The average speed of CH_4 molecules at 300 K | vedclass

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Question

(C) The formula for the average speed $(v_{avg})$ of gas molecules is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since the average speeds are equal

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(C) The formula for the average speed $(v_{avg})$ of gas molecules is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.Since the average speeds are equal,we have $\sqrt{\frac{8RT_1}{\pi M_1}} = \sqrt{\frac{8RT_2}{\pi M_2}}$.This simplifies to $\frac{T_1}{M_1} = \frac{T_2}{M_2}$.Given $T_1 = 300 \ K$,$M_1 (CH_4) = 16 \ g/mol$,and $M_2 (O_2) = 32 \ g/mol$.Substituting the values: $\frac{300}{16} = \frac{T_2}{32}$.$T_2 = \frac{300 	imes 32}{16} = 300 	imes 2 = 600 \ K$.

The average speed of $CH_4$ molecules at $300 \ K$ is equal to the average speed of $O_2$ molecules at ............ $K$.

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