Sol. When the radius R of a drop of water with surface tension γ (also equal to the energy per unit surface area) decreases by R, the energy of the surface decreases. It changes by

E surface = 4 πγ [R 2 – (R – R) 2 ] ≈ – 8 π R γΔ R.

At the same time, the volume of the drop decreases by 4 π R 2 R. For this quantity of water to evaporate, energy

E evaporation = 4 π L ρ R 2 Δ R

has to be supplied. Here ρ is the density of water and L its latent heat of evaporation.

The decrease in the surface energy could provide the evaporation energy of the drop if | E surface | > E evaporation i.e. if

R < ≈ 7 × 10 –11 m.

Since this radius is of the same order of magnitude as the size of one water molecule, a drop of water with this radius cannot exist. Therefore, there is no water drop that can evaporate without absorbing heat, or losing internal energy. However, the above reasoning can be used to estimate molecular sizes using macroscopic properties.