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The average velocity of carbon dioxide at the temperature   $\small T_1$ and maximum probable velocity of carbon dioxide at the temperature $\small T_2$ is $\small 9\times 10^4cms^{-1}$ . Calculate the values of   $\small T_1$ and   $\small T_2$ .
Option: 1
$T_1=1682.5K$ and   $T_2=2143.4K$

Option: 2
$T_1=1600K$ and   $T_2=2000K$

Option: 3
$T_1=21343K$ and   $T_2=1682.5K$

Option: 4
$T_1=168.2K$ and   $T_2=2143.4K$

Answers (1)

best_answer

We have given
Average velocity = Most probable speed $=9\times10^4cm/s$ or $=9\times10^2m/s$

We know:

$Average \: \: velocity=\sqrt{\frac{8RT}{\Pi M}}$

$Thus,\: 9\times10^2=\sqrt{\frac{8\times 8.314\times T_1}{3.14\times44\times10^{-3}}}$

Thus, $T_1=1682.5K$

Again, we know:

$Most\: \: probable\: \: velocity=\sqrt{\frac{2\times 8.314 \times T_2}{44\times 10^{-3}}}$

$Thus, \: T_2=2143K$

Therefore, Option(1) is correct

Posted by

seema garhwal

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