Q 1.64 - An Introduction to Thermal Physics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Given Information

We have to estimate the thermal conductivity of a gas such as helium using the approximate formula.

We know that,

$k_t=\frac{Cv}{2V}l\overline{v}$

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

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

substitute $k$$=$$1.38\times {10}^{-23}{m}^2$ ,$T=300 K$ , $m=4\times 1.66\times {10}^{-27} kg$

The average molecular velocity is therefore:

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

mean free path l is,

$$\begin{gathered} l=\frac{1}{4{\mathrm{\pi r}}^2}\left(\frac{N}{V}\right)=\frac{1}{4{\mathrm{\pi r}}^2}\left(\frac{kT}{P}\right) \\ where r=1.4\times {10}^{-10} m \end{gathered}$$

Step 2 : Explanation

Given a free path is:

$$\begin{gathered} l=\frac{1}{4\mathrm{\pi }{\left(1.4\times {10}^{-10}\right)}^2}\left(\frac{1.38\times {10}^{-23}\times 300}{101325}\right) \\ l=1.66\times {10}^{-7}m \end{gathered}$$

The heat capacity is:

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

Where f is the number of degree of freedom of the molecule.

From ideal gas law $PV=NkT$

Therefore heat capacity is ;

$$\begin{gathered} C_v=\frac{f}{2}\frac{PV}{T} \\ \frac{C_v}{V}=\frac{f}{2}\frac{P}{T} \end{gathered}$$

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

$\frac{C_v}{V}=\frac{3}{2}\frac{101325}{300}=506.625 J.m^{-3}.K^{-1}$

$$\begin{gathered} k_t=\frac{1}{2}\times \left(506.625\right)\times \left(1.66\times {10}^{-7}\right)\left(1367.65\right) \\ {k}_t=0.0575 W.m^{-1}.K^{-1} \end{gathered}$$