The average speed of H_2 at T_1 K is equal t | vedclass

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(D) The formula for average speed is $C_{av} = \sqrt{\frac{8RT}{\pi M}}$.Given that $(C_{av})_{H_2} = (C_{av})_{O_2}$,we have:$\sqrt{\frac{8RT_1}{\pi

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(D) The formula for average speed is $C_{av} = \sqrt{\frac{8RT}{\pi M}}$.Given that $(C_{av})_{H_2} = (C_{av})_{O_2}$,we have:$\sqrt{\frac{8RT_1}{\pi M_{H_2}}} = \sqrt{\frac{8RT_2}{\pi M_{O_2}}}$Squaring both sides:$\frac{T_1}{M_{H_2}} = \frac{T_2}{M_{O_2}}$$\frac{T_1}{T_2} = \frac{M_{H_2}}{M_{O_2}}$Substituting the molar masses ($M_{H_2} = 2 \ g/mol$ and $M_{O_2} = 32 \ g/mol$):$\frac{T_1}{T_2} = \frac{2}{32} = \frac{1}{16}$Thus,the ratio $T_1: T_2$ is $1: 16$.

The average speed of $H_2$ at $T_1 \ K$ is equal to that of $O_2$ at $T_2 \ K$. The ratio $T_1: T_2$ is

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