There are several different equations that better approximate gas behaviour than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules (correcting $V$, shown in red below) and another for the attractive forces between them (correcting $P$, shown in blue below).
$$\color{blue}{P}\color{red}{V} = nRT\\ \color{blue}{\left(P + \frac{an^2}{V^2}\right)}\color{red}{\left(V – nb\right)}=nRT$$The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant b corresponds to the size of the molecules of a particular gas. Both a and b are determined experimentally.
Note that when V is relatively large and n is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, PV = nRT. This corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure (i.e. our “ideal” conditions!).
At low pressures, the correction for intermolecular attraction, a, is more important than the one for molecular volume, b. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume.
Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case.
Comparison of Ideal Gas Law and van der Waals Equation
A 4.25-L flask contains 3.46 mol CO2 at 229 °C. Calculate the pressure (in bar) of this sample of CO2:
(a) from the ideal gas law
(b) from the van der Waals equation
the factor a for CO2 is 3.59 $L^2 bar / mol^2$, and b is 0.0427 $L/mol$.
(c) Explain the reason(s) for the difference.
Solution
(a) From the ideal gas law: $$P=\frac{nRT}{V}=\frac{3.46\;\require{enclose}\enclose{horizontalstrike}{mol}\times 0.08314510\;\enclose{horizontalstrike}{L}\;bar\;\enclose{horizontalstrike}{mol^{-1}\;K^{-1}}\times 502\;\enclose{horizontalstrike}{K}}{4.25\;\enclose{horizontalstrike}{L}}=\mathbf{34.0\; bar}$$
(b) From the van der Waals equation:
$$(P+\frac{an^2}{V^2})\times (V-nb)=nRT⟶P=\frac{nRT}{(V-nb)}-\frac{n^2a}{V^2}\\ P=\frac{3.46\;mol\times 0.0831451\;L\;bar\;mol^{-1}\;K^{-1}\times 502\;K}{(4.25\;L-3.46\;mol\times 0.0427\;L\;mol^{-1})}-\frac{(3.46\;mol)^2\times 3.59\;L^2\;bar\;mol^2}{(4.25\;L)^2}\\ P=\mathbf{32.8 bar}$$(c) The value calculated by the van der Waals equation is somewhat different from the ideal gas law because CO2 molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions. The difference is however not very large, because the pressure is not very high and the temperature is not very low.
Check your Learning
A 560-mL flask contains 21.3 g N2 at 145 °C. Calculate the pressure (in bar) of N2:
(a) from the ideal gas law
(b) from the van der Waals equation
the factor a for N2 is 1.39 $L^2 bar / mol^2$, and b is 0.0391 $L/mol$.
(c) Explain the reason(s) for the difference.
Answer
(a) 47.178 bar
(b) 47.210 bar
(c) As in the example above, the values are slightly different because the van der Waals equation takes into account the attractive forces and size effects that the ideal gas law does not. The difference between the two values is even smaller this time, since the corrective factors for N2 are quite small compared to CO2 – it behaves more similarly to an ideal gas in general.