enter image description here

In the above graph,the minima of the curve for methane is more than that of nitrogen. Also, for a given value of pressure, the value of $Z$ for methane is less than that of nitrogen. They seem to meet at high pressure. Why do some gases have lower value of Z?

The same thing is observed if you compare methane and carbon dioxide. But they don't seem to meet at high pressure. This is because the $Z$ for methane increases at a faster rate. Why does Z increase faster for some gases?

The real gas equation is $$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT$$ I tried expressing $Z$ in terms of $P$ and $T$ by eliminating $V$.

$$V=\frac{ZRT}{P}$$ $$\left(P+\frac{aP^2}{Z^2R^2T^2}\right)\left(\frac{ZRT}{P}-b\right)=RT$$ $$ZRT-Pb+\frac{aP}{ZRT}-\frac{aP^2b}{Z^2R^2T^2}=RT$$ Multiplying by $Z^2$ on both sides, $$Z^3RT-Z^2(Pb-RT)+\frac{ZaP}{RT}-\frac{aP^2b}{R^2T^2}=0$$ $$f(Z,P)=0$$ (I am assuming constant temperature)

Should I find $\frac{\partial f(Z,P)}{\partial P}$ and make it equal to zero to get information about the minima?


2 Answers 2


The trend you've noticed here is a very interesting way of studying the interactions of a gas with itself.

You are certainly right about how to find the minimum of these curves, but the important thing here is what that minimum actually means.

$$Z=\frac{PV}{RT}$$By analyzing this, we expect that $Z$ will tell us something useful about the way the volume of a gas changes as pressure changes (holding $T$ constant just gives better information on a graph). Specifically, we see that if pressure increases but volume decreases more, we will have a number less than 1 and if volume decreases less than pressure increases, we will have a number greater than 1. So what does that number tell us?

Well, if pressure increases and the volume gets even smaller than we would expect for a real gas, these gases must be attracted to each other. Then, we can interpret the minimum of this graph as being the pressure (for a given temperature) at which a gas molecule is attracted to itself most strongly. Beyond this point, repulsive interactions become more and more important. Notice that there is also no maximum. Or, there is no point where attractive interactions dominate repulsive interactions again which is what we would expect.

Now, to explain why we see the behaviors we do from certain molecules, we have to imagine what is causing there to be attractive and repulsive interactions at all. This is due to polarizability and the amount of van der waal's forces.

Indeed, if we look up value of the parameter $a$ for the molecules present (I have them in a textbook) we find:$$a_{\ce{H2}}=.24646$$$$a_{N_2}=1.3661$$$$a_{\ce{CH4}}=2.3026$$$$a_{\ce{CO2}}=3.6551$$where $a$ is a measure of how attracted the molecules are to themselves (in the horrible unit of $\frac{dm^6*bar}{mol^2}$).

These values do indeed follow the trend we see in the graph above. That is, larger value for $a$ has a larger attractive deviation initially. $\ce{H2}$ is something of a special case because it is effectivly unpolarizable, so we see a linear increase in the compressibility factor with pressure. This basically means that hydrogen will deviate from the ideal gas precisely the same at all pressures.

My best guess at why the slopes then increase at different rates has to do with the size of the actual molecule. Again, looking up the values for $b$ which represents the excluded volume of a molecule, we have:$$b_{\ce{H2}}=.026665$$$$b_{N_2}=.038577$$$$b_{\ce{CH4}}=.043067$$$$b_{\ce{CO2}}=.042816$$with $b$ in units of $\frac{dm^3}{mol}$.

So, this predicts methane will have a slope increasing more quickly than both nitrogen and carbon dioxide which is what we see and that hydrogen will have the shallowest slope which is again what we see.

Hope that helps.

  • $\begingroup$ So the equation $f(Z,P)=0$ will not work properly because it will give a Maxima and a minima. $\endgroup$
    – Aditya Dev
    Commented Feb 26, 2016 at 11:04
  • $\begingroup$ I assume there's one compound which reaches the lowest value of Z at a given temperature. I wonder what it is, and what makes it so. Wikipedia has a list which suggests iodobenzene is up there, which I didn't quite expect. $\endgroup$ Commented Feb 26, 2016 at 12:23
  • $\begingroup$ @NicolauSakerNeto cubic functions can also have a Maxima. But Z vs P curves don't have a Maxima. Is there a way to get the curve equation using my approach? $\endgroup$
    – Aditya Dev
    Commented Feb 26, 2016 at 13:34
  • $\begingroup$ Does $f(Z,P)=0$ even make sense? It was derived from the real gas equation. $\endgroup$
    – Aditya Dev
    Commented Feb 26, 2016 at 13:36
  • $\begingroup$ Can you comment on the role of a temperature in this graph? If the temperature is changed, keeping P constant, what changes will be observed? And will the line for dihydrogen remain straight? $\endgroup$ Commented Jan 14, 2018 at 11:16

Considering Z for the van der Waals equation (upper part of the image), deriving this equation with respect to Pressure(P), the slope at P = 0 (initial slope of the curve) can be negative or positive. If "b" is greater than the term "a / RT", the effect of molecular size governs the behavior of the system, the slope is positive and Z>1. If b is less than a / RT, the attractions prove that the gas has a volume less than that calculated through the ideal gas equation, the slope is negative and Z<1.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.