Q: 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)$\large \frac{\rho L}{T}$

(b) $\large \sqrt{\frac{T}{\rho L}}$

(c) $\large \frac{T}{\rho L}$

(d) $\large \frac{2 T}{\rho L}$

Ans: (d)

Sol: Decrease in Surface Energy = Heat required in vaporization

$\large T(dS) = (dm)L$

$\large 4\pi(r+dr)^2 – 4\pi r^2 = (4\pi r^2 dr)\rho L $

$\large T[ 4\pi(r^2 + 2 r dr + dr^2 ) – 4\pi r^2 ] = (4\pi r^2 dr)\rho L $

Neglecting dr^{2} being very small

$\large 8\pi r dr \times T = (4\pi r^2 dr)\rho L $

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