Question

Assume that a drop of a liquid evaporates by a decrease in its surface energy so that its temperature remains unchanged. The minimum radius of the drop for this to be possible is. (The surface tension is $T$ , the density of the liquid is $\rho $ and $L$ is its latent heat of vaporisation.)

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

Answer 1

Answer: (B)

Let the radius of the drop at time $t=t$ be $r$ and at an instant $t=t+dt$ be $r-dr$

As surface area is given by, $A=4\pi r^{2}$ ,
$\therefore $ decrease in surface area is,
$dA=4\pi \left(2 r d r\right)$
$\therefore $ Decrease in surface energy during time $dt$ is,
$dU=TdA=T\cdot 8\pi rdr...\left(1\right)$
Decrease in volume during time $dt$ is,
$dV=4\pi r^{2}dr$
$\therefore $ Decrease in mass during time $dt$ is,
$dm=\rho dV=4\pi \rho r^{2}dr$
$\therefore $ Heat required in vaporisation is,
$dQ=Ldm=4\pi \rho r^{2}drL...\left(2\right)$
Apply conservation of energy,
decrease in surface energy $=$ heat required in vaporisation.
$\Rightarrow dU=dQ$
$\Rightarrow 8T\pi rdr=4\pi \rho r^{2}drL$
$\Rightarrow 2T=\rho rL$
$\Rightarrow r=\frac{2 T}{\rho L}$
Hence, the minimum radius is $\frac{2 T}{\rho L}$