Expert: Philip A. Stahl - 10/8/2007

Question
How do I show that:

i) the mean molecular weight of a pure hydrogen gas in a star is 1/2?

ii) the mean molecular weight of a pure helium gas in a star is 4/3?

iii) the change in the mean molecular weight for a solar-type star which evolves from hydrogen abundance 73% and helium 25% to hydrogen abundance 70% and helium 28% (heavier elements essentially unchanged in proportions)

iv) the electron molecular weight for a pure hydrogen gas

Answer
Hello, and thanks for a genuine astrophysics question!

In obtaining the mean molecular density we assume three components: X(fraction of hydrogen abundance), Y (fraction of helium abundance) and Z (heavier elements’ abundance).

This will entail a condition such that: X + Y + Z = 1

To proceed, we assume the ideal gas law applies (P = nkT) and full ionization of the plasma, so we add the numbers of ions and electrons for each category (X, Y and Z) in turn.

For hydrogen this will yield:

No. of ions =  X(rho)/ m(H)

where m(H) denotes the mass of a hydrogen atom and rho is the mass density.

No. of electrons = X(rho)/ m(H)

WHY the same? Because when you ionize hydrogen you get one ion (proton) and one electron each time.  One to one matching.

For helium this will yield:

No. of ions =   Y (rho)/ 4 m(H)

No. of electrons =  Y (rho)/ 2 m(H)

An He atom contains two protons, two neutrons and two electrons so you should be able to use this fact to explain why the two quantities above differ. (Hint: What IS the helium ion? It is none other than He ++, or the helium nucleus!)

For the heavier elements we include the quantity <A> - or the mean atomic mass for the heavier elements – for the ion enumeration.  Now, since the latter No. of ions = (Z (rho)/ <A>m(H) is negligible by comparison to the other contributions we may ignore it. Meanwhile the number of electrons will be:

Z (rho)/  2m(H)

the mean molecular weight (u(m))  can then be found by first adding all the contributions (for both ions and electrons) to first get n, the number density of particles:.

Thus, n = { X(rho)/ m(H) + X(rho)/ m(H)} + { Y (rho)/ 4 m(H) + Y (rho)/ 2 m(H)} + Z (rho)/  2m(H)}

This can be simplified by factoring out all the quantities except X, Y and Z to obtain:

n = rho/ m(H) [2X  +  Y/4 + Y/2  +  Z/2]

whence:

u(m) =   1/n (rho/ m(H))    =  1/  [2X  +  3Y/4 + Z/2]

For a pure hydrogen gas obviously Y= 0, Z = 0 and X = 1 so that:

u(m) = 1/ 2X  = 1/2

Pure helium will have: X= 0, Y= 1 and Z = 0  so:

u(m) = 1/ 3Y/4  = 1/ (3/4)  =  4/3

You can obtain the changing u(m) for the evolving star by using the same method, and simply substituting the fractional abundances (in decimals) for regular fractions.

Meanwhile, the electron molecular weight (for pure hydrogen gas)  will be:

u(e) = 2/ (1 + X) =   2/ 2 = 1

Hope this helps!