The radii of two soap bubbles are R₁ and R₂ respectively. The ratio of masses of air in them will be

To find the ratio of the masses of air inside two soap bubbles of radii $R_1$ and $R_2$, we can use the ideal gas equation and the concept of excess pressure inside a soap bubble.

According to the ideal gas equation:
$PV=nRT$
where:
- $P$ is the absolute pressure of the air inside the bubble,
- $V$ is the volume of the bubble,
- $n$ is the number of moles of air, which is proportional to the mass $m$ ($n=m/M$, where $M$ is the molar mass of air),
- $R$ is the universal gas constant, and
- $T$ is the absolute temperature.

Since the temperature $T$ and the molar mass $M$ of air are the same for both bubbles, the mass $m$ of air inside a bubble is directly proportional to the product of its absolute pressure and volume:
$m\propto PV$

For a soap bubble of radius $R$, surface tension $T$, and atmospheric pressure $P$ outside:
The excess pressure inside the soap bubble is given by:
$\Delta P=\frac{4T}{R}$
Thus, the absolute pressure $P_{\text{in}}$ inside the bubble is:
$P_{\text{in}}=P+\frac{4T}{R}$

The volume of a spherical bubble of radius $R$ is:
$V=\frac{4}{3}\pi R^3$

Substituting the absolute pressure and volume into the relation for mass, we get:
$m\propto (P+\frac{4T}{R})·\frac{4}{3}\pi R^3$
Since $\frac{4}{3}\pi$ is a constant, we have:
$m\propto (P+\frac{4T}{R})R^3$

Therefore, the ratio of the masses of air in the two bubbles is:
$\frac{m_1}{m_2}=\frac{(P+\frac{4T}{R_1}){R_1}^3}{(P+\frac{4T}{R_2}){R_2}^3}$

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