Comparing weights of balloons

[Thamska]

Homework Statement

You have two balloons with the same amount of air. Then you put one in the freezer so it's volume decreseas because of pressure. Then you put both of them on a scale, and then compare which one weighs more, according to the scale? My teacher said that the frozen balloon will weigh more because it has less volume (archimedes principle). I think the balloons will weigh equally as much since there will be a opposite force (Newtons third law) from archimedes principle from both of the balloons on the scale so it evens out. Who is right?

Relevant Equations

Archimedes principle: F = density of medium * gravitation constant * volume

.

 

[Orodruin]

Your teacher is right.

The weight of both balloons is the same but the scale will measure (weight - buoyancy).

 

[phinds]

The weight of both balloons is the same but the scale will measure (weight - buoyancy).

Seems to me that that would mean that the MASS is the same but the weight is different.

 

[Charles Link]

If you put them on a scale that was in a vacuum, they would both weigh the same. Introduce air to the scale, and the balloon with the larger volume gets more buoyancy from the air.

 

[Orodruin]

Seems to me that that would mean that the MASS is the same but the weight is different.

No, the weight is $mg$ and $g$ is the same.

 

[erobz]

My teacher said that the frozen balloon will weigh more because it has less volume (archimedes principle).

Your teacher is right.

The weight of both balloons is the same but the scale will measure (weight - buoyancy).

I guess if to weigh something is not intended to yield its weight as an output?

EDIT: I guess when we weigh anything of finite volume in a fluid, we really never get the weight as an output.

 

[Orodruin]

I guess if to weigh something is not intended to yield its weight as an output?

EDIT: I guess when we weigh anything of finite volume in a fluid, we really never get the weight as an output.

For most things you would like to weigh the buoyancy will not be that relevant, usually within error bars. However, when the density of the object approaches the density of the fluid you cannot really ignore it.

 

[Charles Link]

For most things you would like to weigh the buoyancy will not be that relevant, usually within error bars. However, when the density of the object approaches the density of the fluid you cannot really ignore it.

See also my post 4. It seems on this one, both the teacher and the OP are correct, but it matters in this case whether you weigh the balloons in an air environment, or in a vacuum.

 

[erobz]

See also my post 4. It seems on this one, both the teacher and the OP are correct, but it matters in this case whether you weigh the balloons in an air environment, or in a vacuum.

I personally see your point, but I'm leaning to the fact that weighing the ballons in a vacuum would be problematic as an application of Archimedes Principle.

 

[Charles Link]

We used balances with weights in a chemistry class in college. I think it was called a Mettler balance. There was always a slight error introduced when there was an unbalanced buoyancy when the density of the substance being weighed was different from that of the weights on the other side of the balance. See https://www.mt.com/us/en/home/perm-lp/product-organizations/balances_scales.html?cmp=sea_56010127&SE=GOOGLE&Campaign=MT_IND-LAB_EN_US&Adgroup=Balances_Generic_Branded_Exact&bookedkeyword=mettler balance&matchtype=e&adtext=606083278760&placement=&network=g&kclid=_k_EAIaIQobChMI0Mnuu4ioiwMV2Uf_AR3tKyjzEAAYASAAEgJj4_D_BwE_k_&gad_source=1&gclid=EAIaIQobChMI0Mnuu4ioiwMV2Uf_AR3tKyjzEAAYASAAEgJj4_D_BwE

Yes, you put the balloons in a vacuum and they would both expand considerably. I'm just considering this hypothetically.

 

[erobz]

In any case the OP should probably try to demonstrate a FBD of the ballon and one of a "typical" scale. They did seem to have some misunderstanding about how a scale "weighs" in my opinion.

 

[kuruman]

I think the balloons will weigh equally as much since there will be a opposite force (Newtons third law) from archimedes principle from both of the balloons on the scale so it evens out. Who is right?

Your argument is specious because the scale measures the normal force on its surface, not the weight (or mass) of the object. If you push the scale with your finger it measures the force applied by your finger, not your weight. The scale pushes back with an equal and opposite force and this is always the case. Newton's 3rd law has nothing to say about the comparison of the readings.

Here, the normal force on the scale is the weight of the balloon plus the air in it minus the buoyant force on the balloon, $BF=\rho_{\text{air}}~g~V_{\text{balloon}}.$ If you decrease the buoyant force by shrinking the balloon, you decrease the term that is subtracted from the weight which means that the measurement is closer to the actual weight.

 

[Charles Link]

Reading the OP again, the buoyant force is the weight of the displaced air and is clearly larger for the warmer (larger) balloon. Thereby the OP needs to study this, especially archimedes, in a little more detail.

 

[kuruman]

Yes, you put the balloons in a vacuum and they would both expand considerably. I'm just considering this hypothetically.

It doesn't have to be hypothetical. I used to do this demo to the delight of my students. In the end I asked them, "if the entire assembly were placed on a scale, as the volume of the funny-looking balloon increases, will the reading on the scale increase, decrease or stay the same?"

Spoiler

Decrease.

 

[Charles Link]

I thought in any case, it is a very good post by the OP @Thamska . It makes for some good physics discussion. :)

 

[erobz]

It doesn't have to be hypothetical. I used to do this demo to the delight of my students. In the end I asked them, "if the entire assembly were placed on a scale, as the volume of the funny-looking balloon increases, will the reading on the scale increase, decrease or stay the same?"

Spoiler

Decrease.

Not certain about that as demo. The air in the bell jar is being removed from the scale?

 

[kuruman]

th certain about that as demo. The air in the bell jar is being removed from the scale?

Of course, otherwise the condom will not expand. The answer is a no-brainer. You remove the weight of the air, you reduce the reading on the scale.

 

[erobz]

Of course, otherwise the condom will not expand. The answer is a no-brainer. You remove the weight of the air, you reduce the reading on the scale.

I thought we were talking about the buoyant force. As far as the Buoyant force is concerned (I see) the density of the displaced fluid is dropping to nothing while the volume of the balloon is increasing.

 

[Charles Link]

Of course, otherwise the condom will not expand. The answer is a no-brainer. You remove the weight of the air, you reduce the reading on the scale.

Suggest you try to describe in more complete detail. I think you know what you are trying to say, but it isn't completely clear with the above statement. You introduce outside air to the balance, and the apparent weight of the balloon is less, because of the buoyant force, with the buoyant force increasing with the volume of the balloon. Meanwhile there is always air inside the balloons which provide for the same weight, regardless of temperature.

Edit: I see now=you are referring to the weight of the air inside the bell jar, (but outside the condom.) My mistake. :)

This complicates matters in a way, because you now have the buoyant force of the whole bell jar in the picture, (with the bell jar on the scale), when there is a vacuum in the bell jar. I think most of us are on the same page though, and we understand the concepts. :)

 

[kuruman]

This complicates matters in a way, because you now have the buoyant force of the whole bell jar in the picture, when there is a vacuum in the bell jar. I think most of us are on the same page though, and we understand the concepts. :)

The buoyant force on the whole bell jar is the same whether there is air in it or vacuum. That's because its external volume is the same.

I thought we were talking about the buoyant force. As far as the Buoyant force is concerned (I see) the density of the displaced fluid is dropping to nothing while the volume of the balloon is increasing.

Correct, but how does that change the reading the scale that supports the bell jar and everything in it when the air is taken out? Answer: The reading is reduced because a bell jar will always weigh less when you take some mass out of it, in this case air. It has nothing to do with buoyancy.

 

[erobz]

Suggest you try to describe in more complete detail. I think you know what you are trying to say, but it isn't completely clear with the above statement. You introduce outside air to the balance, and the apparent weight of the balloon is less, because of the buoyant force, with the buoyant force increasing with the volume of the balloon. Meanwhile there is always air inside the balloons which provide for the same weight, regardless of temperature.

Edit: I see now=you are referring to the weight of the air inside the bell jar, (but outside the condom.) My mistake. :)

This complicates matters in a way, because you now have the buoyant force of the whole bell jar in the picture, (with the bell jar on the scale), when there is a vacuum in the bell jar. I think most of us are on the same page though, and we understand the concepts. :)

I was bringing because we haven't heard from the OP yet, so we don't know how they feel about it. They have already been kind of gifted the answer without effort beyond the opening post.

 

[erobz]

It has nothing to do with buoyancy.

The effect of the buoyancy is increasing the reading on the scale on as the air is being removed. However, the removed air apparently outpaces the buoyancy if the scale reading indeed decreases. My point was this problem is about the buoyant force effect, it didn't seem like an effective demonstration of that.

 

[kuruman]

I was bringing because we haven't heard from the OP yet, so we don't know how they feel about it. They have already been kind of gifted the answer without effort beyond the opening post.

The answer was given by the teacher, which the OP didn't buy. I pointed out OP's misapplication of th 3rd law in post #12.

 

[erobz]

The answer was given by the teacher, which the OP didn't buy. I pointed out OP's misapplication of th 3rd law in post #12.

Normally we make them work for it. I'm not saying that was on you, we all jumped in too soon in that respect (IMO). "my teacher says this, and I say this" could be the problem statement for all we know!

 

[Charles Link]

The buoyant force on the whole bell jar is the same whether there is air in it or vacuum. That's because its external volume is the same.

Correct, but how does that change the reading the scale that supports the bell jar and everything in it when the air is taken out? Answer: The reading is reduced because a bell jar will always weigh less when you take some mass out of it, in this case air. It has nothing to do with buoyancy.

When the air pressure is the same inside the bell jar as outside, everything balances, and you needn't consider the buoyancy of the bell jar. Take out the weight of the air inside, and now you introduce the buoyancy of the volume of the bell jar into the picture. The weight goes down because you removed the inside air, or alternatively you can say that you now introduced a buoyant force that you previously didn't need to consider. LOL

 

[Charles Link]

The effect of the buoyancy is increasing the reading on the scale on as the air is being removed.

No, the reading on the scale will decrease as the air is removed from the bell jar. The bell jar is completely self-contained.

If you removed some of the outside air without a bell jar, then the scale reading would increase, provided the condom or cold balloon didn't expand too much.

 

[erobz]

No, the reading on the scale will decrease as the air is removed from the bell jar. The bell jar is completely self-contained.

If you removed some of the outside air without a bell jar, then the scale reading would increase, provided the condom or cold balloon didn't expand too much.

So are you saying the buoyant force acting on the balloon in the jar becomes larger as the air is evacuated from around it?

 

[kuruman]

Take out the weight of the air inside, and now you introduce the buoyancy of the volume of the bell jar into the picture.

I don't see why that is the case. The buoyant force on the bell jar is equal to the weight of the air it displaces. Whether the bell jar is filled with air or vacuum or water, the external volume is the same so the buoyant force is the same.

 

[erobz]

I think we need a diagram!

 

[Orodruin]

When the air pressure is the same inside the bell jar as outside, everything balances, and you needn't consider the buoyancy of the bell jar. Take out the weight of the air inside, and now you introduce the buoyancy of the volume of the bell jar into the picture.

Buoyancy is always in the picture. It is even in the picture when you have a volume of gas with no container. The buoyant force is the resultant force of external pressure on the contraption. Then this may or not be balanced out by the weight of the air inside. If it is balanced and then you take out the air inside you are not introducing buoyancy, you are removing the weight of the contained gas.

The weight goes down because you removed the inside air, or alternatively you can say that you now introduced a buoyant force that you previously didn't need to consider. LOL

This would be the case if there was vacuum everywhere and you then introduced pressure outside the contraption.

 

[Orodruin]

So are you saying the buoyant force acting on the balloon in the jar becomes larger as the air is evacuated from around it?

Nobody is talking about the buoyant force on the balloon. The question is about the weight of the entire contraption, jar and all.

 

[jbriggs444]

I don't see why that is the case. The buoyant force on the bell jar is equal to the weight of the air it displaces. Whether the bell jar is filled with air or vacuum or water, the external volume is the same so the buoyant force is the same.

I think the idea is that as long as the bell jar contains air of the same composition, temperature and pressure as the ambient atmosphere then a scale will accurately weigh the jar and its contents except for any contained air. [One assumes that the non-air contents are enough more dense than air so that the buoyant discrepancy due to their volume may be ignored. In the same way we ignore the buoyancy of any standard balance weights]

But if we modify the temperature, pressure or composition of the contained air then our measured force no longer accurately reflects the weight of the bell jar plus non-air contents. There will be a discrepancy due to buoyancy.

 

[erobz]

Can I put some forces on these?

I think the force on the scale is initially

$$W_j+ W_{air}+W_{balloon} - F_b$$

Where:

$W_j$ weight of bell jar
$W_{air}$ weight of air surrounding ballon
$W_{balloon}$ weight of ballon material and air inside balloon
$F_b$ Buoyant force from initial displaced volume of balloon.


In the case after the air is removed the $F_b = 0$ because the density of the fluid displaced is zero ( Vaccuum), and $W_{air} = 0$?

 

[jbriggs444]

$$W_j+ W_{air}+W_{balloon} - F_b$$

Where:

$W_j$ weight of bell jar
$W_{air}$ weight of air surrounding ballon
$W_{balloon}$ weight of ballon material and air inside balloon
$F_b$ Buoyant force from initial displaced volume of balloon.


In the case after the air is removed the $F_b = 0$ because the density of the fluid displaced is zero ( Vaccuum), and $W_{air} = 0$?

The buoyancy of the balloon in the atmosphere inside the bell jar is a purely internal force pair. Up on the balloon and down on the inside-the-jar air. It can have no effect on the reading of a scale outside the jar.

A scale positioned inside the jar and supporting the balloon only will not see a contributions from $W_\text{j}$ and $W_\text{air}$

An unusual scale design where the scale is hermetically sealed and covers exactly the bottom of the jar would see a contribution from $W_\text{air}$. But only if we covered the top of the jar with an identical hermetically sealed scale and calibrated the two to measure the force difference. That would be an entirely unexpected situation.

Normally we set scales on legs so that air is allowed underneath.

 

[erobz]

Ok, I got to think for a bit I see.

$$ P_{bottom}A - P_{top}A = F_b $$

 

[jbriggs444]

Ok, I got to think for a bit I see.

$$ P_{bottom}A - P_{top}A = F_b $$

You seem to be computing the weight of the volume of air that would be present if the jar were a perfect cylinder.

Or the buoyancy of a perfect cylinder in said air. Is $A$ the horizontal area of the jar or of the balloon?

 

[erobz]

You seem to be computing the weight of the volume of air that would be present if the jar were a perfect cylinder.

Or the buoyancy of a perfect cylinder in said air. Is $A$ the horizontal area of the jar or of the balloon?

If there is a helium balloon in the jar will it be given by $\rho g h$?

 

[jbriggs444]

If there is a helium balloon in the jar will it be given by $\rho g h$?

The difference in pressure between top and bottom will be given by $\rho g h$ regardless of the presence of a helium balloon floating in the bell jar, yes.

If the helium balloon is resting at the top of the jar on a contact point of negligible surface area then the total force at the top of the jar will have one contribution from $A P_\text{top}$ and a second contribution from the buoyancy of the balloon minus the weight of the balloon. The support force from the bottom of the jar minus the net force at the top of the jar will turn out to match the weight of the air in the jar plus the weight of the balloon floating in that air. As it must -- that's a second law force balance.

 

[Charles Link]

So are you saying the buoyant force acting on the balloon in the jar becomes larger as the air is evacuated from around it?

The bell jar and its contents which includes the air inside the bell jar is what is getting weighed when the bell jar, totally enclosed, is on the scale. The bell jar does displace a volume of ambient air around it, so the buoyant force from the bell jar becomes part of the picture to be considered.

What the balloon does inside the bell jar is no longer of any bearing in the calculation.

Simply its weight plus the air inside the balloon, plus the weight of the air inside the bell jar, plus the weight of the glass of the bell jar minus the buoyant force of the whole bell jar from the outside air is what the scale will read.

 

[Charles Link]

Perhaps it might be worthwhile at this point to mention why archimedes principle works like it does, that the buoyant force is the weight of the air or fluid that is displaced. What actually occurs is the pressure is just ever so slightly larger as you go further down and thereby the submerged object, when all the external pressure forces are summed, experiences a buoyant force that mathematically is equal to the weight of the displaced air or fluid.

The actual mathematics of this is exact, and follows from a form of Gauss' law.

On the item of a helium balloon inside the jar, the helium balloon does not create any magical upward force of its own. If it is inside the jar, the weight of the balloon plus the helium will add to the weight of the jar.

Helium is less dense than air, so that there is a net upward force on a helium balloon in air when the weight of the helium plus the balloon is less than the buoyant force from the weight of the air that is displaced by the helium balloon.

You may ask what happens if the helium balloon pushes against the underside of the top of the jar? When that occurs, the push comes from the underside of the balloon, and any such pressure/force will be balanced by an equal downward force/pressure against the bottom of the jar (on the inside). That's where it really doesn't matter what happens inside the jar. The scale will see the weight of the contents inside the jar, regardless of what the contents are doing.

 

[erobz]

What does the scale measure in this scenario with the scale completely inside bell jar?

 

[Charles Link]

The weight of the blue egg including its contents of air or fluid minus a buoyant force from the weight of the displaced air inside the bell jar.

 

[erobz]

The weight of the blue egg including its contents of air or fluid minus a buoyant force from the weight of the displaced air.

ok, then when we decrease then amount of air in the jar, the ballon expands the displaced air volume goes up, but the density of the displaced air goes down i.e. there is less air in the same jar. So what is $F_b$ doing, I think its declining in magnitude, because I expect there to be no buoyant force in a perfect vacuum - thus, increasing the readout on the scale?

 

[Charles Link]

ok, then when we decrease then amount of air in the jar, the ballon expands the displaced air volume goes up, but the density of the displaced air goes down i.e. there is less air in the same jar. So what is $F_b$ doing, I think its declining in magnitude, because I expect there to be no buoyant force in a perfect vacuum - thus, increasing the readout on the scale?

What the buoyant force does in that case would be hard to predict, and it would depend upon just how much the balloon expands as the external pressure is lowered. If the balloon expands enough, the buoyant force could increase. If you make a vacuum, yes, then the buoyant force would be zero.

 

[erobz]

What the buoyant force does in that case would be hard to predict, and it would depend upon just how much the balloon expands as the external pressure is lowered. If the balloon expands enough, the buoyant force could increase. If you make a vacuum, yes, then the buoyant force would be zero.

So then it's fair to say that there is something complicated occurring if you think it won't necessarily monotonically decrease toward zero. I'm a little bit -

For a "rigid balloon" (I expect) it decreases monotonically as the air is evacuated. What is it about the rubber?

 

[Charles Link]

So then its fair to say that there is something complicated occurring if you think it won't necessarily monotonically decrease toward zero. I'm a little bit -

You've greatly changed the scenario from what @kuruman has with his isolated bell jar sitting on top of the scale. His case is simple. His balloon is inside an isolated bell jar. It doesn't matter for his case how the balloon responds.

When you have a balloon, basically in ambient air on a scale, and you change the air pressure, (e.g. lowering it) the balloon could expand an unpredictable amount. That makes for a case that doesn't have the same simple answer for what the buoyant force will do. If the balloon suddenly expands to a much larger volume, the larger volume could more than offset the lower density of the ambient air.

 

[erobz]

You've greatly changed the scenario from what @kuruman has with his isolated bell jar sitting on top of the scale. His case is simple. His balloon is inside an isolated bell jar. It doesn't matter for his case how the balloon responds.

When you have a balloon, basically in ambient air on a scale, and you change the air pressure, (e.g. lowering it) the balloon could expand an unpredictable amount. That makes for a case that doesn't have the same simple answer for what the buoyant force will do. If the balloon suddenly expands to a much larger volume, the larger volume could more than offset the lower density of the ambient air.

Have I though? So…that whole assembly is now on a scale of its own. The scale reading inside is unpredictable, and the scale reading outside is predictable. That’s kind of interesting isn’t it?

 

[Charles Link]

You could have the balloon inside the bell jar be a helium balloon that originally floats in the bell jar. It doesn't matter at all to the scale outside what the helium balloon is doing. It will read the weight of the contents of the bell jar, minus the buoyant force of the entire bell jar from the outside air.

The scale inside the jar will read zero though if the helium balloon is floating in the air or rises to the top and pushes against the top of the bell jar. If you then empty the air out of the jar, the scale outside the jar will read less, but the scale inside will now have a helium balloon resting on it and it will read its weight. (Note: there is zero buoyant force inside the jar). The outside scale reading went down, and the inside scale reading went up. Yes, it's very interesting. :)

 

[Thamska]

The answer was given by the teacher, which the OP didn't buy. I pointed out OP's misapplication of th 3rd law in post #12.

Let's say you have a bowl of water on a scale. Then you submerge your hand in the bowl without touching the bowl so the hand is freely in the water not moving. This will result in the scale registrering more weight since there will be additional force from somewhere on the scale. I just thought that the additional force was a opposite force from the buoyancy force on my hand, but it seems you think I am wrong?

If you apply this to the balloons then you could say the scale readings will both be the same? Their mass will always be the same.

 

[Charles Link]

@erobz addition to post 48: The scenario mentioned at the bottom of this post with the helium balloon also applies to @kuruman 's condom demonstration of post 14. The outer scale will show less weight, but a scale inside the bell jar would show an increase as the air inside the bell jar is removed.

to answer @Thamska of post 49, your first paragraph is correct, but I don't see how you get the result of the second paragraph from that, and the second paragraph is incorrect if air is present to supply a buoyant force in the scale readings to the balloons.

Note: Even the bowl of water (no hand in it) appears to weigh less than it really does because there is a buoyant force from the outside air. The scale would show a slight increase if you were to remove the air.