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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 $$P_{real}V_{ideal}=nRT$$ ??

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  • $\begingroup$ I think he could do that. But the data will not agree with test data. $\endgroup$ – user115350 Dec 21 '16 at 6:35
  • $\begingroup$ That sounds so wrong, i am sorry to say. He says that the real pressure is less than ideal gas pressure and then actually proceeds to INCREASE it by $\frac{an^2}{V^2}$ instead of decreasing ?! If the real volume available and the real pressure both are lesser than the ideal counterparts, then there should be a minus sign for both ..$(P_{ideal}-\frac{an^2}{V^2})(V-nb)=nRT$ $\endgroup$ – Prasad Mani Dec 21 '16 at 7:07
  • $\begingroup$ I think you misunderstood ideal and real here. $P_{ideal}=\frac{nRT}{V}$ and $P_{real}=P_{ideal}-\frac{an^2}{V^2}$. But because of his modification to the volume, the real pressure becomes $P_{real}=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$. van der Waals was not happy with ideal gas state of equation and modified it. The state of equation is used for real gas, though we also know it is nowhere near exact. $\endgroup$ – user115350 Dec 22 '16 at 0:53
  • $\begingroup$ I corrected it; it was a silly mistake. I actually am pretty clear what real and ideal pressure and volume are. My question was about the order in which he performed the correction which seemed arbitrary. If you correct for the pressure FIRST and then, the volume, then you get the last equation i wrote in my question. $\endgroup$ – Prasad Mani Dec 22 '16 at 2:34
  • $\begingroup$ If you change the order, it shouldn't be different. You go first by pressure correction $P_{real}=\frac{nRT}{V}-\frac{an^2}{V^2}$ and then go second by correct the volume. I note that in your last equation, you are using ideal pressure that's not intended by van der Waals. $\endgroup$ – user115350 Dec 22 '16 at 3:53
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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 $\Omega = 2b$.

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  • $\begingroup$ Wow that escalated quickly. But thankfully i understood it fully. Thank you. Just for future reference, i shouldn't be using the derivation i described in the question right? Because it seems to raise more questions than it answers. $\endgroup$ – Prasad Mani Dec 21 '16 at 13:25
  • $\begingroup$ @PrasadMani There's nothing wrong with trying to reason about it the way you did, but I prefer this approach for three reasons: 1) From the mathematical expansion we understand why it is limited to low densities and high temperatures. 2) The approach is generalisable easily. 3) It gives a relation between the coefficients and the potential itself. $\endgroup$ – JamalS Dec 21 '16 at 13:27
  • $\begingroup$ I actually did not reason it that way; it was taught to me, long back. The stat mech derivation seems like the right way to go since the derivation i was taught is riddled with the questions i highlighted. And did he use only uptil the first order correction? $\endgroup$ – Prasad Mani Dec 21 '16 at 13:32

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