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The answers to Gravitational collapse and free fall time (spherical, pressure-free) describe the calculation of collapse time for a gas. For the case of neutral hydrogen gas with number density $10^4$/cm$^3$, how to find the collapse time, $t$. Also, how to find the collapse time for the case of 70% hydrogen and 30% helium gas (by mass).

I got $10^6$ years for the neutral hydrogen gas by using $d= (M*n)/N$, where $N$ is the Avogadro number, $n$ is the number density and $d$ is the mass density.

How do we do the calculation for the mixture of the two gases hydrogen and helium?

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Under the assumption that the particles do not interact, you can treat the gases separately, leading to the non-physical result, that the gas with the higher density has collapsed earlier than the other one. You could circumvent this problem by assuming that the density is for both gases the same, and given by their mass ratios.

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As given in the answer you link to, the collapse time for a homogeneous sphere of gas where we neglect pressure reads $$t = \sqrt{\frac{3\pi}{32 G \rho_0}}$$ Now we know that the gas has a certain constant number density $n_0$ and atomic mass $m_{\rm A}$. Then we simply substitute $\rho_0 = m_{\rm A} n_0$.

If the gas has more than one components, it will have an average atomic mass $\bar{m}_{\rm A} = \kappa_1 m_{\rm A 1} + \kappa_2 m_{\rm A 2}$, where $\kappa_1, \kappa_2$ are the proportions of the components per mass ($\kappa_1+\kappa_2 = 1$). Then you can use $\rho_0 = \bar{m}_{\rm A} n_0$. I am sure you can now run the numbers to find the right answer.

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