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

Question

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 $A$:

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

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

....

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.