_{o }+ N

_{e}) k T

Where N

_{o}is the number of atoms of all kinds, both neutral (unionized) and ionized and N_{e}is the number of free electrons (the greater this number, the higher the degree of ionization). The total gas pressure therefore is the sum of the two pressures:
P = N

_{o }k T_{ }+ N_{e}k T
Then:

N

_{e}k T = (1.45 x 10^{ 19})( 1.38 x 10^{-23}J/K) (5650 K) = 1.13 Pa
But P, the total gas pressure is 8.3 x 10

^{3}Pa, so:
P = 8.3 x 10

^{3}Pa = N_{o }k T_{ }+ N_{e}k T = N_{o }k T_{ }+ 1.13 Pa
Then: N

_{o }k T_{ }= 8.3 x 10^{3}Pa - 1.13 Pa » 8.3 x 10^{3}Pa
N o = (8.3 x 10

(8.3 x 10

^{3}Pa ) / k T =(8.3 x 10

^{3}Pa ) / (1.38 x 10-23 J/K) (5650 K)
N

_{o }= 1.06 x 10^{ 23}
Where N

_{o }is the total number of neutral and ionized atoms.2) We assume negligible amounts of heavier elements so:

m = 4/ (6X + Y + 2)

Where X = 0.9 and Y = 0.1. Substituting those values:

m = 4/ (6(0.9) + 0.10 + 2)

= 4/ (0.54 + 0.10 + 2) = 4/ 2.64

= 4/ (0.54 + 0.10 + 2) = 4/ 2.64

(Note that this molecular weight is in proton masses)

3) We have:

1/ m = (2X + 3Y/4 + ½Z)

But the no. of electrons is:

N

_{e }= r /H [X + (Y + Z)/ 2] » r [1 + X]/ 2H

1/ m = H N

_{e / }r = (1 + X)/ 2

m = 2/ (1 + X)

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