At what temperature will the average speed of | vedclass

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(B) The formula for average speed is $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.For the average speeds of $CH_4$ and $O_2$ to be equal: $\sqrt{\frac{8RT_{CH_

Vedclass · Accepted Answer

$150$

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(B) The formula for average speed is $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.For the average speeds of $CH_4$ and $O_2$ to be equal: $\sqrt{\frac{8RT_{CH_4}}{\pi M_{CH_4}}} = \sqrt{\frac{8RT_{O_2}}{\pi M_{O_2}}}$.This simplifies to $\frac{T_{CH_4}}{M_{CH_4}} = \frac{T_{O_2}}{M_{O_2}}$.Given $T_{O_2} = 300 \ K$,$M_{O_2} = 32 \ g/mol$,and $M_{CH_4} = 16 \ g/mol$.Substituting the values: $\frac{T_{CH_4}}{16} = \frac{300}{32}$.$T_{CH_4} = \frac{300 	imes 16}{32} = 150 \ K$.

At what temperature will the average speed of $CH_4$ molecules have the same value as $O_2$ has at $300 \ K$ ........... $K$

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