Physics - Fluid Question with Solution | TestHub

Let $P_1$ and $P_2$ be the pressures inside the two soap bubbles, and $V_1$ and $V_2$ be their respective volumes.

Since the bubbles are in a vacuum, the external pressure is zero. The pressure inside a soap bubble of radius $r$ is given by $P=\frac{4T}{r}$, where $T$ is the surface tension of the soap solution.

The volume of a bubble of radius $r$ is $V=\frac{4}{3}\pi r^3$.

Assuming the gas inside behaves ideally, the number of moles of gas is $n=\frac{PV}{R_g{T}_0}$, where $R_g$ is the universal gas constant and $T_0$ is the absolute temperature.

When the two bubbles coalesce under isothermal conditions, the total number of moles of gas is conserved:

$n_1+n_2=n$

$\frac{P_1{V}_1}{R_g{T}_0}+\frac{P_2{V}_2}{R_g{T}_0}=\frac{PV}{R_g{T}_0}$

$P_1{V}_1+P_2{V}_2=PV$

Substituting the expressions for pressure and volume:

$\left(\frac{4T}{r_1}\right)\left(\frac{4}{3}\pi r_1^3\right)+\left(\frac{4T}{r_2}\right)\left(\frac{4}{3}\pi r_2^3\right)=\left(\frac{4T}{R}\right)\left(\frac{4}{3}\pi R^3\right)$

Simplifying the equation, we get:

$r_1^2+r_2^2=R^2$

$R={\left(r_1^2+r_2^2\right)}^{\frac{1}{2}}$

Answer: ${\left(r_1^2+r_2^2\right)}^{\frac{1}{2}}$