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The Volume of the average adult lung when expanded is about $6L$ at $98.4^{\large\circ}F$. If the pressure of oxygen in inhaled air is $168mm$ of $Hg$ , calculate the mass of $O_2$ required to occupy the lung at $98.4^{\large\circ}F$.

$(a)\;2.3g\qquad(b)\;0.5g\qquad(c)\;1.67g\qquad(d)\;2.1g$

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$98.4^{\large\circ}F$ can be converted to $^{\large\circ}C$ as
$\large\frac{F-32}{9} = \large\frac{C}{5}$
$\large\frac{98.4 - 32}{9} = \large\frac{C}{5}$
$C = 36.88^{\large\circ}C$ or 309.88K
Then using,
$PV = \large\frac{w}{m}RT$
$\Rightarrow\large\frac{168}{760}\times6 = \large\frac{w}{32}\times0.0821\times309.88$
$\Rightarrow$ w = 1.67g
Hence answer is (c)
answered Feb 20, 2014 by sharmaaparna1
edited Mar 18, 2014 by mosymeow_1
 

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