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.

• A. $\rho L/T$
• B. $\sqrt{T/\rho L}$
• C. $T/\rho L$
• D. $2T/\rho L$

Answer 1

The correct option is D $2T/\rho L$

surface Tension $=T$ Radius $=R$

density $=\rho$

Latent heat $=L$

Let for $R,$ small decrease on $R$ is $△R$

$\therefore$ change in surface energy

$\begin{array}{cc}\Delta E=& [4\pi R^2-4\pi (R-\Delta R{)}^2]\times T\\ & =4\pi (R-R+\Delta R)(R+R-4R)\times T\\ & =4\pi \cdot \Delta R\cdot 2R\cdot T\therefore △R\approx 0\\ & =8\pi R\cdot \Delta R\cdot T\end{array}$

Change in mass

$\begin{array}{cc}\Delta m& =9/3\pi (R^3-(R-\Delta R{)}^3)\times \rho \\ & =\frac{4}{3}\pi (R-R+\Delta R)(R^2+(R-\Delta R{)}^2+R(R-\Delta R))\rho \\ & =\frac{4}{3}\pi \cdot \Delta R(R^2+R^2+R^2-3R\Delta R+(\Delta R{)}^2)\rho \\ & =4\cdot \pi \cdot \Delta R\cdot R^2\cdot \rho \end{array}$

$\therefore$ Energy for vaporization

$\Delta V=4\pi \Delta R\cdot R^2\cdot L\rho$

$8\pi \cdot R\cdot \Delta R\cdot T=4\cdot \pi \cdot \Delta R\cdot R^2\cdot L\cdot \rho$

$or2T=RL\rho$

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

Correct option is (D).