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According to Wikipedia:

An ideal gas is a theoretical gas composed of many randomly moving point particles whose only interactions are perfectly elastic collisions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.

If the atoms are point particles the probability of their collisions is zero.

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    $\begingroup$ You may either add a radius to the point particle or include an electromagnetic field associated with each particle and include a charge distribution. $\endgroup$
    – JQK
    May 5 '19 at 14:39
  • $\begingroup$ Related: physics.stackexchange.com/questions/292058/… $\endgroup$
    – valerio
    May 8 '19 at 11:57
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My understanding of this matter is the following: the basic and defining characteristic of the ideal gas is that there are no interactions (neither attractive nor repulsive) between the particles. If there are no interactions there can be no collisions between the particles. This does not include collisions with the walls, which are allowed.

The problem that arises from the lack of interactions is that it does not ensure random motion (and thus ergodic mixing flow). So either one uses the extra assumption that the motion of the ideal gas particles is random or that there are collisions that are elastic.

p.s. The ideal gas is different from the hard-sphere gas. These two types of gas are discribed by different equations of state.

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  • $\begingroup$ The ideal gas law still allows for elastic collisions between particles in the gas. $\endgroup$
    – JMac
    May 8 '19 at 12:53
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According to http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/idegas.html the atoms are hard spheres so they collide with non-zero probability.

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  • $\begingroup$ Hard spheres of what radius? If you make the radius arbitrarily small, the collision probability becomes negligible. If you make the radius large enough to make a difference, you're now dealing with a gas that isn't ideal, since it has significant excluded volume and must be treated by the van der Waals equation. $\endgroup$ May 8 '19 at 12:07

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