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Here is the problem I am working on

For what temperatures are the atoms in an ideal gas at pressure P quantum mechanical ? Hint: use the ideal gas law P V = N kT to deduce the interatomic spacing (the answer is $T < (1/k)(h^2/3m)^{3/5} P^{2/5})$. Obviously we want m to be as small as possible and P as large as possible for the gas to show quantum behavior. Put in numbers for helium at atmospheric pressure. Is hydrogen in outer space (interatomic distance ≈ 1 cm and temperature ≈ 3K) quantum mechanical?

According to the answer key, to find the interatomic spacing, we need to find the size of a single gas particle. One gas particle corresponds to N=1, and the volume is V = d^3. This leads to

$d = \left( \frac{kT}{P} \right)^{1/3}$

I have two objections to this, for which I hope you provide correction. Firstly, assigning the volume V=d3 implies that we are assuming that the atoms are square? Secondly, how does finding the size of a single gas particle provide us with the interatomic spacing. It would seem that the most we could deduce from such information is, that closest two gas particles could get. Are we to assume that the gas particles are this closely packed? Wouldn't the gas solidify at this point?

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The idea of the question is to find the temperature at which the average interparticle spacing is equal to the average de Broglie wavelength. Both of these are averages because the atoms of the ideal gas are not evenly spaced and the velocity (and therefore de Broglie wavelength) of the ideal gas atoms follows the Maxwell-Boltzmann distribution. So this is going to be a very rough calculation.

Given how rough the calculation is, the approximation that the gas atoms are on average equally spaced seems a reasonable one. In that case our $N$ atoms are distributed in a volume $V$, so the average spacing $d$ is:

$$\begin{align} d &= \left( \frac{V}{N} \right)^{1/3} \\ &= \left( \frac{\frac{NkT}{P}}{N} \right)^{1/3} \\ &= \left( \frac{kT}{P} \right)^{1/3} \end{align}$$

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"The size of a single gas particle" is a bad term for what you calculate. A better term would be:

"The volume one of the gas particles would occupy, if the total volume were distributed equally among all gas particles"

And this translates loosely to "The volume, in which you find one gas particle on average"

If you then imagine every particle sitting at the centre of their volumes, then two neighbouring particles have indeed the distance $d$. This whole argument is not rigorous, but rather heuristic.

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