# How is the Pressure defined in the Van der Waals' equation?

I have some doubts regarding the Van der Waals' equation which relates the Pressure, volume and temperature of real gases.

It is given by : $$(P_{real}+\frac{an^2}{V^2})(V-nb) = nRT$$

$$P_{ideal}=P_{real}+\frac{an^2}{V^2}$$

I know that the term $$(V-nb)$$ represents the volume remaining for a real gas. So in my opinion, the Pressure term in the above equation should also be the Pressure of the real gas on the walls.

But the second mentioned equation doesn't indicate this and clearly contradicts my intuition.

So my question is What is the term $$P$$ in the Van der Walls' equation ?

If it is the pressure of the ideal gas then why should we include Pressure of ideal gas in the equation for real gases ? Shouldn't the equation be $$(P-\frac{an^2}{V^2})(V-nb)=nRT\;?$$

Also if this is the pressure in the middle of a real gas then why should we include this pressure to indicate the pressure at the walls ?

If someone has doubt with the last paragraph of this question, please see the comments to the Garf's answer.

• from around 25:10 of this video
– Babu
Commented Jan 13, 2021 at 15:57
• @Buraian my doubt is that the pressure which we are using is the pressure that an ideal gas will exert and the volume we are using is the volume occupied by the real gas. Isn't it strange ? Commented Jan 13, 2021 at 18:04
• Apologies that I couldn't reply earlier @A Student, it's kinda hard to difficult to explain here, but if you see the book of physical chemistry by atkin's then in around the first chapter where they introduce the ideal gas law, they talk about how the ideal gas law defines a surface in the P,V,T coordinates. So, I think an intuition for this would be that, this new van der waal equation also gives a surface, but this surface has more of an 'overlap' with the behaviour of a 'real gas'
– Babu
Commented Jan 21, 2021 at 17:13

What is the term $$P$$ in the Van der Walls' equation?

It's Pressure! First, It's better to first go through the origin of this term!

The number of nearest neighbors is proportional to $$n/V$$, and so attractive intermolecular interactions lower the total potential energy by an amount proportional to the number of atoms multiplied by the number of nearest neighbors so that the energy change as $$\frac{an^2}{V}$$ Hence, if you change $$V$$, the energy changes by an amount $$-\frac{an^2dV}{V^2}$$ but this energy change can be thought of as being due to an effective pressure $$p_\mathrm{eff}$$ so that the energy change would be $$-p_{\mathrm{eff}}dV$$. Hence $$p_{\mathrm{eff}}=-\frac{an^2}{V^2}$$ The pressure $$p$$ that we measure is the sum pressure $$p_\mathrm{ideal}$$ where we neglect the interaction and $$p_\mathrm{eff}$$. And So $$p_\mathrm{ideal}=p-p_{\mathrm{eff}}=p+\frac{an^2}{V^2}$$

If it is the pressure of the ideal gas then why should we include the Pressure of ideal gas in the equation for real gases?

The ideal gas equation is given by $$p_\mathrm{ideal}V=nRT$$ As we have discovered, The real pressure (pressure of real gas ) in terms of ideal gas term and an extra term due to interactions we can write $$\left(p+\frac{an^2}{V^2}\right)V=nRT$$ What you have right down is wrong because we know that for the real gas $$pV\not=nRT$$ If it does follow, then we don't have to do all that crap in the first place.

In the last, I don't understand what do you mean by the pressure in the middle and in the wall. It doesn't make any difference pressure is what it is.

• for the last paragraph please see comments below Garf's answer.. Commented Jan 12, 2021 at 19:27
• The energy change that you mentioned , shouldn't it be equal to -pdV (p is the real pressure). Commented Dec 24, 2022 at 16:23

I think it's very easy to get confused here. Let's remind ourselves why this correction is needed...

It's to do with the fact that when we measure the pressure of the gas (call that $$P_\text{real}$$), our measurement will be slightly less than what's going on in the middle of the gas (call that $$P_\text{ideal}$$ - the "true" pressure of our gas). The reason for this is that we can never measure pressure of "bulk" gas, because we have to put some device somewhere and then define an edge to our "box" of gas. Then we argue that gas particles at the edge of the box feel a net force of attraction away from the edge, because of unbalanced van der Waals forces from the other particles in the bulk. Hence the pressure measured ($$P_\text{real}$$) is less than the pressure in the bulk ($$P_\text{ideal}$$).

I would personally call these things slightly different things to avoid confusion. I think generally when we see $$P$$, we assume it's a pressure that can be measured - after all physics equations should be written in terms of measurable things!

• As a side note - I suppose one assumption in the vdW formulation is that in the bulk gas, all vdW forces are equal and in all directions, so have no net effect on the pressure in the bulk. This is probably why it is referred to as $P_\text{ideal}$ here, because we would expect the pressure in the middle to be the same as the pressure in an ideal gas (ignoring the volume effects) but I agree this is confusing.
– Garf
Commented Jan 12, 2021 at 13:34
• so the term $(P+\frac{an^2}{V^2})$ is the pressure in the middle or bulk ? Commented Jan 12, 2021 at 13:40
• That term will be pressure in the bulk
– Garf
Commented Jan 12, 2021 at 13:40
• But that number is less meaningful than $P$, because we could never actually measure that.
– Garf
Commented Jan 12, 2021 at 13:40
• A probably equivalent description of the effect of the Van-der Waals interaction is that the molecules spend more time in each other's potential wells, so that the number of collisions with the walls is lower than without such interactions.
– user137289
Commented Jan 12, 2021 at 19:13