In monoatomic fluids the atoms can move quite freely around each other. Is there any thermodynamic/statistical mechanic equation how much free space there is between the atoms? This has to be proportional to temperature.

  • $\begingroup$ Are you asking what is the dependence of density on temperature? en.wikipedia.org/wiki/Thermal_expansion $\endgroup$
    – pentane
    Commented Apr 23, 2015 at 19:51
  • $\begingroup$ Seems like this must be related to density, or maybe look into the molecular basis of 'mean free path'? $\endgroup$
    – Time4Tea
    Commented Apr 24, 2015 at 3:36
  • $\begingroup$ It is related to mean free path. If in solids the distance between atoms is 1X, I am interested if it is in fluids something like 1.3X or 2X or whatever. $\endgroup$
    – Oseania
    Commented Apr 24, 2015 at 5:11

1 Answer 1


Here's a way to get a decent distance estimate for solids, liquids or gases.:

for solids or liquids, you can get the number density, $n$, of atoms or molecules (as needed), from the density, Avogadro's number ($6.02E23$) and the molecular weight ($\rho,N_a,W$: $$n = \frac{\rho N_a}{W}$$ for a gas with pressure, temperature and the boltzmann constant ($P,T,k_B$): $$n = \frac{P}{k_B T}$$ Then, $n^{-1}$ gives a volume per molecule. Take that volume to be spherical solve for the radius ($r$). $$r=\left(\frac{3}{4n}\right)^{1/3}$$ There are probably arguments for some kind of coefficient that would make this a bit more accurate, based on packing efficiency or somesuch, but this will basically get you the answer.

For example, with iron ($\rho = 7874 \, kg/m^3,W=0.056 \, kg/mol$) you get $r=207 \, nm$, while the atomic diameter is $252 \, nm$. For aluminum ($\rho = 2700 \, kg/m^3,W=0.027 \, kg/mol$), you get $r=232 \, nm$while the atomic diameter is $286 \, nm$.

For air at $101325 \, kPa, 298 \, K$ you get $3.12 \, nm$ while the mean free path is $\sim 68 \, nm$.


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