The RMS velocity of oxygen molecules at NTP i | vedclass

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(B) The formula for the $RMS$ velocity of gas molecules at a given temperature $T$ is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the unive

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$2$

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(B) The formula for the $RMS$ velocity of gas molecules at a given temperature $T$ is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant and $M$ is the molar mass of the gas.Since $T$ is constant at $NTP$,$v_{rms} \propto \frac{1}{\sqrt{M}}$.For oxygen $(O_2)$,$M_1 = 32 \,g/mol$ and $v_1 = 0.5 \,km/s$.For hydrogen $(H_2)$,$M_2 = 2 \,g/mol$ and let the velocity be $v_2$.Using the ratio: $\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}$.Substituting the values: $\frac{0.5}{v_2} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Therefore,$v_2 = 0.5 	imes 4 = 2 \,km/s$.

The $RMS$ velocity of oxygen molecules at $NTP$ is $0.5 \,km/s$. The $RMS$ velocity for the hydrogen molecule at $NTP$ is (in $\,km/s$)

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