The root mean square speed of hydrogen molecu | vedclass

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(B) The root mean square speed $(v_{rms})$ is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.For two different gases,the ratio of their $v_{rm

Vedclass · Accepted Answer

$836$

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(B) The root mean square speed $(v_{rms})$ is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.For two different gases,the ratio of their $v_{rms}$ speeds is: $\frac{(v_{rms})_{O_2}}{(v_{rms})_{H_2}} = \sqrt{\frac{T_{O_2}}{T_{H_2}} 	imes \frac{M_{H_2}}{M_{O_2}}}$.Given: $T_{H_2} = 300 \ K$,$T_{O_2} = 900 \ K$,$M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$,and $(v_{rms})_{H_2} = 1930 \ m/s$.Substituting the values: $\frac{(v_{rms})_{O_2}}{1930} = \sqrt{\frac{900}{300} 	imes \frac{2}{32}} = \sqrt{3 	imes \frac{1}{16}} = \frac{\sqrt{3}}{4}$.Therefore,$(v_{rms})_{O_2} = 1930 	imes \frac{\sqrt{3}}{4} \approx 1930 	imes \frac{1.732}{4} \approx 836 \ m/s$.

The root mean square speed of hydrogen molecules at $300 \ K$ is $1930 \ m/s.$ Then the root mean square speed of oxygen molecules at $900 \ K$ will be ....... $m/s$.

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