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):


*

*The relation between average speed, molar mass and absolute temperature is given by the following equation:
$$\begin{equation}
v_{rms} = \sqrt{\frac{3RT}{M_{H_2O}}} \tag{1}
\end{equation}$$
where $M_{H_2O}$= 18.02 g/mol, $T=373K$, and $R=8.314 K^{-1} \cdot mol^{-1} \cdot J$


*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:


$$\begin{equation}
\mathbb{E}[E_f] = \frac{1}{2} \cdot m_{H2O} \cdot v_{rms}^2 \tag{2}
\end{equation}$$
where $m_{H2O}\approx 3 \cdot 10^{-23} g$ is the mass of a single water molecule and $v_{rms} \approx 718 \space m \cdot s^{-1}$


*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: 
$$\begin{equation}
\rho_{g} = \frac{m}{V} = M_{H_2O} \cdot \frac{P}{RT}= 18.02 g \cdot mol^{-1} \frac{0.0230 atm}{0.0821 \frac{atm \cdot L}{mol \cdot K} \cdot 373 K} = 0.0135 \space kg \cdot m^{-3} \tag{3}
\end{equation}$$

*From $\rho_g$ and $\rho_l$ we can infer that the ratio of average inter-molecular distances is approximately:
$$\begin{equation}
\frac{D_l}{D_g} \approx \big(\frac{\rho_g}{\rho_l}\big)^{\frac{1}{3}}\approx 2.4 \% \tag{4}
\end{equation}$$
where I suspect that if the molecules may be modelled as hard spheres there's probably some kind of hexagonal lattice structure. 


*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:


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


*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:


$$\begin{equation}
N^{\frac{1}{3}} = (\frac{\rho_l}{M_{H20}} \cdot 6 \cdot 10^{23})^{\frac{1}{3}} \approx 6.4 \cdot 10^{10} \tag{6}
\end{equation}$$
and from this we deduce that the inter-molecular distance is approximately $0.016 \cdot nm$


*It follows that the vdW force is on the order of:


$$\begin{equation}
\mathbb{E}[F_{vdw}] = \frac{\mathbb{E}[W]}{\mathbb{E}[D]}=\frac{7.73 \cdot 10^{-18} \cdot J}{.016 \cdot 10^{-9} \cdot m} \approx 5 \cdot 10^{-7} \cdot N \tag{7}
\end{equation}$$
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. 
 A: There are some fundamental shortcomings to this kind of approach.
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.
Secondly, there is nothing special about the boiling point of water at one atmosphere pressure. In principle, one could apply your argument to any temperature, up to the critical point temperature where the distinction between liquid and vapour disappears. Indeed, since the same argument should apply to the sublimation of a solid, you could choose to consider equilibrium between the vapour phase and a solid, right down to absolute zero (because that is what happens to the solid-vapour coexistence line, as the pressure approaches zero). Obviously, if you feed in a different temperature, you will get a different estimate out for the strength of the intermolecular forces.
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.
