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(B) For an ideal gas, the pressure $P$ is given by $P = \frac{nRT}{V} = \frac{m}{MV} RT$, where $m$ is the mass of the gas and $M$ is the molar mass.F

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$3 m_A = 2 m_B$

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(B) For an ideal gas, the pressure $P$ is given by $P = \frac{nRT}{V} = \frac{m}{MV} RT$, where $m$ is the mass of the gas and $M$ is the molar mass.For vessel $A$, the initial pressure is $P_{A,i} = \frac{m_A RT}{MV}$ and the final pressure is $P_{A,f} = \frac{m_A RT}{M(2V)}$.The change in pressure is $\Delta P = P_{A,i} - P_{A,f} = \frac{m_A RT}{MV} - \frac{m_A RT}{2MV} = \frac{m_A RT}{2MV}$.Similarly, for vessel $B$, the change in pressure is $1.5 \Delta P = \frac{m_B RT}{2MV}$.Dividing the expression for $1.5 \Delta P$ by the expression for $\Delta P$, we get:$\frac{1.5 \Delta P}{\Delta P} = \frac{m_B RT / 2MV}{m_A RT / 2MV} = \frac{m_B}{m_A}$.Therefore, $1.5 = \frac{m_B}{m_A}$, which implies $\frac{3}{2} = \frac{m_B}{m_A}$, or $3 m_A = 2 m_B$.

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