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(D) The root mean square velocity $(c_{rms})$ of gas molecules is given by the formula: $c_{rms} = \sqrt{\frac{3RT}{M}}$.Since the temperature $(T)$ i

Vedclass · Accepted Answer

$0.46$

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(D) The root mean square velocity $(c_{rms})$ of gas molecules is given by the formula: $c_{rms} = \sqrt{\frac{3RT}{M}}$.Since the temperature $(T)$ is the same for both gases at $NTP$,we have $c_{rms} \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the velocities is given by: $\frac{(c_{rms})_{H_2}}{(c_{rms})_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.Given $(c_{rms})_{H_2} = 1.84 \, km/s$,$M_{H_2} = 2$,and $M_{O_2} = 32$.Substituting the values: $\frac{1.84}{(c_{rms})_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.Thus,$(c_{rms})_{O_2} = \frac{1.84}{4} = 0.46 \, km/s$.

If the root mean square velocity of the molecules of hydrogen at $NTP$ is $1.84 \, km/s$,calculate the root mean square velocity of oxygen molecule at $NTP$. The molecular weights of hydrogen and oxygen are $2$ and $32$ respectively.

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