In van der Waals equation for a real gas, the equation is

$$\left(P + a\frac{n^2}{V^2}\right)(V - nb) = nRT$$

where $P$ is the pressure of the real gas, $n$ is the number of moles of the gas, $R$ is the ideal gas constant, $T$ is the temperature, $a$ is a constant to correct for the intermolecular forces of attraction, and $b$ is a constant to correct for the size of the molecules.

It follows that gases with large intermolecular forces have large $a$ values and gases with large molecular size have large $b$ values.

Below is a table of the $a$ and $b$ values of the ideal gases:

\begin{array}{ccc} \hline \text{gas} & a & b \\ \hline \ce{He} & 0.0341 & 0.0237 \\ \ce{Ne} & 0.211 & 0.0171 \\ \ce{Ar} & 1.35 & 0.0322 \\ \ce{Kr} & 2.32 & 0.0398 \\ \ce{Xe} & 4.19 & 0.0511 \\ \hline \end{array}

Why is the $b$ value for helium larger than the $b$ value for neon even though helium's atomic radius is smaller than that of neon? Does it maybe have to do with helium's small size, and thus higher electron density surrounding the nucleus, creating more repulsions with other helium atoms?

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    $\begingroup$ What's the source? Also did you check other sources? $\endgroup$
    – Mithoron
    Nov 28, 2019 at 22:52
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    $\begingroup$ This table is invalid, it lacks the temperature/pressure range in which the equation was fitted for each gas. If they used the same range for all of them, that would definitively make the results incomparable. $a$ and $b$ are not universal constants for each gas, and generally vdW is not a good model for any gas over a wider p/T range. $\endgroup$
    – Karl
    Nov 28, 2019 at 22:53
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    $\begingroup$ @Mithoron from the Wikipedia page: en.wikipedia.org/wiki/Van_der_Waals_constants_(data_page) $\endgroup$ Nov 28, 2019 at 23:33
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    $\begingroup$ @Karl Why do you say the table is invalid when almost all of the chemical data pages for van der waals constants give only "a" and "b", as exemplified here: engineeringtoolbox.com/… and here: en.wikipedia.org/wiki/Van_der_Waals_constants_(data_page) and here: www2.ucdsb.on.ca/tiss/stretton/database/… $\endgroup$ Nov 28, 2019 at 23:34
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    $\begingroup$ I know, its terrible. a and b are just empiric parameters, valid for a limited p/T range. If you make the range too big, you get a bad fit and nothing makes sense, if you make it small enough, you still cannot compare between different gases. Giving vdW parameters without a T range is simply nonsense. $\endgroup$
    – Karl
    Nov 29, 2019 at 16:12

1 Answer 1


The Quantum Gases

I will venture a guess, and hopefully a rational explanation. As discussed in the comments, the van der Waals equation parameter can be derived from mathematical considerations at the critical point and can be calculated as $\displaystyle b = \frac{RT_\mathrm{crit}}{8P_\mathrm{crit}}$.

The critical temperatures of the first three entries from your table are

$$\begin{array}{cr} \text{Gas} & T_\mathrm{crit}/\pu{K} \\ \hline \text{He} & 5.2 \\ \text{Ne} & 44.5 \\ \text{Ar} & 150.7 \end{array}$$

From these, I will offer hand-waving "quantum effects" as the culprit affecting the parameters of this classical equation of state. At the critical point of helium, quantum effects dominate. As things warm up to neon's critical temperature, quantum effects become less important, and by the time we hit argon's, many (if not most) quantum things can be ignored. Because the quantum effects will also scale inversely with mass, they are again important in the order: helium, neon, argon.

So I imagine because hydrogen, helium, and to some extent neon are "quantum gases" their van der Waals equation parameters will not be directly comparable to those of "classical gases".


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