Probabilities or pressures in the Saha equation?

[lampCable]

Homework Statement

Consider a system consisting of a single hydrogen atom/ion, which has two possible states: unoccupied (i.e., no electron present) and occupied (i.e., one electron present, in the ground state). Calculate the ratio of the probabilities of these two states, to obtain the Saha equation. Treat the electrons as a monatomic gas, for the purpose of determining the chemical potential. Neglect the fact that an electron has two independent spinstates.

Homework Equations

The Saha equation $$\frac{P_H}{P_{H^+}}=\frac{kT}{P_e}\bigg(\frac{2\pi m_e kT}{h^2}\bigg)^{3/2}e^{-I/kT},$$ where $P_H$, $P_{H^+}$ and $P_e$ are the partial pressures for the hydrogen, ionized hydrogen and the electrons, respectively. The equaition can also be written in terms of number of particles using $$P/kT=N/V.$$

The Attempt at a Solution

I believe that I successfully derived the equation, as I found that $$\frac{P(H)}{P(H^+)}=\frac{kT}{P_e}\bigg(\frac{2\pi m_e kT}{h^2}\bigg)^{3/2}e^{-I/kT},$$ where the $P$'s on the left hand side are the probabilities of finding the particle in one of the two states (hydrogen in the numerator and ionized hydrogen in the denominator). But I do not understand why the ratios of the probabilities is equal to the ratio of pressures. Could someone explain this to me?

 

[TSny]

Can you express your $\frac{P(H)}{P(H^+)}$ in terms of the number of H atoms, $N(H)$, and the number of H ions, $N(H^+)$?

 

[lampCable]

So the system is found either occupied or unoccupied with the probabilities $P(H)$ or $P(H^+)$, respectively. If we then consider $N$ such systems we should find that there are $N(H) = P(H)N$ occupied states and $N(H^+)=P(H^+)N$ unoccupied states, so that $\frac{N(H)}{N(H^+)}=\frac{P(H)}{P(H^+)}$. Is it so simple?

 

[TSny]

I think that's all there is to it. You can then relate the N's to the partial pressures.