Animated Solution for Chemistry - States of Matter: The compressibility factor for a real gas at high pressure is

Imagine you are trying to squeeze a gas into a tiny, microscopic box. The Ideal Gas Law, $pV=nRT$, tells a beautiful, simple story: gases are just point masses bouncing around with no volume and no feelings for each other. But in reality, when you crank up the pressure, gases start to rebel. They take up physical space, and they push back.

To measure this rebellion, we use the Compressibility Factor, denoted by $Z$. It is defined as $Z=\frac{pV}{RT}$ for one mole of gas. For a perfectly ideal gas, $Z$ is exactly $1$ under all conditions. But for real gases, $Z$ deviates from $1$, and the van der Waals equation is our master key to understanding why.

The van der Waals equation for one mole of a real gas is written as:

Here, the term $\frac{a}{V^2}$ accounts for the intermolecular attractive forces (which effectively pull molecules together and reduce the pressure they exert on the walls). The constant $b$ accounts for the excluded volume—the actual physical space the gas molecules occupy.

The question specifically asks us to investigate the gas at high pressure. When the pressure $p$ is extremely high, the gas is compressed into a very small volume. You might think both correction terms become huge. However, the external pressure $p$ is so overwhelmingly massive that the internal pressure correction $\frac{a}{V^2}$ becomes negligibly small in comparison.

Mathematically, we can say $p\gg \frac{a}{V^2}$. Therefore, we can safely approximate the pressure term:

But wait, why don't we ignore $b$? Because at high pressure, the total volume $V$ is very small. The physical volume of the molecules, $b$, now makes up a significant fraction of the container's volume. We absolutely cannot ignore it!

Let's substitute our high-pressure approximation back into the van der Waals equation. The equation simplifies elegantly:

Now, let's expand the bracket by multiplying $p$ through:

We want to find the compressibility factor $Z$, which requires the term $pV$. Let's isolate $pV$ by moving $pb$ to the right side:

Finally, divide the entire equation by $RT$ to construct $Z$:

And there is our answer! At high pressures, $Z$ is strictly greater than $1$ and increases linearly with pressure. Physically, this means the repulsive forces (due to the finite size of molecules) dominate, making the real gas harder to compress than an ideal gas.