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(D) Since the separator is diathermic,the system is in thermal equilibrium,so the temperature $T$ is the same for both gases.The $rms$ speed of molecu

Vedclass · Accepted Answer

$\frac{3\pi}{8}$

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(D) Since the separator is diathermic,the system is in thermal equilibrium,so the temperature $T$ is the same for both gases.The $rms$ speed of molecules in the $L$ part is given by $v_{rms, L} = \sqrt{\frac{3kT}{m_L}}$,where $m_L$ is the mass of a molecule in the $L$ part.The mean speed of molecules in the $R$ part is given by $v_{mean, R} = \sqrt{\frac{8kT}{\pi m_R}}$,where $m_R$ is the mass of a molecule in the $R$ part.Given that $v_{rms, L} = v_{mean, R}$,we have:$\sqrt{\frac{3kT}{m_L}} = \sqrt{\frac{8kT}{\pi m_R}}$Squaring both sides:$\frac{3kT}{m_L} = \frac{8kT}{\pi m_R}$$\frac{3}{m_L} = \frac{8}{\pi m_R}$Rearranging to find the ratio $\frac{m_L}{m_R}$:$\frac{m_L}{m_R} = \frac{3\pi}{8}$

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