The Van Der Waals Forces in Gases

A calculation of van der Waal's potential of two atoms at large separation has been carried out for hydrogen and helium. The method depends upon a representation of the perturbed wave function of the system as $\ensuremath{\psi}={\ensuremath{\psi}}_{0} (1+vR)$ where ${\ensuremath{\psi}}_{0}$ is the unperturbed wave function, $v$ the perturbing potential and $R$ is a function of the radial coordinates of the electrons. The method is equally well adapted to the calculation of polarizabilities. A computation of the mutual energy of two hydrogen atoms confirms the results of Eisenschitz and London. The polarizability of helium is calculated as 0.210\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}24}$ cc which agrees well with the experimental value, 0.205\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}24}$. The mutual energy of two helium atoms is found to be -3.18 $\frac{{E}_{0}}{{(\frac{R}{{a}_{0}})}^{6}}$. A correlation between the mutual energy of the two molecules, $\ensuremath{\epsilon}$, and the polarizability, $\ensuremath{\alpha}$, is obtained: $\ensuremath{\epsilon}=\frac{\ensuremath{-}1.36{{\ensuremath{\nu}}_{0}}^{\frac{1}{2}}{{a}_{0}}^{\frac{3}{2}}{a}^{\frac{3}{2}}{E}_{0}}{{R}^{6}}$ where ${\ensuremath{\nu}}_{0}$ is the number of electrons in the highest quantum state in the molecule, ${E}_{0}$ the energy of the hydrogen atom in the normal state, and $R$ is the separation of the molecules. By means of this formula, the van der Waals cohesive pressure constant is calculated for Ne, A, ${\mathrm{N}}_{2}$, ${\mathrm{H}}_{2}$, ${\mathrm{O}}_{2}$, and C${\mathrm{H}}_{4}$.