Van der Waals Equation of State

Johannes van der Waals earned the 1910 Nobel Prize for two corrections to the ideal gas law. The $a n^2/V^2$ term adds back the pressure that molecular attraction steals, and the $nb$ term subtracts the space the molecules themselves occupy. Put one mole of carbon dioxide in a one-litre vessel at 300 K and the ideal gas law promises 2494 kPa. Van der Waals, with $a = 0.3640$ and $b = 4.267\times 10^{-5}$, says 2242 kPa. Attraction alone accounts for 364 kPa of that, and the real measured value sits near the van der Waals figure.

Expand the equation and it is a cubic in volume, which is the whole reason there is no clean formula for $V$. Below the critical temperature the cubic has three real roots. The largest is the saturated vapour, the smallest the saturated liquid, and the middle one is physically unstable and corresponds to nothing you can put in a vessel. This page's volume brain brackets the largest root and bisects for it, so the answer it gives is the vapour branch. If you are working near or below the critical point and need the liquid root, the equation is telling you something the page cannot: use a proper equation of state.

Watch the units on $a$ and $b$, because handbooks quote them in at least four conventions. This page wants $a$ in Pa·m⁶/mol² and $b$ in m³/mol. The common table entry of 3.640 bar·L²/mol² for CO₂ is the same number as 0.3640 Pa·m⁶/mol², and 0.04267 L/mol is 4.267×10⁻⁵ m³/mol. The factor of ten between 3.640 and 0.3640 has ruined a great many homework sets.