Question

How many moles of $S{O}_2$ will occupy a volume of $10$ litre at a pressure of $15atm$ and temperature $624K$? $(a=6.71atm{L}^{2}mo{l}^{-2};b=0.0564litremo{l}^{-1})$.

Answer 1

First assume ideal behavior.
$n=\frac{PV}{RT}$
$n=\frac{15atm\times 10L}{0.08206Latm/molK\times 624K}$
$n=2.9$ moles.
Now assume non ideal behavior.
$(P+\frac{a{n}^2}{V^2})(V-nb)=nRT$
Neglect nb as $V\left(10L\right)>>>b\left(0.0564L/mol\right)$
$(P+\frac{a{n}^2}{V^2})V=nRT$
$(15atm+\frac{6.71atm{L}^{-2}mo{l}^{-2}\times n^2}{(10L{)}^2})\times 10L=n\times 0.08206Latm/molK\times 624K$
$150+0.671n^2=51.2n$
$0.671n^2-51.2n+150=0$
This is the quadratic equation with solutions
$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$n=\frac{-(-51.2)\pm \sqrt{(-51.2{)}^2-4(0.671)\times 150}}{2\left(0.671\right)}$
$n=\frac{51.2\pm 47.1}{1.342}$
$n=3.05$ or $n=73.25$
The value 73.25 is discarded as it is significantly different from the value obtained from ideal behavior.
Hence, $n=3.05$