At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as $$p = \frac{RT}{V - b}$$ at T. Here, b is the van der Waals constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs p?

We recall that the compression factor is defined as $$Z=\dfrac{pV}{RT}.$$

For the gases in the problem the experimentally observed equation of state neglects the attraction term and is written as $$p=\dfrac{RT}{V-b},$$ where $$b$$ is the van der Waals excluded-volume constant.

We first manipulate this equation to obtain $$V$$ in terms of $$p$$.

Starting with $$p(V-b)=RT,$$ we multiply out to get $$pV-pb=RT.$$

Now we isolate $$V$$. Adding $$pb$$ to both sides and dividing by $$p$$, we obtain $$V=b+\dfrac{RT}{p}.$$

We substitute this expression for $$V$$ in the definition of the compressibility factor: $$Z=\dfrac{pV}{RT}= \dfrac{p\left(b+\dfrac{RT}{p}\right)}{RT}.$$

We separate the two terms in the numerator: $$Z=\dfrac{pb}{RT}+\dfrac{p}{RT}\cdot \dfrac{RT}{p}.$$

The second term simplifies immediately to 1, giving the very convenient linear relation $$Z=1+\dfrac{b}{RT}\,p.$$

Thus, at the fixed temperature $$T$$ of the experiment, the plot of $$Z$$ versus $$p$$ is a straight line with slope $$\text{slope}= \dfrac{b}{RT}.$$

Because $$R$$ and $$T$$ are common to all the gases, the steepness of the line is governed solely by the magnitude of $$b$$. The larger the value of $$b$$, the steeper the increase of $$Z$$ with pressure.

The van der Waals constant $$b$$ reflects the effective molecular size; for the noble gases it increases with atomic size in the order $$\text{Ne} < \text{Ar} < \text{Kr} < \text{Xe}.$$

Therefore, xenon (Xe) has the largest $$b$$ and consequently the largest value of $$\dfrac{b}{RT}$$, giving the greatest slope in the $$Z$$ vs. $$p$$ plot.

Hence, the correct answer is Option D.