# How much energy does a particle need to evaporate?

If i had a container of water at approximately room temperature, how would I work out how much energy an individual molecule needs in order to evaporate?

Obviously water is far too complicated to analyse all the hydrogen bonds and induced dipoles and such.

I though about looking at the Enthalpy of vaporisation $H$, however this is for a large ensemble of molecules and so I doubt the energy of an individual molecule to vaporise is merely $E=H/N_A$ where $N_A$ is Avagadro's number.

• Single molecule is not specific enough. There's a huge difference between a surface water molecule and an interior water molecule. Cf. surface tension. – Zhe Aug 1 '17 at 20:24
• What is wrong with $H/N_A$? Why do you think this is not a reasonable answer for the average molecule? – sammy gerbil Aug 1 '17 at 20:31

## 1 Answer

The problem with your question is that "evaporation" is a statistical process - and that is the reason we can only really describe the average energy over a large ensemble. That makes the calculation $E=H/N_A$ appropriate.

However - if this is the average, we know that the "least amount of energy" you need has to be less than that: some molecules will escape with "more than enough" energy.

That being the case, we could try to estimate the least energy needed from surface tension (which has units of $N/m$ or $J/m^2$). The energy "per molecule" will be the energy per unit area, divided by the number of molecules per unit area.

To make any kind of progress we would have to model the water molecules as "something". Let's (wrongly) model them as spheres. We know that 1 mole of water occupies 18 ml, so the volume occupied by one molecule is $3\cdot 10^{-29}~\rm m^3$. This makes the radius $r=4.1~Å$.

When we pack spheres tightly on a surface, we get a fill factor of about 91%. With a radius of 4.1 Å we can fit $n=1.7\cdot 10^{18}$ molecules.

Dividing the surface tension $\sigma$ by $n$ gives us the minimum energy:

$$E = \frac{\sigma}{n} = \frac{73\cdot 10^{-3}}{1.7\cdot 10^{18}} = 4.3\cdot 10^{-20} ~\rm J/molecule$$

Comparing that with the "simple" calculation:

$$\rm 44 ~kJ/mol = 7.3 \cdot 10^{-20}~ J / molecule$$

Getting within a factor of 2 with such a simplistic approximation (tight packed spheres) seems reasonable. And at any rate we would expect this argument to give us a lower value than the H/A number for the reason I outlined above.