Thermal conductivity of helium

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.64.

We can estimate the thermal conductivity of a gas such as helium using the approximate formula

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

where {\bar{v}} is the average molecular velocity, which we can approximate by the rms speed, which is

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

The mean free path {\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} \ \ \ \ \ (3)


where {r} is the radius of the molecule. The heat capacity is

\displaystyle C_{V}=\frac{1}{2}Nfk=\frac{f}{2}\frac{PV}{T} \ \ \ \ \ (4)

where {f} is the number of degrees of freedom of the molecule.

At room temperature and pressure we can work out {k_{t}} for helium by looking up a few properties using Google. The effective radius of a helium atom seems to depend a lot on which web page you find. It seems that the van der Waals radius is defined as half the distance between two nuclei when two non-bonded atoms are at their closest possible approach. The most commonly quoted value for helium is

\displaystyle r=1.4\times10^{-10}\mbox{ m} \ \ \ \ \ (5)

This gives a mean free path of

\displaystyle \ell=\frac{1}{4\pi r^{2}}\frac{V}{N}=\frac{1}{4\pi r^{2}}\frac{kT}{P}=1.68\times10^{-7}\mbox{ m} \ \ \ \ \ (6)

The mass of a helium-4 atom is about 4 atomic mass units or

\displaystyle m=4\times1.66\times10^{-27}=6.64\times10^{-27}\mbox{ kg} \ \ \ \ \ (7)

The rms velocity at {T=300\mbox{ K}} is therefore

\displaystyle v_{rms}=\sqrt{\frac{3\left(1.38\times10^{-23}\right)\left(300\right)}{6.64\times10^{-27}}}=1.37\times10^{3}\mbox{ m s}^{-1} \ \ \ \ \ (8)

Since helium is monatomic, it has only 3 degrees of freedom so {f=3} and

\displaystyle \frac{C_{V}}{V}=\frac{3}{2}\frac{P}{T}=\frac{3}{2}\frac{10^{5}}{300}=500\mbox{ J m}^{-3}\mbox{K}^{-1} \ \ \ \ \ (9)

Putting all this together gives an estimate of {k_{t}}:

\displaystyle k_{t}=\frac{1}{2}\left(500\right)\left(1.68\times10^{-7}\right)\left(1.37\times10^{3}\right)=0.057\mbox{ W m}^{-1}\mbox{K}^{-1} \ \ \ \ \ (10)

This is only about half the measured value of around 0.142. Using a radius of around {0.95\times10^{-10}\mbox{ m}} gives a better result, and this value is given on a few web sites so who knows? In any case, we’d expect {k_{t}} for helium to be higher than air, since the lower mass of the molecule (it being a single atom) gives it a higher speed so it will transport energy faster.

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