Question

At what pressure and ${127}^{0}C$, the density of $O_2$ gas becomes 1.6g/L? [a=4.0 atm $L^{2}mo{l}^{-2}$,0.4$Lmo{l}^{-1}$, R=0.08 L.atm/K.mol]

Answer 1

$(P+\frac{a{n}^2}{V_{m^2}})(V-nb)=nRT$

let $n=1$ then for $d=1.6,V=20$

$\Rightarrow (P+\frac{4}{20\times 20})(10-0.4)=0.08\times 400$

$\Rightarrow (P+\frac{1}{100})\times 9.6=8\times 4$

$\Rightarrow P+0.01=\frac{82\times 10}{9.6}$

$\Rightarrow P=3.33-0.01$

$=3.32atm$