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(D) Let the mass of a molecule in the $L$ part be $m_1$ and in the $R$ part be $m_2$. Since the separator is diathermic,the temperature $T$ of both ga

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$3\pi/8$

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(D) Let the mass of a molecule in the $L$ part be $m_1$ and in the $R$ part be $m_2$. Since the separator is diathermic,the temperature $T$ of both gases is the same.The rms speed of molecules in the $L$ part is $v_{rms} = \sqrt{\frac{3 k_B T}{m_1}}$.The mean speed of molecules in the $R$ part is $v_{mean} = \sqrt{\frac{8 k_B T}{\pi m_2}}$.Given that $v_{rms} = v_{mean}$,we have:$\sqrt{\frac{3 k_B T}{m_1}} = \sqrt{\frac{8 k_B T}{\pi m_2}}$Squaring both sides:$\frac{3 k_B T}{m_1} = \frac{8 k_B T}{\pi m_2}$$\frac{3}{m_1} = \frac{8}{\pi m_2}$Rearranging to find the ratio $\frac{m_1}{m_2}$:$\frac{m_1}{m_2} = \frac{3\pi}{8}$

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