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.

Short 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-by-step solution

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 localid="1650283569119" $\overline{v}$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 $T={300}^{\circ }\mathrm{K}$

To find CVV

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 ($4$)

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 $r=1.4\times {10}^{-10}\mathrm{m}$.