Average molecular distance between hydrogen gas [closed]








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The whole question goes like this:
Calculate the de Broglie wavelength of hydrogen molecule at 300K (Take the translational energy to be 1.5kT) Compare this to the average intermolecular spacing for hydrogen gas at 300K and 1 bar.


I know how to do the first part easily. I don't know how to approach the second part. Is it supposed to be a known fact or is there some way to calculate it?




From the ideal gas state equation, applied to hydrogen, the molar volume is:


$$V\_\mathrm{m} = \frac{RT}{p}$$


Using the Avogadro constant $N\_\mathrm{A}$ as number of molecules in $\pu{1 mol}$, we get the volume, statistically containing in average $\text{1 molecule}$ of $\ce{H2}$:


$$V=\frac{RT}{pN\_\mathrm{A}}$$


The reciprocal value $\frac{pN\_\mathrm{A}}{RT}$, used below, is the molecular density in $\pu{m-3}$.


The cube root then gives the length of the edge of the respective cube:


$$L=\sqrt[3] {\frac{RT}{pN\_\mathrm{A}}}$$


that gives the mean distance of centers in the tightest case of 2 facing cubes.


For $R=\pu{300 K}$ and $p = \pu{e5 Pa}$, $\sqrt[3]{\frac{RT}{pN\_\mathrm{A} }} \approx \pu{3.46 nm}$


For the most distant case of 2 cubes touching by the corner, it is


$$L=\sqrt{3} \cdot \sqrt[3] {\frac{RT}{pN\_\mathrm{A}}} \approx 1.73 \cdot \sqrt[3] {\frac{RT}{pN\_\mathrm{A}}}$$


So the true mean will lay between these two extreme values.


More exact value would be integration over randomly placed molecules.




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Another approach is to use the molecular density to calculate radius where are 2 molecules.


$$2 = \frac{pN\_\mathrm{A}}{RT} \cdot \frac 43 \pi r^3 $$


$$6RT = pN\_\mathrm{A} \cdot 4 \pi r^3 $$


$$ r = \sqrt[3]{\frac{3RT}{2 \pi pN\_\mathrm{A} }}= \sqrt[3]{\frac{3}{2 \pi}} \sqrt[3]{\frac{RT}{pN\_\mathrm{A} }} \approx 0.782 \cdot \sqrt[3]{\frac{RT}{pN\_\mathrm{A} }}$$


and then


$$ D \approx 1.564 \cdot \sqrt[3]{\frac{RT}{pN\_\mathrm{A} }}$$


This is the maximal value, the more proper one would be to compute the mean distance of two molecules randomly place in a sphere of a given diameter. But this step is already beyond my skills.


But, by stochastic Monte Carlo analysis, it looks like the mean distance of two randomly placed objects in a sphere of radius $r$ is $$L\_\mathrm{mean} = \frac{4}{3}r = \frac{4}{3} \sqrt[3]{\frac{3}{2 \pi}} \sqrt[3]{\frac{RT}{pN\_\mathrm{A} }}= \sqrt[3]{\frac{32}{9 \pi}} \sqrt[3]{\frac{RT}{pN\_\mathrm{A} }} \approx 1.042 \cdot \sqrt[3]{\frac{RT}{pN\_\mathrm{A} }} = 1.042 \cdot \pu{3.46 nm} \approx \pu{3.61 nm}$$


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