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(A) From the ideal gas equation, $pV = nRT = (m/M)RT$, where $m$ is the mass of the gas and $M$ is the molar mass. Since $\rho = m/V$, we have $p = (\

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$\rho V \Delta p / p$

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(A) From the ideal gas equation, $pV = nRT = (m/M)RT$, where $m$ is the mass of the gas and $M$ is the molar mass. Since $\rho = m/V$, we have $p = (\rho/M)RT$, or $\rho = pM / (RT)$.Initially, the mass of the gas in the vessel is $m_1 = \rho V$.After some gas is released, the pressure becomes $p' = p - \Delta p$. Since the volume $V$ and temperature $T$ remain constant, the new density $\rho'$ is given by $\rho' = p'M / (RT) = (p - \Delta p)M / (RT)$.The new mass of the gas in the vessel is $m_2 = \rho' V = \frac{(p - \Delta p)M}{RT} V = \frac{(p - \Delta p)}{p} \rho V$.The mass of the released gas $\Delta m$ is $m_1 - m_2 = \rho V - \frac{(p - \Delta p)}{p} \rho V$.$\Delta m = \rho V \left(1 - \frac{p - \Delta p}{p}\right) = \rho V \left(\frac{p - p + \Delta p}{p}\right) = \frac{\rho V \Delta p}{p}$.

$A$ vessel of volume $V$ contains an ideal gas having mass density $\rho$ at temperature $T$ and pressure $p$. After a portion of the gas is let out, the pressure in the vessel is decreased by $\Delta p$. The mass of the released gas is

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