Question

Pressure inside two soap bubbles is $1.01$ and $1.02$ atmosphere. What is the ratio between their volume?

Answer 1

Step 1. Determine the ratio between their radius.

Excess Pressure in the bubble one is $△P_1$ $=1.01\mathrm{P}-\mathrm{P}=0.01\mathrm{P}$

Similarly, excess pressure in the bubble two is $△P_2$ $=1.02\mathrm{P}-\mathrm{P}=0.02\mathrm{P}$

The excess pressure of the bubble one of radius $R_1$ with surface tension $T$ is given as, $△P_1=\frac{4T}{R_1}$

The excess the pressure for the bubble two of radius $R_2$ with surface tension $T$ is given as, $△P_2=\frac{4T}{R_2}$
$\begin{array}{rcl}\frac{R_2}{R_1}& =& \frac{△P_1}{△P_2}\\ & =& \frac{0.01}{0.02}\\ & =& \frac{1}{2}\end{array}$

Step 2. Determine the ratio between their volume.

The volume of soap bubble $V$ is calculated by $V=\frac{4}{3}{\mathrm{\pi R}}^3$ where $R$ represents the radius of soap bubble.

$\begin{array}{rcl}\frac{V_1}{V_2}& =& \frac{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}_1^3}{\frac{4}{3}{\mathrm{\pi R}}_2^3}\\ & =& \frac{R_1^3}{R_2^3}\\ & =& {\left(\frac{1}{{\displaystyle \frac{1}{2}}}\right)}^3\\ & =& 8\\ V_1:V_2& =& 8:1\end{array}$
Hence, the ratio between their volumes is $8:1$