The temperature at which the r.m.s. velocity | vedclass

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(B) The r.m.s. velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Let $T_H$ be the temperature of hydrogen and $T_O$ be the tempe

Vedclass · Accepted Answer

$132$

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(B) The r.m.s. velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Let $T_H$ be the temperature of hydrogen and $T_O$ be the temperature of oxygen $(T_O = 47 + 273 = 320 	ext{ K})$.Given: $v_{H} = 4.5 	imes v_{O}$.Substituting the formula: $\sqrt{\frac{3RT_H}{M_H}} = 4.5 	imes \sqrt{\frac{3RT_O}{M_O}}$.Squaring both sides: $\frac{T_H}{M_H} = (4.5)^2 	imes \frac{T_O}{M_O}$.Given $M_H = 2$ and $M_O = 32$: $\frac{T_H}{2} = 20.25 	imes \frac{320}{32}$.$\frac{T_H}{2} = 20.25 	imes 10 = 202.5$.$T_H = 405 	ext{ K}$.Converting to Celsius: $405 - 273 = 132^{\circ}C$.

The temperature at which the r.m.s. velocity of hydrogen molecules is $4.5$ times that of an oxygen molecule at $47^{\circ}C$ is (molecular weights of hydrogen and oxygen are $2$ and $32$ respectively). (in $^{\circ}C$)

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