# 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.