The surface tension and vapour pressure of water at $20^{°}$ $\mathrm{C}$ is $7.28 \times {10^{ - 2}}\,{\rm{N/m}}$ and $2.33 \times {10^3}\,{{\rm{P}}_{\rm{a}}}$ respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^{°}$ $\mathrm{C}$ ?
The surface tension and vapour pressure of water at $20^{°}$ $\mathrm{C}$ is $7.28 \times {10^{ - 2}}\,{\rm{N/m}}$ and $2.33 \times {10^3}\,{{\rm{P}}_{\rm{a}}}$ respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^{°}$ $\mathrm{C}$ ?
Surface tension of water $\mathrm{S}=7.28 \times 10^{-2} \mathrm{~N} / \mathrm{m}$
Vapour pressure $\mathrm{P}=2.33 \times 10^{3} \mathrm{P}_{a}$
The drop will evaporate if the water pressure on liquid is greater than vapour pressure above the surface of liquid.
Vapour pressure without evaporation = excess pressure in a drop
$\mathrm{P}=\mathrm{P}_{i}-\mathrm{P}_{0}$
$\mathrm{P}=\frac{2 \mathrm{~S}}{\mathrm{R}} \quad\left(\because\right.$ For drop $\left.\mathrm{P}_{i}-\mathrm{P}_{0}=\frac{2 \rho}{\mathrm{R}}\right)$
$\therefore \mathrm{R}=\frac{2 \mathrm{~S}}{\mathrm{P}}$
$\therefore \mathrm{R}=\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^{3}}$
$\mathrm{R}=6.25 \times 10^{-5} \mathrm{~m}$
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