# Does the van der Waals equation remain valid when repulsive intermolecular forces dominate?

The van der Waals equation for a real gas is:

$$RT =\left(p+\frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m}-b)$$

We have understood this formula by saying that $a$ is the term which is for force of attraction between the gas molecules and thus the pressure of a real gas will be less than that of an ideal gas.

Therefore some factor must be added to the pressure of a real gas to make it equal to ideal gas pressure so that it can be used in $p_\mathrm{ideal}V_{\mathrm{ideal}}= nRT$.

Also, if we see the compressibility factor graph , suppose for $\ce{N2}$: (Source: TutorVista)

If $Z < 1$, then attractive forces are dominating. We see that after a particular temperature, the value of $Z$ becomes greater than $1$, implying that the repulsion forces are dominating in $\ce{N2}$.

Now, since repulsive force dominates, the pressure of the real gas will increase and be greater than that in ideal case.

To find the $p_{\mathrm{ideal}}$ now, we will need to deduct some value from the $p$ of real gas, which means $a$ should be negative. However, this is not so as $a$ can never be negative.

No value can ever be deducted from $p$ in the van der Waals equation.

Then how does the van der Waals equation still work and give the correct data?

• If you think too hard about it you will get a headache. The real truth is that such equations provide extra coefficients to fit the experimental data and can thus be used to adjust the relationships. Wikipedia lists 10 different equations and depending on the range for P, number of moles and V (and how the coefficients were fit), some equations work better than others. en.wikipedia.org/wiki/Real_gas
– MaxW
Jan 8, 2016 at 19:29
• Have you tried plotting up a z factor chart using the Van der Waals equation, in terms of reduced temperature and reduced pressure to see how it compares to the chart for real gases? You might find it very instructive. Jan 9, 2016 at 14:27

You need to consider the $b$ term, as well as $a$.
Rearranging the van der Waals equation to solve for $p$ gives
$$p = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2}$$
You are right that a positive $a$ will always decrease the pressure, compared to the ideal gas result.
But a positive $b$ will act to increase the pressure compared to the ideal gas result.
At high pressures (small molar volumes), the effect from finite molecular volume ($b$) outweighs the effect of intermolecular attraction ($a$).