91Ó°ÊÓ

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

$k_t=\frac{C_V}{2V}\mathrm{ℓ}\stackrel{¯}{v}$let be equation ($1$)

where localid="1650283569119" $\stackrel{¯}{v}$is the average molecular velocity, from which we can find the approximate using RMS speed, which is:

$\stackrel{¯}{v}≈v_{rms}=\sqrt{\frac{3kT}{m}}$

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

$\stackrel{¯}{v}=\sqrt{\frac{3×1.38×{10}^{−23}×300}{6.64×{10}^{−27}}}=1367.65\mathrm{m}â‹\dots {\mathrm{s}}^{−1}$

$\stackrel{¯}{v}=1367.65\mathrm{m}â‹\dots {\mathrm{s}}^{−1}$Equation ($2$)

The mean free path $\mathrm{â„“}$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:

$\mathrm{â„“}=\frac{1}{4\mathrm{Ï€}{r}^2}\left(\frac{N}{V}\right)=\frac{1}{4\mathrm{Ï€}{r}^2}\left(\frac{kT}{P}\right)$

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

This gives a mean free path of:

$\mathrm{ℓ}=\frac{1}{4\mathrm{π}{\left(1.4×{10}^{−10}\right)}^2}\left(\frac{1.38×{10}^{−23}×300}{101325}\right)$

$\mathrm{ℓ}=1.66×{10}^{−7}\mathrm{m}$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:

$\frac{C_V}{V}=\frac{3}{2}\frac{101325}{300}=506.625\mathrm{J}â‹\dots {\mathrm{m}}^{−3}â‹\dots {\mathrm{K}}^{−1}$

$\frac{C_V}{V}=506.625\mathrm{J}â‹\dots {\mathrm{m}}^{−3}â‹\dots {\mathrm{K}}^{−1}$Let be Equation ($4$)

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

$k_t=\frac{1}{2}×\left(506.625\right)×\left(1.66×{10}^{−7}\right)\left(1367.65\right)=0.0575\mathrm{W}â‹\dots {\mathrm{m}}^{−1}â‹\dots {\mathrm{K}}^{−1}$

$k_t=0.0575\mathrm{W}â‹\dots {\mathrm{m}}^{−1}â‹\dots {\mathrm{K}}^{−1}$

This is only regarding half the measured value of around $0.142$. Using a radius of around $0.95×{10}^{−10}\mathrm{m}$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 $k_t=0.0575\mathrm{W}â‹\dots {\mathrm{m}}^{−1}â‹\dots {\mathrm{K}}^{−1}$with the effective radius of a helium atom at $r=1.4×{10}^{−10}\mathrm{m}$.