Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene? Shouldn't the former have a higher boiling point because it possesses a dipole moment? I think the van der Waals forces should be equal in both cases.
The difference between the melting points is small: ortho 180 °C, meta 172 °C, para 174 °C, so they are basically the same. It is just not possible to argue bigger or smaller with hand-waving arguments about dipoles or polarity etc. with something so complicated. However, when the thermal energy is increased, whatever intermolecular interactions exist become less important relative to the average thermal energy, and which now has a dominating and disruptive effect. Thus I would expect all isomers to have a similar boiling point.
This is a guess, but maybe the spread of the chlorine atoms, which has high electronegativity, for the para-dichlorobenzene causes it to have a higher local polarity, causing it to pull the electron cloud more towards each chlorine atom, which may cancel out in total for the atom but when regarding it locally, it has a higher polarity, whereas if you place the two chlorine atoms closer together, the dipole spread is actually smaller, therefore causes it to have actually a smaller effect.
You may argue that m-dichlorobenzene has a larger polarity, but I think the effect of the larger dipole moment is smaller than it seems. The atoms is pretty big.
Disclaimer: I am not sure about the following reasons but they might have a certain amount of merit in them.
p-Dichlorobenzene essentially has no dipole moment whereas the resultant dipole moment vector comes out to be 1.72D in m-dichlorobenzene. Hence certain dipole-dipole interactions are generated in m-dichlorobenzene compared to the London dispersion forces generated in its p- counterpart. It is known that dipole-dipole interactions are stronger than the London dispersion forces thereby making the boiling point of m- dichlorobenzene greater than that of p- dichlorobenzene.