A container holds 6 g of oxygen at a pressur | vedclass

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(B) Using the ideal gas equation,$PV = \frac{m}{M}RT$,where $m$ is the mass of the gas,$M$ is the molar mass,$R$ is the gas constant,and $T$ is the te

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$2$

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(B) Using the ideal gas equation,$PV = \frac{m}{M}RT$,where $m$ is the mass of the gas,$M$ is the molar mass,$R$ is the gas constant,and $T$ is the temperature.Since the volume $V$ and molar mass $M$ are constant,we have $P \propto \frac{mT}{V} \implies P \propto mT$.Let $m_1 = 6 \ g$,$P_1 = P$,$T_1 = 400 \ K$ and $m_2$ be the final mass,$P_2 = P/2$,$T_2 = 300 \ K$.Using the ratio $\frac{P_1}{P_2} = \frac{m_1 T_1}{m_2 T_2}$,we get $\frac{P}{P/2} = \frac{6 	imes 400}{m_2 	imes 300}$.$2 = \frac{2400}{300 m_2} \implies 2 = \frac{8}{m_2}$.$m_2 = 4 \ g$.The amount of oxygen leaked is $\Delta m = m_1 - m_2 = 6 \ g - 4 \ g = 2 \ g$.

$A$ container holds $6 \ g$ of oxygen at a pressure $P$ and temperature $400 \ K$. $A$ small hole is made in the container so that oxygen can leak out. How much oxygen (in $g$) will leak out by the time the pressure becomes $P/2$ and the temperature becomes $300 \ K$?

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