The temperature at which oxygen molecules wil | vedclass

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(C) The root mean square (r.m.s.) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the r.m.s. speeds of oxyg

Vedclass · Accepted Answer

$2640$

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(C) The root mean square (r.m.s.) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the r.m.s. speeds of oxygen $(O_2)$ and helium $(He)$ are equal,we have:$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$Squaring both sides and simplifying,we get:$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{He}}{M_{He}}$Given $T_{He} = 57^{\circ} C = 57 + 273 = 330 \ K$,$M_{O_2} = 32$,and $M_{He} = 4$.Substituting the values:$\frac{T_{O_2}}{32} = \frac{330}{4}$$T_{O_2} = \frac{330 	imes 32}{4} = 330 	imes 8 = 2640 \ K$.

The temperature at which oxygen molecules will have the same r.m.s. speed as helium molecules at $57^{\circ} C$ is (molecular masses of oxygen and helium are $32$ and $4$ respectively.) (in $K$)

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