Fluid Mechanics: Surface Tension

Question

Assume that a drop of liquid evaporates by decrease 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,

$T/\rho L$

$2T/\rho L$

$\rho L/T$

$T/\rho L$

The solution given for it is like,

Heat of vaporization = Surface energy released $Ldm=TdA$

$4L\rho \pi r2dr=T8\pi rdr$

$r=2T/\rho L$

My question is should not heat of vaporization be equal to $L\times \textrm{density}\times4/3~P~R^3$ .Why in the above solution in calculating heat of vaporization change is mass is taken instead of total mass.

Answer 1

But $(4/3)L\rho \pi r^3$ is exactly what the answer is giving. If the drop radius changed from $r\to r+dr$ $dr$ then $$ d\left( \frac 43 \pi r^3 L\rho\right)= 4\pi r^2 L\rho dr, $$ which is what you have in your penultimate line (if I assume that by "r2" you mean $r^2$).