Answer
$\mathrm{a}.\quad \mathrm{K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}=\mathrm{K}_{\mathrm{C}}$
$\mathrm{b}.\quad \mathrm{v}_{\mathrm{C}}\lt \mathrm{v}_{\mathrm{B}}\lt \mathrm{v}_{\mathrm{A}}$
Work Step by Step
Kinetic theory relates the average kinetic energy of the molecules in a gas
to the Kelvin temperature of the gas, $T$:
$(\displaystyle \frac{1}{2}mv^{2})_{\mathrm{a}\mathrm{v}}=K_{\mathrm{a}\mathrm{v}}=\frac{3}{2}kT \qquad (17- 15)$
The rms (root mean square) speed of the molecules in a gas at the Kelvin temperature $T$ is
$v_{\mathrm{r}\mathrm{m}\mathrm{s}}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}}\qquad( 17- 13 )$
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$a.$
Interpreting formula 17-15, we see that kinetic energy depends on temperature only.
Thus, $\mathrm{K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}=\mathrm{K}_{\mathrm{C}}$
$\mathrm{b}.$
Interpreting formula 17-13,
$v_{\mathrm{r}\mathrm{m}\mathrm{s}}$ is inversely proportional to the square root of mass.
Thus the ranking of $v_{\mathrm{r}\mathrm{m}\mathrm{s}}$ is the opposite to the ranking of masses
$\mathrm{v}_{\mathrm{C}}\lt \mathrm{v}_{\mathrm{B}}\lt \mathrm{v}_{\mathrm{A}}$
