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

Question

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

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

I also read that

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

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 ?

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

Answer 1

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 $$p_\mathrm{ideal}=p-p_{\mathrm{eff}}=p+\frac{an^2}{V^2}$$

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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.

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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.

Answer 2

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!