# Weighing Air

1. Apr 25, 2005

### Jayse_83

Hi guys, i was hoping to get an explanation for something that is bugging me. QUESTION - An evacuated flask weighs 350g and with air present it weighs 356.3g ... now i understand that the weight will increase due to the fact that air weighs something, but the part that confuses me is if we consider the fact that the particles are moving around inside te flask and not touching the flask.
What would be a good way of explaining this ?? im thinking along the lines of the flask weight is the sum of the flask + contents ... but its far from an ellegant explanation.
Thanks
Jayse

2. Apr 25, 2005

### dextercioby

You have to be kidding,right...?Is there any sign that the air molecules (mainly $N_{2}$ and $O_{2}$) could pass through the amorphous structure of glass,so that the "mass conservation" would not be possible...?

Daniel.

3. Apr 25, 2005

### Jayse_83

Hey, Im not sure if i explained the problem to well...
The flask does weigh more (its not a shock ... i hope :S) but the way i intially thought of a solution was to consider the gas to be solid or liquid and imagine the constituents of air neatly sat at the bottom of the flask. But the problem states that the air moves does not touch the flask (or if it does we can assume that the collisions with each dA (unit area) of the flask will be equally balanced with a pressure force in the opposite direction. Its got me all muddled!

4. Apr 25, 2005

### whozum

The air molecules inside are still being pulled by gravity, the pressure on the bottom of the flask is responsible for the added weight, as the particles are getting pulled down (I believe).

5. Apr 25, 2005

### Jayse_83

Yeh that makes sense, like atomospheric pressure pushing on us ... on the side what would be the difference between a sealed box filled with a balloon ... and the same scenario but this time the balloon is attached to a weight which is being suspended.
Would the second box weigh more ... and will it weigh box + balloon + weight ?

6. Apr 25, 2005

### DaveC426913

I don't understand how the particles can move around inside the glass without touching it. There must be some physical mechanism, even if it's indirect (such as a second container). If there isn't any transfer of force from air (ultimately) to glass, then, yes the air would effectively be weightless!(which won't happen)

7. Apr 25, 2005

### Integral

Staff Emeritus
In reality the molecules of gas do collide with the sides of the flask. But this is not the reason the flask weighs more. Were it collisions with the sides of the container causing the added mass then the mass would be temperature dependent. This is not the case (lets leave relativity out of this!)

It is not so clear to me that you can ignore what is happening on the BOTTOM of the container. Suppose that we where to suspend (meaning that it is externally supported by something other then the outter flask) a flask full of air inside of larger evacuated flask. would the weight of the suspended flask contribute to the weight of the outer?

Unless the outer flask is some how supporting the inner the answer must be no.

8. Apr 25, 2005

### SpaceTiger

Staff Emeritus
I think a better way to think about this is to say that air is actually pushing "up" on the evacuated flask. Let's look at the equation of hydrostatic equilibrium:

$$\frac{dP}{dr}=-\rho g$$

If g is assumed to be roughly constant, this can be simplified to the Physics 101 equation:

$$\Delta P=-\rho g \Delta r$$

Now, for simplicity, let's imagine a cylindrical vial or cup, closed on the top. The particles outside the cup will be pushing down on the lid and up on the base (yes, there's air underneath the cup). The difference between the force pushing up and that pushing down is given by:

$$\Delta F=-\rho ghA$$

where A is the area of the lid/base and h is the height of the cup. In other words, there is a net force upwards. Since the sides of the cup are distributed identically in height, there will be no net lateral force from the air outside the cup.

What about inside the cup? Well, if there's air in it, then the internal pressure will push up on the lid and down on the base with the same forces that the outside air is pushing, so it will exactly balance the force given above. In other words, the air won't add to the cup's weight. If the cup is evacuated, however, there will be no pressure from inside the cup to balance the air pressure outside. How does all this relate to the weight of the air? Well, we can rewrite the equation above:

$$\Delta F = -\rho ghA=-\rho gV=-m_{air}g$$

where V is the volume of the cup. This is just the "missing" weight of the air that was inside the cup!

So, long story short, if there's no air inside, then the atmosphere will exert a net upward force on the flask/cup/vial. If there is air inside, then its pressure will cancel that of the air outside and you will get only the weight of the solid material. What we've just done is inadvertantly rederived a special case of Archimedes' Principle, which says that a body immersed in a fluid will experience an upward force equal to the weight of the displaced fluid.

9. Apr 25, 2005

### Staff: Mentor

So to boil it all down: the pressure in a closed container will be very slightly higher at the bottom of the container than at the top and that difference is equal to the weight of the air inside the flask (pressure times surface area, actually).

10. Apr 25, 2005

### jdavel

here's a related question.

If an inflated balloon is weighed in air, does it weigh more, less, or the same as it weighs when its deflated?

11. Apr 25, 2005

### SpaceTiger

Staff Emeritus
If the air inside the balloon is the same as outside, it will weigh approximately the same in either case...though the stretching of the balloon material may complicate things a bit.

12. Apr 25, 2005

### whozum

I think you were referring to my post, which was supposed to address this exact concept. The added weight is due to the air pressure on the bottom of the flask.

13. Apr 26, 2005

### Mk

Well, it depends on what the question is asking, it can weigh more or stay the same: If you mean the weight of the balloon, then it will of course stay the same. If you mean the weight of the balloon and its contents, then it will weigh more, because there is more matter there.

14. Apr 26, 2005

### jdavel

Mk,

"....then it will weigh more, because there is more matter there."

Careful! If it's full of He there's more matter, but it weighs less. Remember, we're weighing the balloon in air.

15. Apr 26, 2005

### Jayse_83

wow! thanks for all the replies, u've cleared that problem up for me :D

16. Apr 27, 2005

### Mk

Weight is the force exerted upon an object by virtue of its position in a gravitational field. It is equal to the mass of the object multiplied by the magnitude of the gravitational field. The balloon still weighs more, just because it's floating, or even going up doesn't mean it doesn't have weight! Clouds have weight, dust has weight. Though the balloon's apparent weight is ≤0.

Last edited: Apr 27, 2005
17. Apr 27, 2005

### jdavel

So an object in free fall isn't weightless? How about an object in orbit around the earth? Aren't astronauts weightless in space?

I think a better definition of weight is what a scale reads when you stand on it!

18. Apr 27, 2005

### Mk

The things you describe have ≤0 apparent weight. And your definition is the definition of apparent weight. In space there is microgravity, so they also have apparent weight.

19. Apr 28, 2005

### MR. P

hey! this was just starting to get interesting! don't stop now!

20. Apr 28, 2005

### MR. P

lets go back to the balloons... and evualate three senerios...three balloons of equal volume balloon 1 is filled with He2
balloon 2 is filled with H2
balloon 3 is unique and retains the same volume as 1 &2 but holds a perfect
vacuum
all these balloons are at STP what would be the difference in their 'bouancy' or rather their lifting capability?

thanks

frank MR, P

21. Apr 28, 2005

### MR. P

Spacetiger you seem quite adept at these types of abstracts....in the scenario above if all three balloons were restrained at sea level what would be the rate of acceleration of each balloon and what would be the ceiling of each also assuming the mass of the balloons are equal.....I am curious about this relationship ...I know intuitively how the balloons would respond however , mathematically i'm an idiot.....also intuitively doesn't help when considering balloon 3 since its' volume will not change through the ascension phase also i find it hard to evaluate its' final velocity since the impeding forces diminish with altitude and it would appear that we have discovered a means of achieving NEO without anything.....

.where physics starts and intuition ends

frank MR. P

22. Apr 28, 2005

### SpaceTiger

Staff Emeritus
This all boils down to the ideal gas law:

$$PV=N_mRT$$

where Nm is the number of molecules in the balloon. I hate that form of it, however, so I'm going to define:

$$n=\frac{N_mN_A}{V}$$

where n is the number density and NA is Avogadro's number. So we have:

$$P=nkT$$

where k is Boltzmann's constant. Anyway, let's look at this. In the hydrogen and helium balloons, we know that the internal pressure must equal the outside pressure. If it didn't, then the molecules would push on the balloon, making it expand or contract until the pressures were equal. Likewise, we can treat the temperatures as being equal because we'll assume that the gaseous contents were exposed to the external environment and reached thermal equilibrium with it. This means:

$$n=\frac{P}{kT}$$

is a constant as well. However, any number of hydrogen or helium molecules has a smaller total mass than the same number of air molecules (mainly diatomic nitrogen). To see this, just look on the periodic table. Atomic hydrogen and helium have masses of about 1 amu and 4 amu, respectively, while nitrogen has a mass of about 14 amu. By Archimedes principle, the buoyancy force is given by:

$$F=-m_{air}g$$

where mair is the mass of air that would be in the balloon if it weren't filled with hydrogen or helium. The total weight of the balloon is given by:

$$F=m_{balloon}g$$

If we neglect the weight of the elastic balloon material, this gives a total force of

$$F=(m_x-m_{air})g=-m_{air}(1-\frac{m_x}{m_{air}})g$$

For the cases you listed:

$$\frac{m_{He_2}}{m_{air}}\sim\frac{4}{14},\ \frac{m_{H_2}}{m_{air}}\sim\frac{1}{14},\ \frac{m_{vac}}{m_{air}}=0$$

Thus, the ratios of the upward forces are:

$$\frac{F_{He_2}}{F_{H_2}}\sim\frac{10}{13},\ \frac{F_{H_2}}{F_{vac}}\sim\frac{13}{14}$$

This was done on-the-fly and I have no means of checking it, so please let me know if I made any errors.

NOTE: If the balloon were made of an elastic material, it couldn't maintain a vacuum at the same volume (the external pressure would squash it), but the above calculation assumes that it does anyway. For this case, one can instead imagine that the "balloon" is made of a light rigid material.

Last edited: Apr 28, 2005
23. Apr 28, 2005

### MR. P

Tiger thank you are a genius!!!!

indeed if volume 3 were an ideal restraint of equal mass as 1& 2 what would be the final velocity of 3.

in real life balloons are allowed to expand until they pop or leak from their confining medium however 3 is still purplexing in that its' density or volume to mass ratio would or should be lower than 1 & 2 but the ascension or acceleration from restraint should be quite powerfull in as much the entirety of the earth's atmosphere would be acting upon it with a 2500 lb./ft^2 pressure so perhaps it would be easier to evualate assuming a 1 ft^3 volume . evualating this with the above insight is still beyond my ability ....i can ask the question maybe even build the restraint but analyze beyond me ......
thank you tiger that was quick, too.....sorry I couldn't come up with something that actually challenged you

humbled

frank mr. p

24. Apr 28, 2005

### SpaceTiger

Staff Emeritus
I would imagine that the primary reasons they sink are

1) The balloon material still adds some weight that is much denser than air.
2) The outside air is leaking in to the balloon, increasing its interior mass.

The above treatment is idealized in the fact that it neglected the weight of the elastic material. A truly weightless balloon filled with air should feel no net force from pressure + gravity. Of course, you would still have wind and such.

25. Apr 28, 2005

### MR. P

one cubic foot and a mass of one hundredth of a pound. also I can see why 3 ==0 however the real world sees that mass and acts upon it vacuum or not