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(B) For an isothermal process, the ideal gas law gives $PV = nRT$. Since $T$ is constant, $P = \frac{nRT}{V}$.For container $A$, the change in pressur

Vedclass · Accepted Answer

$3m_{A} = 2m_{B}$

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(B) For an isothermal process, the ideal gas law gives $PV = nRT$. Since $T$ is constant, $P = \frac{nRT}{V}$.For container $A$, the change in pressure is $\Delta P_A = P_{initial} - P_{final} = \frac{n_A RT}{V} - \frac{n_A RT}{2V} = \frac{n_A RT}{2V}$.Given $\Delta P_A = 2\Delta P$, so $2\Delta P = \frac{n_A RT}{2V}$ ... $(1)$.For container $B$, the change in pressure is $\Delta P_B = P_{initial} - P_{final} = \frac{n_B RT}{V} - \frac{n_B RT}{2V} = \frac{n_B RT}{2V}$.Given $\Delta P_B = 3\Delta P$, so $3\Delta P = \frac{n_B RT}{2V}$ ... $(2)$.Dividing equation $(1)$ by $(2)$:$\frac{2\Delta P}{3\Delta P} = \frac{n_A RT / 2V}{n_B RT / 2V} = \frac{n_A}{n_B}$.Thus, $\frac{n_A}{n_B} = \frac{2}{3}$.Since $n = \frac{m}{M}$ (where $M$ is molar mass), $\frac{m_A / M}{m_B / M} = \frac{2}{3}$, which implies $\frac{m_A}{m_B} = \frac{2}{3}$.Therefore, $3m_A = 2m_B$.

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