Does dissociation of molecules into atoms change the volume of mixture?

For example, if $\ce{H2}$ is converted to two $\ce{H},$ does the volume of the mixture change?


If it doesn't, does it imply that the work done is $0$ because work is $P\,\mathrm dV?$




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. 


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In the dissociation of $H\_2$ into $2$ $H$, one mole of something yields $
2$ moles of something else. The number of moles increases. As $V$ is proportional to $n$, at constant pressure, the volume $V$ of the gas must increase.


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