The surface tension and vapour pressure of water at 20^{\circ}C is 7.28 \times 10^{-2} \frac{N}{m} and 2.33 \times 10^{3} Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20^{\circ}C?

Answers (1)

In case when the water pressure in the liquid is greater than the surface above the liquid, the drop will evaporate. We assume R to be the radius of the droplet before evaporation.

We know that pressure of vapour = pressure which becomes excess inside the drop

P=\frac{2\sigma}{R}

R = 2 \times 7.28 \times 10^{-2}/P  =\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 1000 }= \frac{14560}{233}\times 100000

Hence, R= 6.25 \times 10^{-5} m

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