# Evaporation of liquid drop due to change in surface energy

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,

1. $T / \rho L$
1. $2T / \rho L$
2. $\rho L / T$
3. $\sqrt{T / \rho L}$

The solution given for it is like,

Heat of vaporization = Surface energy released $$L dm = T dA$$ $$4L \rho \pi r^2 dr = T 8 \pi r dr$$ $$r = 2T/{\rho L}$$

My confusion, I know that the excess pressure of a drop is $P_{excess}=2T/r$ and that the work done is $T \delta A$ where $\delta A$ represents change in A. So if the whole drop is evaporating why would we take that $dr$ case. And if I were to attempt it then I would do something like $L \rho 4/3 \pi r^3 = T \delta A$ just not sure what $\delta A$ value would be.

I don't know why they have taken $dA=8 \pi r$

And also, why would that be the minimum value of radius. Could someone please explain or give me a hint how to proceed?

Let me know if I am unclear

I have a couple of points to make.

• The reason you have to take a small change in radius, instead of the total surface energy change of the whole drop is that the entire surface energy need not go entirely into the latent heat of vaporization, but could also go into other processes. The only requirement is that the latent heat of vaporization required be less than the change in surface energy. Since we are investigating the extreme case, we have to take it equal. This leads to my $2^{nd}$ point.

• The requirement that the required latent energy be less than the change of surface energy leads to:
$$Ldm \le TdA$$ $$\Longrightarrow r \le \frac{2T}{\rho L}$$ Thus, $\frac{2T}{\rho L}$ is in fact the maximum radius of the drop of liquid and that there is no minimum radius for the drop if it has to evaporated due to the change in surface energy. If there were a minimum radius of the drop, then the drop would never evaporate and would remain in that radius!

• Now, since the surface area of the drop $A=4\pi r^2$, differentiating it we get $dA=8\pi rdr$

Hope you understood.

• why do we need differentials in the first place? Shouldn't it be like $L\times \frac 43 \pi r^3 \rho = T\times 4\pi r^2$ Mar 24, 2018 at 16:35
• @SmarthBansal evaporation is not an instantaneous process, it happens on the surface layer by layer. Even though the bulk of the liquid enough surface energy to evaporate, there might not be enough for the surface to start vapourising away. Mar 24, 2018 at 16:41
• so it should be like $L\times \frac 43 \pi \rho (3r^2)dr = T\times 4\pi (2r)dr$. As we're considering evaporation in a small dt time interval? Mar 24, 2018 at 16:52

The fault in your method is that you are only considering the initial and final states whereas the temperature of the drop has to remain constant throughout the process.

Hence we analyse the situation in a differential amount of time.

The change in volume in this differential time is $4πr²\,{\rm d}r$ and change in area is $8πr\,{\rm d}r$ (the differential of $4πr²$).

• Note that we have an equation editor built into the site which helps for readability of your post. Apr 30, 2018 at 10:18
• Latent heat of vaporisation = 4/3*PR^3 * density * L Change in surface energy = TdA = T*8*PRdR Thus it should not it be, T*8*PRdR= 4/3*P*R^3 * density * L @prathyush poduval Jan 14, 2020 at 5:18