91Ó°ÊÓ

The $rms$speed formula is given by

$v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$

The yield is

$v_{\mathrm{rms}}^2=\frac{3RT}{M}$

Solving for $T$we find

$T=\frac{M{v}_{\mathrm{rms}}^2}{3R}$

The molar mass of Nitrogen diatomic molecular gas $N_2$is $28\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$. The escape velocity for Earth is $11.2\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. Setting $V_{rms}$equal to this value we find for the temperature

$T=1.4â‹\dots {10}^5\mathrm{K}.$

The $rms$speed is given by

$v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$

The yield is

$v_{\mathrm{rms}}^2=\frac{3RT}{M}$

Solving for we find

$T=\frac{M{v}_{\mathrm{rms}}^2}{3R}$

The molar mass of Hydrogen diatomic molecular gas $H_2$is $2\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$. The escape velocity for Earth is $11.2\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. Setting $V_{rms}$ equal to this value we find for the temperature

$T={10}^4\mathrm{K}$.

The average kinetic energy is proportional to temperature. As a result, the temperatures for Nitrogen and Hydrogen should be within a few degrees of each other. $1\%$of the values calculated in parts $a$. and $b$. i.e.

$T_{{\mathrm{N}}_2}≤1400\mathrm{K},T_{{\mathrm{H}}_2}≤100\mathrm{K}$

This criterion is met for nitrogen but not for hydrogen because the real temperature of the atmosphere is approximately.$300\mathrm{K}>100\mathrm{K}$.