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(D) The $rms$ velocity of the gas on the left side is given by: $\upsilon_{rms} = \sqrt{\frac{3KT}{m_L}}$.The average velocity of the gas on the right

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$3\pi/8$

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(D) The $rms$ velocity of the gas on the left side is given by: $\upsilon_{rms} = \sqrt{\frac{3KT}{m_L}}$.The average velocity of the gas on the right side is given by: $\upsilon_{avg} = \sqrt{\frac{8KT}{\pi m_R}}$.According to the problem,the $rms$ velocity of the left gas equals the average velocity of the right gas:$\sqrt{\frac{3KT}{m_L}} = \sqrt{\frac{8KT}{\pi m_R}}$.Squaring both sides:$\frac{3KT}{m_L} = \frac{8KT}{\pi m_R}$.Canceling $KT$ from both sides:$\frac{3}{m_L} = \frac{8}{\pi m_R}$.Rearranging for the ratio of masses $\frac{m_L}{m_R}$:$\frac{m_L}{m_R} = \frac{3\pi}{8}$.

$A$ tube is divided into two parts,containing two different ideal gases,$L$ and $R$. If the $rms$ velocity of the gas on the left side is equal to the average velocity of the gas on the right side,what is the ratio of the masses of the molecules in $L$ and $R$?

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