# When is it a good aproximation to consider a star to be an ideal gas?

I am currently taking a first course on stellar astrophysics, and I noticed that in some cases we use the ideal gas equation of state for stars, so we also use $\gamma =5/3$. Of course it can only be applied where there is no nuclear reaction, so there is a limit in the temperature.

Also, if the radiation pressure if relevant, you have to consider the parameter $\beta$ to calculate the total pressure and the adiabatic coefficient $\gamma$. It is also incorrect (I think) if there are degeneration or relativistic considerations.

I don't know if I have to consider other factors before I can use this equation of state and value for $\gamma$, maybe it is important if it is a convective or radiative area, or other factors I didn't consider.

My question is: can anybody tell me the limits of the ideal gas approximation? (quantitative better than qualitative but any help will be well received)

This means you can take the mean separation ($\sim n^{-1/3}$) and compare that with the size of the particles - it should be much bigger. You can also compare the particle kinetic energies with the potential energy of their interactions at typical separations. eg The ratio of $kT$ to the Coulomb potential energy should be large.
• Hi, could you maybe elaborate a bit on the last part? In fact I'm one of those ppl confusing it. In my mind there is only one relation for the pressure towards the other fluid variables needed (fluid closure relation = equation of state?), so I'm strugglin a bit with the understanding how $P= n k_B T$ and $P = \rho^{\gamma}$ can be true at the same time. – AtmosphericPrisonEscape Dec 3 '15 at 13:58
• @AtmosphericPrisonEscape The equation of state for a perfect gas is $P = nkT$. However, an adiabatic change in a gas will follow an "adiabat" defined by $PV^{\gamma} = const$ (the temperature will of course change). It is conceptually no different to saying an isothermal change implies $PV = constant$ (Boyle's law). In each case $P=nkT$ remains true. – ProfRob Dec 8 '15 at 16:22