Editing Van der Waals radius (section)

=== Van der Waals equation of state ===
{{main|Van der Waals equation}}

The van der Waals equation of state is the simplest and best-known modification of the [[ideal gas law]] to account for the behaviour of [[real gas]]es:
<math display="block">\left (p + a\left (\frac{n}{\tilde{V}}\right )^2\right ) (\tilde{V} - nb) = nRT,</math>
where {{mvar|p}} is pressure, {{mvar|n}} is the number of moles of the gas in question and {{mvar|a}} and {{mvar|b}} depend on the particular gas, <math>\tilde{V}</math> is the volume, {{mvar|R}} is the specific gas constant on a unit mole basis and {{mvar|T}} the absolute temperature; {{mvar|a}} is a correction for intermolecular forces and {{mvar|b}} corrects for finite atomic or molecular sizes; the value of {{mvar|b}} equals the van der Waals volume per mole of the gas. 
[[van der Waals constants (data page)|Their values]] vary from gas to gas.

The van der Waals equation also has a microscopic interpretation: molecules interact with one another. 
The interaction is strongly repulsive at a very short distance, becomes mildly attractive at the intermediate range, and vanishes at a long distance. 
The ideal gas law must be corrected when attractive and repulsive forces are considered. 
For example, the mutual repulsion between molecules has the effect of excluding neighbors from a certain amount of space around each molecule. 
Thus, a fraction of the total space becomes unavailable to each molecule as it executes random motion. 
In the equation of state, this volume of exclusion ({{math|''nb''}}) should be subtracted from the volume of the container ({{mvar|V}}), thus: ({{math|''V'' - ''nb''}}). 
The other term that is introduced in the van der Waals equation, <math display="inline">a\left (\frac{n}{\tilde{V}}\right )^2</math>, describes a weak attractive force among molecules (known as the [[van der Waals force]]), which increases when {{mvar|n}} increases or {{mvar|V}} decreases and molecules become more crowded together.

{| class="wikitable floatright" align=right
|-
! Gas
! ''d'' ([[Ångström|Å]])
! ''b'' (cm{{sup|3}}mol{{sup|–1}})
! ''V''{{sub|w}} (Å{{sup|3}})
! ''r''{{sub|w}} (Å)
|-
| [[Hydrogen]]
| 0.74611
| align=center | 26.61
| 34.53
| 2.02
|-
| [[Nitrogen]]
| 1.0975
| align=center | 39.13
| 47.71
| 2.25
|-
| [[Oxygen]]
| 1.208
| align=center | 31.83
| 36.62
| 2.06
|-
| [[Chlorine]]
| 1.988
| align=center | 56.22
| 57.19
| 2.39
|-
| colspan=5 | <small>van der Waals radii ''r''{{sub|w}} in Å (or in 100 picometers) calculated from the [[van der Waals constant]]s<br/>of some diatomic gases. Values of ''d'' and ''b'' from Weast (1981).</small>
|}

The [[van der Waals constant]] ''b'' volume can be used to calculate the van der Waals volume of an atom or molecule with experimental data derived from measurements on gases.

For [[helium]],<ref name="CRC">{{RubberBible62nd}}, p. D-166.</ref> ''b''&nbsp;= 23.7&nbsp;cm{{sup|3}}/mol. Helium is a [[monatomic gas]], and each mole of helium contains {{val|6.022|e=23}} atoms (the [[Avogadro constant]], ''N''{{sub|A}}):
<math display="block">V_{\rm w} = {b\over{N_{\rm A}}}</math>
Therefore, the van der Waals volume of a single atom ''V''{{sub|w}}&nbsp;= 39.36&nbsp;Å{{sup|3}}, which corresponds to ''r''{{sub|w}}&nbsp;= 2.11&nbsp;Å (≈ 200 picometers). 
This method may be extended to diatomic gases by approximating the molecule as a rod with rounded ends where the diameter is {{math|2''r''{{sub|w}}}} and the internuclear distance is {{mvar|d}}. 
The algebra is more complicated, but the relation
<math display="block">V_{\rm w} = {4\over 3}\pi r_{\rm w}^3 + \pi r_{\rm w}^2d</math>
can be solved by the normal methods for [[cubic functions]].