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Can the interior of the Sun be described as an ideal gas?

From my knowledge, to describe a body of gas as an ideal gas, the separation between the particles must be much greater than the size of the actual particles.

How could one justify whether the Sun fits this?

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  • $\begingroup$ In the very core of the sun, particles are so close together that they fuse. Do you think this fulfils the ideal gas criterion you mention? $\endgroup$
    – astronat
    Commented Apr 14, 2018 at 21:02
  • $\begingroup$ I'm trying to figure a more quantitative way of reasoning. How could one find the distance between these particles in the Sun? Surely it would involve the mean temperature and densities of the star? $\endgroup$ Commented Apr 14, 2018 at 21:27
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    $\begingroup$ The ideal gas approximation of point-like particles that either don't interact or only interact elastically is a very good approximation in the solar interior. $\endgroup$
    – ProfRob
    Commented Apr 15, 2018 at 6:34
  • $\begingroup$ If your definition of ideal gas is just the ideal gas law, PV=nRT, then plasma is almost ideal. $\endgroup$ Commented Apr 15, 2018 at 10:45

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Density of the solar core is 150 g/cc and at a temperature of $1.5\times 10^7$K.

For a rough calculation assume everything is ionised hydrogen (protons and electrons).

The mass is all in protons, with a number density of $1.5\times 10^5/1.67\times10^{-27}=9\times 10^{31}$ m$^{-3}$, with an equal number of electrons.

The average particle separation is roughly the inverse cube root of the number density (imagine each particle in a cube), so is $1.8\times 10^{-11}$m. The "size" of a proton is $10^{-15}$m, so the approximation of point-like particles is satisfied.

However, that is insufficient. It also needs to be the case that the particles are "non-interacting" or at least only inelastically interacting. Fusion is a rare process, so inelastic collisions are rare. That the particles have little interaction can be shown by comparing the Coulomb energy at the average separation with the thermal energy. $e^2/4\pi\epsilon_0 kT \sim 0.06$. Thus the Coulomb interactions are small compared with the thermal energy and the particle motion is not greatly affected by the particles around them.

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  • $\begingroup$ That's a perfect answer! $\endgroup$ Commented Apr 15, 2018 at 12:57

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