91Ó°ÊÓ

(A) The density of the chemical $m$and velocity $v$seems to have an average translational kinetic energy, which may be calculated using equation $(20.19)$in the form

${\mathrm{Ï\mu }}_{\mathrm{avg}}=\frac{1}{2}m{v}_{\mathrm{rms}}^2$...........(3)

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

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

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

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

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 molecular mass of cesium is $m=133\mathrm{Ï\dots }$. Converting this to $kg$, we get the mass of one atom of cesium by

$$\begin{gathered} m=133\mathrm{u}×\frac{1.66×{10}^{-27}\mathrm{kg}}{1\mathrm{u}} \\ =221×{10}^{-27}\mathrm{kg} \end{gathered}$$

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

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

$=\sqrt{\frac{3\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(100×{10}^{-9}\mathrm{K}\right)}{\left(221×{10}^{-27}\mathrm{kg}\right)}}$

$=0.0043\mathrm{m}/\mathrm{s}$

$=4.3\mathrm{mm}/\mathrm{s}$

The$rms$speed of$v_{\mathrm{rms}}=4.3\mathrm{mm}/\mathrm{s}$