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(D) The root mean square speed of molecules in part $L$ is given by: $v_{rms, L} = \sqrt{\frac{3KT}{m_L}}$The average speed of molecules in part $R$ i

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$3\pi/8$

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(D) The root mean square speed of molecules in part $L$ is given by: $v_{rms, L} = \sqrt{\frac{3KT}{m_L}}$The average speed of molecules in part $R$ is given by: $v_{av, R} = \sqrt{\frac{8KT}{\pi m_R}}$Given that $v_{rms, L} = v_{av, R}$,we equate the two expressions:$\sqrt{\frac{3KT}{m_L}} = \sqrt{\frac{8KT}{\pi m_R}}$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 to find the ratio $\frac{m_L}{m_R}$:$\frac{m_L}{m_R} = \frac{3\pi}{8}$

$A$ container is divided into two equal parts $L$ and $R$. If the root mean square (rms) speed of the molecules in part $L$ is equal to the average speed of the molecules in part $R$,find the ratio of the mass of a molecule in part $L$ to the mass of a molecule in part $R$.

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