Chemistry - Why is the pressure correction in the van der Waals equation proportional to (n/V)^2?
Solution 1:
Summing up the argument from the Wikipedia derivation, the reason it is proportional to $(\frac{1}{V_\mathrm m})^2$ where $V_\mathrm m=\frac Vn$ (the molar volume) comes from how the molecules experience the attractive force. The only molecules that experience a net attractive force are those close to the edge of the container because as we move toward the center of the container, the attractive forces on either side of a molecule cancel out. The force toward the center experienced by a molecule at the edge should be proportional to $(\frac{1}{V_\mathrm m})$. But we can also say that the number of molecules at the edge of the container should be proportional to $(\frac{1}{V_\mathrm m})$. So, by combining the proportionality constants into a single constant $a$, we obtain that the correction to the pressure should be $(\frac{a}{V_\mathrm m^2})=\frac {an^2}{V^2}$ .
Solution 2:
Pressure correction term depends upon:
Number of molecules attracting the molecules which comes to strike the wall and as such it is proportional to density of gas i.e. proportional to $n/V$ where $n$ is the number of moles of gas and $V$ is the volume of the container.
It also depends upon number of molecules which has a strike the unit area of the wall and is therefore proportional to the total number of molecules per unit volume, i.e. proportional to the density again (i.e. proportional to $n/V$).
So the pressure correction term is jointly proportional to $n/V$ for factor 1 and again proportional to $n/V$ for factor 2. Hence the pressure correction term is proportional to
$$\left(\frac{n}{V}\right)\left(\frac{n}{V}\right),$$
i.e. proportional to
$$\frac{n^2}{V^2}.$$