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While going through the derivation of how to calculate the pressure of an ideal gas we consider only the translation of molecules. Why don't we take into account that the molecules may rotate and vibrate which may change the pressure of the gas under consideration? Here is the link of the derivation

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    $\begingroup$ How those motions should transfer momentum? $\endgroup$
    – Alchimista
    Commented Aug 19, 2020 at 9:26
  • $\begingroup$ By hitting the wall. $\endgroup$
    – Kashmiri
    Commented Aug 19, 2020 at 9:34
  • $\begingroup$ which derivation? calculating pressure from what? $\endgroup$
    – Umaxo
    Commented Aug 19, 2020 at 9:50
  • $\begingroup$ I've added the link. $\endgroup$
    – Kashmiri
    Commented Aug 19, 2020 at 10:20

2 Answers 2

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"Ideal gas" is jargon for "mathematical model involving point particles with mass and collisions," so it kind of defines away your question as "not askable," but that's an unsatisfying answer.

For a real gas, you get to include other stuff, like the volume of the particles, or (for a molecular gas) rotations and vibrations. Those count as additional "degrees of freedom" in statistical mechanics, and can be considered independently of the components of the velocity along the x, y, and z axes.

For molecular gasses, as you add energy, temperature increases (and along with it, the velocity of the molecules), and at certain temperatures, those rotational or vibrational modes become important.

Here's a graph of the specific heat* (i.e., change in temperature when you add energy) of diatomic hydrogen:

C_v vs. T Source

Note that, at low temperatures, rotational and vibrational modes are not important, and the ideal gas law works surprisingly well (especially if you modify it a bit for the volume of the hydrogen). At higher temperatures, the ideal gas law will lead you astray. The surprising thing is how well it works at low temperatures.

* Technically the specific heat at constant volume

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    $\begingroup$ Thank you brother. $\endgroup$
    – Kashmiri
    Commented Aug 19, 2020 at 16:11
  • $\begingroup$ Sources online say that the ideal gas law breaks down for low temperatures, rather than high ones. Also, doesn't ideal gas law work decently for, say, O2 and N2 around room temperature? Based on that graph in your answer, rotational energy should already be relevant at those temperatures. I'm still lost here $\endgroup$
    – nog642
    Commented Sep 24, 2021 at 15:58
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For an ideal gas, e.g. point like masses, rotation and vibration cannot store energy. For low temperatures and symmetric molecules this is also true for real gases as an observational fact.

For some real gases, however, rotation and vibration must be taken into account., see e.g.

https://en.m.wikipedia.org/wiki/Heat_capacity_ratio

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