Background
Generally, all phase transitions require some input energy in order for the transition to occur. For instance, the transition from solid-to-liquid or vice versa requires what is called the enthalpy of fusion or latent heat of fusion. This is the amount of energy needed to change the total interal energy (i.e., enthalpy) of a substance in order to produce the phase transition. The amount of energy necessary to produce sublimation is, not too surprisingly, called the enthalpy of sublimation.
For example, imagine we had water ice at $-10^{\circ}$ C and we wanted to convert this to a gas. We would need to add energy, but it would occur in the following steps:
- add enough energy to change the temperature of the ice from $-10^{\circ}$ C to $0^{\circ}$ C;
- add enough energy to overcome the latent heat of fusion;
- add enough energy to change the temperature of the ice from $0^{\circ}$ C to just below $100^{\circ}$ C;
- add enough energy to overcome the latent heat of vaporization; and finally
- add more energy to increase the temperature of the steam (if one so desires, so long as the steam is contained).
The process would occur in these steps under STP if we added the energy continuously in at a slow rate (i.e., supply slighly more energy than is lost through radiation and/or conduction). It is possible to sublimate ice into a vapour but this generally occurs through ablation by UV light (and/or higher energy photons).
Why do some substances undergo sublimation while others do not?
The answer lies in thermodynamics, specifically the triple point of the substance. If this occurs at a large pressure compared to STP, then it is possible for a material to sublimate. For instance, if you look at the phase diagram image below you can see that the sublimation point for CO$_{2}$ occurs at a lower temperature and pressure than its triple point. Whereas, water has a triple point at a much lower pressure than standard atmospheric pressure, so sublimation is less likely to occur under normal conditions (and in the absence of high energy photons or particles that can induce ablation).
Thus, if you continously add energy to solid CO$_{2}$, its first transition will be to a gaseous form not a liquid form under STP. If you increased the pressure on the system and then added energy, so long as the pressure is high enough then you can produce a liquid CO$_{2}$. The substance's chemical properties determine its critical and triple points and these have been well documented for most substances.
as there must be a state between solid-gas in which particles have greater velocity then solids but less than gaseous particle...
No, there must not. If you add enough energy to induce a phase transition and it is easier for the substance to change into a gas than a liquid, it will go straight to the gaseous form, not though the liquid.
so why do we use this term for some substances at some temperatures and for some we do not...if we know that the transition is always happening continuously.?
A phase transition is, by definition and occurrence, a discontinuous change. The problem is, while you can change the temperature of a substance in a semi-continuous manner when it is far away from any phase transition lines, once at a phase transition line you can add energy but see no change in temperature until the substance undergoes the phase transition. Then, once in the new state, the substance can change temperature again (assuming the transition was not bounded by another phase transition line) in a semi-continuous manner by adding more energy.
You can try this at home with a standard cooking thermometer. Start with near-ice-cold water and then raise it to a boiling temperature. Make sure the thermometer is not touching the metal pot/pan in which the water lies. If your oven surface elements provide a roughly constant energy to the pan, then you should see a semi-continuous change in temperature up until just under $100^{\circ}$ C. If you keep the heat low enough, you can sit at this point, just below boiling, for a relatively long time. This is because the rate at which you add energy may be only just overcoming radiative and conductive heat losses to the surrounding room. If you crank up the heat, the time spent just below the $100^{\circ}$ C mark will be much shorter as your energy input can vastly exceed radiative and conductive losses.
Caveats and Notes
Note that phase transitions are defined within the context of thermodynamics, which is, by construction, a fluid model (i.e., meant for macroscopic parameter space) not a kinetic model. Note that kinetic models can describe macroscopic parameters as well, but they do so by finding the ensemble average in velocity space of the distribution function to treat a large number of particles as if they exhibit a bulk behavior similar to a fluid.
The differences between fluid and kinetic models can be subtle, but you can think of them in the following ways.
- We treat fluids as a continuous blob occupying a unit volume that can deform under stress but the initial fluid element, if incompressible, will occupy the same volume.
- For kinetic models, we assume a model distribution function that most closely describes the discrete particles in the system. In many cases, this function is continuous in the mathematical sense and the use of ensemble averages can yield bulk properties similar to those in thermodynamics (e.g., temperature). However, the interpretations are often different and the limitations and assumptions required in kinetic models are generally fewer than fluid models.
Continuous vs. Discontinuous
When I use the term discontinuous above, I am referring to a change that occurs on a smaller scale than the resolution of the specific observation. For instance, we assume that shock waves contain a discontuous jump in density, pressure, etc. in a region called the ramp. This region is generally assumed to be "infinitely" thin when looking at the asymptotic state of the fluid/kinetic gas on either side of the ramp. However, it is well known that the ramp has a finite thickness of the order of $\sim 1 \ \mu m$ in Earth's atmosphere at STP. For most fluid models, this spatial scale is so small we can approximate it as being infinitesimal and neglect it. This approximation greatly simplifies many of the equations we would use to try and model such phenomena, even though the transition from upstream to downstream is not truly discontinuous. We define the transition as being discontinuous because it is comparable to or smaller than the smallest relevant scale lengths considered for the problem at hand (i.e., in this case, the inter-particle mean free path).
In nature, there are few things that could be truly called discontinuous (I actually know of none, but some of the quantum whisperers on this site might know of some. Thus I am trying to be careful in this statement.). However, that some phenomena changes continuously on the smallest scales may not matter for the macroscopic dynamics where we assumed a discontinuous change. As in the shock wave example above, that the ramp region has a finite thickness does not render the conservation relations used to model most shocks (i.e., Rankine-Hugoniot relations) useless. The assumption that the ramp is discontinuous works because the transition is faster than the scales (i.e., fluid) considered in this specific problem. Thus, the definition of continuous vs. discontinuous depends upon the problem being addressed!
So in the purest sense, yes, a phase transition is close to (not exact because particles are discrete) a continuous transition if we could measure things "infinitely" fast and on an "infinitely" small scale.
Interesting Side Note: The use of a model distribution function generally inserts irreversibility into any model one would evolve dynamically from this point forward.
Reference
The phase diagram image was taken from Wikipedia, courtesy of Cmglee - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=29176053
Best Answer
A safer solution - buy everclear.
1) Mix one part ethanol to three parts water. The mixture freezes at around -15C. Make ice cubes out of most of this mixture, cool the rest. Your freezer should be able to just freeze this ethanol+water mixture, if not add some water.
2) Replace the water in your computer cooler with a 30% ethanol - 70% water mixture. This won't freeze and jam your impeller, and it wont burn (the rubber might dissolve though)
3) Immerse your radiator into a bucket of full of chilled 25% ethanol and pour the ice cubes into it.
4) As your ice cubes melt, replenish them. As the ethanol evaporates out occasionally add a shot of ever-clear.
Now you switched the unpleasantness of CO2 intoxication, for the pleasure of ethanol intoxication!
(As noted in the comments, with such low temperatures you'll have condensation problems. So buy a bunch of silica packets and dump them into your case and seal it up. Or blow dry hot air on exposed electronic parts. Or whatever)