Real Gas Equation

When a real gas has a pressure lower than the expected value from ideal gas (because of the intermolecular force of attraction) , we have -

$$P_{\text{ideal}}= P_{\text{real}} + \frac{an^2}{V^2}$$ where a is the Van der Waal's constant for intermolecular forces of attraction between a gas.

Similarly, a real gas has a lower volume than the calculated volume in ideal gas. Here, when we say volume , we mean the empty space around molecules or atoms. So, for volume, we have - $$V_{\text{real}}=V_{\text{ideal}}-nb$$ where b is the Van der Waal's constant for the volume of a molecule or atom.

Having said this, when we plug in these values into the ideal gas equation to get the real gas equation.

We put $\rightarrow$

$$\left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT$$

But shouldn't this be wrong? Because for pressure, we put $P_{\text{ideal}}$ but for Volume we put $V_{\text{real}}$. Shouldn't we put both $P_{\text{ideal}}$, $V_{\text{ideal}}$ so as to satisfy ideal gas equation?

Why do we not do so?

Please give a theoretical explanation along with a mathematical one if possible.

You have the volumes wrong. The ideal gas law is: $$\mathrm{ P_{ideal}V_{ideal} = nRT}$$
So: $$V_{\text{ideal}}=V_{\text{real}}-nb$$ when compared to Van der Waal's equation. Thus the "ideal" volume is less than the "real" volume because the molecules take up space. At least that is the rationale.
• I was regurgitating the assumptions behind Van der Waal's equation. An ideal gas is supposed to contain gas molecules which are points in space with no volume, not molecules of finite volume. So an "ideal gas" is supposed to occupy 22.4 liters at STP. But the "real gas" molecules occupy some of that volume, so the "empty" space is less than 22.4 liters. Look at a quick dimensional analysis. In the correction the factor $n$ is the number of moles of gas and $b$ is the volume per mole of the molecules. – MaxW Mar 16 '16 at 3:29