma=2 \times\left(4 \pi r_1^2\right) \sigma+2 \times\left(4 \pi r_2^2\right) \sigma\)
or \(R=\sqrt{r_1^2+r_2^2}\)