91Ó°ÊÓ

$T$The gravitational potential between atoms is 0 in the micro system, and the total energy of a monatomic gas equals translational kinetic energy.. As a result, a monatomic gas's thermal energy is $N$atoms is given by equation $(20.27$in the form

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

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 related to the temperature per molecule in the form.

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

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

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

Knowing 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 (1) to get an expression for $v_{\text{avg}}^2$

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

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

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

Solve equation

(3) for $v_{avg}$and use the values for$M$and $E_{\text{th}}$

$v_{\mathrm{avg}}=\sqrt{\frac{2E_{\mathrm{th}}}{M}}$

$=\sqrt{\frac{2\left(1700\mathrm{J}\right)}{0.01\mathrm{kg}}}$

$=580\mathrm{m}/\mathrm{s}$