Real gas equations and liquefaction.

If you have ever studied thermodynamics, you might recognize this equation:

If you were to draw a graph of vs assuming constant temperature for several temperatures, you would obtain a very familiar set of curves known as isotherms:

In real life, however, we know that gases deviate from this behavior at low temperatures and high pressures due to the interaction between molecules/atoms which are not accounted for in the ideal gas model. Furthermore, ideal gases would never become liquid at a given temperature just by applying pressure. This process is known as liquefaction and, if you come from a country like mine, you would know that this doesn’t make sense, considering that we buy gas for our kitchens in containers like this one and you can even hear the “liquid gas” inside.

Liquefaction of gases is a very common process and, in fact, all real gases can be liquefied if the temperature is low enough. Interestingly, for each gas you can find a specific temperature above which it becomes impossible to liquefy it just by increasing the pressure. This is known as the critical temperature of that gas. Some gases such as helium have a very low critical temperature (5.19 K) whereas substances that are liquid at room temperature show rather high ones (water, 647.27 K). If we think about it considering that helium is the smallest monoatomic gas, it makes sense that the interaction between helium atoms is almost negligible unless the temperature is extremely small, which cannot be said about water molecules.

Since we know now that our magical ideal gas equation doesn’t work in real life, what can we do? Well, luckily (or not, depending on whether you have to memorize them for your next mid-term) we have better approximations for real gases. There are many analytical ways to fit experimental data into an ideal gas equation using a certain correction factor and simply collect a bunch of data into different tables for each gas, but that is a very unsatisfying answer for a scientist. We want a more general equation. One of the most famous ones is the van der Waals equation, which is:

Van der Waals won the Nobel Prize in Physics in 1910 for his work on gases. We can qualitatively describe his equation as a real gas equation with corrections for pressure, thinking that the measured pressure would be equal to the ideal pressure minus a factor that accounts for the interaction between molecules that reduces this pressure and is proportional to the square of the concentration (my personal interpretation of this is that interactions occur mostly between no more than two molecules/atoms and that’s why it depends on the square of the concentration only), and volume, considering that the ideal volume is the total volume of the container minus the volume occupied by the atoms/molecules. a and b depend on each species.

All right, we have a real gas equation. Does it account for liquefaction? Let us see the isotherms for a generic van der Waals equation:

It turns out that the isotherms on the upper right side of the graph tend to fit experimental results very nicely, but we get this interesting phenomenon: below a certain temperature, we can observe some oscillations (the pressure goes up and down as the volume becomes bigger). This does not make any physical sense because volume cannot become both bigger and smaller as pressure increases (in other words, we cannot have several possible volumes for the same species in the same phase at the same pressure). We need to correct this. And that is what Maxwell (yes, the famous Maxwell whose name you’ll keep finding everywhere the more you learn Physics) did with the so-called Maxwell construction. The idea is that a behavior in which a substance can have multiple volumes at a certain pressure can only be observed during a phase change. We know experimentally that during a liquid-gas phase change carried out by expansion pressure always remains constant so we would observe a straight line. Therefore, if we draw a straight line that cuts our van der Waal isotherms in three points, we can account for liquefaction in our model. The Maxwell construction consists in drawing the line in such a way that the area under and above the line are equal, based on the idea that since the hypothetical van der Waals equation isotherm should be mathematically accurate, going throw the straight line and the oscillation should give us the same amount of work (remember that work is the area under the P-V curve).

This also explains perfectly the idea of critical temperature, since we can see that above a certain temperature the oscillations don’t take place anymore. We can calculate it considering that an oscillation contains an inflection point (at which the second derivative becomes zero) so the temperature at which the first derivative is also zero during that inflection point would be the critical temperature (red isotherm).

But why exactly does the van der Waals equation behave like this? This can be easily understood if we turn it into a polynomial in V:

This is a cubic equation, which means that, for a given P, we should have either three real values (as in the oscillations) or one real and two imaginary values of V that satisfy the equation, and that will depend on the temperature (I will not solve the cubic equation, but remember that every polynomial or order n has n complex roots, and if n is odd at least one of them is real).

Let us generalize this behavior to other gas models. Recently a friend asked in a study group linked to this blog at Nagoya University a question from an exam which asked whether gases ruled by the following two equations can be liquefied (i.e. whether they have a critical temperature):

You can try to solve the first and second derivative in both cases and conclude whether there are any inflection points, but an easier approach is to realize that the first equation is quadratic whereas the second is linear. In the case of A, we can always find two different possible values for volume for a given pressure, but if you solve the quadratic equation, you would realize that one of the values should be positive and the other negative, so we ignore the negative one and there should be no ups and downs in our graph on the positive volume side. That should be the same for the linear equation since it should be a bijective relation. Therefore, neither of the gases could be liquefied and they don’t have any critical temperature.

Update (Nov 19th, 2016): The fact that the quadratic equation gives one positive and one negative solution is easy to see by solving the quadratic equation only if we assume that b is a positive constant (this is an exercise for the readers). I did it assuming that b is a physical constant related to the volume of the molecules/atoms, but that was not stated in the problem. Instead, we can see that the plot P-Vm should have an asymptote only in Vm=0, so it should be continuous on the positive side, but if b was negative we would have that the pressure becomes zero when the volume becomes as small as b, and, unlike the van der Waals equation, the pressure just tends to minus infinity as volume decreases, which makes no physical sense. See an example of a plot like that: