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

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(B) The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given,$T_{O_2} = 47^{\circ} C = 47 + 273 = 320 \ K$.Let $V_H$

Vedclass · Accepted Answer

$132$

Vedclass · Answer

(B) The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given,$T_{O_2} = 47^{\circ} C = 47 + 273 = 320 \ K$.Let $V_H$ be the $r.m.s.$ velocity of hydrogen and $V_O$ be that of oxygen.According to the problem,$V_H = 4.5 	imes V_O$.Substituting the formula: $\sqrt{\frac{3RT_H}{M_H}} = 4.5 	imes \sqrt{\frac{3RT_O}{M_O}}$.Squaring both sides: $\frac{3RT_H}{M_H} = (4.5)^2 	imes \frac{3RT_O}{M_O}$.Canceling $3R$ from both sides: $\frac{T_H}{M_H} = 20.25 	imes \frac{T_O}{M_O}$.Given $M_H = 2$ and $M_O = 32$: $\frac{T_H}{2} = 20.25 	imes \frac{320}{32}$.$\frac{T_H}{2} = 20.25 	imes 10 = 202.5$.$T_H = 202.5 	imes 2 = 405 \ K$.Converting to Celsius: $T_H = 405 - 273 = 132^{\circ} C$.

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 weight of hydrogen and oxygen molecules are $2$ and $32$ respectively). (in $^{\circ} C$)

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