Animated Solution for Chemistry - States of Matter: One mole of a monoatomic real gas satisfied the equation $p(V-b)=RT$ where $b$ is a constant. The relationship of interatomic potential $V(r)$ and interatomic distance $r$ for the gas is given by –

To truly understand the behavior of gases, we must look beyond the ideal gas law, $PV=nRT$. Real gases don't behave perfectly because their molecules take up space and exert forces on one another. This is beautifully captured by the van der Waals equation for one mole of a gas:

Here, we have two crucial correction terms. The parameter $a$ accounts for the intermolecular attractive forces that pull molecules together, effectively reducing the pressure they exert on the container walls. The parameter $b$ accounts for the excluded volume—the actual physical space the molecules occupy, which reduces the free volume available for them to move.

Now, let's look at the specific equation provided in the problem:

By comparing this directly with the standard van der Waals equation, a glaring difference emerges: the pressure correction term, $\frac{a}{V^2}$, is completely missing.

Mathematically, this implies that $a=0$.

What does $a=0$ mean in the physical world? It means that these gas molecules do not attract each other at all. There is no long-range pull, no dipole interactions, and no London dispersion forces. Because there is no attraction, the potential energy curve will not dip below zero. There is no "attractive well" that molecules fall into as they approach each other.

However, the constant $b$ is still present ($b eq 0$). This tells us that the molecules are not mere point masses; they have a definite physical size. They act like tiny, impenetrable billiard balls. They can move freely until they touch, but they absolutely refuse to overlap.

Let's translate this physical behavior into an interatomic potential energy graph, $V(r)$ versus the interatomic distance $r$.

When the molecules are far apart ($r>\sigma$, where $\sigma$ is the collision diameter), they feel absolutely no force. Therefore, the potential energy remains exactly zero. This is represented by a flat horizontal line along the $r$-axis.

But what happens when they get too close? The exact moment they touch ($r=\sigma$), the repulsive force becomes infinite because they cannot occupy the same space. To prevent overlap, the potential energy shoots straight up to infinity.

This sudden, dramatic jump from zero to infinity is known in physics as the Hard Sphere Potential.

If we examine the given options, we are looking for a graph that shows $V(r)=0$ for large distances, and a vertical wall at a specific short distance.

Graph (A) shows the Lennard-Jones potential with an attractive well, which is incorrect since $a=0$. Graph (B) shows an ideal gas with no forces at all, which is incorrect since $b eq 0$. Graph (D) shows a square well potential, which implies some constant attraction over a range, also incorrect.

Graph (C) perfectly captures the hard-sphere behavior: a flat line at zero, followed by an abrupt vertical wall. Therefore, (C) is the correct answer.