0
$\begingroup$

Why is it necessary that all inter-molecular collisions in an ideal gas be elastic?

My understanding is that a gas behaves ideally so long as the potential energy arising from inter-molecular interactions is negligible compared to the kinetic energy. Collisions are necessary to establish equilibrium, but the nature of the collisions seems to me irrelevant.

Further, it seems to me that, following a transfer of heat to the gas, inelastic collisions are necessary in order to properly distribute energy among the internal molecular degrees of freedom (vibration, rotation).

$\endgroup$
2
$\begingroup$

inelastic collisions are necessary in order to properly distribute energy among the internal molecular degrees of freedom (vibration, rotation).

They are, but the ideal gas (nominally) does not have such internal degrees of freedom.1 Allowing them is a simple extension that is often made without announcing it2 and with that extension you get some degree of inelasticity for exactly the reason you state.


1 For instance see the part of the Wikipedia page where it says

*The ideal gas model depends on the following assumptions [...] The molecules of the gas are indistinguishable, small, hard spheres".

2 When we introduce the effects of molecular structure on the heat capacity for instance. The Wikipedia article does that and the authors/editors appear to have failed to notice the conflict between the two parts of the article.

$\endgroup$
  • $\begingroup$ If the equation of state $PV=NkT$ still holds for polyatomic molecules, then why initially restrict analysis to particles with only the three external translational degrees of freedom? What do we gain? $\endgroup$ – creillyucla Jul 3 '14 at 4:07
  • $\begingroup$ Without internal degrees of freedom you have a simpler model and can develop the kinetic theory of gasses without distractions. Indeed this is usually done in a intro-level course. $\endgroup$ – dmckee Jul 3 '14 at 4:17
1
$\begingroup$

A reason from statistical physics point of view is, that with hard spheres the ensembles are equivalent. This is what one is often interested in. If you want to consider other possible interactions it is in general necessary to check whether this is true. For a detailed consideration you may have a look in the textbook Statistical Mechanics Rigorous Results, David Ruelle, 1969.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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