Question

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remain 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. $\sqrt{T/\rho L}$
• B. $T/\rho L$
• C. $2T/\rho L$
• D. $\rho L/T$

Answer 1

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

given that,

surface tension =$T$

density of liquid = $\rho$

latent heat =$L$

We know that,

Decrease in surface energy = heat required in vaporization

Now,

Decrease in surface energy

$=T\times \Delta A$

$=2T\times \left(4\pi rdr\right).....\left(I\right)$

Now, heat required in vaporization = latent heat

$=ML$

$=V\rho L$

$=4\pi r^2\rho L....\left(II\right)$

Now, from equation (I) and (II)

$2T\left(4\pi rdr\right)=4\pi r^2\rho L$

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

Hence, the radius is $\frac{2T}{\rho L}$