Working out liquid to gas pressures

[Jolly Roger]

I'm curious as to how to work out the pressure of a vessel after a liquid such as oxygen or nitrogen etc. is heated beyond it's boiling point inside it.

I am led to believe that oxygen cannot be contained as a liquid once it goes beyond it's boiling point, not matter what the pressure, so I'm wondering what happens if you poured 1 litre of oxygen into a 1 litre high pressure vessel, and let it sit at room temperature. What would be it's final pressure after all the liquid has boiled off and converted to a gas inside the vessel?

Don't ask me why I started thinking about it, even I don't know

[jackssmirkingrevenge]

I though we had this conversation before

Boiling point depends on pressure. One very common thing to do when you're concentrating a liquid in a chemical process is to put the container under vacuum, because this lowers the pressure above the liquid and therefore means that it boils at a lower temperature.

If you put 1 litre of liquid oxygen at -200°C in a container really quickly and closed the lid, it would never evaporate because there simply is too much pressure preventing most of the liquid from boiling.

[Jolly Roger]

Yeah that's what I thought, however after some research I found that they have to store oxygen in dewar that a vaccum flasks, and eventually the liquid boils off through the pressure release valves as it can never keep the liquid 100% isolated from outside heat.

They said that oxygen, once it reaches it's boiling point, will turn into a gas. However if contained it will reach it's critical point where the fluid is neither a gas or liquid.

That's where I got confused.

[Zeus]

Works the same for almost everything.

[Jolly Roger]

So 1 litre of liquid oxygen, in 1 litre vessel @ 30 degrees centigrade = ?? atm

[Zeus]

You'd need to find that out yourself, that chart shows you the relationship of pressure and temperature for water.

Wikipedia might be a good place to start, look for evaporation, boiling point, triple point, and oxygen/liquid oxygen.

[Jolly Roger]

Yeah I have been for a while but haven't come up with anything so far apart from graphs that go up to its critical points, but no further. Haven't been able to extrapolate the graphs either

[ramses]

If I recall, 30 degrees is above the critical point of oxygen, so you will have an EXTREMELY high pressure. Tens of thousands of PSI would not surprise me. You could find the number of moles of O2 in 1 L of liquid oxygen and use pv=nRt, but that would tend to overestimate the pressure.

[D_Hall]

I don't recall the exact number off the top of my head, but yes, the critical point for O2 is safely in the realm of cryogenic.

If you put 1 litre of liquid oxygen at -200°C in a container really quickly and closed the lid, it would never evaporate because there simply is too much pressure preventing most of the liquid from boiling.

Not so.

What you end up with is a supercritical fluid. The pressure is extremely high, and the density is unchanged, but it behaves like a gas, not a liquid. For a rough guesstimate, of the pressure, just use ye ol' ideal gas law, set density equal to that of LOX, and the temperature to something that approximates room temperature.



For the classic thought experiment on this (I'm probably going to screw this up, but here goes....):

Start with a pressure vessel half full of a liquid and the other half full of the same substance but in a gas form. Let's say water and steam. As you increase the temperature more water turns to steam (ie, the water boils). The steam's density increases as there is more steam occupying essentially the same volume. At the same time, the density of the water decreases slightly (hot things expand, remember?).

Heat it some more and the steam continues to get more dense while the water continues to get less dense.

Keep increasing the temperature until the density of the steam equals the density of the water.

Congratulations, you're now at the critical point.

As long as the volume of the pressure vessel remains unchanged the density of either phase can no longer change. So what does it mean to increase the temperature further? It means that your liquid phase completes it's transition to gas phase (ie, it still boils in a sense) and pressure starts increasing rapidly... But it also means that while your density is very nearly that of the room temperature liquid, you no longer have a liquid. You have a gas...but one that is referred to as a supercritical fluid for clarity.

For water, this happens at roughly 700 F, and [uh...I forgot the pressure].

[DYI]

D_Hall has it. To give a little background, it's probably worthwhile to elaborate on the boiling process.

As you're probably aware, boiling temperature of a liquid is pressure dependent (in the case of water evaporating into air, the water evaporates because its temperature is above the boiling temperature at the partial pressure of water vapor in the air). Typically, boiling temperature increases with the pressure.

After the liquid reaches the boiling temperature, it needs further energy input to change phase from liquid to gas, called the heat (or enthalpy) of vaporization. In water at standard conditions this is about 2.3kJ/g, compared to the ~0.5kJ/g to raise it from the freezing point to the boiling point. The heat of vaporization is also temperature dependent, with the 2.3kJ/g figure being at roughly 100C. Heat of vaporization decreases with increasing temperature, until, at the critical point, it reaches zero. At this point there is no energetic distinction between the phases and, as Dave's thought experiment correctly implies, the densities are the same.

The most basic, hand-wavy sort of definition for the transition is that, above the critical point, the substance can no longer slosh. Dioxygen's critical point is at -120C and about 50atm, so the oxygen tanks you use for welding contain supercritical oxygen when they're full.

In the case of your question the density it probably too high for the ideal gas law to give great results, but it'll still be adequate for rough calculation. LOX is at 1.14g/cc at its boiling point, giving an approximate pressure at 295K of P=ρRT = (1141)*(8.31/0.032)*295 = 87.4MPa, or about 12.5kpsi. Actual pressure will be higher due to reduction in compressibility.

[Jolly Roger]

Yeah cheers guys, that helps a lot.

I only started thinking about it after trying to work out how many shots you could get out of a gas oxygen tank supplying a small hybrid at 200+ mixes. Wasn't a whole lot given the preignition pressures are around 3000psi, almost the same pressure as a full oxy bottle.

So I started thinking, if one was able to safely release a small amount of LOX from a dewar into a larger tank, they would have a concentrated source of oxygen and they could control the pressure to provide for well over 200x mixes.

Obviously some hurdles to cover there in regards to materials etc. but in theory, could you see any reasons as to why it wouldn't work?

[D_Hall]

Obviously some hurdles to cover there in regards to materials etc. but in theory, could you see any reasons as to why it wouldn't work?

No reason why it wouldn't work.... But be advised that LOX is one hell of a tiger to grab by the tail.

[Jolly Roger]

Hopefully there will be no grabbing of it by the tail

Cheers all

[Jolly Roger]

BTW I'm getting around 25,000 psi for any concealed amount of liquid oxygen when the temperature reaches 25 degrees celcius, or room temperature.

[jackssmirkingrevenge]

Not so.

What you end up with is a supercritical fluid. The pressure is extremely high, and the density is unchanged, but it behaves like a gas, not a liquid. For a rough guesstimate, of the pressure, just use ye ol' ideal gas law, set density equal to that of LOX, and the temperature to something that approximates room temperature.

Apologies for the oversimplification