Question

The average velocity at $T_{1}K$, and the most probable velocity at $T_{2}K$ of $C{O}_2$ gas is $9.0\times {10}^4{\text{cm sec}}^{-1}$. The value of $T_1$ is K and $T_2$ is K.

Answer 1

Average velocity $=\sqrt{\frac{8RT}{\pi M}}$
and most probable velocity $=\sqrt{\frac{2RT}{M}}$
Given - for $C{O}_2$
Average velocity at $T_1=$ Most probable velocity at $T_2$
$9\times {10}^4\text{cm/sec}=\frac{9\times {10}^4}{100}\text{m/sec}$
$=9\times {10}^2\text{m/sec}$

$\therefore 9\times {10}^2=\sqrt{\frac{8\times 8.314\times T_1}{3.14\times 44\times {10}^{-3}}}\dots \left(A\right)$ [Average velocity at $T_{1}K$]

and

$9\times {10}^2=\sqrt{\frac{2\times 8.314\times T_2}{44\times {10}^{-3}}}\dots \left(B\right)$ [Most probable velocity at $T_{2}K$]

On solving, $T_1=1682.5\text{K},T_2=2143.4\text{K}$