Question

The pressure exerted by $1mol$ of $C{O}_2$ at 273 $K$ is 34.98 $atm$. Assuming that volume occupied by $C{O}_2$ molecules is negligible, the value of van der Waals' constant for attraction of $C{O}_2$ gas is :

• A. $3.59d{m}^{6}atmmo{l}^{-2}$
• B. $2.59d{m}^{6}atmmo{l}^{-2}$
• C. $1.25d{m}^{6}atmmo{l}^{-2}$
• D. $1.59d{m}^{6}atmmo{l}^{-2}$

Answer 1

The correct option is A $3.59d{m}^{6}atmmo{l}^{-2}$

$[P+\frac{a}{V^2}][V-b]=RT$

$\therefore [P+\frac{a}{V^2}]V=RT$

$V^{2}P-RTV+a=0$

$\therefore V=\frac{+RT\pm \sqrt{R^2{T}^2-4Pa}}{2P}$

Since,$V$ is constant at given $P$ and $T$, $V$ can have only one value or discriminant $=0$

$\therefore R^2{T}^2=4Pa$ or $a=\frac{R^2{T}^2}{4P}$

$=\frac{(0.821{)}^2\times (273{)}^2}{4\times 34.98}$

$3.59d{m}^{6}atmmo{l}^{-2}$

Hence, the correct option is $A$