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(A) The droplet will be in equilibrium when the excess pressure due to surface tension is equal to the vapour pressure $P$. The excess pressure inside

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$64$

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(A) The droplet will be in equilibrium when the excess pressure due to surface tension is equal to the vapour pressure $P$. The excess pressure inside a spherical droplet is given by $\Delta P = \frac{2S}{R}$,where $S$ is the surface tension and $R$ is the radius of the droplet. For the droplet to exist without evaporating,the pressure inside must balance the vapour pressure $P$. Thus,$P = \frac{2S}{R}$. Rearranging for $R$,we get $R = \frac{2S}{P}$. Substituting the given values: $R = \frac{2 	imes (8 	imes 10^{-2} 	ext{ Nm}^{-1})}{2.5 	imes 10^3 	ext{ Pa}}$. $R = \frac{16 	imes 10^{-2}}{2.5 	imes 10^3} = 6.4 	imes 10^{-5} 	ext{ m}$. Converting to micrometers: $R = 64 	imes 10^{-6} 	ext{ m} = 64 \mu m$.

The surface tension and vapour pressure of a liquid at $25^{\circ} C$ are $8 \times 10^{-2} \text{ Nm}^{-1}$ and $2.5 \times 10^3 \text{ Pa}$ respectively. The radius of the smallest spherical water droplet which can form without evaporating at $25^{\circ} C$ is (in $\mu m$)

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