At what temperature is the R.M.S. velocity of | vedclass

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(B) The root mean square ($R$.$M$.$S$.) velocity of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $R$.$M$.$S$

Vedclass · Accepted Answer

$20$

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(B) The root mean square ($R$.$M$.$S$.) velocity of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $R$.$M$.$S$. velocities of hydrogen $(H_2)$ and oxygen $(O_2)$ are equal,we have $v_{H} = v_{O}$.Substituting the formula: $\sqrt{\frac{3RT_H}{M_H}} = \sqrt{\frac{3RT_O}{M_O}}$.Squaring both sides and simplifying: $\frac{T_H}{M_H} = \frac{T_O}{M_O}$.Given $T_O = 47^{\circ} C = 47 + 273 = 320 \ K$,$M_H = 2$,and $M_O = 32$.Rearranging for $T_H$: $T_H = T_O 	imes \frac{M_H}{M_O}$.Substituting the values: $T_H = 320 	imes \frac{2}{32} = 320 	imes \frac{1}{16} = 20 \ K$.

At what temperature is the $R$.$M$.$S$. velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} C$ (in $K$)? (Molecular weight of hydrogen $= 2$,molecular weight of oxygen $= 32$)

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