# Ideal Gas: Equation Of State in Diatomic Gases: still valid?

1. Sep 6, 2006

### Clausius2

That came out in my qualifying exam.

The question was if a gas with rotational and vibrational energies enhanced, does obey the Ideal Gas Equation PV=NKT.

You can answer with all the tools you have at hand. My comitee member didn't agree with me, but anyway I persisted in my answer and I said: yes it does, assuming the energetics of the rotation and vibration are uncoupled.

I based my answer on the fact that the only partition function which depends on volume is the Translational partition function. And as we know the pressure can be found as:

$$P=NKT\frac{\partial (ln Q)}{\partial V}$$

if you substitute there the partition function, the only contribution comes from the translation one, so the only thing that is producing pressure is the translational motion of the molecules of the gas, no matter if they are vibrating or rotating.

What is your opinion?

PS: God knows that I will remember my persistance in the exam about this fact the rest of my life. I kinda challenged a faculty. But I justified my answer adecuately I think.

Last edited: Sep 6, 2006
2. Sep 6, 2006

### Andrew Mason

The problem is that when you introduce vibrational energy you change the (average) moment of inertia of the molecules. As temperature increases, proportionately more energy will go into rotational energy so the relationship between translational energy (pressure) and temperature will not be linear.

AM

3. Sep 6, 2006

### Bystander

Recall the "properties of an ideal gas:" no intermolecular potential; means non-interacting particles; means no translational coupling with rotational or vibrational energies; there are no restrictions regarding angular moments, degrees of freedom within molecules, or couplings among intra-molecular degrees of freedom.

4. Sep 6, 2006

### Gokul43201

Staff Emeritus
I'm not following this argument. How does the second clause follow from the first?

My current opinion is that Clausius is right. The existence of rotational/vibrational states does not alter the volume dependence of the total number of available states, $\Omega$ by more than a scaling factor. Take the logarithm and find the volume derivative; the constant coefficient disappears and you're left with the same equation of state.

5. Sep 6, 2006

### Galileo

I agree with Clausius as well. The energy equation of state will change, but not the pressure equation of state.
The internal degrees of freedom have no effect on the pressure equation for an ideal gas, which is natural since the pressure is due to momentum transferred to the wall by the rebounding particles and only center of mass momentum is involved in that transfer.

6. Sep 6, 2006

### Clausius2

Thanks folks. It seems that all the business of statistical mechanics is built over the fact that there is no atomic interaction between particles, so I would think that a ideal gas is still ideal under vibration and rotational motion of its particles under the assumption that the gas behaves as ideal, as the boltzmann statistics does, doesn't it?. Therefore the theory is consistent with itself, as no other way could be.

7. Sep 7, 2006

### Andrew Mason

If it is correct that the proportion of rotational energy to the total kinetic energy of the gas molecules increases, then necessarily a smaller proportion of the total kinetic energy would be in the form of translational and vibrational energy. Since T is a (statistical) measure of the total kinetic energy of the gas molecules, this would mean that the ratio of translational energy to T changes with temperature. Since P is proportional to translational energy, the relationship between P and T would not be linear (as in P = nRT/V) so it would not behave as an ideal gas.

I may be wrong that more energy will go into rotational energy as temperature increases. I'll have to think a bit more about that.

My thinking at at this point is that if you take a spinning polyatomic molecule and increase the amplitude of vibration of a vibrational mode, some of the vibrational energy is transferred into additional rotational energy (because angular momentum is conserved, when the moment of inertia changes, the rotational energy changes).

AM

8. Sep 7, 2006

### Bystander

Equipartition principle --- Cv, Cp are greater for polyatomic gases than for monatomic. Translational energies of all are equal for a given T.

9. Sep 7, 2006

### Andrew Mason

Well, I think the whole issue here is whether the equipartition principle applies to polyatomic molecules with significant vibrational modes. Obviously, if the equipartition principle applies, translational energy will always maintain the same proportion of total energy. The real question is whether Cv varies with temperature.

AM

10. Sep 7, 2006

### Clausius2

You are talking about the coupling of rotational and vibrational energies. Re read the answer I provided (the first post). I said that under the assumption that both are uncoupled the equation of state is valid. I don't mean in real world they are uncoupled, but what I mean is that my answer was consistent with the assumptions. In other words, I used the accurate words.

11. Sep 7, 2006

### Andrew Mason

Ok. I guess the question then is whether their is any physical reason that the equipartition principle would not to apply to all degrees of freedom of the molecule. The only thing I can think of would be some restriction on the vibrational frequencies (eg. some sort of resonance effect within the molecule that would depend on the frequency of impact on the container walls).

Just out of curiosity, what was your examiner's argument?

AM

12. Sep 8, 2006

### Bystander

Each mode has a "characteristic temperature," theta, for which kT is equal (give or take powers of pi, and other assorted constants) to the ground state energy. For translational modes, in ordinary systems, V large compared to molecule, theta is quite small; rotational modes, again, small; vibrational modes depend on the molecule; excited electronic states, again on the molecule. Heat capacities are functions of temperature and thetas and the spacing of the energy levels for a particular mode. You get into a temperature range where T ~ theta of a mode, the calculation of heat capacities gets messy --- if you're far above, fully excited mode, or far below, unexcited mode, you simply add up the number of active modes to calculate heat capacity.