Understanding the Problem

We are given the equation of state for a monoatomic real gas: $P(V-b)=RT$, where $b$ is a constant. We need to find the correct graph that shows the relationship between the interatomic potential $V\left(r\right)$ and the interatomic distance $r$ for this gas.

Step 1: Analyze the Equation of State

The given equation, $P(V-b)=RT$, is a modified version of the ideal gas law. The term $V-b$ accounts for the excluded volume. This means the gas molecules have a finite size and cannot occupy the same space. The constant $b$ is related to the volume occupied by one mole of the gas molecules themselves.

Key Point: This equation only corrects for the repulsive forces between atoms. It does not account for any attractive forces. Attractive forces are typically represented by a term like $\frac{a}{V^2}$ in equations like the van der Waals equation.

Step 2: Relate the Equation of State to Interatomic Forces

The equation of state of a gas is directly related to the intermolecular potential, $V\left(r\right)$.

• Repulsive Forces: These are very strong, short-range forces that dominate when atoms are very close to each other. They prevent atoms from overlapping. In a potential energy graph, this is represented by a very steep (almost vertical) rise in potential energy as $r$ decreases below a certain point.
• Attractive Forces: These are weaker, longer-range forces that pull atoms together. In a potential energy graph, this is represented by a dip or a "well" in the curve, where the potential energy is negative.

Since our equation of state $P(V-b)=RT$ only has the excluded volume term $b$, it implies that the attractive forces between the atoms are negligible for this gas. The interatomic potential will only show the effects of repulsion.

Step 3: Visualize the Correct Interatomic Potential

For a gas with only repulsive forces:

• When atoms are far apart ($r\to \infty$), the potential energy $V\left(r\right)$ is zero. They do not interact.
• As they get closer, nothing happens until they are almost touching.
• When the distance between their centers $r$ becomes less than or equal to their diameter ($r\le \sigma$), the strong repulsive force dominates. The potential energy $V\left(r\right)$ rises very sharply to positive infinity.

Therefore, the graph of $V\left(r\right)$ vs. $r$ will be a horizontal line at $V\left(r\right)=0$ for most $r$, with a vertical "wall" at a specific value of $r$ (the atomic diameter). There is no attractive well in the graph.

Step 4: Match the Graph to Our Conclusion

We need to find the graph among the options that shows:

1. A potential of zero for large $r$.
2. A steep, repulsive wall at a small, finite value of $r$.
3. No dip below zero, indicating no attractive forces.

The graph that fits this description is the one that looks like a hard-sphere potential: a flat line that jumps vertically to infinity at $r=\sigma$.

Final Answer

The correct relationship is given by the graph that shows a purely repulsive potential with no attractive well. This corresponds to the first option (top-left image).

Related Topics & Formulae

Equations of State for Real Gases

• Ideal Gas Law: $PV=nRT$
• van der Waals Equation: $(P+\frac{a{n}^2}{V^2})(V-nb)=nRT$
  Here, $a$ corrects for attractive forces and $b$ corrects for repulsive forces (excluded volume).
• Given Equation (Repulsive only): $P(V-b)=RT$

Intermolecular Potentials

• Lennard-Jones Potential: A common model that describes both attraction and repulsion. $$V\left(r\right)=4\epsilon \left[{\left(\frac{\sigma }{r}\right)}^{12}-{\left(\frac{\sigma }{r}\right)}^6\right]$$ The term with exponent 12 represents repulsion, and the term with exponent 6 represents attraction.
• Hard-Sphere Potential: A simplified model with only repulsion, which matches our given equation. $$V\left(r\right)=\{\begin{array}{ll}\infty & r\le \sigma \\ 0& r>\sigma \end{array}$$