Question

Assume that a drop of liquid evaporates due to decrease in its surface energy so that its temperature remains unchanged. What should be the minimum radius of drop for this to be possible?

Surface tension = T; density of liquid = ρ , Latent heat of vaporation = L.

(1) ρL T

(2) \(\sqrt{\frac{T}{pT}}\)

(3) \(\frac{T}{pL}\)

(4) \(\frac{2T}{pL}\)

Answer 1

 (4) \(\frac{2T}{pL}\)

The liquid drop evaporates if energy released due to decrease in surface area equals latent heat of vaporisation. Consider a drop of radius r. Let radius decreases by dr.

Surface energy released = T(ds)

The change in mass dm of drop due to change dr in radius

The latent heat of vaporisation = (dm)L

From Eqns. (1) and (2) we have