Two soap bubbles of radii $r_1$ and $r_2$ in vacuum coalesce under isothermal conditions. The resulting…
Two soap bubbles of radii $r_1$ and $r_2$ in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to
$\sqrt{r_1^2+r_2^2}$
$\frac{r_1+r_2}{2}$
$r_1+r_2$
$\frac{r_1 r_2}{r_1+r_2}$
Solution
In Iso-thermal process: $p V=\mathrm{C}$. The pressure inside bubble is: $p \propto 1 / r$ and volume is: $V \propto r^3$.
$\therefore p V \propto r^2$
And $n=\frac{p V}{R T}$, therefore number of moles of gas inside the bubble is proportional to the square of its radius
$\therefore n \propto r^2$
On combining bubbles, the total number of moles is conserved: Therefore, $n_1+n_1=N$
$\therefore r_1^2+r_2^2=R^2$
The above result can also be obtained by total surface energy conservation:
$2 \times\left(4 \pi R^2\right) \sigma=2 \times\left(4 \pi r_1^2\right) \sigma+2 \times\left(4 \pi r_2^2\right) \sigma$
or $R=\sqrt{r_1^2+r_2^2}$