Excess Pressure In A Soap Bubble

Question

A soap bubble of radius 2 cm is blown using a soap solution whose surface tension is $0.03\,	ext{N/m}$. What is the excess pressure inside this bubble compared with the surrounding air?

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A soap bubble differs from a liquid drop because it has two thin liquid surfaces in contact with air, an outer and an inner one, each contributing to the inward pull of surface tension. The excess pressure inside is therefore $\Delta P = \frac{4S}{r}$, twice the value for a single-surface drop. Substituting $S = 0.03\,	ext{N/m}$ and $r = 2	imes10^{-2}\,	ext{m}$ gives $\Delta P = \frac{4	imes0.03}{2	imes10^{-2}} = \frac{0.12}{0.02} = 6\,	ext{Pa}$. The value 3 Pa uses the drop formula $2S/r$ and overlooks the second surface. The value 1.5 Pa applies only $S/r$. The value 12 Pa wrongly doubles the radius effect or uses $8S/r$. This distinction between a two-surface bubble and a one-surface drop is emphasised in the NCERT treatment of surface tension, where a real soap film is so thin that its inner and outer radii are taken to be essentially equal. The same reasoning explains why two bubbles of different sizes, when joined, push air from the smaller into the larger: the smaller bubble holds the greater internal pressure. As a plausibility check, the excess pressure increases as the bubble shrinks, so a centimetre-scale bubble showing only a few pascals of excess pressure is entirely reasonable, and it exceeds what an identical-radius single drop would show by exactly a factor of two.

Excess Pressure In A Soap Bubble

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