# Real Gases , Van der Waal Equation , Solved Examples

The hypotheses dealing with negligible volume and intermolecular forces are clearly approximations and it is not surprising that gas under conditions such as high pressure and low temperature will exhibit behaviors that violate the ideal gas laws.

The simplest of the equations to try to treat real gases was developed by Johannes van der Waals.

Based on experiments on pure gases in his laboratory, van der Waals recognized that the variation of each gas from ideal behavior could be treated by introducing two additional terms into the ideal gas equation.

These terms account for the fact that real gas particles have some finite volume, and that they also have some measurable intermolecular force.

The two equations are presented below:

$\large P V = n R T$

$\large (P + \frac{a n^2}{V^2}) (V – n b) = n R T$

In the van der Waals equation the constant “a” is a correction term for intermolecular force and “b” is a correction for the real volume of the gas molecules.

Intermolecular forces are often lumped together and called van der Waals’ forces in recognition of his being the first to recognize their existence.

The “a” term is a correction for intermolecular forces and by itself is quite simple. The magnitude of “a” is reflective of the intermolecular forces between molecules. Thus the increase in “a” going down the noble gases reflects the increasing London forces between these atoms.

Molecules that are polar typically have larger values of “a” than do nonpolar molecules. The surprisingly large value of a molecule like carbon tetrachloride reflects the London forces as well as a property called Polarizability.

The portion of the equation containing the “a” term is somewhat more confusing that the term itself and requires a bit of explanation.

First, we note that the volume comes into play even though this is a term correcting for intermolecular force. The volume term is necessary because we must have a way of recognizing that intermolecular forces will be negligible at long distances, but significant at small distances.

The bigger the volume (notice that it’s on the bottom) the smaller the correction term because the bigger the volume the further apart the molecules are and the less their intermolecular forces are likely to come into play.

Next we note that the term containing “a” is added to the pressure P.

In other words the corrected pressure is slightly larger than the observed pressure. How does this make any sense ? If we go back to the origins of pressure itself, we remember that pressure arises when molecules strike the walls (not Waals) of the container.

As our molecule moves toward the wall all of its gaseous neighbors are behind it, thus any intermolecular forces acting on the molecule are slowing it as it moves toward the wall. The molecule striking the wall does so at a velocity that is slightly less (determined by the size of the intermolecular forces) than it would without these forces.

Or, the observed pressure is actually slightly less than the pressure would be without the intermolecular forces.

To correct for this, a small positive pressure must be added to the observed pressure, hence the term with “a” in it is positive. The “b” term is relatively easy to understand. Our gas is in a container with some volume “V”.

If the container is sufficiently large, the volume of the gas molecules is negligible, but if the volume of the gas molecules themselves starts to become significant with respect to the volume, we have a problem.

Under these conditions, we may wish to think of the actual free volume in the container as being the measured volume, V, minus the volume occupied by the molecules.

We calculate the volume occupied by the molecules by multiplying the number of moles of molecules , n , by their effective molar volume , b.

Let’s think about the values of “b” for a moment. The volume of one mole of water molecules condensed into liquid water would be 18 ml/mole (water is 18 g/mol and the density of water is by definition 1 g/ml).

The value of “b” for water, however, is 30.5 ml/mole (notice I’ve taken the table value which is in liters/mol and converted it to ml/mol since that’s the easier number to think about). Why the difference?

In the gas phase, our water molecules spin around rapidly so that they occupy a larger volume than they do when they are relatively fixed in their positions in liquid water.

This is like comparing the area around you if you were to stand still and hold your arms out compared to the area you occupy if you were to spin around.

All of the values for “b” will be somewhat larger than the volume of one mole of that compound in its liquid form.

Example : 2 moles of NH3 occupied a volume of 5 litres at 27°C. Calculate the pressure if the gas obeyed Vander Waals equation. Given a = 4.17 litre2 atm mole−2 , b = 0.037 litre/mole.

Solution : Applying Vander Waal’s equation for n moles

$\large (P + \frac{a n^2}{V^2}) (V – n b) = n R T$

$\large (P + \frac{4.17 \times 2^2}{5^2}) (5 – 2 \times 0.037) = 2 \times 0.082 \times 300$

P = 9.33 atm