Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is `rho` and L is its latent heat of vaporization.

Options

A. `rhoL//T`
B. `sqrt(T//rhoL)`
C. `T//rhoL`
D. `2T//rhoL`

Answer

D

Explanation

If a layer of thickness dr is evaporated, then
Change in surface energy
= Change in surface area × T
`=[4pi(r+dr)^(2)-4pir^(2)]T`
`=[4pir^(2)+4pidr^(2)+8pirdr-4pir^(2)]T`
As dr is very small , so `dr^(2)` neglected.
So change in surface energy `=8pirdrT`
Energy required to evaporate layer of thickness `dr=(4pir^(2)dr)rhoL`
The processes of evaporation just starts, if change in surface energy is just sufficient to evaporate the water layer.
`therefore(4pir^(2)dr)rhoL=(8pirdr)TrArrr=(2T)/(rhoL)`