Ok, I believe I figured this out. The pressure of each gas is proportional to the force with which the collide and the frequency of each collision. The force with which they collide is also proportional to the momentum of the particles. Therefore, the pressure is proportional to the collision frequency and momentum. The frequency of collision for the light particles is: $ {f_L} = \sqrt{x} {f_H} $, and the ratio of the pressures is $ {P_L} / {P_H} = (\sqrt{x}/x * \rho_H * \sqrt{x}{f_H}) / \rho_H * {f_H} = 1/1 = 1 $, therefore the pressures must be equal.
The more formal derivation of the van der Waals equation of state utilises the partition function. If we have an interaction $U(r_{ij})$ between particles $i$ and $j$, then we can expand in the Mayer function,
$$f_{ij}= e^{-\beta U(r_{ij})} -1$$
the partition function of the system, which for $N$ indistinguishable particles is given by,
$$\mathcal Z = \frac{1}{N! \lambda^{3N}} \int \prod_i d^3 r_i \left( 1 + \sum_{j>k}f_{jk} + \sum_{j>k,l>m} f_{jk}f_{lm} + \dots\right)$$
where $\lambda$ is a convenient constant, the de Broglie thermal wavelength and this expansion is simply obtained by the Taylor series of the exponential. The first term $\int \prod_i d^3 r_i$ simply gives $V^N$, and the first correction is simply the same sum each time, contributing,
$$V^{N-1}\int d^3 r \, f(r).$$
The free energy can be derived from the partition function, which allows us to approximate the pressure of the system as,
$$p = \frac{Nk_B T}{V} \left( 1-\frac{N}{2V} \int d^3r \, f(r) + \dots\right).$$
If we use the van der Waals interaction,
$$U(r) = \left\{\begin{matrix}
\infty & r < r_0\\
-U_0 \left( \frac{r_0}{r}\right)^6 & r \geq r_0
\end{matrix}\right.$$
and evaluate the integral, we find,
$$\frac{pV}{Nk_B T} = 1 - \frac{N}{V} \left( \frac{a}{k_B T}-b\right)$$
where $a = \frac23 \pi r_0^3 U_0$ and $b = \frac23 \pi r_0^3$ which is directly related to the excluded volume $\Omega = 2b$.
Best Answer
There are two terms in the Van der Waals correction:
$$\left(P + a\left(\frac{n}{V}\right)^2\right)\left(\frac{V}{n}-b\right)=RT$$
The first of these ($a$) is an attraction term: the molecules attract each other, which makes the pressure lower. It's the sum of the actual pressure and the attraction term that gives you the "ideal gas pressure".
The second term ($b$) is a volume term: the "apparent volume" that the molecules can move in is smaller than the volume of the container, so we need to subtract a number from the size of the container to get the "ideal volume".
And this is why you have the two terms, with opposing signs.