I understand that the behavior of ideal gases deviates largely from that of real gases in terms of pressure exerted by the gas molecules on the container in which it is present, space available for each gas molecule to move, and other factors. I know the van der Waals equation includes the necessary correction terms in pressure and volume so that one could fit the behavior of a real gas using ideal gas equation.

In all the examples given in my textbook,for a real gas,the ideal gas equation would be:

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

where $n$ is the amount of gas, $T$ the temperature, $R$ the gas constant, and $(a, b)$ van der Waals' constants.

  1. What exactly is $V$ in the equation? $P$ in the equation has to be the pressure exerted by the gas molecules in an nonideal state since due to attraction of gas molecules in real state, the pressure exerted by the gas would be lesser than what it would be in the case of an ideal gas. Hence we are adding $an^2/V^2$.

    But, the $V$ is representative of the space available for gas molecules to move. In an ideal case, since the size of the molecules is insignificant, we take the space available for the gas molecules to be that of the volume of the container. However, in a nonideal situation, the gas molecules occupy a significant portion of the container, and hence the available space for them to move is less than the volume of the container.

  2. Is $V$ the volume of the container? If so, then why are we even including a correction term to the volume part of the equation? Doesn't the van der Waals equation attempt to fit the behaviour of a real gas using an equation similar to the ideal gas law?

  • 2
    $\begingroup$ Useful links for text and formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial. $\endgroup$
    – Poutnik
    Aug 17, 2022 at 17:18
  • $\begingroup$ i accidentally typed "large" for "real".apologies! $\endgroup$ Aug 17, 2022 at 17:18
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    $\begingroup$ Please consider reading the links in Poutnik's comment. They will let you typeset maths professionally. Also, please, put spaces after punctuation marks. Spaces after commas, spaces after full stops. Lastly, I took the liberty of correcting your equations from $an/V^2$ to $an^2/V^2$: I assume what you really meant was $a(n/V)^2$, but if you want to write it like that, you need the parentheses. $\endgroup$ Aug 17, 2022 at 17:35
  • $\begingroup$ @Poutnik thank you, I was looking everywhere for that and just gave up ! $\endgroup$ Aug 18, 2022 at 21:08

1 Answer 1


The van der Waals equation is represented as such:

$$\left(P + \frac{an^2}{V^2}\right)(V-nb) = nRT$$

Here, $P$ is the pressure of the container and $V$ is indeed the volume of the container. All the symbols that come from the original ideal gas law maintain their definitions. The best way to conceptualize the van der Waals equation is as a perturbation of the ideal gas law that brings it closer to real gas behavior. Let's start with the ideal gas law:

$$PV = nRT$$

This equation simply states that pressure and volume are inversely proportional to each other at a constant number of molecules $n$ and a constant temperature $T$; that their product is directly proportional to number of molecules $n$ and absolute temperature $T$; and that these parameters are the only considerations to model gases' behaviors. It is not simply that the equation conforms to evidence from the world (empirical) but also that it applies in the fictional but reasonably approximate case that molecules 1) do not interact and 2) do not occupy space. It is often more useful to conceptualize an ideal gas as a gas where gas particles pass through each other; this is equivalent both to excluding particle interactions and to allowing each gas particle the entire volume of the container (because nothing gets in the way). If gas particles all passed through each other, then the ideal gas law would be true, but alas this is not the case.

Since this is not the case, we need to add parameters that account for particle interactions. An entirely new equation is not necessary though: instead of starting from scratch, we can add a term to the pressure and volume variables that corrects for the existence of particle interactions. It is simplest to demonstrate with the case of the $(V-nb)$ term:

Let's say that particles do not interact except they collide with each other and bounce off elastically like billiard balls. The volume term in the ideal gas law $V$ stands for the volume of the entire container; however, if two molecules simply collide with each other, then for every molecule added to the container, there is a molecule's volume less space for every other molecule to be. So, while we still see the volume of the container as $V$, from the molecule's perspective there isn't a full $V$ volume worth of space to go, because other molecules are in the way.

For an analogy, if you walk into an empty room, you can navigate every square foot of the room easily. However, if you invite a bunch of friends over, suddenly there is much less space to navigate even though the room did not literally shrink in size—you effectively have less space to move. We'll call the amount of space that the molecule can actually move "effective volume" or $V_\text{eff}$.

We expect for every additional molecule we add that the volume $V$ stays the same, but the effective volume $V_\text{eff}$ decreases a little bit. So, our effective volume should be 1) linearly related to the number of molecules proportional to the molecule's volume and 2) smaller than the original volume. We can use this to make an equation, where $b$ depends on the volume of the molecule:

$$V_\text{eff} = V - nb$$

A similar argument is made to account for molecular attractions with the pressure parameter, but this is much more complex and includes a squared volume term for complicated reasons. Regardless, we can construct an effective pressure $P_\text{eff}$ which is just the original pressure term with an additional correction for particle interaction included. If we let:

$$P_\text{eff} = P + \frac{an^2}{V^2}$$


$$V_\text{eff} = V - nb$$

and simply let $a$ and $b$ be corrections to the original pressures and volumes, then we get the van der Waals equation looking like this:

$$P_\text{eff}V_\text{eff} = nRT$$

which looks exactly like the ideal gas law, but corrected for additional factors.

  • $\begingroup$ so the P in van der Waal's equation is actually the pressure exerted by the gas molecules without taking into account the mutual interactions between other molecules?and not the pressure that the gas molecules exert on the container in a non ideal situation? $\endgroup$ Aug 18, 2022 at 4:32
  • $\begingroup$ @anotherhyooman P is pressure on container walls. Internal gas pressure is higher by an^2/V^2. (according to the van der Waals real gas model) $\endgroup$
    – Poutnik
    Aug 18, 2022 at 7:46
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    $\begingroup$ @anotherhyooman yes—by "effective pressure" I mean it, similar to effective volume, as pressure from the molecular perspective. you can ~sorta~ conceptualize P_eff as the average amount of forces experienced by a molecule, which are for essentially all purposes attractive forces. If you rearranged the equation to solve for P, you'd see that P is LESS than the term linearly related to temperature. More attractive forces means the molecule experiences more forces, but that also means a smaller portion of the total forces goes to hitting the container. $\endgroup$ Aug 18, 2022 at 21:19

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