Q. 1.64 - An Introduction to Thermal Physics

Question

Make a rough estimate of the thermal conductivity of helium at room temperature. Discuss your result, explaining why it differs from the value for air.

Step-by-Step Solution

Answer

It is solved that the Thermal conductivity of helium $k_t=0.0575\mathrm{W}\cdot {\mathrm{m}}^{-1}\cdot {\mathrm{K}}^{-1}$ with the effective radius of a helium atom at $r=1.4\times {10}^{-10}\mathrm{m}$

Step 1: Estimate thermal conductivity

The approximation formula can be used to calculate the thermal conductivity of a gas such as helium.

$k_t=\frac{C_V}{2V}\ell \overline{v}$ let be equation ($1$)

where $\overline{v}$ is the average molecular velocity, from which we can find the approximate using RMS speed, which is:

$\overline{v}\approx v_{rms}=\sqrt{\frac{3kT}{m}}$

substitute $k=1.38\times {10}^{-23}{\mathrm{m}}^2\cdot \mathrm{kg}\cdot {\mathrm{s}}^{-2}\cdot {\mathrm{K}}^{-1}$, and at room temperature $T={300}^{\circ }\mathrm{K}$, and $m$ is the mass of helium which is about $4$ atomic mass units or $m=4\times 1.66\times {10}^{-27}=6.64\times {10}^{-27}\mathrm{kg}$, so the average molecular velocity is therefore:

$\overline{v}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 300}{6.64\times {10}^{-27}}}=1367.65\mathrm{m}\cdot {\mathrm{s}}^{-1}$

$\overline{v}=1367.65\mathrm{m}\cdot {\mathrm{s}}^{-1}$ Equation ($2$)

The mean free path $\ell$ is based on the idea that the length of a cylinder with a radius equal to the molecule's diameter and volume equal to the average volume per molecule is equal to the length of a cylinder with a radius equal to the molecule's diameter and volume equal to the average volume per molecule$\frac{V}{N}$, so that:

$\ell =\frac{1}{4\pi r^2}\left(\frac{N}{V}\right)=\frac{1}{4\pi r^2}\left(\frac{kT}{P}\right)$

where $r$ is the effective radius of a helium atom, $r=1.4\times {10}^{-10}\mathrm{m}$. substitute with $k=1.38\times {10}^{-23}{\mathrm{m}}^2\cdot \mathrm{kg}\cdot {\mathrm{s}}^{-2}\cdot {\mathrm{K}}^{-1}$, at atmospheric pressure $P=1\mathrm{atm}=101325\mathrm{Pa}$, and at room temperature $T={300}^{\circ }\mathrm{K}$

Step 2: To find C V V

This gives a mean free path of: 

$\ell =\frac{1}{4\pi {\left(1.4\times {10}^{-10}\right)}^2}\left(\frac{1.38\times {10}^{-23}\times 300}{101325}\right)$

$\ell =1.66\times {10}^{-7}\mathrm{m}$ Equation($3$)

The heat capacity is:

$C_V=\frac{f}{2}Nk$

where $f$ is the number of degrees of freedom of the molecule. from the ideal gas law $PV=NkT$, the heat capacity is therefore:

$C_V=\frac{f}{2}\frac{PV}{T}$

$\frac{C_V}{V}=\frac{f}{2}\frac{P}{T}$

Since helium is monatomic, it has only $3$ degrees of freedom so $f = 3$, so:

$\frac{C_V}{V}=\frac{3}{2}\frac{101325}{300}=506.625\mathrm{J}\cdot {\mathrm{m}}^{-3}\cdot {\mathrm{K}}^{-1}$

$\frac{C_V}{V}=506.625\mathrm{J}\cdot {\mathrm{m}}^{-3}\cdot {\mathrm{K}}^{-1}$ Let be Equation ($4$)

Step 3: Substituting

Putting all together, equations ($2$),($3$) and ($4$) into equation ($1$), gives an estimate of $k_t$ :

$k_t=\frac{1}{2}\times \left(506.625\right)\times \left(1.66\times {10}^{-7}\right)\left(1367.65\right)=0.0575\mathrm{W}\cdot {\mathrm{m}}^{-1}\cdot {\mathrm{K}}^{-1}$

$k_t=0.0575\mathrm{W}\cdot {\mathrm{m}}^{-1}\cdot {\mathrm{K}}^{-1}$

This is only regarding half the measured value of around $0.142$. Using a radius of around $0.95\times {10}^{-10}\mathrm{m}$ gives a better result. In all cases, we'd expect $k_t$ helium to be higher than air since the lower mass of the molecule (it is a single atom) gives it a higher speed so it will transport energy faster.

Thus, the Thermal conductivity of helium $k_t=0.0575\mathrm{W}\cdot {\mathrm{m}}^{-1}\cdot {\mathrm{K}}^{-1}$ with the effective radius of a helium atom at $r=1.4\times {10}^{-10}\mathrm{m}$.