One mole of a monoatomic real gas satisfies 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

Gaseous State PYQ — JEE Advanced 2015 · Shift 2 · Q18 with Solution

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Correct answer: C

For one mole of gas, $p(V-b)=RT$ so $p=\frac{RT}{V-b}.$

This is the van der Waals equation with only excluded volume correction and no attractive term.

The usual van der Waals form is $\left(p+\frac{a}{V^2}\right)(V-b)=RT.$ Here, effectively, $a=0.$

So the gas behaves like a system of hard spheres:

At very small separation, particles cannot overlap, so potential is effectively infinite repulsion.
At larger separation, there is no interaction, so potential is zero.

Hence the interatomic potential $U(r)$ versus distance $r$ must be:

\begin{cases} \infty, & r<\sigma \\ 0, & r>\sigma \end{cases}$$ where $\sigma$ is the hard-sphere diameter. This is the **hard-sphere potential**. Graphically: - a vertical rise to $\infty$ at some finite distance, - zero potential for all larger distances, - **no negative well**. --- 4. **Choosing the correct option** Therefore, the correct graph is the one representing **hard-sphere repulsion only**. So the correct option is: $$\boxed{\text{C}}$$ --- 5. **Comparison with stored answer** Stored correct answer: **C** My derived answer: **C** They agree.