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.

Step 1: Express the surface energy of the drop.

The surface energy $E_s$ of a spherical drop of radius $r$ with surface tension $T$ is given by the product of surface tension and surface area.

$$E_s = 4\pi r^2 T$$

Step 2: Relate the change in surface energy to a change in radius.

Let the radius of the drop decrease by $dr$ during evaporation.

$$dE_s = 8\pi r T \, dr$$

Step 3: Relate the mass of the evaporated liquid to the change in radius.

The mass $dm$ of the evaporated liquid is the product of density $\rho$ and the change in volume $dV$.

$$dm = 4\pi r^2 \rho \, dr$$

Step 4: Set up the energy conservation condition.

The released surface energy must be at least equal to the latent heat needed for vaporization.

$$dE_s \ge L \, dm$$

Step 5: Solve for the radius $r$.

Substitute the expressions for $dE_s$ and $dm$ into the inequality.

$$8\pi r T \, dr \ge 4\pi r^2 \rho L \, dr$$

Simplify the inequality to find the condition for $r$.

$$2T \ge r \rho L$$ $$r \le \frac{2T}{\rho L}$$

The critical radius for this process is the limiting value.

$$r = \frac{2T}{\rho L}$$

Answer: The minimum radius of the drop is $\dfrac{2T}{\rho L}$.