Two identical containers A and B with frictio | vedclass

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Question

(C) The initial pressure in the two containers is given by the ideal gas law $PV = nRT = (m/M)RT$,where $M$ is the molar mass.$P_A = \frac{m_A RT}{MV}

Vedclass · Accepted Answer

$3m_A = 2m_B$

Vedclass · Answer

(C) The initial pressure in the two containers is given by the ideal gas law $PV = nRT = (m/M)RT$,where $M$ is the molar mass.$P_A = \frac{m_A RT}{MV}$ and $P_B = \frac{m_B RT}{MV}$.After isothermal expansion to volume $2V$,the final pressures are:$P'_A = \frac{m_A RT}{M(2V)}$ and $P'_B = \frac{m_B RT}{M(2V)}$.The change in pressure for each container is:$\Delta P_A = P_A - P'_A = \frac{m_A RT}{MV} - \frac{m_A RT}{2MV} = \frac{m_A RT}{2MV}$.$\Delta P_B = P_B - P'_B = \frac{m_B RT}{MV} - \frac{m_B RT}{2MV} = \frac{m_B RT}{2MV}$.Given that $\Delta P_A = \Delta P$ and $\Delta P_B = 1.5 \Delta P$,we take the ratio:$\frac{\Delta P_A}{\Delta P_B} = \frac{\Delta P}{1.5 \Delta P} = \frac{1}{1.5} = \frac{2}{3}$.Substituting the expressions for $\Delta P_A$ and $\Delta P_B$:$\frac{m_A RT / 2MV}{m_B RT / 2MV} = \frac{2}{3} \implies \frac{m_A}{m_B} = \frac{2}{3}$.Therefore,$3m_A = 2m_B$.

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