Van der waals equation derivation? He assumed that the intermolecular forces result in a reduced pressure on the walls of the container which has a real gas in it. Also that the molecules are finite in size which means they do not have the entire volume of the container to themselves; something less than that. So when he accounted for the reduced volume by $V-nb$, why did he not do $P-\frac{an^2}{v^2}$ and instead, did the below:
He accounted for the reduced volume first with $V-nb$, then he used $$P(V-nb) = nRT$$ and then $$P=\frac{nRT}{V-nb},$$ then said that the real pressure is less than the ideal gas pressure by an amount $\frac{an^2}{V^2}$ from which follows the below $$P_{real}=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$$and therefore $$(P_{real}+\frac{an^2}{V^2})(V-nb)=nRT.$$
My question is: what is the logic behind this? What if he did the other way around? meaning corrected for the reduced pressure first and then corrected for the reduced volume which would have given the following steps
Correction for the pressure FIRST (reducing the ideal pressure by an amount$\frac{an^2}{V^2}$) $$V=\frac{nRT}{(P_{ideal}-\frac{an^2}{V^2})}$$
Then correcting the volume by reducing it by an amount $nb$, giving
$$V=\frac{nRT}{(P_{ideal}-\frac{an^2}{V^2})}-nb$$
giving
$$(P_{ideal}-\frac{an^2}{V^2})(V+nb)=nRT$$
Should the equation of state be $$P_{real}V_{real}=nRT$$ or $$P_{ideal}V_{real}=nRT$$ or $$P_{real}V_{ideal}=nRT$$
 ??
 A: 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 $\Omega = 2b$.
