The van der Waals equation is:

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

I understand the volume term, it basically reduces the avalaible volume because the particles have volume themselves. In a sense it is the volume occupied by all the particles if we squeeze them all together.

However I don't understand the pressure term. It is because of intermolecular attraction. So I understand that if you have more particles, there is more intermolecular attraction and the pressure is reduced.

But why exactly is this term squared? I don't see why this isn't proportional. There are a lot of derivations to find, but it is commonly skimmed over.

I am specifically talking about:

$$ (p + \frac{n^2 a}{V^2})$$

My proposed heuristic solution:

  1. If you increase the number of particles with $N$ the attractive force on one particle doubles. When this one particle would be on the wall to exert a force $F$ this force would thus be reduced by $F_{\text{net,1}}=F_{i,1}-aNF_1$.

  2. Since the pressure is created by $N$ particles, to get the total force on the wall:

$$ F_{\text{total}} = NF_1 $$

$$ F_{\text{total}} = N(F_{i,1}-aNF_1 )=NF_{i,1}-a N^2F_1 $$

Dividing by the area $A$:

$$ \frac{F_{\text{total}}}{A} = P_{\text{total}}$$

$$ F_{\text{total}} = P_{i}-a N^2P_1 $$



1 Answer 1


Short Answer: Every particle attracts every other, so the total attraction (which is the correction to pressure) has to be proportional to $N^2$.

Long Answer: If you want to rigorously derive the VdW equation of state, you should look into the Cluster expansion. It is covered in most textbooks on statistical mechanics. For a good introduction, have a look at David Tong's lecture notes.

More intuitively, you can view the VdW equation as a mean field theory of a weakly interacting gas. Take a look at the partition function $$ Z = \frac{1}{N! \lambda^{3N}} \int e^{-\beta \sum U(\Delta x_{ij})} dx^{3N} \,, $$ with the momentum integration already performed and put into the thermal wavelength $\lambda$.

For an arbitrary potential $U$, this integral is impossible to do exactly. But you can do a mean field approximation, where you assume that all particles feel the same 'mean field potential' $U_{\rm MF}(r)$. The partition function then factorizes into $$ Z = \frac{1}{N! \lambda^{3N}} \left[\int e^{-\beta U_{\rm MF}(x)} dx^3\right]^N \,. $$

What can you say about $U_{\rm MF}(r)$? You know two things:

  1. Particles can't overlap, so that $U_{\rm MF}(r < R) = \infty$. You therefore have an exlcluded volume of size $nb$, where $b$ is you particle volume. This gives the correction to the $V$ term in the VdW equation.
  2. The potential will be slightly attractive at long range. Since every particle attracts every other, $U_{\rm MF}(r >> R)$ should be proportional to the particle density, say $U_{\rm MF} = -a N / V$.

Since our simplified $U_{\rm MF}$ does not depend on position anymore, we can pull it out of the integral. $$ Z = \frac{(e^{\beta a N / V})^N}{N! \lambda^{3N}} \left[\int dx^3\right]^N = \frac{1}{N!} \left[\frac{e^{\beta a N / V} (V- Nb)}{\lambda^3}\right]^N\,. $$ Here you can see that the $N^2$ dependence comes from the overall exponent of $N$: $$ \left(e^{\beta a N / V}\right)^N = e^{\beta a N^2 / V} $$ You can derive the VdW equation of state by calculating the Gibbs free energy $G$ and looking at its derivatives. Along the way you will pick up another $V$ to give you $a N^2 / V^2$.

If you want to know more I recommend the book Introduction to the Theory of Soft Matter by Jonathan Selinger.

  • $\begingroup$ Thank you for your very indepth comment however I am afraid it is a bit too complicated for me. Do you have a more easy to understand version? $\endgroup$ Jun 6, 2022 at 18:29
  • 2
    $\begingroup$ Think of it this way: The pressure of the interacting gas is reduced because the molecules attract each other. Say there are $N$ molecules in total. If you pick a random one, it will be attracted to all others, which means it will feel a potential that looks like $U = -a N/V$ (proportional to $N$ and negative because it is attractive). But we this is just the potential that a single molecule is subjected to. There are $N$ of these molecules, so that the total interaction energy is $\sum U = N \times U = -a N^2 / V$. If you divide by volume to get an energy density, you find $u = - aN^2 / V^2$. $\endgroup$
    – leapsheep
    Jun 7, 2022 at 9:46
  • $\begingroup$ Thank you. that last one is easier to understand. Maybe it is a good suggestion to add it as an alternative less rigorous derivation to your more rigorous derivation for future readers of my level $\endgroup$ Jun 7, 2022 at 12:15

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.