Question

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

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

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
$P=\frac{4 T}{R} ;$
where $T$ is surface tension and $R$ is radius.
Also an isothermal process obeys Boyle's law.

Hence, $P V=$ constant
Let $R$ be the radius of coalesce system then
$P_1 V_1+P_2 V_2=P V$
$ \left(\frac{4 T}{R_1}\right)\left(\frac{4}{3} \pi R_1^3\right)+\left(\frac{4 T}{R_2}\right)\left(\frac{4}{3} \pi R_2^3\right)=\left(\frac{4 T}{R}\right)\left(\frac{4}{3} \pi R^3\right) $
$\Rightarrow r_1^2+r_2^2=R$
Given, $R_1=3 cm$ and $R_2=4\, cm$
$\therefore R =\sqrt{3^2+4^2}$
$R =5\, cm$