# Can the van der Waals coefficients be negative in the van der Waals equation for real gases?

A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally. True or false?

### My approach

$$P(\text{ideal}) = P(\text{real}) + P(\text{changes due to intermolecular forces})$$

So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.

### Questions

1. The answer to the question is given as True. So I want to understand what is wrong about my approach.

2. Since the intermolecular forces depend on $$a$$, for attractive nature $$a$$ is positive. Is $$a$$ negative for repulsive nature of forces?

• Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt. – Groverkss Apr 22 '19 at 17:53
• Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent. – Poutnik Apr 22 '19 at 18:22
• @Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify? – Groverkss Apr 22 '19 at 18:30
• No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions. – Poutnik Apr 22 '19 at 18:36
• – Alchimista Apr 23 '19 at 8:19

This question requires a simplistic notion of real gas behavior.

The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $$b$$ is negative in the equation and the pressure correction, $$a$$ is positive. The formula is

$$(P + a/V_\mathrm{m}^2)(V_\mathrm{m} -b) = RT$$

If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.

### The rest of the story...

The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)

• Is it correct to say that if we ignore 'a' then we are ignoring all the attraction forces and if we ignore 'b' then we are ignoring all the repulsive forces? – Groverkss Apr 23 '19 at 5:31
• If "a" is ignored then the attraction forces within the gas are ignored. If "b" is ignored then the actual volume of the gas molecules is ignored. – MaxW Apr 23 '19 at 5:34
• Do we account for repulsive forces in the van der Walls equation then? – Groverkss Apr 23 '19 at 5:37
• You have to understand that the ideal gas model and the van der Walls model are supposed to work from a temperature of zero to infinity and pressure from zero to infinity. That is nuts of course. If you fit data from just a small pressure and temperature range to get the best fit "a" and "b" constants for the van der Walls equation then you could easily get a negative "a" term. – MaxW Apr 23 '19 at 6:00
• The repulsive forces might be simply reflect in a further V reduction. If the molecules repel each others, there is apparently less Volume for them to span, ie collisions happen at a longer distance. They will have the apparent effect of increasing the proper volume of the molecules. – Alchimista Apr 24 '19 at 6:51

The constant $$a$$ only describes attractive forces while the constant $$b$$ describes repulsive forces. So by definition $$a$$ cannot be negative. Can anyone tell me if this is correct?

It is correct.

Neither parameter $$a$$ or $$b$$ is regarded as negative, because a negative parameter would be opposite to its original meaning within the van der Waals theory. In agreement with this meaning, as pointed out by MaxW, when the theory is compared to data for real gases, the parameters are consistently positive.

As you correctly point out, the $$b$$ parameter encodes the repulsive interaction when gas particles come into close contact. Excluded volume refers to the same thing but can be encoded in different mathematical language by different models. The parameters in the van der Waals equation can be related to the second virial coefficient $$B_2$$ as follows: $$B_2 = b - \frac{a}{R T} \tag{1}$$ as can the parameters describing a square-well, Lennard-Jones, or other similar potential. All of these models can generate equivalent virial equations of state to first order, provided appropriate values are chosen for the coefficients (in some cases both attractive and repulsive terms are specified in the potential function). Note that equation (1) includes a second term, $$-a/RT$$ , that reflects the attractive nature of the interaction implied by the parameter $$a$$. When $$B_2>0$$ repulsions dominate. When $$B_2<0$$ attractions dominate.

So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.

Absolutely. The vdW equation can be written as
$$P = \frac{RT}{(V_\mathrm{m} -b)}- \frac{a}{V_\mathrm{m}^2}$$

Therefore for constant temperature and volume, the repulsive parameter $$b$$ results (intuitively) in a larger pressure, whereas the attractive parameter $$a$$ results in a reduced pressure. Equation (2) can be expanded as follows:

\begin{align} P &= \frac{RT/V_\mathrm{m}}{(1 -b/V_\mathrm{m})}- \frac{a}{V_\mathrm{m}^2}\ \\ &\approx \frac{RT}{V_\mathrm{m}}\left(1+\frac{b}{V_\mathrm{m}}\right)- \frac{a}{V_\mathrm{m}^2} \\ &= \frac{RT}{V_\mathrm{m}} + \frac{RTb-a}{V_\mathrm{m}^2} \end{align}

which shows that (to first order) when $$RTb=a$$ the gas behaves as if ideal. Consistently, in that case you also have that the second virial coefficient $$B_2=0$$. Depending on the relative magnitudes of $$a$$ and $$b$$ either attractive or repulsive interactions will predominate.

The answer to the question is given as True. So I want to understand what is wrong about my approach.

Nothing is wrong with your approach except. It is more usual that the repulsive term associated with $$b$$ is dominant, so that the pressure in a real gas is typically greater than in an ideal gas at the same $$T$$ and $$V_m$$. It is not impossible for the opposite to occur, though (see this problem, for instance, but beware the different naming convention of the parameters used there).