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. `rho L//S`
B. `sqrt(S//rho L)`
C. `S//rho L`
D. `2S//rho L`

Answer

D

Explanation

Conside a liquid drop of radius r. When a layer of lilquid drop of thickness dr is evaporated and temperature remains unchanged, then change in surface energy = change in surface area x surface tensiion `=d(4 pi r^(2))xxS=(8pi r dr)S`
Energy required to evaporate the liquid drop layer of thickness dr `=(4pi r^(2))xxS=(4pi r^(2) dr)rho L`
The process of evaporation only start if change in surface energy is just sufficient to evaporate the layer of liquid drop. Therefore,
`(4pi r^(2) dr)rho L=(8pi r dr)S` or `r=(2S)/(rho L)`