Two identical containers A and B with frictio | vedclass

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(C) The process is isothermal,therefore $T = 	ext{constant}$.Since $PV = \mu RT$,we have $P = \frac{\mu RT}{V}$.For chamber $A$,the change in pressur

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$3{m_A} = 2{m_B}$

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(C) The process is isothermal,therefore $T = 	ext{constant}$.Since $PV = \mu RT$,we have $P = \frac{\mu RT}{V}$.For chamber $A$,the change in pressure is $\Delta P = P_i - P_f = \frac{\mu_A RT}{V} - \frac{\mu_A RT}{2V} = \frac{\mu_A RT}{2V} \dots (i)$.For chamber $B$,the change in pressure is $1.5 \Delta P = P_i - P_f = \frac{\mu_B RT}{V} - \frac{\mu_B RT}{2V} = \frac{\mu_B RT}{2V} \dots (ii)$.Dividing equation $(i)$ by $(ii)$,we get $\frac{\Delta P}{1.5 \Delta P} = \frac{\mu_A}{\mu_B} \implies \frac{1}{1.5} = \frac{\mu_A}{\mu_B} \implies \frac{\mu_A}{\mu_B} = \frac{2}{3}$.Since $\mu = \frac{m}{M}$,where $M$ is the molar mass,we have $\frac{m_A/M}{m_B/M} = \frac{2}{3} \implies \frac{m_A}{m_B} = \frac{2}{3}$.Therefore,$3m_A = 2m_B$.

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