Chemistry - Van der Waals real gas equation

Solution 1:

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

Solution 2:

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.