91Ó°ÊÓ

In , the potential energy between the bonds is zero, and the kinetic energy of a monatomic gas is translational. As a result, a monatomic gas's thermal energy is$N$atoms is given by equation $(20.27)$in the form

$E_{\text{th}}=N{\mathrm{Ï\mu }}_{\mathrm{avg}}$

Where ${\mathrm{Ï\mu }}_{\text{avg}}$is the average energy. The molecule with mass $m$and velocity $v$has a translational kinetic energy on average 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

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

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

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

The mass $M$and the molar mass $m$, we can get the number of molecules by

$N=\frac{M}{m}$

Use this expressions of $N$and ${\mathrm{Ï\mu }}_{\text{avg}}$into equation $\left(1\right)$to get an expression for $v_{\mathrm{rms}}^2$

$E_{\mathrm{th}}=N{\mathrm{Ï\mu }}_{\text{avg}}$

$=\left(\frac{M}{m}\right)\left(\frac{1}{2}m{v}_{\mathrm{avg}}^2\right)$

$=\frac{1}{2}M{v}_{\mathrm{rms}}^2$

Now, we plug the values for $M$and $v_{\text{rms}}$into equation $E_{th}$

$E_{\mathrm{th}}=\frac{1}{2}M{v}_{\mathrm{rms}}^2$

$=\frac{1}{2}\left(2×{10}^{−3}\mathrm{kg}\right)(700\mathrm{m}/\mathrm{s}{)}^2$

$=490J$