Two identical vessels A and B with frictionle | vedclass

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

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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 $P = \frac{nRT}{V} = \frac{m}{M} \frac{RT}{V}$,where $m$ is the mass,$M$ is the molar mass,$R$ is the gas constant,and $T$ is the temperature.Initially,for vessel $A$,$P_{A,i} = \frac{m_A RT}{MV}$. After isothermal expansion to volume $2V$,$P_{A,f} = \frac{m_A RT}{M(2V)}$.The change in pressure for vessel $A$ is $\Delta P = |P_{A,f} - P_{A,i}| = \frac{m_A RT}{2MV}$.Similarly,for vessel $B$,the change in pressure is $1.5 \Delta P = |P_{B,f} - P_{B,i}| = \frac{m_B RT}{2MV}$.Dividing the two equations:$\frac{1.5 \Delta P}{\Delta P} = \frac{\frac{m_B RT}{2MV}}{\frac{m_A RT}{2MV}}$$1.5 = \frac{m_B}{m_A}$$\frac{3}{2} = \frac{m_B}{m_A}$$3 m_A = 2 m_B$.

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