Re: Van der Waals constant (b) for neon

Date: Fri Feb 13 13:30:14 1998
Posted By: Dan Berger, Faculty Chemistry/Science, Bluffton College
Area of science: Chemistry
ID: 886784809.Ch

Message:

While teaching the Van der Waal's equation for real gases, I came across something i did not completely understand: If the "b" constant is correlated to molecular volume, why would the b value for neon be smaller than the value for H or He? I checked the Handbook of Chemistry and Physics, and the closest I could come to an explanation was the fact that the b value was also correlated to a compressibility factor.

By the way, the van der Waals equation is

According to P.W. Atkins (Physical Chemistry, 3^d Edition), a relates to the density of the gas and b to the total volume occupied by the gas molecules. It is important to recognize that these constants are derived from experiment, that is, they are empirical.

The first thing I did was to check my handy Sargent-Welch periodic table. It gives atomic and covalent radii. A few calculations (and a check of the Handbook of Chemistry and Physics) gave the following information:

Element	Atomic Radius	Molecular Volume	van der Waals a constant*	van der Waals b constant*
Hydrogen	0.79	3.3	0.24	0.027
Helium	0.49	0.49	0.03	0.024
Neon	0.51	0.55	0.21	0.017
* van der Waals constants taken from the Handbook of Chemistry and Physics, 61^st Edition

For hydrogen (a diatomic molecule), we need the covalent radius (0.32) to convert the information into a molecular volume; I assumed a cylinder, with hemispheric ends, with radius 0.79 and length 2.22 (= 2*0.32+2*0.79). The volume is then given by: where r_cov is the covalent radius of hydrogen. (The two hemispheres add to the volume of a sphere of radius r, and the remainder of the volume is a cylinder with radius r and height 2r_cov.)
For helium and neon, (monatomic) molecular volume is just .

These results explain the difference with hydrogen (which is, after all, H₂) by showing that a hydrogen molecule will be rather larger than a neon atom. (In fact, hydrogen's b constant is much smaller than one might expect, given its molecular volume. Of that, more anon.) However, the neon atom will be slightly larger than a helium atom, so that volume cannot be the whole story.

We must again be reminded that the van der Waals constants are empirical. Thus, they reflect many real-world variables, such as "compressibility." Compressibility ought to be affected by how well the molecules can interact with each other; the better the interactions, the higher the compressibility.

You see, the stronger the non-bonded intermolecular (that is, the van der Waals) forces, the more closely the molecules will be able to approach each other and the lower the value of the b constant. One source of van der Waals interactions is thought to be "induced-dipole/induced-dipole" interactions, in which a temporary dipole in one molecule induces an opposite dipole in a neighbor. The two temporary dipoles then attract each other.

However, the more tightly electrons are held within a molecule, the harder it will be to induce a dipole (this is called polarizability) and the weaker the van der Waals interactions. Indeed, the volume occupied by the molecules will go up because of electron-electron repulsions. Therefore, I think a clue to the van der Waals b constant may be found in ionization potentials, which measure how tightly electrons are held.

Looking at the three elements (and the same periodic table), we find:

Element	Ionization Potential
Hydrogen	13.6
Helium	24.6
Neon	21.6

Here, at last, we find an explanation not only for neon but also for hydrogen. Both neon and hydrogen are more polarizable than helium -- hydrogen very much so -- and thus their van der Waals b constants are lower than one would expect from volumes alone.

I am afraid that I have been carried away!

	Dan Berger
	Bluffton College
	http://cs.bluffton.edu/~berger

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