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 $\Omega = 2b$.
This feels a bit like a homework question, so you will only get half of the answer and need to do the final part. Make sure you understand what is going on though, so you can finish fast!
You need to try and write the VdW equation in the form suggested, and then find terms $\sim n/V$ or $\sim n^2 / V^2$ and identify the coefficients ($B_1$ and $B_2$ in this case).
You can start by rewriting
$$(p+{an^2 \over V^2})(V-nb) = nRT$$
as
$$pV-pnb+{an^2\over V}-{abn^3\over V^2} = nRT$$
so that
$$p(V-nb)=nRT+{an^2\over V}+{abn^3\over V^2}$$
Notice that we had to keep the $p$-dependet terms on the left, because pressure has to only appear on the left. On the other hand, $V$ can appear on the right but only in powers of $n/V$. So we need to decouple things.
To do that, we first collect all things on the right:
$$p(V-nb)=nRT\left(1+{a \over RT} {n\over V}+{ab\over RT}{n^2\over V^2}\right)$$
and you see we only get powers of $n/V$ inside the parenthesis.
We would be done if we did not have that annoying $p(V-nb)$ term that is keeping an extra $..-nb$ on the left.
To get rid of that, we can use the fact that
$$p(V-nb)=pV(1-b{n\over V})$$
and we can now isolate $pV$ and divide by the remaining term
$$pV = nRT {1\over 1-b{n\over V}}\left(1+{a \over RT} {n\over V}+{ab\over RT}{n^2\over V^2}\right)$$
And you see that we have $pV$ on the left (good) and only powers of $n/V$ on the right (good). Unfortunately, one $n/V$ is at the denominator still...
So all that is left - and I leave it to you so you can finish the exercise - is to expand ${1\over 1-b{n\over V}}$ in series (because we assume $nb\ll V$ of course, as it is a correction) and again collect terms in order $\sim n / V$, $\sim n^2 / V^2$, $\sim n^3 / V^3$ etc.
Best Answer
If I understand your question, then the following might help. Solve the van der Waals equation for p and you get: $$p=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$$ So note that the pressure is reduced from what it would be if the factor $a$ was 0.