The van der Waals gas: two corrections to the ideal gas that predict liquids

Real molecules take up room and tug on each other, so the ideal gas law needs two fixes

Intuition

An ideal gas is a crowd of ghosts: they pass through each other and never feel each other. Real molecules are more like people on a train platform. They cannot stand in the same spot, and they are mildly drawn to a friend a few steps away. Those two facts are the whole story of the van der Waals gas.

Real gases involve interactions between molecules. The oldest real-gas equation of state is the van der Waals equation, and importantly it predicts the gas to liquid transition. The lecture derives it with mean-field theory.

Molecules interact in two ways. They repel when very close, because they cannot overlap. They attract at medium distances, which is the van der Waals attraction. At low pressure the molecules are far apart, so the interaction is weak and the gas becomes ideal. The slide plots potential energy against separation: steep repulsion at short range, a shallow minimum below zero, and an attractive tail that fades to zero.

Fix one: molecules have size, so the free volume is V minus Nb

The exact form of the interaction is not important, and usually not known. The lecture approximates it with a square well: an infinite wall at small separation, a shallow attractive well out to a cutoff, and zero beyond. The infinite wall at small $r$ is why the van der Waals equation replaces $V$ with $V - Nb$. $Nb$ is the excluded volume of all the molecules, the room they take up themselves. It leaves less space to move in, which raises the pressure.

Fix two: attraction pulls molecules together, which lowers the pressure

Adding up every pairwise attraction is too hard. So the lecture uses the mean-field idea. Look from the point of view of one molecule and assume the others are evenly spread through the remaining volume. Its potential energy is then proportional to the density of nearby molecules (how many are in its radius of attraction).

$U_{\text{pot,mol}} = -a\,\frac{N}{V}$

The minus sign means attraction lowers the energy. $a$ is the attraction constant. Multiply by $N$ molecules for the total:

$U_{\text{pot}} = -a\,\frac{N^2}{V}$

To turn an energy into a pressure, use the first law, $dU = T\,dS - p\,dV$. At constant entropy $p = -(\partial U/\partial V)_S$. So

$p_{\text{pot}} = -\frac{d}{dV}\left(-a\frac{N^2}{V}\right) = -\frac{aN^2}{V^2}$

This pressure is negative. Add it to the non-interacting pressure $NkT/(V - Nb)$.

$p = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2} \qquad\text{or}\qquad \left(p + \frac{aN^2}{V^2}\right)(V - Nb) = NkT$

Try it: worked example

Derive the van der Waals equation from the ideal gas law in the order the lecture uses.

Show answer and steps

AnswerReplace V by V - Nb for the excluded volume, then add the attractive pressure -aN^2/V^2 found from U_pot = -aN^2/V.

1. Start from the ideal gas law $p = NkT/V$.
2. Molecules cannot overlap, so only $V - Nb$ is free. Replace $V$: $p = \dfrac{NkT}{V-Nb}$.
3. For attraction, mean-field says one molecule feels energy $-aN/V$. For $N$ molecules, $U_{\text{pot}} = -aN^2/V$.
4. Pressure comes from energy: $p = -(\partial U/\partial V)_S$. Differentiate $-aN^2/V$ with respect to $V$: the derivative of $1/V$ is $-1/V^2$, so $\dfrac{d}{dV}\left(-\dfrac{aN^2}{V}\right) = +\dfrac{aN^2}{V^2}$.
5. Apply the minus sign in front: $p_{\text{pot}} = -\dfrac{aN^2}{V^2}$.
6. Add it to the pressure from step 2: $p = \dfrac{NkT}{V-Nb} - \dfrac{aN^2}{V^2}$.
7. Move the attraction to the left: $\left(p + \dfrac{aN^2}{V^2}\right)(V - Nb) = NkT$.

Exam detail

Learn both forms of the van der Waals equation and which term is which. $b$ and $Nb$ are excluded volume (from repulsion at short range). $a$ and $aN^2/V^2$ are attraction (mean-field at medium range).

Below T_c the isotherms have a loop, and the real gas skips across it

Plot pressure against volume and you get the isotherms of slide 18. Above $T_c$ the curves fall smoothly. At $T_c$ there is a flat inflection point. Below $T_c$ each curve has a minimum followed by a maximum, called the van der Waals loop. In that stretch it seems, unphysically, that compressing the gas would lower its pressure.

The true equilibrium at a given $T$ and $p$ is decided by the Gibbs free energy. In reality the system goes straight from point 2 (gas) to point 6 (liquid), an abrupt change in volume. Between those volumes the liquid and gas coexist. The right-hand plot on slide 19 shows $G$ against $P/P_c$ and a common tangent between 6 and 2 that gives the equilibrium.

Slides 19 and 20 show real CO$_2$ numbers read from a plot. The isotherms are at 243 K, 258 K, 274 K, 304.12 K (the critical isotherm, peak near 74 bar), 334 K and 365 K, with a dashed ideal-gas curve at 334 K, and the two-phase region shaded. Lecture 22 (slide 19) shows the same plot with coexistence plateaus of roughly 36 bar at 243 K, 47 bar at 258 K and 60 bar at 274 K. These are read off the plot and are approximate.

Compare

Strengths of the vdW model

Discontinuous transition from gas to liquid below $T_c$. Coexistence of liquid and gas at certain volumes. A supercritical fluid above $T_c$ (more next lecture).

Limits of the vdW model

Not particularly accurate, and better equations exist for real gases. Mean-field is an approximation, and real molecules tend to clump at high pressure.

Where the Gibbs energy comes from (appendix)

The appendix, taken from Schroeder, shows how to get $G$ for a vdW fluid. The lecture marks it as a supporting derivation.

Try it: worked example

Show how G for a van der Waals fluid is found, starting from dG = -S dT + V dP + mu dN.

Show answer and steps

AnswerAt fixed N and T, dG = V dP. Differentiate the vdW pressure, multiply by V, then integrate in V to get G = -NkT ln(V - Nb) + NkT Nb/(V - Nb) - 2aN^2/V + c(T).

1. At fixed $N$ and $T$, $dT = 0$ and $dN = 0$, so $dG = V\,dP$. Dividing by $dV$ gives $\left(\dfrac{\partial G}{\partial V}\right)_{N,T} = V\left(\dfrac{\partial P}{\partial V}\right)_{N,T}$.
2. Differentiate the vdW pressure $P = \dfrac{NkT}{V-Nb} - \dfrac{aN^2}{V^2}$: $\left(\dfrac{\partial P}{\partial V}\right) = -\dfrac{NkT}{(V-Nb)^2} + \dfrac{2aN^2}{V^3}$. This step is not printed on the slide.
3. Multiply by $V$: $\left(\dfrac{\partial G}{\partial V}\right) = -\dfrac{NkTV}{(V-Nb)^2} + \dfrac{2aN^2}{V^2}$ (eq. 5.55).
4. To integrate the first term, write the $V$ on top as $(V-Nb) + Nb$ and integrate the two pieces separately.
5. The first piece gives $-NkT\ln(V-Nb)$. The second gives $+\dfrac{(NkT)(Nb)}{V-Nb}$.
6. The attraction term integrates to $-\dfrac{2aN^2}{V}$.
7. Add the constant of integration, which can depend on $T$ but is unimportant here: $G = -NkT\ln(V-Nb) + \dfrac{(NkT)(Nb)}{V-Nb} - \dfrac{2aN^2}{V} + c(T)$ (eq. 5.56).

Pitfall

The vdW loop is not a real state of the system. The lecture says it is unphysical. The real path skips from the gas end to the liquid end, with coexistence in between, and the Gibbs free energy decides where.

Recall

Why does the van der Waals equation use V minus Nb and a minus aN squared over V squared?

$V - Nb$ removes the volume the molecules occupy themselves (repulsion at short range). The $-aN^2/V^2$ term is the pressure drop from average attraction between molecules (mean-field at medium range).