Animated Solution for Chemistry - States of Matter: If $Z$ is a compressibility factor, van der Waals' equation at low pressure can be written as

The behavior of real gases is a fascinating departure from the perfect, idealized world we often study. While the Ideal Gas Law ($pV=nRT$) assumes gas molecules have zero volume and exert no forces on each other, reality is quite different. Enter the van der Waals equation, a brilliant mathematical model that brings us back to the real world.

In this problem, we are asked to find the expression for the compressibility factor ($Z$) of a real gas at low pressure using the van der Waals equation. Let's embark on this derivation and uncover the physical meaning behind the math.

For one mole of a real gas, the van der Waals equation is written as:

This equation introduces two crucial corrections: 1. Pressure Correction ($\frac{a}{V^2}$): Accounts for the intermolecular forces of attraction that pull molecules together, effectively reducing the pressure they exert on the container walls. 2. Volume Correction ($b$): Accounts for the actual, finite volume occupied by the gas molecules themselves, which reduces the "free" volume available for them to move.

The problem specifically dictates a low pressure scenario. Imagine a gas in a highly expanded state. When the pressure is low, the volume ($V$) of the container is exceptionally large.

Because the container's volume is so massive, the tiny volume occupied by the gas molecules themselves ($b$) becomes completely negligible in comparison. Mathematically, we can say:

But what about the pressure correction term? You might be tempted to ignore $\frac{a}{V^2}$ since $V$ is large. However, there is a catch here! At low pressure, the pressure $p$ itself is very small. The term $\frac{a}{V^2}$ is actually comparable in magnitude to this small $p$. Therefore, the intermolecular attractions are still significant relative to the external pressure, and we cannot ignore the $\frac{a}{V^2}$ term.

Now, let's substitute our assumption ($V-b\approx V$) back into the master equation. The equation simplifies beautifully:

Next, we expand the bracket by multiplying $V$ with both terms inside:

Our goal is to find the compressibility factor, $Z$, which is defined as $Z=\frac{pV}{RT}$. To get there, we first need to isolate $pV$. Let's move the $\frac{a}{V}$ term to the right side of the equation:

To transform this equation into an expression for $Z$, we divide the entire equation by $RT$:

Simplifying this, we arrive at our final answer:

This elegant result perfectly matches option (b). But more importantly, it tells a physical story. Notice the negative sign? It indicates that at low pressures, $Z$ is less than 1. This mathematical truth perfectly explains the initial downward dip we see in the $Z$ versus $p$ graph for real gases. It proves that at low pressures, the attractive forces (represented by '$a$') dominate, making the gas more compressible than an ideal gas!