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

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[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.