Boiling point of liquids under negative pressure

[Sillychris]

So, if you spin the hell out of a liquid you can generate strong negative pressures in the center.

It's fairly common knowledge that boiling points of liquids reduce with pressure. What I am wondering is the behaviour of this trend below 0 pressure. Does the boiling point reach a minimum, or does it continue to decrease with decreasing pressure?

[MBobrik]

there won't be negative pressure. at sufficient speed you would just create a bubble filled with diluted fluid vapor.

[K^2]

So, if you spin the hell out of a liquid you can generate strong negative pressures in the center.

No, you cannot. For the same reason that gravity pulling down on gas particles never creates negative pressure in space. Vacuum is the most you can do.

[Guest]

Actually, negative pressure is a thing. Though I don't think you can get it by spinning a liquid. Now, the source I have won't be much use for you, being in Polish, but I'm sure that you can find materials in English about that. It's not exactly like positive pressure, though, and it definitely behaves rather oddly.

[K^2]

Things like capilary pressure and osmotic pressure can sometimes work as negative pressure. But none of these things are going to affect boiling point as if pressure is negative. In fact, I'm pretty confident nothing will.

[Sillychris]

No, you cannot. For the same reason that gravity pulling down on gas particles never creates negative pressure in space. Vacuum is the most you can do.

Yeah I get that by definition of pressure it's not formally pressure, but for mathy purposes it behaves as though it's negative pressure. Here are some papers exploiting the concept:

Pless, I. A. & Plano, R. J. (1956). Negative pressure isopentane bubble chamber. Rev. Sci. Intsr, 27(11), 935-937

Hahn, B. (1961). The fracture of liquids under stree due to ionizing particles. Il Nuovo Cimento, 22(3)

[Sillychris]

Things like capilary pressure and osmotic pressure can sometimes work as negative pressure. But none of these things are going to affect boiling point as if pressure is negative. In fact, I'm pretty confident nothing will.

That's good news. Can you supply me with a simple argument for why "negative pressure" should not lower the boiling point of a liquid past the vacuum boiling point? I'm doing my senior research project presentation on friday and I don't want to stumble on this if someone asks me.

[K^2]

In your case, centrifugal potential is no different from gravitational potential. It's really that simple. Another way to think of it, imagine a centrifuge that's already spinning. It starts out completely empty, vacuum, and you start filling it with liquid until it's almost completely full. Yes, the centrifugal force is going to pull the liquid to the sides, but at the center, you still just have vacuum.

[Sillychris]

mmk, that poked me in the right direction. I think I have my final answer ready:

Boiling point is related to vapor pressure which is related to atmospheric pressure. Since the vapor pressure can never get below zero, the boiling point can never get below the vacuum boiling point.

[Kermunist]

For a senior project, you might as well settle for "negative pressures are impossible". You are also correct with your above point that if any vapor is formed, it cannot have a negative pressure, so any hypothetical negative pressure in the liquid will be relieved by vapor formation.

That said, I don't think negative pressure is outright impossible. In a solid, negative uniaxial pressure is easy to achieve: just stretch your sample with a piezo. The amount of negative pressure you can achieve is very small, because the sample will break. Negative hydrostatic/multiaxial pressure is more difficult, but by no means impossible, just use more piezos.

In a liquid, in theory the same thing is possible. But void formation would kick in even sooner, because liquids are usually held together by weaker forces. To allow negative pressure you would need the energy of 'stretching' the liquid to be less than the energy of the void formation (whether that is a vacuum void or a void containing vapor). The former depends on the compressability and is going to be a large. The only energy cost I can think of for the void is surface tension, which is normally a much weaker thing. Worse, the compressability energy scales as the whole volume of the liquid (l^3), whilst the surface tension scales as the surface area of the void necessary to relieve the stretching (l^2). I don't know of any theoretical upper limits on the energies of surface tension, nor any lower limits on the compressability of a liquid, and I would be surprised if no such liquid was possible.

Edit: further thought. I rather doubt that any macroscopic-sized equilibrium system can show negative pressure in a liquid, as it will always be unstable to formation of a single large void. I do think that a non-equilibrium state might be metastable, with no void formation unless nucleated. Because of the scaling laws above, it should be possible to achieve a system where small voids are unstable, and decay, but voids over a critical size grow rapidly and release the negative pressure. If the critical size can be made large enough, voids will not occur unless nucleated by an external peturbation (such as a high energy particle strike).

The above mentioned paper by Hahn indicates that this is certainly possible in a quasi-1D system, where the scaling laws will be different: l^1 and l^0 probably. I bet a literature search would show up 3D theoretical work at least. But it's too late at night in my timezone to start looking.

Edited April 9, 2014 by Kermunist

[Mattasmack]

Liquids can indeed exist in metastable states where they should be vapor (at thermodynamic equilibrium). I'm more familiar with heating a liquid at constant pressure above its boiling point, but it can also be achieved by dropping pressure at constant temperature. The liquid needs a nucleation site to start the boiling process: a scratch on a solid surface that trapped some atmospheric gasses, a suspended dust particle with gas absorbed on its surface, etc. In the absence of a nucleation site, any bubble embryos that form in the liquid due to fluctuations quickly collapse again due to surface tension -- since the pressure difference across a surface due to surface tension goes as the inverse of the radius of curvature, the pressure difference pulling these nanometer-scale embryos closed is enormous. This situation can persist (again in the absence of nucleation sites) up until the limit of superheat, which can be predicted using an appropriate thermodynamic equation of state (thermodynamic limit of superheat) or kinetics (kinetic limit of superheat).

That was a bit to the side of the original question asked, sorry. But it's kind of related an area I used to do research in so I have it on the brain. Metastable liquids and the limits of superheat have been studied experimentally for decades. Usually they're studied by heating liquids at constant pressure; a common apparatus is a column of heavy oil that is heated unevenly so that its temperature rises steadily from bottom to top. Then you can introduce a droplet of a lighter, low-boiling-point liquid at the bottom, and as it rises through the oil it gets hotter and hotter until it reaches its limit of superheat, and goes pop! The beauty of the apparatus is that the droplet doesn't ever touch a solid wall when its superheated, so it doesn't come into contact with any nucleation sites there. (See Blander, Hengstenberg, and Katz's paper in J Phys Chem 1971, for one example. Katz and Blander had a number of papers on the subject.)

Anyway, if you have a liquid that isn't anywhere close to its critical temperature (water at room temperature, say), you can reduce its pressure quite a ways before reaching the limit of superheat. In many cases the limit is at a negative pressure; the liquid is in tension. This situation wouldn't last for long, because the liquid will find a nucleation site on the walls of the container. But the situation of liquid in tension has been produced, I believe using acoustic waves (which don't last for very long, and can be arranged to focus on a region in the middle of a container).

Finally getting to the original question, talking about the boiling point in this situation doesn't really make sense in the first place! The boiling point is the dividing line between where a fluid at thermodynamic equilibrium will be liquid and where it will be vapor. But the fluid at negative pressure cannot be at thermodynamic equilibrium; while it is liquid it is in a metastable state, and as a vapor it must have a positive pressure. So the concept of boiling point loses its meaning.

Or at least that's my take on it, FWIW.

Edited April 10, 2014 by Mattasmack

[lajoswinkler]

Negative pressures exist in certain situations. If you had a syringe full of oil and you tried to stretch the plug, you're applying negative pressure which can, by its absolute value, go below vacuum. It happens with water in the trees, and that's how they manage to pull water above some 10 metres of height. It works as long as the xylem column is voidless. If a void appears, the water column breaks.

[Nao]

Negative pressures exist in certain situations. If you had a syringe full of oil and you tried to stretch the plug, you're applying negative pressure which can, by its absolute value, go below vacuum. It happens with water in the trees, and that's how they manage to pull water above some 10 metres of height. It works as long as the xylem column is voidless. If a void appears, the water column breaks.

Oh, I remember a Veritasium channel had a video on how very tall trees get water up to the top. They concluded that in the tallest trees negative pressure can reach -15 atm, it blew my mind! Trees are awesome

[Z-Man]

Negative pressures exist in certain situations. If you had a syringe full of oil and you tried to stretch the plug, you're applying negative pressure

That is not negative absolute pressure, that is just pressure that is lower than the pressure of the surrounding air.

[Nao]

That is not negative absolute pressure, that is just pressure that is lower than the pressure of the surrounding air.

And what if the pressure inside the synergine is for example 2 atm lower than pressure of surrounding air? (1atm)

Also the same experiment can be done in space, where you take a volume of liquid at zero pressure and stretch it to higher volume.

[lajoswinkler]

That is not negative absolute pressure, that is just pressure that is lower than the pressure of the surrounding air.

You can apply even lower pressure. As I said, trees do it.