Question

A soap bubble in vacuum has a radius $3\text{cm}$ and another soap bubble in vacuum has a radius $4\text{cm}$. If both bubbles coalesce under isothermal condition, then the radius of the new bubble will be

• A. $6\text{cm}$
• B. $5\text{cm}$
• C. $4\text{cm}$
• D. $3\text{cm}$

Answer 1

The correct option is B $5\text{cm}$

In vacuum, the pressure inside the soap bubble is equal to the excess pressure.
$\Rightarrow P=\Delta P=\frac{4T}{r}$
Let the radius of the soap bubble be $r_1$ and $r_2$ before coalescing and $r$ be the radius of the single soap bubble after coalescing.
At isothermal condition, $T=$constant
Applying conservation of moles,
$n_1+n_2=n$
$\Rightarrow \frac{P_1{V}_1}{RT}+\frac{P_2{V}_2}{RT}=\frac{PV}{RT}$
$\Rightarrow P_1{V}_1+P_2{V}_2=PV$
$\Rightarrow \left(\frac{4T}{r_1}\right)\left(\frac{4}{3}\pi r_1^3\right)+\left(\frac{4T}{r_2}\right)\left(\frac{4}{3}\pi r_2^3\right)=\left(\frac{4T}{r}\right)\left(\frac{4}{3}\pi r^3\right)$
$\Rightarrow r_1^2+r_2^2=r^2$
$\Rightarrow r^2=\sqrt{3^2+4^2}=5\text{cm}$

Hence the radius of the new bubble is 5 cm.