I have a problem in understanding the van der Waals real gas equation and it goes like this. $$ nRT=\left(P+a\frac{n^2}{V^2}\right)(V-nb) $$ Do the variables $P$ and $V$ on the RHS have the same physical meaning as the one which appear in ideal gas equation? If yes then pressure of a real gas must always be more than that of ideal gas which actually is the opposite. Secondly, the volume of the real gas must always be less than ideal (since there is a negative term). But this is also not true since there is both positive and negative deviation.

As to what I am interpreting the meaning of the two terms on RHS also seems wrong which besides having spoken above, I am also trying to explain below.

  1. Pressure term

Here $P$ stands for the pressure had the gas been ideal. Since the gas isn't ideal molecules experience mutual attractive force which accounts for the pressure correction term.By this very logic the correction term should have been negative to support the claim that pressure of an ideal gas is less than that of real.

  1. Volume term

Here $V$ accounts for the volume had the gas been ideal that is the volume of the container. Since the gas isn't ideal and molecules of real gases are not point sized the volume correction term is introduced which excludes the volume of the gas molecules and ensures that the volume accounted by the second term is only the one in which gas molecules can have free movement.

This was my understanding about real gases which I think is wrong for obvious reasons. Any help would be greatly be appreciated. Thanks in advance.


2 Answers 2


Firstly, $P$, $V$ and $T$ in the van der Waals equation of state have exactly the same meanings as they have elsewhere in thermodynamics: they represent the pressure, volume, and temperature of a sample of the material being studied.

Secondly, this equation of state is one of many approximate equations of state which attempt to model the values measured in experiment, especially (in this case) non-ideal gases. There are many other, more accurate, equations of state which are empirical fits to the measured data. The van der Waals equation is a useful one to study because it is simple, and one can understand in fairly simple terms where it comes from. It has some basis in statistical mechanics, which also helps explain its form.

But remember, it is just an approximation. For example, although it predicts the existence of a critical point, it continues in an unphysical way below the critical temperature into the two-phase region where liquid and gas are in equilibrium with each other.

The equations of state page actually gives a handwaving explanation of the two modified terms. I'll paraphrase here.

Indeed $(V-nb)$ attempts to account for the excluded volume of the atoms. If we ignore all the attractions, we are essentially modelling the atoms as hard spheres, and writing the equation of state as $$ \frac{PV}{nRT} = \left( \frac{V}{V-nb}\right) = \left( \frac{1}{1-b\rho}\right) $$ where $\rho=n/V$. This means that, for a given density $\rho$ and temperature $T$, the pressure is higher than we might have expected if we had assumed that the gas were ideal. Also, it recognizes that there is a minimum possible value of $V/n$ or a maximum density $\rho$, because at some point the atoms will overlap. (We don't take the precise value of the maximum density too literally; in practice, the system will crystallize to a solid, which the equation does not describe). Nowadays, we know a lot more about the hard sphere fluid, and have more accurate equations of state such as Carnahan-Starling for this system.

The effects of attractions are then treated as a weak perturbation, superimposed on the above equation. The gas surrounding a typical atom is treated as a uniform attractive background: a kind of mean field theory. This reduces the pressure, compared with what we might have thought on the basis of the ideal gas equation. The equations of state page gives a way to think of this:

While ideal gas molecules do not interact, we consider molecules attracting others within a distance of several molecules' radii. It makes no effect inside the material, but surface molecules are attracted into the material from the surface. We see this as diminishing of pressure on the outer shell (which is used in the ideal gas law), so we write ( $P$ + something) instead of $P$. To evaluate this "something", let's examine an additional force acting on an element of gas surface. While the force acting on each surface molecule is $\sim\rho$, the force acting on the whole element is $\sim\rho^{2}$ ....

In fact, this can be made more rigorous: it can be shown using that, for thermodynamic consistency, if we add a uniform attractive background to a system, the effect on the equation of state must take the form $$ P = P_0 - a\rho^2 $$ where $P_0$ is the pressure of the system without the attractions (for example, $P_0=1/(1-b\rho)$ as above, or a more accurate equation such as Carnahan-Starling) and $a$ is a positive constant. This is the form appearing in van der Waals' equation.

[EDIT following OP comment]

If the van der Waals equation describes the real gas accurately, then $P$ is the pressure of the real gas (not an ideal gas) and likewise $V$, $T$, $n$ are properties of the real gas. If you want to compare the properties of a real gas and an ideal gas, then it is necessary to specify what is kept the same, and what is to be compared. I believe that it is clearest to keep $n$, $V$ and $T$ the same, and compare pressures. Then, rearranging the equation of state \begin{align*} P &= \frac{nRT}{V-nb} - a\frac{n^2}{V^2} \\ \frac{PV}{nRT} = \frac{P}{P_{\text{id}}} &= \frac{V}{V-nb} - \frac{a}{RT}\frac{n}{V} = \frac{1}{1-\rho b} - \frac{a}{RT}\rho \end{align*} where I have introduced the density in moles per unit volume, $\rho=n/V$, and $P_{\text{id}}=nRT/V$ which is the ideal gas pressure at the same density and temperature. At low density, $\rho\rightarrow 0$, $P\rightarrow P_{\text{id}}$, as we would expect. The equation makes clear that the $b$ term (excluded volume due to short range repulsive interactions) tends to increase $P/P_{\text{id}}$ and the $a$ term (attractive long range interactions) tends to decrease $P/P_{\text{id}}$. Depending on the temperature and the density, it is possible for $P$ to be higher, or lower, than $P_{\text{id}}$.

  • $\begingroup$ My argument is that P+an^2/V^2 is the effective pressure of the real gas.This we know is less than the pressure of the ideal gas(P) but then adding an^2/V^2 makes it more than P. $\endgroup$ Oct 13, 2018 at 12:01
  • 1
    $\begingroup$ $P+a\rho^2=nRT/(V-nb)$ is greater than the pressure of an ideal gas at the same $\rho$ and $T$, which would be $nRT/V$. The repulsive short range interatomic forces increase the pressure, relative to an ideal gas at the same $\rho$ and $T$. As I have explained, the attractive term involving $a$ then reduces the pressure relative to the ideal gas. $\endgroup$
    – user197851
    Oct 13, 2018 at 12:20
  • 1
    $\begingroup$ I have edited my answer in an attempt to clarify further. $\endgroup$
    – user197851
    Oct 13, 2018 at 13:29

Simply put, you are basically correct that real gasses have a slightly lower pressure than ideal gasses at the same temperature. Your basic understanding of volume seems correct as well - the $n_b$ term is an attempt to correct for the incredibly-small-but-not-zero size of the atoms.

With regard to your attempt to "interpret" the $P$ and $V$ variables and assign "meaning" to them, I prefer that people think of $P$ and $V$ as variables that attempt to predict the readings of pressure gauges and volume gauges and nothing more. Attempts to do anything more usually result in over-analysis and over-thinking.

If you read your own logic very carefully, I think you will find that you are thinking in doubles. The $P'$ of the Van der Waals equation will be slightly less than the $P$ of the ideal gas equation. Note my use of $P'$ in the Van der Waals equation vs. $P$ in the ideal gas equation. They don't necessarily have the same numerical values.

Additionally, in the Van der Waals equation, I think of $(P' + a^2/V^2)$ as corresponding to plain ol' $P$ in the ideal gas equation. This is all consistent with having the $a^2/V^2$ be a positive correction factor rather than a negative one.

Another thought about volume: Perhaps we all should think of $V$ as the total amount of free space inside the container that the gas molecules have to fly around. When the molecules have non-zero size, we have to deduct this small volume from the total volume of the container.

Lastly, I admit that the analysis I have given above does not address every aspect of your question - it is merely intended to set you in the right direction about rethinking the plus and minus issues you present.

  • $\begingroup$ so that means (P+a^2/V^2) as a whole is numerically less than the P appearing in ideal gas equation. $\endgroup$ Oct 13, 2018 at 8:40
  • $\begingroup$ Also V-nb is the total space for movement of molecules.V is the volume of the container??I f yes can it be found out using ideal gas equation given we use the 'real pressure' $\endgroup$ Oct 13, 2018 at 8:43
  • $\begingroup$ They prof gave a very intuitive proof of the equation and from what I could comprehend he meant that P is pressure due to collision of the molecules with the walls of the container while n^2/V^2 is the pressure due to mutual attraction between gas molecules.P in ideal gas accounts for only the pressure due to collision with the walls of the container since attractions are absent.Plz check. $\endgroup$ Oct 13, 2018 at 8:48

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