Vacuum inside barometer

[Sisyphus]

I was wondering something today. Given a simple liquid barometer, the idea is to have two connected surfaces of a fluid, one exposed to vacuum, the other to air. The weight of the air column above the air-exposed surface forces the other upwards until the weight of the difference in height is equal to the air pressure. Right? Right. Now what I'm wondering is, won't there always be vapor in the vacuum chamber as some of the liquid boils when exposed to vacuum? If so, how negligible is the pressure exerted by said vapor, and how does it relate to the exterior air pressure? Does it in any way depend on the liquid used?

[J.C.MacSwell]

I was wondering something today. Given a simple liquid barometer, the idea is to have two connected surfaces of a fluid, one exposed to vacuum, the other to air. The weight of the air column above the air-exposed surface forces the other upwards until the weight of the difference in height is equal to the air pressure. Right? Right. Now what I'm wondering is, won't there always be vapor in the vacuum chamber as some of the liquid boils when exposed to vacuum? If so, how negligible is the pressure exerted by said vapor, and how does it relate to the exterior air pressure? Does it in any way depend on the liquid used?

 

Right. You would want to choose your liquid wisely or at least calibrate it for temperature.

[Bignose]

One of the fluid mechanics books on my shelf (by Munson, Young, and Okiishi) says that if you use mercury in a barometer, the vapor pressure of mercury at 68 deg F is 0.00023 lbf/in^2. For engineering purposes, this is pretty close to zero.

 

For most any fluid, the vapor pressure versus temperature can be looked up in an appropriate handbook. Most have been fitted to a polynomial:

 

P_vap(T) = A + B*T + C*T^2 + D/T

 

The handbook will have the values of A,B,C & D, as well as the range over which this polynomial fit is valid.

[Sisyphus]

Any particular explanation for why that is?

[swansont]

Any particular explanation for why that is?

 

Why the fit is valid? You can fit anything to a polynomial as long as you have enough terms.

[Sisyphus]

I understand that. I just mean, is there any explanation for why a given substance is a given pressure at a given temperature?

[swansont]

I understand that. I just mean, is there any explanation for why a given substance is a given pressure at a given temperature?

 

 

I imagine that if there were a reasonable physical model, you wouldn't have to fit four parameters to a polynomial.

[Sisyphus]

So nobody actually knows, then. And here I was, trying to figure it out...

[insane_alien]

the liquid exists in equilibrium with is gaseous state with the equilibrium extremely to the liquid side. the vapour pressure is caused by the few atoms/molecules of the liquid that gain enough energy by random collisions to vapourise. this therefore creates a small pressure instead of a true vacuum.

[Sisyphus]

But the pressure at a given temperature for a given substance at a given outside pressure is always the same, apparently. The percentage of vaporizing molecules is predictable. That's why I think it's so strange that there isn't some known law that determines that percentage, that we just have to observe it and fit it to some approximate function. What that tells me is that there is no way to actually predict what that pressure will be for some known substance without actually trying it. Is nobody else bothered by that?

[swansont]

But the pressure at a given temperature for a given substance at a given outside pressure is always the same, apparently. The percentage of vaporizing molecules is predictable.

 

 

Is it? It depends on the freezing point, boiling point and the chemical bond nature (all inter-related). The nature of molecules on the surface is different than for bulk material, too, because of the lack of symmetry and how those unused bonds behave. Factors that would probably effect the result are whether the material forms a monotomic or diatomic gas or some more complex molecule, and whether that molecule is polar.

 

There are a number of different equations of state and I think you still have to have some empirical data for those to predict anything.

[Sisyphus]

you still have to have some empirical data for those to predict anything.

 

I guess that's just indicative of how little we understand about it. Presumably, it must be theoretically possible to construct a model such that given a hypothetical substance, its behavior under given conditions would be entirely predictable.

[Bignose]

There is plenty of understanding... open up an undergraduate level thermodynamics or physical chemistry textbook. It comes down to this: the chemical potential of each phase must be equal. Otherwise, there will be a change. It is not possible at the moment to predict the rate of change, just that there will be a change.

 

The polynomial fit is for ease of use, the complete thermodynamics is a complex issue. Especially with non-ideal mixtures, azeotropes, polar molecules, charged molecules, asymmetrical molecules, etc. Nevertheless, there have been several theories proffered. It comes down to properly describing the chemical potential of these complicated cases.

 

I guess if pressed, there is no complete first principles basis for all of thermodynamics, but it is being worked on pretty fervently. There are several models and ideas today that come pretty close.

[Sisyphus]

But not complete understanding, which is really what I was asking. We still need empirical data, and we still need to fit the equation to that data. Therefore, there are necessarily factors that we do not yet know how to predict. Compare this with throwing a ball for which you know its mass and density distribution, its initial vector, gravitational forces, and the density, temperature, and vectors of the media its travelling thrugh. You don't have to actually throw it to know where it will land.

[Bignose]

I suppose then it's all going to depend on what your definition of complete understanding is. In you example, if the media the object were travelling in were turbulent, by today's knowledge it would be impossible to know what the exact velocity is at every point. In fact, if the flow is complicated, it may be very difficult to even get just the time-average flow.

 

If you want to be super-accurate, you can use a computational fluid dynamics technique called direct numerical simulation, but expect it to take several months to maybe even years.

 

Is this complete understanding? I would rather use correlations and have an answer that is pretty darn close in a matter of minutes than months.

 

It is the same why with the vapor-liquid equilibrium problem above. We know the complete problem -- the chemical potential of each phase must be equal. And there have been several models proposed for many, many situations. These work spectacularly well for the situations they are designed for. The software used by engineers to simulate entire chemical plants use these descriptions all the time. But the complete problem -- simulating each molecule to see how it moves around, what its chance of leaving the liquid to become a vapor, etc? That is not perfectly understood. But, like I said above, there are several models that are coming pretty darn close.

 

So, again, it all comes down to your definition of complete understanding.

[Morbid Steve]

Well, as said before, though, a substance such as mercury, which has a very low vapor pressure at room temperature, is used.

 

It is impossibly to have a complete vacuum in the conditions provided, becuase any of the amount of liquid evaporated will cause the vacuum to seize to exist. As well as the fact that in a vacuum, it is empty space, and if you have a liquid coming in contact with it that changes pressure and volume of the vacuum, it would therefore be impossible to have a complete vacuum, so in the end it's not really even a vaccume.

 

Also, looking here, the picture on the right as you open the page shows how a barometer is constructed. You must have something to compress at the 'top' of the glass. So, also, as I'm seeing there isn't a complete vacuum much at all in the first place, because of the contact with the mercury and the evaporation, there is some evaporation which would make it impossible to get an exact measurement.

 

If I'm wrong (I'd happen to think that I am) please correct me.

[Sisyphus]

It's different from turbulence because that involves a constantly changing situation. You throw the ball at different times and you get a different result. You apply the same pressure to the barometer at different times and you get the same result every time, and yet we still can't explain it, entirely.