Well my question is simple, if two real gas particles are colliding (head on collision) then will the kinetic energy will be conserved i.e. will it be a perfectly elastic collisions?


Well, what is an inelastic collision, really? Suppose you have two balls made of steel; they collide, then fly away with some lasting deformation, so some energy is lost. With molecules, it is not quite like that. You can't leave a dent on a molecule. It has certain discrete energy levels, and that's it. You either excite the molecule to one of these levels, or you don't excite it at all.

To sum it up, some collisions of molecules are perfectly elastic, and others are combined with excitation of some rotational or (more probable at higher temperatures) vibrational mode in one of the molecules, or maybe in both.

Noble gases which have no molecules and hence neither rotational nor vibrational modes may enjoy perfectly elastic collisions up to pretty high temperatures.

  • $\begingroup$ " It has certain discrete energy levels, and that's it. " are you referring to Molecular orbital theory? Because, why would otherwise a molecule have energy levels? $\endgroup$ – Gaurang Tandon Jan 14 '18 at 14:31
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    $\begingroup$ A molecule has discrete energy levels because it is a finite quantum system. Whether you describe it in terms of molecular orbital theory or otherwise is irrelevant. $\endgroup$ – Ivan Neretin Jan 14 '18 at 14:51

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