Firstly, thanks go to Fnord for posting that diagram here, because I was going to link to it in my response otherwise. I'll try to go against the pattern set by two of the three previous posts, and give you some possibly useful information that you might NOT already know from a cursory glance at Wikipedia.
Your idea is achievable, if not particularly helpful. This stems from the fact that, as I tend to state incessantly, CO<sub>2</sub> is a terrible propellant gas, and electrothermal tech is bulky and expensive. If you're going to spend several thousand dollars building a launcher that you can't hold or move, it should at least be made energetic and fast.
Loading the cartridges would be the easy part. Open them up, cool to roughly the sublimation point of CO<sub>2</sub>, and pack them about half full with crushed dry ice (this based on the density of the solid and the liquid). Designing and building the cartridges is the hard part. Having designed (but not implemented) ET and ETC cartridges, I can say that they're
very complicated compared to traditional firearm cartridges. You obviously have your own lathe if you're asking this question, but it's prudent to warn you that you'll be looking at possibly over an hour of machining per shot, assuming that you reload used cartridges with new seals, insulators and capillary tubes.
Now, on to the science. CO<sub>2</sub>'s critical point occurs at 304 K, roughly ten degrees above room temperature. Heated above that temperature it becomes a supercritical fluid. Practically speaking, this means that you can't slosh it around in the container any more, as it has become filled to constant density. For most of our applications, we can treat supercritical fluids as gases. In fact, in some of our modern high pressure launchers (over 500psi), nitrogen is actually a supercritical fluid.
At its critical point, the enthalpy of vaporization of a substance is zero. As such, to "vaporize" the liquid CO<sub>2</sub>, all you need to do is warm it by about ten degrees, which by itself is very easy to achieve (based on cubic fit I have for liquid CO<sub>2</sub>'s heat capacity, it takes about 34J/g). The real issue is with it converting back to a liquid in the barrel. This is where GGDT comes in. Model this hypothetical launcher, and take a look at the pressure-temperature. If the temperature is maintained above 304 K at all times, there's no problem. If it drops below that value, the pressure should be below the line on the phase diagram indicating a liquid state. If it isn't, then you need to make the CO<sub>2</sub> hotter.
This is where the tricky bit comes in - converting the temperature you want to the energy you should input. Take a look at pages two and three of
this for an idea of why. Regrettably, I've been unable to find a polynomial fit, and I have neither the dataset nor the motivation to construct my own. You, however, might. As you can see, CO<sub>2</sub> is a particularly nasty gas to do these calculations for. You're looking for a constant volume heat capacity, not constant pressure, but that's easily enough obtained by looking at the ratio of specific heats, which is much easier to come across data for.
Looking at the 3-D graph in my link, you can imagine the process you're working with taking some diagonal path across from low pressure and low temperature to increased values of each. I'm not you, but I'd settle for approximations and ballpark figures at this point (or I could write code to model it for you, but I'm afraid I'd have to start asking for hourly pay at that point
). So, here's my approximation: imagine a plane corresponding to a roughly "average" of that surface in 3-space. I'll assume a C<sub>p</sub> of 3 kJ/(kg * K). This converts to a C<sub>V</sub> of roughly 2.4 J/(g*K). For example, if you want to heat 1g of CO<sub>2</sub> to 600K (from its critical point of 304), you need to add roughly 710J.
How much energy do you need to actually pump into the cartridge? Well, a good capillary tube plasma generator might be 70% efficient at actually creating the plasma. The mixing process is tremendously complicated (turbulent flow, non-equilibrium chemistry and thermodynamics all jammed together with a healthy dollop of plasma physics on top) and depends on the geometry of the chamber and capillary tube. Some parts of the fluid will be heat to thousands of degrees, some will be almost unaffected, and it's all up to the firing process to determine what happens from there, and how this translates to barrel temperature. One thing that's certain is you'll need a good deal more than (710/0.7) J to make it work the way you hope. Remember though, that's just an example. Fooling with GGDT will tell you more about the temperature you actually need.
Also remember that you can't treat the CO<sub>2</sub> as a perfect gas in your calculations. It's far too dense for that in this case, and you'll need a better equation of state. The only constant here (before the disk ruptures) is the density - i.e., you know the number of entities, the volume they occupy, and the fact that you're dealing with CO<sub>2</sub>. Go find yourself a suitable equation of state for the temperatures and pressures you're looking to work with, because you'll need that to calculate the GGDT inputs.
Finally, if you do decide to try this, good luck, and tell us all about it if you succeed.