Among the following, the incorrect statement is:

We recall that a gas behaves ideally when the van der Waals corrections for intermolecular attraction and molecular volume become negligible. Mathematically, the van der Waals equation is first written

$$\left(P + \dfrac{a}{V_m^{\,2}}\right)\!\left(V_m - b\right) = RT,$$

where $$P$$ is pressure, $$V_m$$ is molar volume, $$T$$ is absolute temperature, $$R$$ is the gas constant, and the constants $$a$$ and $$b$$ account for attraction and finite molecular size, respectively. A gas will approach ideal behaviour when the two correction terms $$\dfrac{a}{V_m^{\,2}}$$ and $$b$$ are very small compared with the other quantities present in the equation.

We now examine each experimental condition in the four options one by one, always asking whether those two corrections really become negligible.

Option A: “at low pressure, real gases show ideal behaviour.”

At low pressure we have $$P \rightarrow 0$$. From the ideal gas relation $$PV_m = RT$$ we see that $$V_m \rightarrow \infty$$ when $$P$$ becomes very small (with $$T$$ fixed). Because $$V_m$$ is now very large, the attraction term

$$\dfrac{a}{V_m^{\,2}} \rightarrow 0$$

and the molecular volume correction $$b$$ is also negligible compared with the huge $$V_m$$. Hence the van der Waals equation reduces to $$PV_m = RT$$, which is the ideal gas law. So Option A is a correct statement.

Option B: “at very low temperature, real gases show ideal behaviour.”

At very low temperature we have $$T \rightarrow 0$$ (in practice, simply “much lower than room temperature”). Substituting a small $$T$$ into the ideal gas law $$PV_m = RT$$ requires either a very small pressure or a very small volume if the product $$PV_m$$ is to stay equal to $$RT$$. In reality, however, molecules at low temperature possess little kinetic energy, so the attractive forces between them (represented by $$a$$ in the van der Waals equation) become more significant. Consequently the correction term $$\dfrac{a}{V_m^{\,2}}$$ cannot be neglected; it in fact dominates the behaviour, often leading to liquefaction. Therefore real gases deviate strongly from ideality at very low temperatures. Option B is therefore an incorrect statement.

Option C: “at Boyle's temperature, real gases show ideal behaviour.”

Boyle’s temperature $$T_B$$ is defined as the temperature at which the second virial coefficient $$B(T)$$ becomes zero, i.e.

$$B(T_B) = 0.$$

The virial expansion $$\dfrac{PV_m}{RT} = 1 + \dfrac{B(T)}{V_m} + \dots$$ then gives $$\dfrac{PV_m}{RT} = 1$$ when $$T = T_B$$, even for a moderate range of pressures. Hence real gases do behave ideally at Boyle’s temperature. Option C is a correct statement.

Option D: “at very large volume, real gases show ideal behaviour.”

Very large molar volume is the same physical situation as very low pressure: molecules are far apart. As already shown for Option A, large $$V_m$$ makes $$\dfrac{a}{V_m^{\,2}} \rightarrow 0$$ and makes $$b$$ negligible. So the gas again obeys $$PV_m = RT$$, i.e. ideal behaviour. Option D is therefore a correct statement.

We find that Options A, C, and D are correct statements, whereas Option B is not. Therefore the one and only incorrect statement is Option B.

Hence, the correct answer is Option B.