Question

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

• A. $3.59 \,dm ^6 atm\, mol ^{-2}$
• B. $2.59 \,dm ^6 atm\, mol ^{-2}$
• C. $1.25 \,dm ^6 atm \,mol ^{-2}$
• D. $1.59 \,dm ^6 atm\, mol ^{-2}$

Answer 1

Answer: (A)

${\left[P+\frac{a}{V^2}\right][V-b]=R T}$
$\therefore\left[P+\frac{a}{V^2}\right] V=R T$
Or $V^2 P-R T V+a=0$
$V=\frac{+R T \pm \sqrt{R^2 T^2-4 Pa }}{2 P}$
Since, $V$ is constant at given $P$ and $T, V$ can have only one value or discriminant $=0$
$\therefore R^2 T^2=4$ Paor $a=\frac{R^2 T^2}{4 P}$
$=\frac{(0.821)^2 \times(273)^2}{4 \times 34.98}$
$=3.59\, dm ^6\, atm\, mol ^{-2}$