At what pressure and $127^{o}$ C, the density of $O_{2}$ gas becomes $1.6$ g/L? [ $a = 4.0$ atm $L^{2}$ $m o l^{- 2}$ , $0.4$ L $m o l^{- 1}$ , R $= 0.08$ L-atm /K-mol]

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Solution

Density $P = 1.5 g / L$ $(P = M / V) \Rightarrow V = (M / P)$
Temp. $T = 127^{0} c = 400 k$
$a = 4 a t m L^{2} {m o l}^{- 2}$ $b = 0.4 L / m o l$
By vander waal's gas equation,
$[P + \frac{n^{2}}{V^{2}} a] (V - n b) = n R T$ (assume $1$ mole $\Rightarrow n = 1$ )
$\Rightarrow [P + \frac{1}{{(32 / 1.6)}^{2}} (4)] (\frac{32}{1.6} - (1) (0.4))$ Mass of $O_{2} = 32$
$= 1 \times 0.08 \times 400$
$\Rightarrow [P + 0.01] (19.6) = 32$
$\Rightarrow P + 0.01 = 1.6326 \Rightarrow P = 1.6227 a t m$

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