Why is a gas of elastically colliding hard balls of finite size not ideal?


Why is it essential that the particles of an ideal gas are point-like?


Which differing thermodynamic properties of a gas of not point-like particles are most striking in comparison with an ideal gas (of point-like particles)?

  • $\begingroup$ Interestingly the wikipedia article on ideal gas contradicts itself. In the introduction it states ideal gases are composed of point particles, whereas later in classical thermodynamic ideal gases it states only that the separation between particles is much greater than the particle size. $\endgroup$ – BMS Mar 27 '14 at 18:12
  • $\begingroup$ Are you sure that non point-like particles does not form an ideal gas? Isn't the presence of an inner structure the reason why the formula for the internal energy, $U = \frac{3}{2} n R T$, has sometime different factors (e.g .5/2)? $\endgroup$ – giulio bullsaver Mar 27 '14 at 18:12
  • $\begingroup$ @user3376924. To be honest: I am not sure. It has been only an impression. (Among others I found this source: sklogwiki.org/SklogWiki/index.php/Hard_sphere_model but could not delve deeper into it.) $\endgroup$ – Hans-Peter Stricker Mar 27 '14 at 18:19
  • 2
    $\begingroup$ @BMS I disagree with your analysis that the two statements are contradictory. The later statement is a condition on which classical thermodynamic ideal gas applies. It is because particle size is much smaller than separation that we can assume zero size for our calculations. It is like saying that for classical Newtonian mechanics velocities are much less than c. $\endgroup$ – Aron Mar 28 '14 at 9:47
  • $\begingroup$ @user3376924 Internal structure and point-like are not mutually exclusive. Take for example Bose-Einstein Condensates/Cooper pairs. When electrons form cooper pairs, it is their internal structure that allows them to have a point-like behavior. $\endgroup$ – Aron Mar 28 '14 at 9:51

If the particles are not point-like, they will take up some volume. As the gas is compressed, the collision frequency will rise more quickly, which will make the pressure-volume curve change. The corrections in the Van der Waals model of a real gas account for the volume of the particles. Also if they have internal structure, that structure can have degrees of freedom that change the specific heat. The fact that nitrogen and oxygen are diatomic leads to the specific heat of air being $\frac 52R$ at normal conditions. There are two rotational degrees of freedom available along with translation.

| cite | improve this answer | |
  • $\begingroup$ An ideal gas obeys Boyles law. Your gas with finite molecules doesn't obey Boyles law because there is a minimum volume (when the gas molecules are close packed) beyond which the gas can no longer be compressed. Therefore the gas cannot be ideal. $\endgroup$ – John Rennie Mar 28 '14 at 10:25

Because if those particles aren't point objects, you must also take into consideration that they take some space in the system and have properties like density and size which have to be taken into consideration when formulating the laws. This makes everything extremely complicated. A huge part of classical mechanics is only true for point objects for the very same reason.

| cite | improve this answer | |
  • $\begingroup$ I wonder: What observable difference does it make whether the particles are point-like or not - assumed they are "hard balls"? $\endgroup$ – Hans-Peter Stricker Mar 27 '14 at 18:47
  • $\begingroup$ They acquire Volume, Size and Density. $\endgroup$ – user42733 Mar 27 '14 at 18:51
  • $\begingroup$ I believe Hans is referring to observable differences according to theory when modeling a gas as small hard spheres vs point particles. $\endgroup$ – BMS Mar 27 '14 at 19:36
  • $\begingroup$ Oh right. I don't think there will be a lot of observable differences except probably the increase in frequency of the collisions and the heat released. $\endgroup$ – user42733 Mar 27 '14 at 19:49
  • $\begingroup$ Ideal gasses only exist in frictionless vacuums. $\endgroup$ – John Mar 28 '14 at 2:59

There are multiple ways of looking at the ideal gas model. One is to say we have point particles colliding elastically (as an ideal scenario) and proceed to obtain the exact equation of state, i.e $pV=nRT$. The other approach, is to state a priori the relevant length scales and time scales at which the system is going to be studied are sufficiently macroscopic, so as to be able to disregard any microscopic structure in the constituent particles. In both cases, the excluded volume effect (which is what the effect of existence of a finite size of particles in a gas is called) is neglected, as a model assumption in the first case and as a practical approximation in the second. The simplest fluid models that can be solved analytically are the so-called hard sphere models which involve the crudest form of binary excluded volume interactions. In such a case, one can in principle write down the entire virial equation for pressure (the virial coefficients being calculable from microscopic parameters of the model, size of the particles included) and hence the equation of state for this fluid is derivable. The Van der Waals equation is a specific mean field limit of such a virial expansion. The effect of excluded volume will be most prominent a low temperatures and high densities and it will have a substantial effect on the condensation of the fluid (the critical temperature and the scaling laws).

| cite | improve this answer | |

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.