# Change in pressure due to dissociation of gas

Let's say I have a container with fixed volume, containing Hydrogen Gas at some predefined temperature. Now I pass an electric spark in the whole container such that this reaction takes place

$$H_2 \rightarrow 2H$$

Now actually the number of moles of Hydrogen(atoms) remain constant, the volume is also constant, would there be an increase in pressure assuming the whole situation to be ideal. I just want to know the contribution in increment due to dissociation and not due to increase in temperature of the gas due to the electric spark. Will the pressure increase, decrease or remain constant?

• The temperature would be lower: it takes a lot of energy to dissociate hydrogen molecules. – Pieter Dec 25 '16 at 10:27
• That would be true if the spark contained exactly enough energy to dissociate the gas. Odds are, this being an electrical spark, this is not the case, and, assuming complete dissociation, we err on the side of too much energy, which would heat the gas back up. – probably_someone Dec 25 '16 at 11:16

$$PV=nRT$$
Here, $n$ is not the number of moles of hydrogen, but rather the number of moles of gas particles, whatever those may be. Before the spark, there are $n$ moles of hydrogen molecules. After the spark, each molecule dissociates into two hydrogen atoms, making $2n$ moles of hydrogen atoms. So, all else being constant, pressure in the chamber should double.
Of course, this is assuming that you can actually keep the gas dissociated into neutral atoms for long enough to reach steady-state conditions. In practice, this is difficult for any case except very dilute gases, as collisions between the neutral hydrogen atoms favor recombination into $H_2$.