91Ó°ÊÓ

The molecule with mass $m$and velocity $v$has an average translational kinetic energy and it is given by equation $(20.19)$in the form

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

The change in the temperature of the molecule affects its average translational kinetic energy, so it is related to the temperature $T$per molecule in the form

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

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

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

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

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

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

$T=\frac{m{v}_{\mathrm{rms}}^2}{3k_{\mathrm{B}}}$..........(3)

The molecular mass of nitrogen is $m=14\mathrm{Ï\dots }$. But the nitrogen is a diatomic gas $N_2$, so we get the mass for two atoms $m=28\mathrm{Ï\dots }$. Converting this to $kg$, we get the mass of one atom of argon by

$$\begin{gathered} m=28\mathrm{u}×\frac{1.66×{10}^{-27}\mathrm{kg}}{1\mathrm{u}} \\ =46.48×{10}^{-27}\mathrm{kg} \end{gathered}$$

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

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

$=\frac{\left(46.48×{10}^{-27}\mathrm{kg}\right)(1.5\mathrm{m}/\mathrm{s}{)}^2}{3\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)}$

$=0.0025K$

The conversion between the Celsius scale and the Kelvin scale is given by equation in the form

$T_{\mathrm{C}}=T_{\mathrm{K}}−273$

$=0.0025\mathrm{K}+273$

$=−{272.99}^{∘}\mathrm{C}$

The temperature of nitrogen molecules is $=-272.99°\mathrm{C}$