[Bidagdha_TADIR]

My Friend and I got into an argument yesterday. The argument was about - if a deflated rubber balloon is weighed in an electronic balance and then the rubber balloon is inflated and weighed again then will the weight change?

I said that, since more air is inside the balloon now, the weight will increase but he says that weight will remain same and explains something that I don't understand. So, what do you think, will the weight change upon inflation (even if it is miniscule; also let's not get into the topic where an inflated balloon can't be properly weighed in an electronic balance)?

[Arkcon]

Remove the technology from the equation.  You have two balloons full of air, balancing each other in a simple lever balance, maybe even a ruler on a string.  Once balanced, you pop one.  What happens to the balance?

https://www.teachengineering.org/view_activity.php?url=collection/cub_/activities/cub_air/cub_air_lesson01_activity2.xml

[Borek]

You have two balloons full of air, balancing each other in a simple lever balance, maybe even a ruler on a string.  Once balanced, you pop one.  What happens to the balance?

https://www.teachengineering.org/view_activity.php?url=collection/cub_/activities/cub_air/cub_air_lesson01_activity2.xml

I don't like the fact they ignore buoyancy. Even if the system behaves the way they describe it, the explanation is IMHO wrong.

[Furanone]

I think about it this way:

If a balloon has hydrogen or helium pumped into it to inflate it relative to the atmospheric gas on the outside (ie ~78% nitrogen 21% Oxygen, 0.5% CO2, 0.5% other gases) then it will weigh lighter as it will float up not even allowing weighing. If you pump it full of argon, a much heavier gas, it will weigh more on the scale. So knowing this what would you think would be the result if you pumped the same exact mix of gases into the balloon as what is the equilibrium atmospheric conditions?

Now what about if you take that balloon with regular atmosphere gases and climb a 5000 metre high mountain, what happens to the balloon? Does it weigh the same, less or more?

[Corribus]

I don't like the fact they ignore buoyancy. Even if the system behaves the way they describe it, the explanation is IMHO wrong.

I have to agree. I read the webpage, and my first thought was, "That's a totally wrong explanation."

[Bidagdha_TADIR]

Today I had a heartier talk with my friend. His logic is that if a water balloon is immersed in water, then it will loose weight equal to the weight of the volume of water replaced by the balloon (which is equal to the volume of water inside the balloon). Since electronic balances determine mass based on weight, if we could somehow determine the weight of that balloon underwater, we would only get the weight of the rubber part.

Similarly, when a balloon filled with air is being weighed in normal environment, the balloon is actually immersed in air and thus Due to that Archimedes principle, weight will be lost and only the rubber part can be weighed by the Electronic balance. This is his logic.

I couldn't disagree and conceded defeat. But how can I determine the mass of air inside a balloon? Also it seems that even if a gas much heavier than air was used to inflate the balloon, we still would not get the actual mass of the gas due to buoyancy.

[Borek]

Today I had a heartier talk with my friend. His logic is that if a water balloon is immersed in water, then it will loose weight equal to the weight of the volume of water replaced by the balloon (which is equal to the volume of water inside the balloon). Since electronic balances determine mass based on weight, if we could somehow determine the weight of that balloon underwater, we would only get the weight of the rubber part.

Yes, that's the buoyancy at work.

Similarly, when a balloon filled with air is being weighed in normal environment, the balloon is actually immersed in air and thus Due to that Archimedes principle, weight will be lost and only the rubber part can be weighed by the Electronic balance. This is his logic.

Yes, that's the buoyancy at work.

If you could find a way of measuring the balloon volume, you can always account for the buoyancy.

But how can I determine the mass of air inside a balloon?

You mean - if you wanted to, for whatever reason? This is a very good question. TBH I can't think of any simple approach. Weighing in vacuum is out of the question. You could try freezing it down and weighing the liquid. You could try to pump the air from the balloon and to weigh it compressed , so that buoyancy doesn't matter much.

Also it seems that even if a gas much heavier than air was used to inflate the balloon, we still would not get the actual mass of the gas due to buoyancy.

Yes.

[Sapperdaddy]

Sorry to jump in so late, but one thing you seem to be overlooking in your use of the Archimedes principle is that you are not compressing the water inside the balloon. It was the same density as the water around it. This isn’t the case with a balloon filled with air. When you blow up the balloon, you are compressing more air into the balloon than would be there at STP. This is very simple to show if you have a balance that can go to enough decimal places. It is best If you have one that has the air shields. Weigh the balloon empty, inflate and weigh again. It will have gained mass. This is why a balloon that you inflate with your lungs falls to the ground; it is denser than the surrounding gas.

[Borek]

When you blow up the balloon, you are compressing more air into the balloon than would be there at STP. This is very simple to show if you have a balance that can go to enough decimal places. It is best If you have one that has the air shields. Weigh the balloon empty, inflate and weigh again. It will have gained mass. This is why a balloon that you inflate with your lungs falls to the ground; it is denser than the surrounding gas.

While it is true that the pressure inside the balloon is slightly higher, you don't need that to explain why it falls down. Weight of the filled balloon is always weight of the trapped air plus weight of the balloon itself. If the pressures inside and outside are identical, weight of the trapped gas and its buoyancy cancel out, but you are still left with the weight of the balloon itself, so it has to fall.

Fact that the air is compressed works in the same direction (makes the balloon slightly heavier), but is not necessary to explain the result.