91Ó°ÊÓ

A box of atomic hydrogen, initially at room temperature and atmospheric pressure. Then raise the temperature, keeping the volume fixed.

The equation is given by:

$\frac{P_p}{P_H}=\frac{kT}{P_e}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}$

Where,

$I$is the ionisation energy

The ratio of partial pressures equals the ratio of ionised hydrogen to non-ionised hydrogen (pressure of protons where the proton is ionised hydrogen). Let N₁be the number of ionised hydrogen and N_Gbe the number of hydrogen in ground state.

$\frac{P_p}{P_H}=\frac{N_I}{N_G}=\frac{kT}{P_e}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}$

By ideal gas law,

$\frac{kT}{P_e}=\frac{V}{N_e}=\frac{1}{n_e}$

Where,

$n_e$is number of density of electrons

Therefore,

$\frac{N_I}{N_G}=\frac{1}{n_e}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}\left(1\right)$

Electron density in terms of number density of hydrogen atoms:

$n_e=\left(\frac{N_I}{N_G+N_I}\right)n$

Substitute this into (1)

$$\begin{gathered} \frac{N_I}{N_G}=\frac{1}{n}\left(\frac{N_G+N_I}{N_I}\right){\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT} \\ \frac{N_I^2}{N_G\left(N_G+N_I\right)}=\frac{1}{n}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}\left(2\right) \end{gathered}$$

Let N_T be the number of ionised atoms plus the number of non ionised atoms, therefore the fraction of the ionised atom can be written as:

$x=\frac{N_I}{N_I+N_G}=\frac{N_I}{N_T}$

where N_T = N_I + N_G. Write (2) in terms of N_T and N_I, to get:

$\frac{N_I}{N_T-N_I}\frac{N_I}{N_T}=\frac{1}{n}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}$

Multiply the first term in the LHS by N_T/N_T to get:

$\frac{N_I/N_T}{1-N_I/N_T}\frac{N_I}{N_T}=\frac{1}{n}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}$

But $x=N_I/N_T$,

$\frac{x^2}{1-x}=\frac{1}{n}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}$

The number of density n can be replaced by $P/kT$

role="math" localid="1647289106044">x21-x=kTP2Ï€mekTh23/2e-I/kT(3)

We need to solve this equation for c let the RHS be C, so:

$$\begin{gathered} x^2=C-xC \\ {x}^2+xC-C=0 \end{gathered}$$

Solving the quadratic equation:

$$\begin{gathered} x=\frac{-CÂ\pm \sqrt{C^2+4C}}{2} \\ x=\frac{\sqrt{C^2+4C}-C}{2} \\ x=C\frac{\sqrt{1+4/C}-1}{2} \\ x=\frac{kT}{2P}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT}\left(\sqrt{1+\frac{4P}{kT}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{-3/2}{e}^{I/kT}}-1\right) \end{gathered}$$

(b)To find at what temperature half of the hydrogen atoms will be ionised, we set x= 1/2 into equation (3), to get:

$$\begin{gathered} \frac{1}{4(1-1/2)}=\frac{kT}{P}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT} \\ \frac{1}{2}=\frac{kT}{P}{\left(\frac{2\mathrm{Ï€}{m}_{e}kT}{h^2}\right)}^{3/2}{e}^{-I/kT} \end{gathered}$$

Substitute with

$$\begin{gathered} 1.0×{10}^5\mathrm{Pa}\text{for}P \\ 9.11×{10}^{-31}\mathrm{kg}\text{for}{m}_e \\ 1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\text{for}k \\ 6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\text{for}h \\ 13.6\mathrm{eV}=13.7×1.6×{10}^{-19}=2.176×{10}^{-18}\mathrm{J}\text{for}I \end{gathered}$$

To get

$\frac{1}{2}=\left(3.33×{10}^{-7}{\mathrm{K}}^{2/5}\right)T^{5/2}{e}^{-1.577×{10}^5\mathrm{K}/T}$

Using python to plot f(T), solve this equation:

$f\left(T\right)=\left(3.33×{10}^{-7}{\mathrm{K}}^{2/5}\right)T^{5/2}{e}^{-1.577×{10}^5\mathrm{K}/T}-\frac{1}{2}$

The code is shown below, and the solution is the intersection point between the curve and the x axis.

By looking at the graph of f(T) we can see that the curve intersect with x axis at temperature of:

$T=25800\mathrm{K}$

(c)Because the x is inversely related to the pressure, raising the pressure will raise the temperature in section (b).

This is because when there are fewer particles, there is less pressure, and increasing the pressure, according to Le Chatelier's principle, prevents the ionisation process.

(d)Now we need to plot the result of part (a), using the values in part (6), so:

$x=\frac{k}{2P}{\left(\frac{2\mathrm{Ï€}{m}_{e}k}{h^2}\right)}^{3/2}{T}^{5/2}{e}^{-I/kT}\left(\sqrt{1+\frac{4P}{k}{\left(\frac{2\mathrm{Ï€}{m}_{e}k}{h^2}\right)}^{-3/2}{T}^{-5/2}{e}^{I/kT}}-1\right)$

Let, $t=\frac{kT}{I}→T=\frac{It}{k}$

so,

$x=\frac{1}{2P}{\left(\frac{2\mathrm{Ï€}{m}_e}{h^2}\right)}^{3/2}(It{)}^{5/2}{e}^{-1/t}\left(\sqrt{1+4P{\left(\frac{2\mathrm{Ï€}{m}_e}{h^2}\right)}^{-3/2}(It{)}^{-5/2}{e}^{1/t}}-1\right)$

Substitute the values from part (b)

$x=\left(1.644×{10}^6\right)t^{5/2}{e}^{-1/t}\left(\sqrt{1+\left(1.216×{10}^{-6}\right)t^{-5/2}{e}^{1/t}}-1\right)$

Plot this function using python, the range of graph is t = 0 to = 0.4, the code is

The graph is: