# Why is the value of "a" is positive instead of negative in Van der Waals equation?

Higher value of a means higher atttraction force between molecules. Attraction force causes the pressure to be less than expected. Now the actual pressure becomes $P + \frac{n^2a}{V^2}$ . Which means if a increases the neat pressure increases too. But isn't it the other way around?
(I assume here P is the pressure exerted on the wall of the gas container considering there is no intermolecular force)

You understand the chemistry: intermolecular attraction forces cause the pressure to be less than expected. The problem here is with the math. Though unintuitive at first sight, a positive $a$ leads to a negative correction to pressure. Let's look at a numerical example.

Assume that the pressure inside a certain container is $50$ atm with the ideal gas approximation and $45$ atm with the van der Waals equation of state. Then $P_\text{ideal} = P_\text{vdW} + 5$.

In other words, the corrected pressure is $P_\text{vdW}$, not $P_\text{vdW} + a\frac{n^2}{V^2}$.

The formula is good because it is correct in most of the regions and the parameters make sense. It is attractive because it both describes behavior of gas and approximate behavior of a liquid.

This formula has a problem in a region that corresponds to phase transition (boiling of a liquid). To account for that region correctly you need a function that is constant within some range and changes within other ranges.

Complex analysis (a branch of mathematical analysis) has a theorem that proves that you cannot have an analytical function that is constant on one interval and changing over the other intervals.

This sort of functions exist, but they are not analytic and behave poorly (hard to work with).

As a free bonus it turns out that the "strange" region describes behavior of overheated liquid and overcooled gas. Those two have different behavior at the same temperature.

• I downvoted this answer because it is irrelevant to the posed question. Commented Jul 23, 2016 at 0:16