Contribution of rotational mode in ideal gas law

Question

The ideal gas law is very accurate for an $N_2$ gas when the temperature is around 300K and the pressure is around 1atm. At these conditions, $N_2$'s compressibility factor is 0.997, which means that for fixed temperature (around 300K) and fixed pressure (around 1atm), we have $$P_{\text{real}}=0.997 \cdot P_{ideal}$$ i.e., the error is 0.03%. This seems very weird to me. Here is why:

The derivation of the ideal gas law assumes spherical molecules, while a diatomic molecule like $N_2$ is not spherical. Furthermore, at room temperature the rotational mode is active, and by equipartition theorem, the average rotational energy is of the same order as the translational (the exact ratio is 2/3). It is intuitive that for fixed temperature and volume, this additional kinetic energy (due to rotation) should affect the pressure: an $N_2$ molecule bounces on the wall of the container due to its translational kinetic energy and also "slaps" the wall due to its rotational. This sounds hand-wavy, so I tested it with a simple two-dimensional model, where the molecule is a weightless rod with identical point-particles at its ends. Suppose this rod is approaching a wall with velocity $v$ while rotating with angular velocity $\omega$, such that the translational and rotational kinetic energies are equal. I simulated this model (the angle of contact at collision is assigned randomly) and found my intuition correct: after the collision, the new translational velocity has (on average) magnitude significantly larger than $|v|$: around 10% increase! This toy model indicates, that for the conditions I mentioned in the beginning, $N_2$'s compressibility factor should be significantly higher than $1$.

I understand that there are more factors at play (like intermolecular forces), but this even adds to the surprise that the ideal gas law is so accurate for $N_2$!

Answer 1

The derivation of the ideal gas law assumes spherical molecules

The ideal gas law assumes zero-volume and zero-interaction molecules. Sphericity doesn't really come into it. Real molecules occupy volume and have interactions, but these are small corrections for $N_2$ at room temperature.

Suppose this rod is approaching a wall with velocity 𝑣 while rotating with angular velocity 𝜔 , such that the translational and rotational kinetic energies are equal. I simulated this model (the angle of contact at collision is assigned randomly) and found my intuition correct: after the collision, the new translational velocity has (on average) magnitude significantly larger than |𝑣| : around 10% increase!

What happens to the ratio of velocity and angular velocity after this? I suspect you will find that that (on average) the rotational velocity has reduced. On some later bounces energy will go back into the rotational mode and on those bounces the translational velocity will show a decrease from what you expected. Rather than look at the average change on a single interaction from one particular state, you should look at how the energy moves over multiple bounces.

Answer 2

Some observations that you will hopefully find helpful:

• There is some randomness in how the molecule is spinning (how fast and which direction).
• There is also some randomness in what direction the molecule is pointing when it hits the wall.
• These variables, as well as the overall velocity of the molecule, are all important in determining how hard the wall gets "slapped".
• It's possible to for molecule to hit the wall both less hard and more hard than a solid ball of the same mass and velocity would. For example, if the molecule isn't rotating much when it hits, and the collision causes it to rotate more, that means there's less kinetic energy left over to go into translational motion. So the impact on the wall is actually less.
• For any given collision with the wall, you can imagine running the movie backwards and the result will also be a physically valid collision.
• At thermal equilibrium, the Boltzmann distribution gives precise predictions for how much translational and rotational energy a particle should have on average.
• If you run a simulation with excess rotational energy compared to this prediction, then colliding with the wall will on average result in some energy being transferred from rotational motion to translational motion.
• Likewise, excess translational energy will tend to get converted to rotational energy. Anyway, the point is you have to match the Boltzmann distribution to get your simulation to predict what actually happens in a gas.