Two spherical soap bubbles formed in vacuum have diameters $3.0 mm$ and $4.0 mm$. They coalesce to form a single spherical bubble. If the temperature remains unchanged, the diameter of the bubble so formed will be 

Since the bubbles are in vacuum, the pressure of air inside them are
$P_1=\frac{4\sigma }{r_1} and P_2=\frac{4\sigma }{r_2}$ where r₁ = 3.0 mm and r₂ = 4.0 mm. Since the temperature remains unchanged, we have from Boyle's law

$$\begin{gathered} P_1{V}_1+P_2{V}_2=PV or \\ \frac{4\sigma }{r_1}.\frac{4}{3}\pi r_1^3+\frac{4\sigma }{r_2}.\frac{4}{3}\pi r_2^3 \left(i\right) \end{gathered}$$

where $r$ is the radius of the single bubble formed. From (i), we get

$r^2 = r_1^2 + r_2^2 or r =\sqrt{ r_1^2 + r_2^2 }=\sqrt{{(3.0)}^2+{(4.0)}^2}=5.0 mm$