Question

Find he volume occupied by $7g$ of $N_2$ under a pressure of $100atm$ at ${27}^{o}C$ $(a=1.39atm{L}^2{mol}^{-2}.b=0.391L{mol}^{-1})$

Answer 1

$(P+a\ast n^2/v^2)(v-n\ast b)=n\ast R\ast T$

no. of moles of Nitrogen$=\frac{7}{28}=0.25$

substituting in the equation:

$\frac{100v^2+1.39\ast 0.0625}{v^2}(v-0.25\ast 0.391)=0.25\ast 0.082\ast 300$

equation 1: $100v^2+(1.39\ast 0.0625)=0.25\ast 0.082\ast 300v^2$

$93.85v^2=-0.086875$

$v^2=-0.0009256792$

$v=-0.03042497$

equation2: $v-0.25\ast 0.391=0.25\ast 0.082\ast 300$

$v=6.15+0.09775=6.24775l$