91Ó°ÊÓ

(a) The molecule with mass $m$and velocity $v$has an average translational kinetic energy. 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 for

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

Where $k_{\mathrm{B}}$is Boltzmann's constant and in SI unit its value is

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

Plug the values for $k_{\mathrm{B}}$and $T$into equation (1) to get the energy for proton

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

$=\frac{3}{2}\left(1.38×{10}^{−23}\mathrm{J}/\mathrm{K}\right)\left(2.0×{10}^7\mathrm{K}\right)$

$=4.1×{10}^{−16}\mathrm{J}$

(b) 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

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

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 $v_{rms}$

$\begin{array}{c}\frac{1}{2}m{v}_{\mathrm{rms}}^2=\frac{3}{2}{k}_{\mathrm{B}}T\\ v_{\mathrm{rms}}^2=\frac{3k_{\mathrm{B}}T}{m}\\ v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m}}\end{array}$

The mass of the proton is $m=1.67×{10}^{−27}\mathrm{kg}$.Now we plug the values for $K_B$and $m_p$into equation (3) to get $v_{rms}$

$=\sqrt{\frac{3\left(1.38×{10}^{−23}\mathrm{J}/\mathrm{K}\right)\left(2.0×{10}^7\mathrm{K}\right)}{1.67×{10}^{−27}\mathrm{kg}}}$

$=7×{10}^5\mathrm{m}/\mathrm{s}$