If the rms speed of the molecules of a diatom | vedclass

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(A) The root mean square (rms) speed of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constan

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hydrogen

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(A) The root mean square (rms) speed of gas molecules is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant $(8.314 \ J \ mol^{-1} \ K^{-1})$,$T$ is the temperature in Kelvin,and $M$ is the molar mass in $kg \ mol^{-1}$.Given: $v_{rms} = 2000 \ m \ s^{-1}$,$T = 322 \ K$.Rearranging the formula to solve for $M$: $M = \frac{3RT}{v_{rms}^2}$.Substituting the values: $M = \frac{3 	imes 8.314 	imes 322}{(2000)^2} = \frac{8031.324}{4,000,000} \approx 0.0020078 \ kg \ mol^{-1} = 2.0078 \ g \ mol^{-1}$.The molar mass of hydrogen gas $(H_2)$ is approximately $2 \ g \ mol^{-1}$.Therefore,the gas is hydrogen.

Column-$I$	Column-$II$
$(a)$ ${v_{rms}} = \sqrt {\frac{3P}{\rho}}$	$(i)$ For $1 \text{ mole ideal gas}$
$(b)$ ${v_{rms}} = \sqrt {\frac{3RT}{M_0}}$	$(ii)$ For one molecule of gas
$(c)$ ${v_{rms}} = \sqrt {\frac{3{k_B}T}{m}}$	$(iii)$ On the basis of kinetic theory

If the rms speed of the molecules of a diatomic gas at a temperature of $322 \ K$ is $2000 \ m \ s^{-1}$,then the gas is

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