A vessel has 8 g of oxygen at pressure P and | vedclass

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(C) The volume $V$ of the vessel remains constant.Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is the mass of gas,$M$ is the m

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(C) The volume $V$ of the vessel remains constant.Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is the mass of gas,$M$ is the molar mass).Since $V$ and $M$ are constant,we have $\frac{P}{mT} = 	ext{constant}$.Therefore,$\frac{P_1}{m_1 T_1} = \frac{P_2}{m_2 T_2}$.Given: $P_1 = P$,$m_1 = 8 \ g$,$T_1 = 400 \ K$,$P_2 = P/4$,$T_2 = 200 \ K$.Substituting the values: $\frac{P}{8 	imes 400} = \frac{P/4}{m_2 	imes 200}$.$\frac{1}{3200} = \frac{1}{800 	imes m_2}$.$m_2 = \frac{3200}{800} = 4 \ g$.The mass of oxygen that leaked out is $\Delta m = m_1 - m_2 = 8 \ g - 4 \ g = 4 \ g$.

$A$ vessel has $8 \ g$ of oxygen at pressure $P$ and temperature $400 \ K$. $A$ small hole is made in it so that oxygen leaks out. How many grams of oxygen leak out if the final pressure is $P/4$ and the temperature is $200 \ K$?

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