# Limits of Real Gas Behavior

1. Mar 10, 2008

### Astronauta

Ok, let's say we have at our disposal a thermometer of constant volume gas.

Inside it, we'll put a real gas, like Helium or Molecular Hydrogen. Ok, the gas compartment is placed inside a mixture of liquid water and ice, so we guarantee the gas inside is at 273.15 K. We measure its pressure (from a Hg column bar), and annotate it. Let's say the value is Pm, pressure for molten ice.

We repeat the process, but this time we place the gas into a mixture of boiling water, so we guarantee the gas will be at 373.15 K. We also register its new pressure, let's call it Pb, pressure for the gas in contact with boiling water.

Ok, after we do that, we are interested in the fraction y = Pb/Pm, for that real gas. Of course this y will vary depending on the number of mols of the real gas in the compartment, let's call it n. But we could, in theory, make a graph of y X n, or maybe y X Pm, which would look similar (the graphs will be equivalent).

We could make this graph for various substances, like H2, O2, N2, He.

Now is the main part: if there were an ideal gas in the compartment, the rate Pb/Pm would be a constant, since the volume is constant, and it would be yIdeal = 1.3661.

The question is: is it possible for any real gas, like H2, in any circumstance, to have Pb/Pm smaller than 1.3661, the value for the perfect gas? Or is it just impossible?

Astronauta

Last edited: Mar 10, 2008
2. Mar 11, 2008

### Astronauta

Ok. For further analyzing the subject, I don't know if it would be acceptable to apply the van der Waals equation for real gases in the preceding situation. It's just an approximation, but let's suppose this equation works for the situation above.

Let's do the calculations:

Van der Waals equation:

$$P = \frac{n R T}{V - b n} - a \frac{n^2}{V^2}$$

In it, a and b and coefficients that vary depending on the real gas.

For H2 we have a = 0.248 L² atm / mol², and b = 2.661e-2 L / mol.

If we are to know the rate Pb/Pm, we could substitute the values of the temperatures in the above equation. The end result would be:

$$\frac{Pb}{Pm} = \frac{373.15 R - a n (1 - b n)}{273.15R - a n (1 - b n)}$$

I'm already assuming the volume of the gas in the thermometer to be fixed in 1 L.

If we suppose n = 0.01 for H2, and use its coefficients, we would find the fraction above to be y = 1.36614, already above that of an ideal gas.

In fact, it seems that it's impossible, if the van der Waals equation is valid in that situation, for any gas to have an smaller value than those of perfect gases.

To see this situation, we could simply force it to be smaller by making an inequation:

$$\frac{373.15 R - a n (1 - b n)}{273.15R - a n (1 - b n)} < \frac{373.15}{273.15}$$

Solving it, we find that either a is negative for a normal value of n, or n has to be really big, which could cause an extremelly high pressure, and therefore would invalidate the van der Waals equation, since it is not supposed to work for pressures that are too high.

The real problem here is that this "mini-proof" that no real gas (i.e., no a and b coefficients) are able to be "bellow" the ideal gas in the Pb/Pm X Pm graph. I would accept it fine if I hadn't seen a graph in which H2 really happens to be bellow that of ideal gas, i.e., it's a straight line going down.

The graph is available in the fifth edition of the Halliday & Resnick & Walker book "Fundamentals of Physics", in the 21th chapter.

How is it possible then for a gas to be bellow the ideal gas in a curve? And why is the preceding equation not valid?

Last edited: Mar 11, 2008
3. Mar 14, 2008

### Astronauta

I talked to a teacher of mine these days, and he argued that the reason H2 is bellow Ideal gas in the Pb/Pm X Pm graph (assuming it does as shown in the Halliday 5th edition book) has something to do with it's low mass and so low ability to transfer momentum between its particles.

He said that since hydrogen has a low "inertial mass" and a greater "electromagnetic independence", Pb is less sensible to the change in mass of the gas than Pm, since hydrogen already has a greater individual independence among its particles. He also said that the only thing able to account for it would be quantum mechanics, and so the van der Waals equation would not work in this case.

In class he also argued about an "aggregation state" of H2 gas in higher pressures, so that when it gets less mols of particles we increase the distance between particles and the aggregation energy is somewhat released.

I don't quite buy that explanation, since it's too much "conceptual based" and cannot be well translated into plotting a graph, at least not without quantum physics. Since I don't know how to work with it, it's like just saying "the explanation is far too advanced", and so I can't argue back.

I guess that maybe some of you may have something to say on this subject? I bet many do have the Halliday book I talked earlier, so are aware of the graph I'm talking about.

Perhaps it would be better to move this topic into the thermodynamics section?

4. Mar 15, 2008

### Staff: Mentor

That would make sense to me. The ideal gas equation is derived assuming non-interacting point particles. A real gas molecule cannot have less volume than a point and it cannot interact less than non-interacting! The only complication I could see is the extra degrees of freedome in a real molecule, and I don't know how that would affect things.