The RMS velocity of dihydrogen (H_2) is sqrt{ | vedclass

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(C) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given: $v_{H_2} = \sqrt{7} 	imes v_{N_2}$.Substituting the formula: $\sqrt{\fr

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$T_{H_2} = \frac{T_{N_2}}{2}$

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(C) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given: $v_{H_2} = \sqrt{7} 	imes v_{N_2}$.Substituting the formula: $\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{7} 	imes \sqrt{\frac{3RT_{N_2}}{M_{N_2}}}$.Squaring both sides: $\frac{3RT_{H_2}}{M_{H_2}} = 7 	imes \frac{3RT_{N_2}}{M_{N_2}}$.Given molar masses: $M_{H_2} = 2 \ g/mol$ and $M_{N_2} = 28 \ g/mol$.Substituting these values: $\frac{T_{H_2}}{2} = 7 	imes \frac{T_{N_2}}{28}$.Simplifying: $\frac{T_{H_2}}{2} = \frac{T_{N_2}}{4}$.Therefore, $T_{H_2} = \frac{T_{N_2}}{2}$.

The $RMS$ velocity of dihydrogen $(H_2)$ is $\sqrt{7}$ times more than that of dinitrogen $(N_2)$. If $T_{H_2}$ and $T_{N_2}$ are the temperatures of dihydrogen and dinitrogen respectively, then the correct relationship between them is:

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