Excess Pressure In A Liquid Drop

Question

A spherical raindrop of radius 1 mm forms in air, and the surface tension of water is $0.072\,	ext{N/m}$. What is the excess pressure inside this drop compared with the surrounding atmosphere?

RankGuru · Accepted Answer

Surface tension makes a liquid surface behave like a stretched elastic membrane, and for a spherical liquid drop with a single curved surface the excess internal pressure is $\Delta P = \frac{2S}{r}$, where $S$ is the surface tension and $r$ the radius. The factor of two arises because the curved surface tends to contract, squeezing the enclosed liquid until the pressure difference balances the surface forces. Substituting $S = 0.072\,	ext{N/m}$ and $r = 1	imes10^{-3}\,	ext{m}$ gives $\Delta P = \frac{2	imes0.072}{10^{-3}} = 144\,	ext{Pa}$. The value 72 Pa drops the factor of two and treats the drop like a flat film. The value 288 Pa uses the soap-bubble formula $4S/r$, which applies only when there are two liquid–air surfaces. The value 36 Pa halves the result without justification. This matches the NCERT derivation distinguishing a single-surface drop from a two-surface bubble. As a check, the excess pressure rises as the drop gets smaller, so a millimetre drop showing only a modest 144 Pa above atmospheric is physically reasonable.

Excess Pressure In A Liquid Drop

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