The boiling points of the chlorinated silanes and methanes are given below:
$$\begin{array}{ccc} \hline \text{Species} & \text{Boiling point (X = Si) / }\mathrm{^\circ C} & \text{Boiling point (X = C) / }\mathrm{^\circ C} \\ \hline \ce{XH3Cl} & -30.4 & -24.2 \\ \ce{XH2Cl2} & 8 & 39.6 \\ \ce{XHCl3} & 31.8 & 69.2 \\ \ce{XCl4} & 57.6 & 76.7 \\ \hline \end{array}$$
The chlorosilanes are expected to have larger dipole moments and stronger dispersion forces. Why do the silanes instead have lower boiling points than the methanes?