If the r.m.s. velocity of hydrogen molecules | vedclass

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(B) The formula for r.m.s. velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given,$v_{H_2} = 4 	imes v_{O_2}$.Temperature of Oxygen $T_{O_2} = 47 + 273 =

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$47$

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(B) The formula for r.m.s. velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given,$v_{H_2} = 4 	imes v_{O_2}$.Temperature of Oxygen $T_{O_2} = 47 + 273 = 320 \ K$.Molar mass of Hydrogen $M_{H_2} = 2 \ g/mol$ and Oxygen $M_{O_2} = 32 \ g/mol$.Using the relation $\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = 4 	imes \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.Squaring both sides: $\frac{T_{H_2}}{M_{H_2}} = 16 	imes \frac{T_{O_2}}{M_{O_2}}$.Substituting the values: $\frac{T_{H_2}}{2} = 16 	imes \frac{320}{32}$.$\frac{T_{H_2}}{2} = 16 	imes 10 = 160$.$T_{H_2} = 320 \ K$.Converting back to Celsius: $T(^{\circ}C) = 320 - 273 = 47^{\circ} C$.

If the r.m.s. velocity of hydrogen molecules is $4$ times that of an oxygen molecule at $47^{\circ} C$,the temperature of the hydrogen molecules is (Molecular weight of Hydrogen and Oxygen are $2$ and $32$ respectively). (in $^{\circ} C$)

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