The N_2 molecule is 14 times heavier than the | vedclass

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(D) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $rms$ speed of $H_2$ at temperature $T$ is equal to the $rms$ speed

Vedclass · Accepted Answer

$21.4 \ K$

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(D) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $rms$ speed of $H_2$ at temperature $T$ is equal to the $rms$ speed of $N_2$ at $T_2 = 27^{\circ}C = 300 \ K$.So,$\sqrt{\frac{3RT}{M_{H_2}}} = \sqrt{\frac{3RT_2}{M_{N_2}}}$.Squaring both sides: $\frac{T}{M_{H_2}} = \frac{T_2}{M_{N_2}}$.Given $M_{N_2} = 14 	imes M_{H_2}$.Substituting the values: $\frac{T}{M_{H_2}} = \frac{300}{14 	imes M_{H_2}}$.$T = \frac{300}{14} \approx 21.4 \ K$.

The $N_2$ molecule is $14$ times heavier than the $H_2$ molecule. At what temperature will the $rms$ speed of $H_2$ molecules be equal to the $rms$ speed of $N_2$ molecules at $27^{\circ}C$?

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