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 ρ and L is its latent heat of vaporization.
A. $\sqrt{T/\rho L}$
B. $T/\rho L$
C. $2T/\rho L$
D. $\rho L/T$

Answer 1

Hint: Generally the property of liquid is surface tension. All the molecules inside the liquid experience force from all the 360 degrees direction. Hence net force will be zero on those molecules. The molecules which are on the liquid surface will be experiencing force only from below molecules as there are no molecules above them. So they have some surface energy.
Formula used:
$\mathrm{\Delta }S=T\mathrm{\Delta }A$
$\begin{array}{rl}& dQ=dmL\\ & dm=\rho dV\end{array}$

Complete answer:
Drop possesses the surface energy and when the surface energy is decreased then the decreased energy is lost in the form of heat. Since it is given that the temperature is not varied that means we can tell it as phase change. During phase change there will be no variation in temperature and the heat liberated or heat taken is called latent heat.
Latent heat is given as a product of mass and latent heat of vaporization.
Change in the latent heat will be given as $dQ=dmL$ where m is the mass and L is the latent heat of vaporization.
Now for a spherical drop
$dm=\rho dV$
$\begin{array}{rl}& \Rightarrow dm=\rho 4\pi r^{2}dr\\ & \Rightarrow dQ=dmL\\ & \Rightarrow dQ=\rho 4\pi r^{2}drL\end{array}$ where ‘r’ is the radius of the drop and $\rho$ is the density of liquid.
For the drop the decrease in surface energy is given as
$\mathrm{\Delta }S=T\mathrm{\Delta }A$
$\Rightarrow dS=T8\pi rdr$
Decrease in surface energy is nothing but the increase in liberated latent heat. So equate both.
$dS=dQ$
$\Rightarrow T8\pi rdr=\rho 4\pi r^{2}drL$
$\Rightarrow r={\displaystyle \frac{2T}{\rho L}}$

Hence option C will be the answer.

Note:
The same question can be easily solved by dimensional analysis. Dimension of latent heat is the same as work dimensions. Excess pressure in the drop is already known. If we multiply excess pressure with the volume, we get the work dimensions. Now equate this result to latent heat and from that we will get the same expression which we had got.