Calculate the pressure exerted by one mole of CO₂ gas at 273 K if the van der Waals' constant a = 3.592 dm⁶ atm mol^-2. Assume that the volume occupied by CO₂ molecules is negligible.

In case of negligible morecular volume, b = 0 and van der Waals' equation reduces to

$$\begin{gathered} \left(p+\frac{n^{2}a}{V^2}\right)V=nRT,p=\frac{RT}{V}-\frac{a}{V^2}(n=1\text{ mole }) \\ =\frac{{\displaystyle 0.082\times 273}}{{\displaystyle 22.4}}-\frac{{\displaystyle 3.592}}{{\displaystyle (22.4{)}^2}}=0.99\mathrm{atm}\approx 1\mathrm{atm} \end{gathered}$$