A vessel has 6 g of oxygen at pressure P and | vedclass

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(C) From the ideal gas equation,$PV = nRT$.Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.This i

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(C) From the ideal gas equation,$PV = nRT$.Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.This implies $m = \frac{PVM}{RT}$,so $m \propto \frac{P}{T}$.Let $m_1 = 6 \ g$,$P_1 = P$,and $T_1 = 400 \ K$.Let $m_2$ be the final mass,$P_2 = \frac{P}{2}$,and $T_2 = 300 \ K$.Using the ratio $\frac{m_2}{m_1} = \frac{P_2}{P_1} 	imes \frac{T_1}{T_2}$,we get:$\frac{m_2}{6} = \frac{P/2}{P} 	imes \frac{400}{300} = \frac{1}{2} 	imes \frac{4}{3} = \frac{2}{3}$.Therefore,$m_2 = 6 	imes \frac{2}{3} = 4 \ g$.The mass of oxygen that leaked out is $\Delta m = m_1 - m_2 = 6 \ g - 4 \ g = 2 \ g$.

$A$ vessel has $6 \ g$ of oxygen at pressure $P$ and temperature $400 \ K$. $A$ small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is $\frac{P}{2}$ and temperature is $300 \ K$ (in $g$)?

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