Avogadro’s number from thermal conductivity in an ideal gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.65.

We can get a rough estimate of Avogadro’s number from the thermal conductivity and other macroscopic quantities measurable for an ideal gas. Schroeder’s derivation of the thermal conductivity formula gives the result

\displaystyle k_{t}=\frac{C_{V}}{2V}\ell\bar{v} \ \ \ \ \ (1)

Suppose we set up the system so that we have a box of volume {V} and cross-sectional area {A}. We know that the average speed {\bar{v}} is approximately the rms speed, which is

\displaystyle \bar{v}\approx v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3NkT}{Nm}}=\sqrt{\frac{3PV}{M}} \ \ \ \ \ (2)

where {M} is the total mass of the gas. If we measure the thermal conductivity and heat capacity, we can then get the mean free path:

\displaystyle \ell=\frac{2Vk_{t}}{C_{V}\bar{v}}=\frac{2k_{t}}{C_{V}}\sqrt{\frac{MV}{3P}} \ \ \ \ \ (3)


Schroeder’s expression for {\ell} is based on the idea that the mean path length is equal to the length of a cylinder of radius equal to the molecule’s diameter and volume equal to the average volume per molecule {V/N}, so that

\displaystyle \ell=\frac{1}{4\pi r^{2}}\frac{V}{N} \ \ \ \ \ (4)


where {r} is the radius of the molecule. To get {N} from this formula, we need to know {r}, but this is a microscopic quantity which we’re assuming we don’t know. I can’t see any way of progressing from here unless we take a different value for {\ell}. Since we’re after only a rough approximation of Avogadro’s number, we can take {\ell} to be the average distance between molecules, rather than the average distance between collisions. That is

\displaystyle \ell\approx\left(\frac{V}{N}\right)^{1/3} \ \ \ \ \ (5)

We can now combine this with 3 to get an estimate for the number of molecules {N}:

\displaystyle N\approx\left(\frac{C_{V}}{2k_{t}}\right)^{3}\left(\frac{3P}{MV}\right)^{3/2}V=\frac{1}{\sqrt{V}}\left(\frac{C_{V}}{2k_{t}}\right)^{3}\left(\frac{3P}{M}\right)^{3/2} \ \ \ \ \ (6)

Assuming we know the gas constant {R}, we can get the number of moles from the ideal gas law

\displaystyle n=\frac{PV}{RT} \ \ \ \ \ (7)

so Avogadro’s number is roughly

\displaystyle N_{A}=\frac{N}{n}\approx RT\sqrt{P}\left(\frac{C_{V}}{2k_{t}}\right)^{3}\left(\frac{3}{MV}\right)^{3/2} \ \ \ \ \ (8)

As a check on this formula, we can verify that it’s dimensionless. {RT} has the dimensions of {PV}, so dimensionally, the quantity is equivalent to

\displaystyle \left(\frac{C_{V}}{k_{t}}\right)^{3}\left(\frac{P}{M}\right)^{3/2}\frac{1}{\sqrt{V}} \displaystyle \rightarrow \displaystyle \left[\frac{\mbox{energy}/\mbox{K}}{\mbox{energy}/\mbox{s m K}}\right]^{3}\left[\frac{\mbox{force}/\mbox{m}^{2}}{\mbox{kg}}\right]^{3/2}\frac{1}{\mbox{m}^{3/2}}\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle \left[\mbox{s}^{3}\mbox{m}^{3}\right]\left[\mbox{m}^{-1}\mbox{s}^{-2}\right]^{3/2}\mbox{m}^{-3/2}\ \ \ \ \ (10)
\displaystyle \displaystyle = \displaystyle 1 \ \ \ \ \ (11)

Thus the dimensions check out.

I don’t know if this is the solution Schroeder had in mind, or whether it’s possible to get {N_{A}} using the formula 4 for the mean free path. If we’re not allowed to know a priori any microscopic quantities, that means we don’t know {k} (Boltzmann’s constant), {m} (the mass of a single molecule) or {r} (the radius of a molecule). Together with {N}, that makes 4 unknowns, so we need 4 independent equations to find them. Comments welcome.

3 thoughts on “Avogadro’s number from thermal conductivity in an ideal gas

  1. snailmecream.ru

    Today, Avogadro’s number is formally defined to be the number of carbon-12 atoms in 12 grams of unbound carbon-12 in its rest-energy electronic state. The current state of the art estimates the value of N

    1. Jesse

      When I worked through this problem, I had the same issue, not knowing of a way to eliminate the microscopic parameters. I’m not sure if this is what Schroeder had in mind, but I ended up deciding that the ‘b’ parameter in the Van der Waal equation of state was suitably macroscopic and measurable. This equation was discussed in problem 1.17 so I figured it’s fair game.

      You can do a quick “proof” that b = V/A, where V is the Van der Waal volume associated with the Van der Was radius and A is Avogadro’s number. Just show that this is the limit of the volume per molecule at high pressure. This lets you eliminate r in the mean free path and then you can solve the ideal gas conductivity equation for Avogadro’s number.

      Plugging in values that I looked up for helium at 20°C and atmospheric pressure, I calculated A=5.23×10^23 using only macroscopic quantities, which seems pretty good for all the approximations involved!


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