At high temperatures the Van Der Waals equation has the form $$P\approx\frac{RT}{V_\mathrm{M}-b}$$

What would happen with the equation if in addition to high T, molar volume of the gas ($V_\mathrm{m}$) will get very large too?

The only sense I was able to make out of this was that the pressure would be reduced, but if $V_\mathrm{m}$ stands for the volume occupied by a mole of a substance, than wouldn't a larger $V_\mathrm{M}$ mean less available void space, less volume, and thus a greater pressure?


1 Answer 1


First, let's look at this from a purely mathematical standpoint. This isn't going to be rigorous at all, but I'm happy to clarify anything if asked.

If we assume we have 1 mole, and we say that $V_\mathrm{m}$ and $T$ become very large, then the question becomes what happens to $P$ (our only remaining variable)? Well as $T$ and $V_\mathrm{m}$ become very large with respect to (w.r.t) our excluded volume parameter, $b$, (also written as $T$, $V_\mathrm{m} >> b$), the -b term in the denominator becomes negligible and we approach the conventional ideal gas law:

$$ P \approx \frac{R T}{V_{\mathrm{m}} - b} \implies P \approx \frac{R T}{V_\mathrm{m}} = \frac{n R T}{V} $$

If this is not clear, try plugging in some numbers to the equation and calculating what $P$ would be for small $P$ and $T$ (same order of magnitude as $b$) vs. large $P$ and $T$ (100 times larger than $b$). I will let you determine if this leads to greater or less pressure.

Now let's switch gears to a chemistry point of view. What Johannes van der Waals did, is add this little $b$ parameter in the denominator of the ideal gas law. This parameter is a correction to account for the actual volume that each mole of particles can occupy. You can think of it equivalently as the actual volume each particle can occupy, since a mole is just a really big number of particles.

This $b$ correction is taking into account the fact atoms can repel each other at a relatively large distance. If we approximate these atoms to be hard sphere bouncing off of each other, the radii of these hard sphere is what we call the van der Waals radii - the distance at which atoms "bounce" off of each other if they come into contact. So the parameter $b$ in the denominator is derived purely from the van der Waals radii of the atoms in question. So in the case of large $T$ and $V_\mathrm{m}$, we find that this correction is no longer necessary to make a reasonable approximation, because its presence makes such a small difference.

I recommend looking at the Wikipedia page for more information.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.