Question

Calculate the pressure exerted by one mole ${CO}_2$ of gas at 273 K, if the van der Waal's constat $"a"=3.502{dm}^{6}atm{}^{-2}$. Assume that the volume occupied by ${CO}_2$ molecules is negligible.

Answer 1

Ssolution:-

According to van der Waals equation of state for $1$ mole of a gas-

$(P+\frac{a}{V^2})(V-b)=0$

Since the volume occupied by gas molecules is negligible, i.e.,

$b=0$

Therefore,

$(P+\frac{a}{V^2})V=RT$

$\Rightarrow P=\frac{RT}{V^2}-\frac{a}{V^2}$

$\Rightarrow P=\frac{0.0821\times 273}{22.4}-\frac{3.502}{{\left(22.4\right)}^2}$

$\Rightarrow P=1.000-0.007=0.993$