What are the $rms$ speeds of (a)argon atoms and (b) hydrogen molecules at $800°C$ ?

(A) The median rotational kinetic energy of a molecule with mass $m$and velocity$v$ is given by equation. $(20.19)$in the form

${ϵ}_{\mathrm{avg}}=\frac{1}{2}m{v}_{\mathrm{rms}}^2$..........(1)

The average translational kinetic energy of a molecule is affected by its temperature, hence it is connected to the temperature. $T$per molecule in the form

${ϵ}_{\mathrm{avg}}=\frac{3}{2}{k}_{\mathrm{B}}T$.....(2)

Where $k_B$is Boltzmann's constant and in $SI$unit its value is

$k_{\mathrm{B}}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

As shown, both equations $\left(1\right)$and $\left(2\right)$have the same left side, so we can use these expressions to get an equation for root mean square velocity $v_{rms}$

$\frac{1}{2}m{v}_{\mathrm{rms}}^2=\frac{3}{2}{k}_{\mathrm{B}}T$

$v_{\mathrm{rms}}^2=\frac{3k_{\mathrm{B}}T}{m}$

$v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m}}$..........(3)

The conversion between the Celsius scale and the Kelvin scale is given by equation $(18.7)$in the form

$T_{\mathrm{K}}=T_{\mathrm{C}}+273$

$=800°\mathrm{C}+273$

$=1073\mathrm{K}$

The molecular mass of Argon is$m=40\upsilon$ . Converting this to $kg$, we get the mass of one atom of argon by

$$\begin{gathered} m=40\mathrm{u}\times \frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}} \\ =66.4\times {10}^{-27}\mathrm{kg} \end{gathered}$$

Now, we plug the values for $k_B$, $T$and $m$into equation (3) to get

$v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m}}$

$=\sqrt{\frac{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(1073\mathrm{K})}{\left(66.4\times {10}^{-27}\mathrm{kg}\right)}}$

$=818\mathrm{m}/\mathrm{s}$

(B) The molecular mass of hydrogen is $m=1.007u$. But the hydrogen is a diatomic gas $H_2$, so we get the mass for two atoms $m=2.015U$. Converting this to $kg$, we get the mass of one atom of argon by

$$\begin{gathered} m=2.015\mathrm{u}\times \frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}} \\ =3.34\times {10}^{-27}\mathrm{kg} \end{gathered}$$

Now, we plug the values for $k_B$, $T$and $m$into equation (3) to get $v_{rms}$

$v_{\mathrm{rms}}=\sqrt{\frac{3k\mathrm{B}T}{m}}$

$=\sqrt{\frac{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(1073\mathrm{K})}{\left(3.34\times {10}^{-27}\mathrm{kg}\right)}}$

$=3600\mathrm{m}/\mathrm{s}$.

The $rms$speed of argon atoms $v_{rms}=818m/s$.

The $rms$speed of hydrogen molecules$v_{rms}=3600m/s$.