I decided to inquire with someone who should have hard data. Below is the email I sent, and the response I received. I think the respondent is an alumnus of Microsoft tech support.
Helium boiling point is 452.1°F (268.9°C) at atmospheric pressure. Liquid Helium is available - although i don’t know if it is available to amateurs. I have used liquid helium before for MRI studies (No not the medical kind but flow visualizations).
Liquid Helium is kept liquid just like most of the other liquified gases by self refrigeration. That is, you flash continously a part of the liquid (and use the resulting refrigeration to cool the remaining liquid) to make up for the heat lost to the surroundings.
For more info - see this -
http://www.airliquide.com/en/business/products/helium/index.asp
This brochure from air products shows some pictures of the containers but is a big file.
http://www.airproducts.com/gases/KeepCold%20BrochureFinal.pdf
D’oh! Which would explain why that diagram earlier only went to that temperature. Some of my old chem profs would sigh in dismay that I didn’t pick up on that.
Sigh… Andy, that’s negative 452.1°F ; and it’s not terribly relevant because the OP asked about gas cylinders.
I still haven’t seen an adequate discussion of buoyancy. The mass of helium in a balloon, plus the balloon, is less than the mass of air displaced by the balloon. An empty ballon has less mass than one full of helium, but an empty balloon is flat and small, and is therefore not [buoyant**.
A gas cylinder containing helium at atmospheric pressure (an “empty”) occupies exactly the same volume as one containing air, a vacuum, or compressed helium, so buoyancy depends entirely on mass. A cylinder conaining air is not buoyant at all; one containing a vacuum is lightest of all. One containing helium at one atmosphere is buoyant, and lighter than one containing compressed helium. However, if you cram enough helium into the cylinder, the resulting compressed gas is heavier than the outside air, and is not at all buoyant. I’m pretty sure that a standard helium cylinder is sufficiently pressurized to eliminate buoyancy. Even at atmospheric pressure, the buoyancy of that amount of helium is only a few ounces.
Damn. That last sentence should read: “Even at atmosheric pressure, the buoyancy of that volume of helium is only a few ounces.”
I think this has more to do with an exhausted Walloon than an exhausted tank. 
Let’s take 2 extremes, a tank that holds 225 liters in volume (about 55 gallons), and weighs 25 kilos with a vacuum inside (lightest possible situation). If you add 225 liters of air, the chem majors out there will know we have about 10 Moles of air (22.4L each), a mole of air is probably about 20 grams each, so that’s .2 kilos of extra weight, not nearly enough to make a difference.
Did I do that right, or have I made a fool of myself?
Ok, so can someone resolve the two confusing subjects we have going here: buoyancy and weight? For an English major.
My elementary physics taught me that weight is mass times gravity. Therefore, it shouldn’t matter that what the buoyancy of a given object is, right? It confused me at first because the way I was taught to “weigh” something was to put it on a scale–pretty impossible for a helium balloon.
If I stand on a scale and mark down my weight, then do the same while holding a helium balloon, the second measurement will be less, correct? Can I get the weight of the balloon by subtracting the second measurement from the first?
A full tank of helium is pressurized at 1800-2400 psi. Atmospheric pressure is about 14.5 psi, so the full helium tank is pressurized about 140 times what normal atmospheric pressure is.
A mole of helium weighs 8 grams, if memory serves, and a mole of air, being mostly nitrogen, weighs 28+ grams.
A large helium tank, I’m guessing, is about 2 cubic ft, which is about 56 liters. That’s 2.5 moles at atmospheric pressure, or 350 moles (2.5*140) at ~2000psi.
Thus:
Tank = vacuum: content mass = 0g.
Tank = atmospheric air: content mass = 2.5 moles28g/mole = 70g.
Tank = atmospheric helium (“empty”): content mass = 2.5moles8g/mole = 20g.
Tank = pressurized helium (“full”): content mass = 350moles*8g/mole = 2800g = 6lbs or so.
If I did the math right. This guy here (warning! pdf file) got an answer of about 3 lbs, so maybe I erred. Point is, though, that the weight of the pressurized helium is not negligible, and the cylinder most certainly does weigh more when pressurized.
Good math, zut; however, a mole of He is 4 grams, not 8. Your estimate is also off (by the same factor) for the “empty” helium cylinder; i.e., the helium weighs 10 grams. This gives two grams of buoyancy = two ounces = not enough to notice.
Chriszarate, bouyancy is not normally an issue, but it is always present when weighing things. Truly accurate weights can only be determined (a) in a vacuum or (b) by applying a “buoyancy correction.” When I get my calibration weights cleaned and recertified, the lab weighs them in a vacuum and weighs them in air to determine buoyancy effects (actually, they use balances, and the “weight in air” represents the difference in density between the tested weight and the reference weight).
Bah! For some reason I thought helium was diatomic. Thanks Nametag for the correction. So my numbers match my second cite, and an “empty” helium cylinder (containing He at atmospheric pressure) weighs 60g less than the same cylinder containing atmospheric air, while a pressurized cylinder is 3 lbs heavier.
This can’t be right; how would the air get into the tank?
I’m tending to agree with the exhaustion theory; you carried the heavier full tank to the store, tiring yourself out so the lighter, empty tank felt subjectively heavier to you on the way back to the vehicle.
Otherwise we’d be saying that helium is a substance which, when compressed, becomes less dense - absurd.
Air can get into the tank through the same valve that got helium out of the tank.
To hell with the “exhaustion” theory, I often compared two tanks side by side – the empty one was heavier than the one with the helium in it.
There’s no good reason why it would; to begin with you have a tank full of helium gas at pressure, when the tank has been used, it will be full of helium gas at the same pressure as the air outside the tank; there’s no reason for the air outside the tank to force its way inside. Certainly diffusion may cause a little of the helium and air to be exchanged over time, but air does not automatically displace the helium as the tank becomes empty.
Well, as I said, for that to be true, you would have discovered a substance that becomes less dense as it is compressed…
Walloon’s claim makes even less sense when you remember that “helium” for party balloons is not pure helium. It’s a helium/air mixture, with just enough helium to lift the balloon. So a full tank is even heavier than the above calculations indicate. I really can’t think of any explanation for Walloon’s observations.
I guess you folks would have to have hands on experience with the tanks, to really challange what Walloon is saying. But then again, his observations could be due to the belief that the tanks that are full of “He” are lighter, a sort of psychological conditioning.
So science is a democratic process now?
Please, does anybody reading this have access to a full cylinder of compressed helium and a set of accurate scales? (if the scales are accurate enough, it shouldn’t be necessary to empty the whole tank, just vent a little of it and measure which direction the weight goes.
I know which way my money is going.
You are wrong. Period. End of statement.
A full helium tank ways more than an empty one. The size of the tank (displacement) does not change as helium or any other gas for that matter is pumped in. It is impossible to increase a container’s buoyancy by filling if it does not change the displacement of the container.
The only way filling the tank with ANY substance could make it lighter is if you were filling it with some strange substance that was repulsed by gravity.
The page you quoted shows a “55 Cubic ft” Helium cylinder. I assume that means it holds 55 ft[sup]3[/sup] (= 1557 liter) of uncompressed Helium? That doesn’t quite work with the rest of your calculations.
Also, that link is talking about laboratory Helium, not Helium for balloons. I don’t think it’ll be quite the same. For one thing, he says that the tanks have a mass of 91 kg!
Mathematically, I can definately see the proof for a tank of pure compressed helium to be heavier than a tank at neutral atmospheric pressure and temperature… On the other hand, my own anecdotal evidence agrees with Walloon… those “empty” tanks sure do feel heavier, although, they were brought in full on a hot summer day, and taken back empty later in the evening. Maybe we can get Cecil to spring for a tank of helium and a scale at a readers party. 