# Van der Waals derivation: How is the pressure term exactly derived?

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.

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

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 $$A$$:

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

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

....

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.

• 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? Jun 6, 2022 at 18:29
• 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 $u = - aN^2 / V^2$. Jun 7, 2022 at 9:46
• 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 Jun 7, 2022 at 12:15