Buoyancy - (helium in a sealed vessel)

[men5j2s]

TL;DR

1atm of air + 1atm of HE in a 17.5ml vessel = a mass gain of ~2.9mg

Hi All,

I'm trying to answer a question once and for all that has caused more debate than it ever should have (talking about an internal debate)...

If I fill a sealed vessel (say an aluminium vessel with appox. 17.5ml of internal volume) with varying amounts of helium ( 1bar, 2bar, 3bar ... 10bar), will I see a net mass gain or loss?

My current understanding is as follows:

• The sealed vessel is displacing a volume of air (assuming a vacuum), and therefore will have some buoyancy! (approx -2.3mg)
• The weight of the vessel is larger, so it won't float, but the buoyancy effect can be measured!
• Adding air to 1atm (~1bar absolute) inside the vessel, means density equilibrium inside and outside the vessel, so no buoyancy effect!
• Adding helium to 2atm (on top of the 1atm of air) will simply add more moles to a closed system that is already overcoming any buoyancy effect, by virtue of the 1atm of air that was in there, therefore there will be a net mass gain.
• I should end up with a weight gain of ~2.9mg

 

[russ_watters]

If I fill a sealed vessel (say an aluminium vessel with appox. 17.5ml of internal volume) with varying amounts of helium ( 1bar, 2bar, 3bar ... 10bar), will I see a net mass gain or loss?

Yes, adding mass to a container increases the mass inside the container and total mass of container and contents.

I have occasionally heard where people think buoyany is itself a property of a gas like helium, so adding helium to a container adds buoyancy. Nope.

 

[RPinPA]

You have an answer, but I'll address your specific points.

My current understanding is as follows:

• The sealed vessel is displacing a volume of air (assuming a vacuum), and therefore will have some buoyancy! (approx -2.3mg)

Correct. The buoyant force is the weight $mg$ of the displaced air, which I'll take your word for it has mass 2.3 mg.

Note that this depends only on the volume of the vessel. It doesn't matter what's inside it: vacuum, helium, air, or solid iron.

• The weight of the vessel is larger, so it won't float, but the buoyancy effect can be measured!

The buoyancy effect can be measured by comparison to the weight measured in other media. In vacuum, you should get a value with is 2.3 mg heavier.

• Adding air to 1atm (~1bar absolute) inside the vessel, means density equilibrium inside and outside the vessel, so no buoyancy effect!

There will always be a buoyancy effect in the sense of the weight when immersed in air appearing to be 2.3 mg less than the weight in vacuum. I think you are thinking about neutral buoyancy, where the buoyant force is equal to the weight of the object. But that won't be true here with a steel vessel. It's still a lot heavier than the displaced volume of air.

• Adding helium to 2atm (on top of the 1atm of air) will simply add more moles to a closed system that is already overcoming any buoyancy effect, by virtue of the 1atm of air that was in there, therefore there will be a net mass gain.
• I should end up with a weight gain of ~2.9mg

Whatever the buoyant force was, adding mass will cause a net mass gain.

 

[gmax137]

I'll take your word for it has mass 2.3 mg.

I'm pretty sure that is off by a factor of 10. Sea level air density is ~1.225 kg/m3 which corresponds to 1.225 mg/ml; times 17.5 ml would be 21.4 mg.

Not that that affects anything said, really.

 

[men5j2s]

I'm pretty sure that is off by a factor of 10. Sea level air density is ~1.225 kg/m3 which corresponds to 1.225 mg/ml; times 17.5 ml would be 21.4 mg.

Not that that affects anything said, really.

1.3 g/L * 0.0175 L (or 17.5ml) = 0.02275g (or 22.75mg)

You are correct, I made an error in transposing (that's why we Verify ), well spotted!

 

[gmax137]

Yeah, I have spent nearly half my working career checking calculations done by colleagues (nuclear power is all about verifying). I've gotten pretty good at picking out "that doesn't look right" values.