Two spherical soap bubbles of radii r₁ and r₂ in vacuum coalesce under isotherm al conditions. The resulting bubble has a radius R such that

 The bubbles coalesce in vacuum and hence there is no change in temperature, So, the surface energy does not change. This means that surface area remaining unchanged. Hence
$$\begin{gathered} 4{\mathrm{\pi r}}_1^2+4{\mathrm{\pi r}}_2^2=4{\mathrm{\pi R}}^2 \\ \mathrm{or} \mathrm{R}={\left({\mathrm{r}}_1^2+{\mathrm{r}}_2^2\right)}^{1/2} \end{gathered}$$