# Van der Waals real gas equation

Van der Waals corrected the ideal gas equation for real gases and gave the following equation:

$$\left(p+\frac{an²}{V^2}\right)\left(V-nb\right)=nRT$$

I’m very confused about one aspect of this equation, which is this: why add the pressure term but subtract the volume term?

Edit: If somebody could please just explain the physical significance of the corrected equation and also, exactly what Van der Waals was trying to do by correcting this equation, it would be tremendously helpful for me.

• Where do you see that you substract the volume?
– G M
Aug 24, 2016 at 10:37
• The equation does not have any wall-related parameters in it. It describes an idealized situation without walls at all. Otherwise things might have been different. Aug 24, 2016 at 11:59
• Think of an infinite expanse of gas at certain pressure and temperature, where such-and-such volume contains the prescribed amount of gas... No, that's too far-fetched. OK, think of a gas in a container with idealized walls which do not interact with anything. Aug 24, 2016 at 12:11
• @IvanNeretin I respectfully disagree with your second comment, the repulsions act over relatively short distances, and the potential on average is attractive at intermediate distances. Aug 25, 2016 at 13:14
• Got it. You were right, and I was not. Of course it is attraction that matters. It is attraction that is responsible for the critical point and the emergence of liquid state after that, even though the liquid itself is not described. Aug 25, 2016 at 13:46

Let me begin my discussion by first introducing the notion, using some intuitive model building, and will try to derive, or perhaps guess the van der Waals equation of state.

The perfect gas equation, $PV = nRT$ doesn’t allow for any interactions between molecules. The van der Waals equation is a refinement of this model, in the sense that it introduces the possibility of both repulsive and attractive interactions.

Imagine the gas molecules are small, hard, impenetrable spheres. This accounts for repulsive interactions, as now they can’t get arbitrarily close to each other. Consequently, they are no longer free to move about in a volume, $V$, and only have access to a much smaller volume $V-nb$ where $nb$ is roughly the volume taken up by the molecules.

The modified equation now becomes: $$P = \frac{nRT}{V-nb}$$

Now to calculate the excluded volume: $$V_\text{molecule} = \frac{4}{3}\pi r^3$$ where $r$ is obviously the radius.

Now note, a particle is surrounded by a sphere of radius $2r$ (two times the original radius) that is forbidden for the centres of the other particles. If the distance between two particle centres were to be smaller than $2r$, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.

The excluded volume for the two particles (of average diameter $d$ or radius $r$) is: $\frac{4}{3}\pi (2r)^3$ or, $8V_\text{molecule}$. Divide this by two to get the volume excluded per molecule and we get $b \approx 4V_\text{molecule}$

Note: Of course, molecules are not infinitely hard, as van der Waals thought; there exists some “softness”. Thus, the factor four yields an upper bound; empirical values for b maybe lower.

Now turn on an attractive force between the particles.

Assume, notwithstanding the existence of this force, the density of the fluid is homogeneous.

And assume that a large majority of the particles are always surrounded by other particles, which attract each other (unless they get too close, when they repel; thus, the bulk of the particles do not experience a net force pulling them to the right or to the left. They feel that the container is infinitely large without any walls.

Now, the surface particles feel a net force from the bulk particles pulling them into the container, because there are no particles on the side where the wall is (and we assume here that there is no interaction between walls and particles, which is not true in reality).

This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density: $$F = \frac{N_\mathrm a}{V_\mathrm m}$$

Assuming homogeneity, the number of surface particles is also proportional to the number density. So the total force on the walls is decreased by a factor proportional to the square of density.

Since pressure is force per unit area, it's decrease proportional to: $$a'C^2 = a'\left(\frac{N_\mathrm a}{V_\mathrm m}\right)^2 = \frac{a}{V_\mathrm m^2}$$

Now, the modified equation becomes: $$P = \frac{RT}{V_\mathrm m-b}- \frac{a}{V_\mathrm m^2}$$

Derivation using Statistical Mechanics

For a system of $N$ particles, in 3 dimensions

The partition function: $$Z = \frac{1}{N!} \int \mathrm dp^{3N}\mathrm dx^{3N}\mathrm e^{-\frac{\beta}{2M} \Sigma p^2_n}\mathrm e^{-\beta u(x)}$$

This integral factorises into two, an integral over $p$, and an integral over $x$, i.e. a kinetic energy contribution and a potential energy contribution.

I write the result as $$Z = \frac{z^N}{N!}$$ for an ideal gas, this result is $$\frac{V}{\Lambda^{3N}}$$ where $$\Lambda = \frac{h}{(2\pi mk_\mathrm bT)^{1/2}}$$ (thermal wavelength)

For a van der Waals gas,

$$U(r) = \begin{cases} \infty, & \text{when }r<d \\ -\epsilon\left(\frac{d}{r}\right)^6, & \text{when }r \ge d \end{cases}$$

basically attractive up to a certain distance, and repulsive when they come too close. r is the distance between the spheres and d is the distance where they touch. Also, each particle moves independently in an average potential field offered by the other particles

Also, for an ideal gas we integrate over all space i.e. the entire volume of the container, however, we cannot do that anymore. Instead our volume of integration is $V - Nb'$ (or $V – nb$, same as what we did before, $n$ is number of moles instead of number of particles). $$b'= \frac{2\pi d^3}{3}$$

Now, assume that in a shell of thickness $\mathrm dr$ there are $\frac{N}{V} 4 \pi r^2\mathrm dr$ particles, i.e. the position of the particles is averaged (reality, like always is different)

So for a single particle, the attraction felt is $$\phi = \int_{d}^{\infty} u(r)\frac{N}{V} 4 \pi r^2\mathrm dr$$ and $$z = \frac{(V-Nb')\mathrm e^{-\phi/2k_\mathrm bT}}{\Lambda^3}$$

I will omit the computations and quote the results,

$$\phi = -2a' \frac{N}{V}$$ where $$a' = \epsilon (2/3) \pi d^3$$

going back and putting things together,

$$\ln Z = N\ln(V-Nb') + \frac{N^2 a'}{Vk_\mathrm bT} - N\ln(\Lambda^3) - \ln(N!)$$

Now, $$P = k_\mathrm bT \frac{\partial \ln Z}{\partial V}$$ which turns out to be,

$$P = \frac{Nk_\mathrm bT}{V-Nb'}- \frac{N^2 a'}{V^2}$$

UPDATE: Physical Interpretation

I am writing this section because I get the impression (from the discussion the in the comments) that my answer hasn't adequately addressed this.

In the limit $P \to 0$ (or $V_m \to \infty$) we can expect perfect gas behaviour, i.e $PV = \text{constant}$

However, say at high pressures, we observe that real gases would have larger molar volumes than expected, so we remove a term from the volume.

Similarly, the pressure is lower than ideal gas pressure, so we add a term to it to compensate.

So, instead of saying $PV_m = \text{constant}$, we say $(P +\text{"something"})(V_m - \text{"something"}) = \text{constant}$

It also accounts for critical behaviour, simply solve the following system to obtain critical constants (since the critical point, is a point of inflection on the PV diagram) $$\frac{\partial P}{\partial V_m} = 0$$ $$\frac{\partial^2 P}{\partial V_m^2} = 0$$

Of course, it would be too optimistic to expect a simple, elegant model such as this to truly determine all properties of gases with complete precision; the van der waals equation of state does have its failings, but if nothing else, it improves at least the qualitative discussion of properties/behaviours of real gases

When you see the ideal gas equation, derieved from the kinetic theorey of gases, there are many assumptions taken for the ideal gases, a few like their Potential energy is 0, the molecules are spherical, and so on. What different real gas equations attempt to do is to reduce the number of assumptions taken by providing "corrections". Van der Waal attempted to correct the pressure and volume terms.

According to Van der Waal, his equation of real gas should be of this format:

{ Pressure on walls of container without any intermolecular attraction }. { Volume available for motion of molecules } = nRT

In ideal gases, we assume that there are no intermolecular attractions between any of the molecules, thus the pressure of the gas is used in the equation. Also, we assume that the volume available for motion of the molecules is the volume of the container itself.

What exactly is the volume available for motion of molecules? It is the space in which the molecule can move in. If two molecules are near each other, then neither can travel into the space occupied by the other molecule. Thus the volume available for motion is actually less than the volume of container.

Thus, Van der Waal introduced two correction terms, $\frac{n^2a}{V^2}$ and $nb$, which are the pressure correction term and the volume correction term respectively.

If we account for intermolecular attractions, the pressure of real gases increases. If we assume molecules to occupy space in real gases, the volume available for motion decreases.

• But are you absolutely sure about the fact that pressure increases? Most texts describe the situation quite differently. Please refer to my response(comment)to Ivan Neretin's comment.
– user33789
Aug 24, 2016 at 11:57
• @Kaumudi, I'm certain that pressure increases. Aug 24, 2016 at 12:03
• But most of the texts seem to point to the contrary. Even Khan Academy!
– user33789
Aug 24, 2016 at 12:07
• @Kaumudi, the corrected pressure we plug in the real gas equation increases. Real gases' molecules have intermolecular attractions, which decrease the overall pressure of the real gas. If we were to treat it as an ideal gas, we would only put the overall pressure. But since it is a real gas and we have to exclude the intermolecular attractions, pressure increases. Aug 24, 2016 at 12:13
• Since it is a real gas, we are required to take the intermolecular forces into consideration! You seem to have got it all wrong, I'm afraid. Read getafix's answer.
– user33789
Aug 24, 2016 at 14:28