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I completely understand that the average energy of each degree of freedom in a thermodynamic system is (1/2)kT and that we do not consider the spin about an axis of symmetry in a polyatomic molecule as the atoms are spread out compared to the size of nucleus so the Moment of Inertia becomes significantly small for the spin about its own axis and that energy would not be comparable with the rotational freedom.

But why would we not consider it for Monoatomic gases as according to the law of equipartition equal energy would be given to each degree of freedom and so that would increase the angular velocity to compensate it?

P.S.: Please try to avoid quantum physics, etc if possible.

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    $\begingroup$ Are you talking about spin of nuclei, or spin of electrons? $\endgroup$
    – Pavlo. B.
    Apr 27 '21 at 23:10
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    $\begingroup$ If you consider the atom as a point particle it can have three different directions for speed (three degrees of freedom). But there is no rotation for point particles (in reality the atom is not point-like, but the energy contained in the rotation can never be as high as that of the linear motion). $\endgroup$
    – Deschele Schilder
    Apr 27 '21 at 23:17
  • $\begingroup$ spin enters in macroscopic ball collisions osapublishing.org/DirectPDFAccess/…., so the answer must be that the size of the atom is such that it can be assumed point like. $\endgroup$
    – anna v
    Apr 28 '21 at 5:05
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    $\begingroup$ the answer here says so physics.stackexchange.com/q/192746 also john's here physics.stackexchange.com/q/168943 $\endgroup$
    – anna v
    Apr 28 '21 at 5:10
  • $\begingroup$ @Pavlo.B. Indeed, even electron orbital momentum could be meant. $\endgroup$
    – my2cts
    Apr 28 '21 at 9:44
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If we avoid quantum mechanics, then spin does not exist hence it need not be considered :)

More seriously, considering spin in the basic statistical physics texts would only unnecesarily complicate the discussion - remember that these texts are designed for teaching students statistical physics rather than presenting a theoretical research in real phenomena.

A good practical reason why spin can be neglected is that in absence of magnetic field the spin states are degenerate, so the repartition of energy between them is trivial, and need not be specially considered. Moreover, many monoatomic gases have zero total spin (the ground state is often a singlet state).

If magnetic field is present however, one may have to account for spin - perhaps, people withe xpertise in astrophysics or plasma could come up with relevant examples.

Update
It has been noted in the comments and clarified in the edited question that the point of the question is not spin of individual particles, but the angular momentum of molecules (which consists of the mechanical angular momentum of a molecules as a whole and the spins of the constituting particles). In this respect classical statistical mechanics treats atoms as point-like objects or, at best, spherically symmetric objects, so that the direction of the angular moment can still be ignored in the thermodynamic calculations.

The existing asymmetry may have to be accounted for at the level where the internal structure of atoms plays role, i.e., where the transitions between atomic energy levels are important. As an example one can consider thermodynamic state of a gas in a discharge lamp or a gas laser. Again, such cases are not considered in basic stat. physics textbooks, and rather easily treated once the basics are mastered.

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  • $\begingroup$ On the first sentence: this is not true. There is a classical model of intrinsic spin, as there is a quantum model of spin. Classically it arises in the considerations of Lorentz-covariant symmetries and leads to a classical spinor (one whose components are all c-numbers and can all be observed simultaneously). $\endgroup$
    – Andrew Steane
    Apr 28 '21 at 12:09
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    $\begingroup$ @AndrewSteane This is technically correct, but this is not the point of the question. Moreover, I don't think OP had in mind relativistic statistical mechanics. $\endgroup$
    – Roger Vadim
    Apr 28 '21 at 12:16
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The point is that for the atoms to acquire a substantial amount of rotational energy they have to collide with other atoms in a very specific way. Of course, all atoms can have the same rotational energy as the linear energy. They can even all have only rotational energy without linear energy. Likewise, they can have linear energy only. It's very unlikely though, due to the Nature of collisions between the atoms where only a substantial amount of rotational energy is exchanged if they collide tangentially, and for the collisions to impart a substantial amount of rotational energy to the atoms the collisions have to be in the same relative (to the direction of rotation) direction every time they collide, which is highly unlikely. The rotational motion isn't taken into consideration for defining the temperature of a monoatomic gas. For a diatomic gas, only the rotational motion associated with the axes perpendicular to the axis connecting the atoms is taken into consideration (you said you understand this). In a way, these two last rotations are connected more to linear motion than with the rotational motion of the atoms themselves.

I completely understand why we do not consider the spin about an axis of symmetry in a polyatomic molecule

For the same reason the rotational energy of individual atoms is not taken into consideration (the rotation of a diatomic around the length axis of a diatomic involves atomic rotation only). Why are these rotations not taken into consideration? For the same reason as for the single atoms.

When atoms have a linear velocity $v$, then the rotational velocity at the surface of the atom can be at most $2v$, the maximum relative velocity between two atoms. In that case the atoms have a rotational energy of $\frac{4}{10}mv^2$, which is almost the same as the kinetic energy. The probability that this situation is realized is very very small though.

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It's because of the symmetry assumption of the monoatomic gas particle.

To put it simply, if a monoatomic gas "rotates", the state after the rotation is indistinguishable from the original state.

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  • $\begingroup$ A single atom may have orbital angular momentum, electron and nuclear spin. $\endgroup$
    – my2cts
    Apr 28 '21 at 9:48
  • $\begingroup$ Yes, but in a classical thermodynamics model, the energy is assumed to be zero. $\endgroup$
    – Stefan Tanuwijaya
    Apr 28 '21 at 12:00
  • $\begingroup$ Are you talking about classical particles or quantum mechanical atoms? In the classical case a rotation has a distinguishable effect. In the quantum case it does not, that is, unless the atom has some kind of angular moment. You should also distinguish the rotation of the gas from the rotation of a particle in it. Your answer needs a few repairs. $\endgroup$
    – my2cts
    Apr 28 '21 at 14:14

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