# Estimating the inter-molecular van der Waals' force from the boiling point of water

Background:

I understand that inter-molecular van der Waals' forces are responsible for maintaining water in the liquid phase. Now, if we suppose that the net van der Waals' force on a given H2O molecule suspended in liquid phase H2O is due to its interaction with a very large number of neighbouring H2O molecules, I wonder whether we can estimate the average van der Waals' force on this molecule given that the boiling point of water is known to be 373 Kelvin.

Method:

Here's what I have tried(although my approach may be incorrect as this is entirely based on independent study):

1. The relation between average speed, molar mass and absolute temperature is given by the following equation: $$$$v_{rms} = \sqrt{\frac{3RT}{M_{H_2O}}} \tag{1}$$$$

where $$M_{H_2O}$$= 18.02 g/mol, $$T=373K$$, and $$R=8.314 K^{-1} \cdot mol^{-1} \cdot J$$

1. From $$(1)$$ we may determine that for a particular molecule at boiling point the expected final kinetic energy, $$\mathbb{E}[E_f]$$, is given by:

$$$$\mathbb{E}[E_f] = \frac{1}{2} \cdot m_{H2O} \cdot v_{rms}^2 \tag{2}$$$$

where $$m_{H2O}\approx 3 \cdot 10^{-23} g$$ is the mass of a single water molecule and $$v_{rms} \approx 718 \space m \cdot s^{-1}$$

1. To obtain the average initial kinetic energy $$\mathbb{E}[E_i]$$ of the average water molecule in liquid phase at boiling point we may assume that this particle would cover the same distance in the same time interval if not for the significantly greater density of liquids:

• Let's suppose that the density of water in liquid phase at boiling point is approximately given by $$\rho_l = 10^3 \space kg \cdot m^{-3}$$
• To determine the density in the gas phase we have: $$$$\rho_{g} = \frac{m}{V} = M_{H_2O} \cdot \frac{P}{RT}= 18.02 g \cdot mol^{-1} \frac{0.0230 atm}{0.0821 \frac{atm \cdot L}{mol \cdot K} \cdot 373 K} = 0.0135 \space kg \cdot m^{-3} \tag{3}$$$$

• From $$\rho_g$$ and $$\rho_l$$ we can infer that the ratio of average inter-molecular distances is approximately:

$$$$\frac{D_l}{D_g} \approx \big(\frac{\rho_g}{\rho_l}\big)^{\frac{1}{3}}\approx 2.4 \% \tag{4}$$$$

where I suspect that if the molecules may be modelled as hard spheres there's probably some kind of hexagonal lattice structure.

1. Using the previous calculations we may estimate the average work done to overcome the van der Waals' forces to change from the liquid to the gas phase as follows:

$$$$\mathbb{E}[W]=\mathbb{E}[E_f]-\mathbb{E}[E_i]= \frac{1}{2} m_{H2O} (v_{rms}^2-(0.024 \cdot v_{rms})^2) \approx 7.73 \cdot 10^{-18} \cdot J \tag{5}$$$$

1. Now, if we assume that on average the particles are approximately uniformly distributed in space in the liquid phase then the average inter-molecular distance in the liquid phase may be determined by calculating the number of particles per meter:

$$$$N^{\frac{1}{3}} = (\frac{\rho_l}{M_{H20}} \cdot 6 \cdot 10^{23})^{\frac{1}{3}} \approx 6.4 \cdot 10^{10} \tag{6}$$$$

and from this we deduce that the inter-molecular distance is approximately $$0.016 \cdot nm$$

1. It follows that the vdW force is on the order of:

$$$$\mathbb{E}[F_{vdw}] = \frac{\mathbb{E}[W]}{\mathbb{E}[D]}=\frac{7.73 \cdot 10^{-18} \cdot J}{.016 \cdot 10^{-9} \cdot m} \approx 5 \cdot 10^{-7} \cdot N \tag{7}$$$$

Any constructive thoughts and comments are welcome.

Note:

All of these calculations appear to make sense to me but it's not yet clear how I can be certain that they aren't completely off. I mean this calculation isn't coupled with a testable hypothesis although I think I can figure out whether this estimate is reasonable by using this estimate of vdW force magnitudes to predict how far a metal ball with mass $$M$$ would travel in a container filled with water after a time interval $$T$$ when released from a given height.

• Unfortunately the main reason why the boiling point of water is at such high temperature compared to many other "similar" molecular compounds is that the main driving force behind the water-water attraction is hydrogen bonding rather than van der Waals attraction. If I recall correctly, in water the strength of hydrogen bonds is ~10 times larger than the VdW one. Another small thing: the microscopic structure of water is a four-coordinated network which has a sensibly lower density than a hexagonal lattice. – lr1985 Jan 1 at 21:52
• @lr1985 Thank you for this clarification. I meant VdW forces in the original sense used by Van Der Waals' which includes any inter-molecular forces. However, I think this only makes sense if I can approximate water molecules as hard spheres. I'm also curious about the microscopic structure which you describe. Might you know any good references on the subject? – Aidan Rocke Jan 1 at 22:15
• Sure, try here or here. There are also countless scientific papers on the matter, mostly published tens of years ago, as today the matter is more or less settled (even though people still discuss some of the details). – lr1985 Jan 2 at 8:36
• @lr1985, I have an unrelated question. How do you add a hyperlink to an individual word, such as the word "here"? – David White Jan 4 at 1:35
• @DavidWhite you just use markdown: put the word between square brackets and the link between regular brackets. When you are writing a comment, click on the "help" link on the right to get to know more about formatting. – lr1985 Jan 4 at 8:02

Firstly, your eqn (1) applies equally well to the liquid and the gas. It's a common misconception that the average kinetic energy of a molecule in the liquid is less than that in the gas, at the same temperature $$T$$. In fact, within the approximation of classical statistical mechanics, they are equal, if the temperatures are equal. So section (3) of your analysis is not applicable.
Naturally, one can estimate the cohesive energy of a solid or a liquid, per molecule, by multiplying the typical molecule-molecule interaction energy by an estimate of the number of neighbours "within range" of a molecule in the condensed phase. It is tempting then to say that a gas will be produced when the temperature is sufficient to make the kinetic energy per molecule outweigh this cohesive energy. But the argument neglects any consideration of entropy, and of course these phase transitions occur when the Gibbs free energies $$G$$, or chemical potentials $$\mu$$, of the two phases become equal (at a given $$T$$ and $$P$$). That means considering both enthalpy and entropy contributions to $$G$$ or $$\mu$$.
From the law of corresponding states, it is generally possible to correlate the strength of the interactions, $$\epsilon$$ say, with the temperature of fixed points, such as the critical point, on the phase diagram (and the diameter $$\sigma$$ of the molecules with the molar volumes of those fixed points, etc). But this only applies when one is comparing substances composed of molecules which interact in a similar way (e.g. noble gases, assuming something like a Lennard-Jones pair interaction). So, it makes sense to express things in terms of reduced variables such as $$k_BT/\epsilon$$, $$N\sigma^3/V$$, $$P\sigma^3/\epsilon$$. For Lennard-Jones, for instance, the critical temperature is at roughly $$k_BT_c/\epsilon\approx 1.3$$. So, knowing $$T_c$$, in Kelvin, for a real gas, and assuming a Lennard-Jones form for the interaction between molecules, one can estimate $$\epsilon$$ in J. But going beyond that kind of calculation is a bit of a stretch.