Does dissociation of molecules into atoms change the volume of mixture?
Short answer is "not necessarily". If we isolate V in the ideal gas law, we have $$V=\frac{nRT}{P}.$$

If $n$ doubles and nothing else changes, then $V$ does also double. But the doubling of $n$ could also be offset by a doubling of $P$ or a halving of $T$, either of which would result in no change in the volume.

And yes, if the volume increases, pressure-volume work has been done, and if the volume remains constant, no pressure-volume work has been done.  That isn't the same as "no work has been done", as there are other types of work besides pressure-volume work.