Estimate the ratio of the thermal conductivity of Xe to that of He.

(a) 0.015; (b) 0.061; (c) 0.10; (d) 0.17.

The thermal conductivity k of a gas depends on its molar heat capacity${\mathit{C}}_{\mathbf{v}}$, molar mass M, and molecular radius r. The dependence on those qualities at a given temperature.

The molar heat capacity for Helium and Xenon, at constant volume, is always the same.

$C_v\left(He\right)=C_v\left(X_e\right)=C_v$

The molar mass of Helium and Xenon are given as

$$\begin{gathered} M_{He}=4.0gm/mol \\ {M}_{Xe}=131gm/mol \end{gathered}$$

The molecular radius of Helium and Xenon are given as

$$\begin{gathered} r_{He}=0.13nm \\ {r}_{Xe}=0.13nm \end{gathered}$$

Hence, the thermal conductivity k of a gas depends on its molar heat capacity$C_v$, molar mass M, and molecular radius r

$k\propto \frac{C_v}{r^2\sqrt{M}}$

Therefore, the ratio of thermal conductivity of Xe to He is given by

$$\begin{gathered} \frac{k_{Xe}}{k_{He}}=\frac{C_v}{r_{Xe}\sqrt[2]{M_{Xe}}}.\frac{r_{Xe}\sqrt[2]{M_{He}}}{C_v} \\ \frac{k_{Xe}}{k_{He}}=\frac{r_{Xe}\sqrt[2]{M_{He}}}{r_{Xe}\sqrt[2]{M_{Xe}}} \\ \frac{k_{Xe}}{k_{He}}=\frac{{\left(0.13nm\right)}^2\sqrt{4.0gm/mol}}{{\left(131nm\right)}^2\sqrt{131gm/mol}} \\ \frac{k_{Xe}}{k_{He}}=0.0610 \end{gathered}$$

Hence, the ratio of thermal conductivity of Xe to He is 0.061.