At some point (probably before ionizing the hydrogen in a plasma) the thermodynamic concept of pressure will stop having a physical meaning. So one can just compute the second velocity moment (e.g., see https://physics.stackexchange.com/a/218643/59023) to get a proxy for the mean kinetic energy density in the particle species rest frame. This is often used as a way to estimate what is called a temperature in plasmas but really has little to do with the typical thermodynamic concept of temperature.
What would an isochoric pressure-temperature graph look like for hydrogen, going from 300K to 10,000K?
Assume we fix the volume of some box containing diatomic hydrogen then dissociate into monatomic hydrogen then ionize into a plasma. That is, the total number of individual subatomic particles is constant and the total mass is constant. Technically the number density will double after total disassociation but we haven't changed the total number of protons in this hypothetical box.
As a zeroth order start, we can assume the pressure, $P_{s}$, is just given by:
$$
P_{s} = n_{s} \ k_{B} \ T_{s} \tag{0}
$$
where $n_{s}$ is the number density of species $s$, $k_{B}$ is the Boltzmann constant, and $T_{s}$ is the temperature of species $s$.
For the moment, let's ignore how we know $n_{s}$ and $T_{s}$ at any given moment in time. Let's also ignore the pesky problem of how to actually increase the temperature. Let's just assume the temperature increases at some rate (e.g., you can assume exponential or linear or whatever). There will be some discontinuous jumps at the point of disassociation if the entire system does this at once (not likely... more likely that $H_{2}$ will gradually start to dissociate rather than all occurring at nearly the same time). Again, let's ignore this complication and assume it's a gradual process. If Equation 0 above represents our pressure and $n_{s}$ does indeed increase every time $H_{2}$ dissociates into $2 \ H$, then during this period of time the pressure should rapidly increase above the rate of temperature increase.
After fully dissociating all the $H_{2}$, the pressure would just increase at the rate you define for the increase in $T_{H}$ since $n_{H}$ no longer changes. There will be a murky period where ionization starts to occur and you have some neutral and some ionized hydrogen with electrons flying around, in which case there will be slighly different $P_{s}$'s for each species $s$. However, once past that point you should have the situation where $n_{p} = n_{e}$, i.e., the system is fully ionized and there are equal numbers of protons and electrons.
If we start with absurdly large values for $n_{H2}$ (relative to typical space plasmas like the solar wind), the particle-particle Coulomb collision rate in this final system should be low. If $n_{H2}$ is comparable to what we have in Earth's atmosphere under typical conditions (i.e., on the order of $10^{19}$ cm-3), then the particle-particle Coulomb collision rate would be very large.
The point of making these two distinctions is that in the weakly collisional case there is no particular reason why $P_{p} = P_{e}$ (e.g., see https://physics.stackexchange.com/a/268594/59023) because the mechanisms that heat a plasma do not act uniformly on the different particle species. In the collisionally mediated case of high densities, then it is likely that the system will be in equipartion and $P_{p} = P_{e}$ is actually close to satisfied. Examples of each are the solar wind (weakly collisional) and the solar photosphere (collisional).
If you heat it further and then start ionizing the H to H+ and e- (from 7000 - 10000K), what will happen to the pressure?
In short, the pressure will rise at the rate of temperature increase so long as we define pressure by the second velocity moment over each particle species (as I said above) it is related to temperature through something like the ideal gas law.
