The temperature at which RMS velocity of SO_2 | vedclass

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(D) The formula for $RMS$ velocity is $U = \sqrt{\frac{3RT}{M}}$.Given that the $RMS$ velocity of $SO_2$ is half of that of $He$ at $300 \ K$,we have

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$1200$

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(D) The formula for $RMS$ velocity is $U = \sqrt{\frac{3RT}{M}}$.Given that the $RMS$ velocity of $SO_2$ is half of that of $He$ at $300 \ K$,we have $\frac{U_{SO_2}}{U_{He}} = \frac{1}{2}$.Substituting the formula: $\frac{1}{2} = \sqrt{\frac{M_{He}}{M_{SO_2}} 	imes \frac{T_{SO_2}}{T_{He}}}$.Substituting the molar masses ($M_{He} = 4 \ g/mol$,$M_{SO_2} = 64 \ g/mol$) and temperature $(T_{He} = 300 \ K)$: $\frac{1}{2} = \sqrt{\frac{4}{64} 	imes \frac{T_{SO_2}}{300}}$.Squaring both sides: $\frac{1}{4} = \frac{4}{64} 	imes \frac{T_{SO_2}}{300}$.$\frac{1}{4} = \frac{1}{16} 	imes \frac{T_{SO_2}}{300}$.$T_{SO_2} = \frac{16 	imes 300}{4} = 4 	imes 300 = 1200 \ K$.

The temperature at which $RMS$ velocity of $SO_2$ molecules is half that of $He$ molecules at $300 \ K$ is ............... $K$.

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