Concept 06 / 06Kinetic Theory of Gases
Real Gases, the van der Waals Equation and Deviations from Ideal Behaviour
Why the ideal gas law fails, how the a and b corrections repair it, the derivation of the critical constants, and why deviations are largest near the critical point.
Every equation in this chapter so far has rested on two idealisations: molecules are points, and they do not attract one another. Both are excellent at low density and both fail as you compress a gas. Where they fail is not a nuisance to be apologised for; it is where liquefaction, critical phenomena and refrigeration live. The van der Waals equation is the simplest correction that captures all of it.
Measuring the failure: the compressibility factor
Define
For an ideal gas at every pressure and temperature, a horizontal line. For real gases, as the graph above shows, wanders. At moderate pressures most gases dip below , some of them dramatically. At high pressures every gas rises above and keeps climbing. Hydrogen and helium at room temperature never dip at all. Any successful theory of real gases must explain all three of those facts, and van der Waals does.
The volume correction
Molecules have size. If each molecule is a hard sphere, the centre of another molecule can never come closer than one diameter, so a portion of the container is permanently unavailable. Replace the free volume by , where is the excluded volume per mole.
The size of is worth deriving because it is a neat counting argument. Two spheres of diameter cannot have their centres closer than , so each pair excludes a sphere of radius , of volume . Since that volume is shared between two molecules, the excluded volume per molecule is . Writing the actual volume of one molecule as , we get . So
The excluded volume is four times the true molecular volume, not one times. Units of : .
The pressure correction
Molecules attract one another, weakly and at short range. A molecule deep in the bulk is pulled equally in all directions and feels nothing. A molecule about to strike the wall, however, has neighbours only on the inward side, so it is pulled back and arrives with less momentum than it otherwise would. The measured pressure is therefore lower than the pressure the molecules would exert if they did not attract.
How large is the deficit? It is proportional to the number of molecules being pulled back near the wall, and also to the number of molecules doing the pulling. Both are proportional to density, so the deficit goes as :
with a constant measuring the strength of the attraction. Units of : .
The van der Waals equation
Putting both corrections into :
or for one mole, with the molar volume,
Notice the physical opposition built into it. The term reduces the available volume, which raises the pressure relative to ideal; the term reflects attraction, which lowers it. Which one wins decides whether is above or below .
Why dips and then rises
Expand for one mole at moderate density. Multiply out and drop the small term :
At low pressure , so , giving
Everything follows from the bracket.
If , that is at low temperature, the bracket is negative and : attraction dominates and the gas is easier to compress than an ideal gas.
If , at high temperature, the bracket is positive and from the start: molecular size dominates and the gas resists compression.
At the temperature where the bracket vanishes, , the first-order deviation disappears entirely and the gas behaves almost ideally over a wide pressure range. is called the Boyle temperature. For hydrogen is only about , far below room temperature, which is exactly why the hydrogen curve in the graph above never dips below .
At very high pressure the approximation above fails and the term always takes over, because approaches and approaches zero, forcing up without limit. That is why every gas eventually has .
Critical constants from the equation
Look at the isotherm diagram above. At high temperature the van der Waals isotherms are smooth and nearly hyperbolic. Below a certain temperature they develop an S-shaped wiggle. Exactly at the boundary, the maximum and minimum of the wiggle merge into a single point of inflection with a horizontal tangent. That is the critical point, and it is found by demanding
and
Write and differentiate:
Dividing the first by the second kills , and at once:
Substituting back into the first condition:
And then from the equation of state itself:
A striking prediction follows. The critical compressibility factor
contains no and no at all, so van der Waals predicts the same for every substance. Real gases give to , so this is both a triumph (a universal number does emerge, which is the origin of the law of corresponding states) and a clear demonstration that the equation is only approximate.
Why deviations are worst near the critical point
At the critical point , so the isothermal compressibility diverges. An infinitesimal pressure change produces a huge volume change; the gas has essentially no resistance to being squeezed. Microscopically this means enormous density fluctuations: regions of gas spontaneously condense and re-evaporate on all length scales, including scales comparable to the wavelength of visible light. Those fluctuations scatter light strongly, and a normally transparent fluid turns milky. This is critical opalescence, and it is a phenomenon no equation built on "molecules are independent" could ever produce. It is exactly where the attraction term and the size term are of comparable importance and neither can be neglected, which is why both the ideal gas law and the van der Waals equation are at their worst there.
Reading the isotherm diagram
Above the isotherms fall monotonically and no amount of squeezing produces a liquid. Andrews established this for carbon dioxide in 1869: above C, cannot be liquefied at any pressure. This is why oxygen and nitrogen, whose critical temperatures are and , resisted liquefaction for decades and were called permanent gases until people learned to cool before compressing.
Below , the van der Waals isotherm develops a loop whose middle branch has , meaning that squeezing the gas would lower its pressure. That is mechanically unstable and cannot exist. What nature does instead is split into two phases at a single pressure, giving the flat horizontal segment drawn in the diagram. The correct pressure is fixed by Maxwell's equal-area rule: choose the horizontal line that cuts off equal areas above and below the loop. The horizontal segment is the liquid-vapour coexistence region, and the locus of its endpoints is the dome, which closes at the critical point.
Key Takeaways and Formulae
, and .
, four times the actual molecular volume; measures attraction; in , in .
Low-pressure behaviour: . Boyle temperature .
Critical constants: , , , , .
Common mistakes. The correction is added to and the correction is subtracted from , never the other way round. Both use and respectively, so watch the mole count. means more compressible than ideal, not less. A gas above its critical temperature cannot be liquefied by pressure alone, no matter how large the pressure. And is not the molecular volume; it is four times it.
Brain Teaser 1
For carbon dioxide, and . Using , predict , and . Compare with the measured values , and , and comment on where the model does well and where it does not.
Solution.
The temperature is accurate to within percent and the pressure is essentially exact. The critical volume, however, is overestimated by about percent.
That pattern of errors is highly informative. Check it with :
, while
Since and both come out right, essentially all of the error in sits in . The reason is that the hard-sphere excluded-volume model, which produced , is the crudest part of the theory: real molecules are soft, they interpenetrate somewhat under pressure, and the effective excluded volume shrinks as the gas is compressed, so the true critical volume is smaller than . Meanwhile and are governed mainly by the balance between attraction and thermal energy, which the term captures rather well. The practical moral: van der Waals is excellent for predicting whether a gas can be liquefied and at roughly what pressure, and poor for predicting liquid densities.
As a bonus, the Boyle temperature: , and indeed , exactly as the formulae demand. Because for is so high, at room temperature is far below its Boyle temperature and shows a deep dip in , which is precisely what the graph above displays.
Brain Teaser 2
A hypothetical gas obeys exactly, at all pressures. (a) Show that it can never be liquefied. (b) Find and show it always exceeds . (c) Does such a gas have a Boyle temperature?
Solution. This gas has the size correction but no attraction, so . That single fact drives everything.
(a) Write and differentiate at constant :
This is strictly negative for every and never zero. A critical point requires a point of inflection with a horizontal tangent, which requires somewhere. Since that never happens, there is no critical point, no S-shaped loop, no coexistence dome, and therefore no liquid phase at any temperature or pressure. You can compress this gas indefinitely and it will simply get denser and stiffer, never condensing.
This is the deep point of the question. Liquefaction is caused by attraction, not by compression. The term alone can make a gas dense; only the term can make it condense.
(b)
Since always (the molar volume cannot fall below the excluded volume), the second term is strictly positive and at every pressure. Such a gas is always harder to compress than an ideal gas. In the language of the graph above, its curve starts at as and rises monotonically thereafter, exactly like hydrogen at room temperature, which is a real gas so far above its Boyle temperature that the attraction term is negligible.
(c) No. The Boyle temperature is where the leading deviation from ideality vanishes, , and with this gives . There is no finite temperature at which the gas mimics an ideal gas over a range of pressures; the positive deviation is there at every temperature, because it comes from geometry rather than from a competition between energy scales.
Brain Teaser 3
Two identical rigid cylinders are filled with the same number of moles of two different gases at the same temperature, well above the critical temperatures of both. Gas A has and in SI units; gas B has and . Without computing any pressure, predict which gas has the lower pressure at moderate density and which has the higher pressure at extreme compression. Justify from the equation.
Solution. Write the van der Waals pressure for one mole explicitly:
The two terms have opposite effects and different dependence on , so they dominate in different regimes. The key is how fast each grows as shrinks.
At moderate density, is still much larger than , so the first term is only slightly enhanced above and the difference between the two gases in (about percent) hardly matters. The subtracted term , however, differs by a factor of about . So gas B, with the much larger , loses more pressure to attraction and therefore has the lower pressure. Its dips further below , which matches the graph above, where the gas with the largest shows the deepest dip.
At extreme compression, approaches . Now approaches zero and the first term diverges like , while the attraction term merely approaches the finite value . A diverging term always beats a finite one, so the repulsive size term wins and both gases end up with . Between the two, the one whose vanishes first at a given , that is the one with the larger , blows up sooner. Gas B has the slightly larger , so at extreme compression gas B reaches the higher pressure.
So the ordering reverses as the gas is compressed. This crossover is the entire content of the versus graph: attraction is a low-and-moderate-density effect because it scales as density squared, while excluded volume is a high-density effect because it produces a divergence. Every real gas shows both, and the only question is which temperature and pressure range you happen to be looking at.
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