How is the Energy of a Star's Atoms Calculated and Related to its Temperature?

[shirin]

Hi
1) As a star is composed of a huge number of atoms and molecules in the sate of gas, how is it near to reality to consider it as an ideal gas?
2) Is the energy of these atoms pure kinetic? I mean is the energy of each atom/molecule 1/2mv^2?
3) how is KT related to the energy of these atoms?
Thanks

 

[Astronuc]

Most of the matter (mostly hydrogen and helium) in stars is plasma, that is the conditions are so hot, and there is so much electromagnetic radiation in the form of gamma and X-rays, that most matter is in the form free nuclei and electrons.

There is a spectrum of kinetic energies.

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html

 

[Ken G]

1) As a star is composed of a huge number of atoms and molecules in the sate of gas, how is it near to reality to consider it as an ideal gas?

The concept of ideal gas actually works better the more molecules and atoms there are, as long as the density does not get too high (and stars usually don't). The two things that break the concept are interparticle forces and having steep "gradients" (meaning, the environment changes dramatically between nearby locations, where "nearby" means on the scale of the distance between particles-- a distance that gets shorter, and less of an issue, if the density of particles is high). It sounds like you are confusing the problem presented by forces between particles, which become an issue if they are important over the scale of the distance between particles (so that's where high density can be a problem, but stars in many cases don't get dense enough for that to matter), with the "modes of energy" that can be held internally to the particles (like the binding and excitation of atoms and molecules).

The issue of internal energy modes does not relate directly to the ideal gas law, because the ideal gas law is fundamentally just a relationship between the temperature and the kinetic energy of each particle on average, and that can be used to determine the pressure simply if one assumes the particles have a much higher rest energy (mc²) than kinetic energy. So that doesn't have anything to do with what might be going on inside that atom or molecule, it's all about the motion of the whole particle, and the reasons why the kinetic energy of that motion will be 3KT/2 in statistical equilibrium at temperature T. Internal energy is something different, and if thermalized and treated classically, it will also be related to the kinetic energy of motion-- for example, a diatomic molecule gets kT of internal energy to go along with its 3kT/2 energy of motion. Atoms and molecules have additional internal energy owing to their electrons, but that must be treated quantum mechanically, and typically only requires that the degree of ionization be tracked, as a function of T, to understand its energy consequences. Again, that does not affect the 3kT/2 energy of motion of the particle as a whole, which is all you need for the pressure contribution, which is all you need for the ideal gas law.

2) Is the energy of these atoms pure kinetic?

It should be clear now that it is not, especially in regions of the star where everything is not fully ionized, or where there are molecules present (stars typically only have molecules very close to the surface, being generally pretty hot places). But the point here is, even though the energy usually is pure kinetic, even in situations where it is not this has nothing to do with the ideal gas law. Internal energies don't affect the ideal gas law, because those deal in intraparticle forces and only interparticle forces ruin that law.

3) how is KT related to the energy of these atoms?

That was answered above, but if you didn't see the link, there is 3kT/2 of kinetic energy in each particle, on average. That fact stems entirely from the absence of interparticle forces, the assumption of statistical equilibrium, the free motion in 3 dimensions, and the meaning of temperature.

 

[shirin]

Thank you very much for your full answer.