Question

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

• A. $7cm$
• B. $5cm$
• C. $4.5cm$
• D. $2.3cm$

Answer 1

Answer: (B)

An isothermal process obeys Boyle's law.
Since process is isothermal the total pressure of air inside the bubble is same as excess of pressure given by
$P=\frac{4T}{R};$
where $T$ is surface tension and $R$ is radius.
Also an isothermal process obeys Boyle's law.

Hence, $PV=$ constant
Let $R$ be the radius of coalesce system then
$P_1{V}_1+P_2{V}_2=PV$
$\left(\frac{4T}{R_1}\right)\left(\frac{4}{3}\pi R_1^3\right)+\left(\frac{4T}{R_2}\right)\left(\frac{4}{3}\pi R_2^3\right)=\left(\frac{4T}{R}\right)\left(\frac{4}{3}\pi R^3\right)$
$\Rightarrow r_1^2+r_2^2=R$
Given, $R_1=3cm$ and $R_2=4cm$
$\therefore R=\sqrt{3^2+4^2}$
$R=5cm$