The v_{rms} value for a hydrogen molecule at | vedclass

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(B) The formula for root mean square speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.For hydrogen at $T_1 = 300 \ K$ and $M_1 = 2 \ g/mol$:$(v_{rms})_H = \s

Vedclass · Accepted Answer

$836$

Vedclass · Answer

(B) The formula for root mean square speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.For hydrogen at $T_1 = 300 \ K$ and $M_1 = 2 \ g/mol$:$(v_{rms})_H = \sqrt{\frac{3RT_1}{M_1}} = 1930 \ m/s \implies 1930 = \sqrt{\frac{3R(300)}{2}} \quad \dots(1)$For oxygen at $T_2 = 900 \ K$ and $M_2 = 32 \ g/mol$:$(v_{rms})_O = \sqrt{\frac{3RT_2}{M_2}} = \sqrt{\frac{3R(900)}{32}}$$(v_{rms})_O = \sqrt{\frac{3R(300) 	imes 3}{16 	imes 2}} = \sqrt{3} 	imes \frac{1}{4} 	imes \sqrt{\frac{3R(300)}{2}}$Substituting equation $(1)$:$(v_{rms})_O = \frac{\sqrt{3}}{4} 	imes 1930 = \frac{1.732}{4} 	imes 1930 \approx 836 \ m/s$.

The $v_{rms}$ value for a hydrogen molecule at $300 \ K$ is $1930 \ m/s$. Then the $v_{rms}$ for an oxygen molecule at $900 \ K$ is . . . . . . $m/s$.

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