Question

A container filled with air under pressure $P_0$ contains a soap bubble of radius $R$. The air pressure has been reduced to half isothermally and the new radius of the bubble becomes $\frac{5R}{4}$. If the surface tension of the soap water solution is $S$, $P_0$ is found to be $\frac{nS}{R}$ SI unit,where $n$ is

Answer 1

let,
$P_1$ be initial pressure inside the soap bubble
$V_1$ be initial volume of the soap bubble
$P_2$ be final pressure inside the soap bubble
$V_2$ be final volume of the soap bubble
Given,
Pressure insides the container reduces from $P_0$ to $P_0/2$ as radius of soap bubble changes from $R$ to $5R/4$
Due to excess pressure inside soap bubble
$P_1=(P_0+\frac{4S}{R})$
$P_2=(P_0+\frac{4S}{5R/4})$
As pressure is reduced isothermally,
$P_1{V}_1=P_2{V}_2$
$\Rightarrow (P_0+\frac{4S}{R})\left(\frac{4}{3}\pi R^3\right)=(\frac{P_0}{2}+\frac{4S}{\frac{5R}{4}})\left[\frac{4}{3}\pi {\left(\frac{5R}{4}\right)}^3\right]$
$\Rightarrow (P_0+\frac{4S}{R})=(\frac{P_0}{2}+\frac{16S}{5R})\left(\frac{125}{64}\right)$
$\Rightarrow P_0\times \frac{3}{128}=\frac{25S}{4R}-\frac{4S}{R}=\frac{9S}{4R}$

$\Rightarrow P_0=\frac{96S}{R}$