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This was recently brought up, and I haven't been able to conclude a solid answer.

Let's say we have two identical boxes (A and B) on Earth, both capable of holding a vacuum and withstanding 1 atm outside acting upon them.

A is holding a vacuum, while B is filled with air (at 1 atm). Shouldn't they weigh the same as measured by a scale?

Current thought process

The following thought experiment suggests they'd have the same weight, but I haven't formulaically shown this — and everyone has disagreed so far.

Take a box like B (so it's full of 1 atm air) and place it on a scale. Here's a cross section:

    +------------+
    |            |
    |            |
    |            |   <-- B box
    |            |
    +------------+
***********************
|                     |    <-- scale

Now, taking note of the scale readings, start gradually pushing down the top "side" (rectangle/square) of the box (assume the air can somehow escape out of the box as we push down)

    |            |
    +------------+
    |            |
    |            |
    |            |
    +------------+
***********************
|                     |

Then

    |            |
    |            |
    +------------+
    |            |
    |            |
    +------------+
***********************
|                     |

etc., until the top side is touching the bottom of the box (so the box no longer has any air between the top and bottom sides):

    |            |
    |            |
    |            |
    |            |
    +------------+
    +------------+
***********************
|                     |

It seems to me that:

1) pushing the top of the box down wouldn't change the weight measured by the scale.

2) the state above (where the top touches the bottom) is equivalent to having a box like A (just a box holding a vacuum).

Which is how I arrived to my conclusions that they should weigh the same.

What am I missing, if anything? What's a simple-ish way to model this?

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    $\begingroup$ Does a box full of water weigh more? Yes, of course. Therefor a box full of air weighs more than a box with a vacuum inside. $\endgroup$
    – Brandon Enright
    Dec 6, 2014 at 23:10
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    $\begingroup$ Hint: does a box full of helium weigh the same as a box full of air? Why or why not? If the box is immersed in water (and sinks to the bottom, where it rests on a scale), does it weigh the same when full of water as it does when full of air? Why or why not? What about a box "full" of vacuum underwater? And what about a box full of vacuum in air? $\endgroup$
    – Steve Jessop
    Dec 7, 2014 at 0:54
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    $\begingroup$ This, incidentally, is a classic experiment. Nowadays done with ballons, but I think one of the oldest versions actually used metal balls (one full of air and the other partially evacuated). $\endgroup$
    – Superbest
    Dec 7, 2014 at 21:48
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    $\begingroup$ You're essentially asking if air has weight. If the answer was no, you would not have survived long enough to type your question. :-) $\endgroup$
    – The111
    Dec 7, 2014 at 22:27
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    $\begingroup$ I don't know what's so complicated here. Air is a gas, made up of a mixture of atoms, 99%+ nitrogen and oxygen. Atoms have mass. They will be pulled down by gravity. Therefore the box with air will weight more. Am I missing something? $\endgroup$
    – tcrosley
    Dec 7, 2014 at 22:36

18 Answers 18

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The buoyant force on a body immersed in a fluid is equal to the weight of the fluid it displaces. In other words,

$$ F_B = -\rho_{\text{fluid}} V_{\text{body}} ~g $$

The force of gravity on the body is equal to $$ F_g = m_{\rm body} ~g $$

The apparent weight of this body will therefore be equal to the sum of these two forces. $$ W_{\rm app} = -\rho_{\rm fluid} V_{\rm body}~g + m_{\rm body} ~g $$

When you add air into a box full of vacuum an evacuated box, the mass of the body (which is now $m_{\rm box} + m_{\rm air}$) increases, but the volume still stays the same. Therefore $W_{\rm app}$ must increase.

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    $\begingroup$ I like "a box full of vacuum" - it's like when a block of ice "radiates cold"! $\endgroup$
    – corsiKa
    Dec 8, 2014 at 22:00
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    $\begingroup$ I prefer one half full of vacuum. $\endgroup$
    – Oldcat
    Dec 8, 2014 at 23:05
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    $\begingroup$ @Oldcat That immediately reminds me of: what-if.xkcd.com/6 For those of you who haven't seen it, read that. $\endgroup$
    – reirab
    Dec 9, 2014 at 6:16
  • $\begingroup$ @Oldcat 0.5 Atm doesn't sound too hard to obtain. Three typical domestic vacuum cleaners in a russian doll formation should be able to cut it. $\endgroup$
    – John Dvorak
    Dec 24, 2014 at 2:06
  • $\begingroup$ I feel like it should be $W_{\rm app} = - \rho_{\rm fluid} V_{\rm body}~g + m_{\rm body} ~g$, i.e. the buoyant force reduces the weight (minus sign)... $\endgroup$
    – KIAaze
    Dec 9, 2019 at 13:54
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Just my two cents to complement the other answers. The mistake in your reasoning is that:

the state above (where the top touches the bottom) is equivalent to having a box like A (just a box holding a vacuum).

is incorrect, it rather equivalent to a box full of air (there is air in between the walls, regardeless of the vertical position of the top). In the vaccumm case there is no air in between. The buoyancy increase when ther is vaccum can be explained intuitively in the following way. See the two boxes in the picture below. When there is air inside the box, the air preasure acts on the two sides of each surface, and thus it cancels. When there is vacuum, there is preasure in only on one side of the surfaces, and the preasure at the bottom is higher than that at the top ($\Delta p =\rho_{air} g h$), so they do not cancell and the net force of the air is upwards. That is why the balance reads less.

enter image description here

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  • $\begingroup$ Your words look better than your pictures. Your words acknowledge that, "the preasure at the bottom is higher than that at the top," But the pictures don't show that. The pictures show the same $P$ pushing upward on the bottom of the boxes as the $P$ pushing downward on the top. $\endgroup$
    – Solomon Slow
    Jul 21, 2020 at 18:54
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1) Technically while you push it down it will cause an increase in pressure (so the weight will change) but assuming the box has a hole and the air can equalise then the weight of your initial and end state will be the same. This is because the air is equalised within the box at all stages and at the end state the air that was in the box is now above it.

2) This is wrong, the collapsed box is not equal to a box with a vacuum in it. This is because there is a pressure difference between the two boxes. One has air above it (and is equal to air pressure) the other has a vacuum in it (and is less than air pressure, that is, absence of preasure).

enter image description here

Note that air has weight but because it is a gas (with molecules travelling in random directions) the force of gravity is spread out over the directions of the molecules and results in an increase in air pressure the lower you are in the atmosphere. So the weight of air does not act directly downwards, it results in higher air pressure (which acts in all directions).

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  • $\begingroup$ I've added a note to my answer, but essentially the weight of air above is spread out as air pressure so is not significant. It's the difference in air pressure within the containers that is important. $\endgroup$
    – Quantumplate
    Dec 7, 2014 at 0:51
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    $\begingroup$ @Quantumplate: but look at Quantum's diagram. Draw a horizontal line level with the top of the box-o-vacuum on the right. The column of air above this line (all the way to the top of the atmosphere) is the same in both cases. So no, that part of the column doesn't contribute to the weight measured by the scale[*]. But on the left, there's a little bit more column. So the one on the left weighs more, the difference depending on the mass of air in that extra little bit of air-column at the bottom, i.e. the volume of the box. $\endgroup$
    – Steve Jessop
    Dec 7, 2014 at 1:22
  • $\begingroup$ * Well, actually, it kind of does. But it's the same contribution in both cases, and it's much more than the mass of a small tin box or whatever. Thing is, a real-life scale is itself full of air throughout its measuring mechanism. If the scale itself contained a vacuum, and you measured "weight" by the stress on its upper surface, then it would be measuring the weight of the whole column of air directly. And the box+column on the right would still be lighter than the box+column on the left. $\endgroup$
    – Steve Jessop
    Dec 7, 2014 at 1:25
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Just consider the same situation in water. A box filled with water, immersed in water will have more weight on a weighing machine than an empty box (with vacuum or air)--which might even float depending on the mass and volume of the box.

The only difference in this case is the density of the medium.

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  • $\begingroup$ My apologies - I didn't read through the comments... However, I do believe my answer is a perfect analogy to the OP question. :) $\endgroup$
    – navigator
    Dec 8, 2014 at 8:18
  • $\begingroup$ The water example definitely leads to the answer however without additional details spelling it all out it isn't an answer. If it were an answer I would have posted it as such. $\endgroup$
    – Brandon Enright
    Dec 8, 2014 at 8:20
  • $\begingroup$ The general idea I had in mind for the water example is to take two fluids, water and oil (less dense that water) and compare the difference in weight. Now (with your imagination) reduce the density of the water and the oil slowly. The whole time you reduced them, the water stays heavier than the oil. At some point the oil gets to a density of zero (vacuum) and the water still has a positive density and weighs more. Air is more dense than a vacuum. It weighs more. The helium example others used is a similar idea to the water idea. $\endgroup$
    – Brandon Enright
    Dec 8, 2014 at 8:22
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    $\begingroup$ @BrandonEnright (1st comment) seems like just a coincidence in this case, but in general, anything you post as a comment is fair game for answerers to use or improve on. If it bothers you to have someone post the content of your comment as an answer, you should have posted it as an answer yourself - assuming it does in fact constitute an answer to the question. $\endgroup$
    – David Z
    Dec 8, 2014 at 16:59
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    $\begingroup$ @DavidZ my initial assumption was that it copied me rather than a coincidence. I felt the answer added words but didn't really expand on the idea so the answer didn't really add any additional value. You're right, I should have taken the time to provide a full answer rather than the kernel of an answer in a comment. $\endgroup$
    – Brandon Enright
    Dec 8, 2014 at 17:03
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Let me highlight lhree points:

1) The mass of a box is the sum of the mass of the box structure, and the mass of the box contents;

2) The force of gravity downward on the box depends only on the mass; (and the local acceleration of gravity)

3) The buoyant force upward depends only on the density of the surrounding medium and the volume of the box.

So, how does the net force change when you add air to an empty (vacuum) box?

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  • $\begingroup$ The net force does change, though, doesn't it? I mean, if it's a weak enough empty (vacuum) box, it would collapse (net force =/= 0). If it has air in it, it wouldn't (net force = 0). Maybe I'm missing something. $\endgroup$
    – Chris
    Dec 6, 2014 at 23:13
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    $\begingroup$ Of course it changes; answering the points above should indicate how it changes... $\endgroup$
    – DJohnM
    Dec 7, 2014 at 5:14
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A box filled with helium would weigh less than a box filled with air, because helium is less dense than air.
A rigid box containing a vacuum would weigh even less than the same box filled with helium, because it is even less dense.

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  • $\begingroup$ Nice and simple! $\endgroup$
    – GreenAsJade
    Dec 8, 2014 at 4:44
  • $\begingroup$ Why though? The why is more often than not a crucial part of the answer, because you want to help someone to follow your thoughts - what's intuitive to you might not be to someone else. And another thing: It doesn't matter if you weigh it in an atmosphere or in a vacuum, as long as you weigh all cases in the same conditions. $\endgroup$
    – pho
    Dec 8, 2014 at 8:07
  • $\begingroup$ @PranavHosangadi Thanks for pointing out the irrelevance of the surrounding medium. I had considered expanding on the density explanation by discussing buoyancy, but that would spoil the simplicity of the answer, and furthermore, the topic has already been covered in depth by other answers. $\endgroup$
    – 200_success
    Dec 8, 2014 at 8:55
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    $\begingroup$ So if you could remove all the air from a balloon while it still maintaining it's shape, would the balloon with a vacuum float like a helium balloon? $\endgroup$
    – Jonathan.
    Dec 9, 2014 at 5:28
  • $\begingroup$ @Jonathan. Yes, the balloon would float. The trick, though, is maintaining the same shape while vacuuming out its contents. Most lightweight materials would collapse due to lack of strength. $\endgroup$
    – 200_success
    Dec 9, 2014 at 6:01
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Here's a way to think about it that avoids assuming the bouyant force equation is correct. All you need to know is that air pressure increases with depth / decreases with height.

First off, before we put anything at all on our scale, there's already a large force pushing down on the scale's plate.

$$F_0 = P_0 A_\mathrm{plate}$$

This force gets cancelled out while calibrating the scale, so we don't typically count it toward the weight of anything.

Now let's put a box full of air on the scale. Assume it has area $A_\mathrm{box} < A_\mathrm{scale}$ on the top and bottom, vertical sides, and thin faces, and total mass $m_\mathrm{box}$. At the top of the box, air pressure is a little bit less by a difference given by the density of air times the local gravitational acceleration times the height of the box, $P_\mathrm{top} = P_0 - \rho g h$. So the vertical forces acting on the box are:

  • $-m_\mathrm{box} g$, downward from gravity
  • $-P_0 A_\mathrm{box}$, downward from inside air pushing on the bottom
  • $+(P_0-\rho g h) A_\mathrm{box}$, upward from inside air pushing on the top
  • $-(P_0-\rho g h) A_\mathrm{box}$, downward from outside air pushing on the top
  • $+F_\mathrm{contact}$, the upward force from the plate scale.

Since the box is not moving,

$$F_\mathrm{contact} = m_\mathrm{box}g + P_0 A_\mathrm{box}.$$

The plate still has some air pressure pushing down on the remaining area, so the total forces downward on the plate are

$$F_\mathrm{total} = P_0 (A_\mathrm{plate} - A_\mathrm{box}) + F_\mathrm{contact}.$$

The weight reading is the difference

$$F_\mathrm{total} - F_0 = m_\mathrm{box} g.$$

(Luckily for most scale applications, air inside a container ends up cancelling out.)

Next let's put the identical box full of vacuum on the scale. The forces from air inside the box disappear, leaving:

  • $-m_\mathrm{box} g$, downward from gravity
  • $-(P_0-\rho g h) A_\mathrm{box}$, downward from outside air pushing on the top
  • $+F_\mathrm{contact}$, the upward force from the plate scale.

Again, the box is not moving when we read the weight, so this time the force between the scale and box is

$$ F_\mathrm{contact} = m_\mathrm{box} g + (P_0-\rho g h) A_\mathrm{box}. $$

We still have

$$F_\mathrm{total} = P_0 (A_\mathrm{plate} - A_\mathrm{box}) + F_\mathrm{contact},$$

So this time

$$F_\mathrm{total} - F_0 = m_\mathrm{box} g - \rho g h A_\mathrm{box}.$$

The airless box results in a lower scale reading, and the difference is exactly the bouyant force $\rho g V_\mathrm{box}$.

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The weight is less with vacuum inside the box.

The forces on the box are the gravity acting on the mass of the box and its contents, and the buoyant force which is equal to the weight of the inner volume of the box when filled with air.

Let's say the box is one cubic meter, then the air content weighs some 1200 grams depending on temperature and local pressure. The box may weigh, say, ten kilos.

(Also, on each side of the box there is one atm of pressure, equal to about ten tons force. When filled with air, the air is compressed by that pressure until it exerts an equal reaction force from the inside, equalizing the pressure and keeping the box from collapsing).

The mass of the empty box is ten kilos, and the box full of air has a mass of 11.2 kilograms - 10 kg of box, 1.2 kg of air at one atmosphere of pressure.

The buoyant force on the box is some 11.7 Newton upward, whether the box is empty or full. All that counts is that the box has a volume.

So the empty box weighs 98.1 Newton, less 11.7 Newton of buoyancy, giving 86.4 Newton total (the scale is in kilograms and will read 8.8). If you make a hole in the box, its mass and weight will increase until the scale reads 10.0 Kg.

(Of course I don't think there's a material strong enough and light enough to hold one atmosphere of pressure - that's ten tons on each meter-square side of our box - while weighing as little as ten kilograms).

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The internal pressure in the top of the box full of air is lower than the internal pressure at the bottom - because of the weight of the column of air. You experience this when you go up in a plane (or even an elevator) and your ears "pop".

Air has mass. The box with air weighs more.

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When you push the lid of the air-filled box down you are doing two things:

  1. Reducing the volume of the box; and
  2. Reducing the mass of the box and its contents (as the air is squeezed out).

The weight measured on the scale is:

"weight of the box and contents" - "buoyancy due to surrounding air"

The "buoyancy due to surrounding air" is the weight of a volume of air equal to the volume of the box.

So the two processes cancel each other out and the weight stays the same as the top of the box is pushed down to the bottom.

The vacuum-filled (i.e. evacuated) box A has the same weight as the "squashed" box, but larger volume: and hence has a lower total weight as measured by the scale.

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The key thing you're forgetting is that the atmosphere also has a vertical pressure gradient, much like any body of water. This means that, if you take your box with the lid pressed down,

         (*)                                         (*)            z=h
    +------------+                     |        |            |
    |            |                     |        |            |
    |            |                     |        |            |
    |            |   <-- B box         |        |   (**)     |      z=0
    |            |                     |        +------------+
    +------------+                     |        +------------+
***********************                |    ***********************
|                     |    <-- scale   |    |                     |

the pressure on the pressed-down lid at $z=0$ (point (**)) is greater than on the unpressed lid, at $z=h$. As with water, the pressure difference is given by $$\Delta p=\rho g h,$$ so the additional force on the empty box is $$\Delta F=A\Delta p=\rho g h A=\rho V g,$$ which is exactly the weight of the air that's been let in.

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I am missing in the answers so far the consideration that air consists of single particles: all the arguments I've read so far treat it as a continuum. Sometimes particles are mentioned, but in the next step already "pressure" which is a resulting statistic phenomenon is mentioned again.

The salient point is that those particles are moving in straight lines between collisions when no outer forces are applied.

Imagine filling a box with ping ping balls. Then you let the box vibrate heavily so that balls are bouncing off walls, floor, and ceiling "indiscriminately". Due to gravity's acceleration between bounces, however, they will tend to hit off the bottom with greater speed than off the ceiling and consequently transfer a greater amount of impulse. Basically they "collect" impulse (force times time) while flying in the gravitational field and give about double of their impulse in the direction of the wall off when bouncing off. So even if they are all in perpetual motion, the net effect is pressing down more than it presses up.

Of course, that's the point behind "pressure" as a macroscopic phenomenon. But I was missing the description of the microscopic cause.

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The way I would think about this just for a quick answer:

A balloon filled with air gradually sinks. Now if you took the same balloon and made it rigid, sucked all the air out of it but it still had the same volume, it would float straight up. So I would say that the balloon filled with air weighs more.

Same would go for boxes.

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The various answers above are, essentially, different ways of describing the same things..

Boyancy explains the difference... However you could also simply note that the box with the pushed-down lid still 'contains' the air -- it's just 'outside' the lid. The weight remains the same. If you managed to add a force-field at the top of the box, and then pumped out the air, the box would be 'lighter' by the volume of the air removed. .. this is a different way of describing the 'boyancy' factor.

If you created a vacuum in the entire room containing the box, then a box with a vacuum would 'weigh' the same as a box full of air would in a room full of air.. and the same box, having one atmosphere would weigh more -- accounted for by the mass of the air contained in the box.

The difference, in both cases (vacuum room, and air-filled room) between the 'full' box and the 'vacuum' box is the same.. but the 'base' weight is a bit less in the atmosphere-filled room, as explained by the boyancy force of the air.

Let's put it another way: If you filled the box with helium, it would be lighter because a given volume of helium 'weighs' less than a given volume of air. That same volume of vacuum will weigh even less than the same volume of wir. The only difference being that it's a lot easier to make a 'box' able to hold a cubic foot of helium at one atmosphere.

If you could make a magical balloon that weighed the same a a rubber balloon but was able to withstand the 15psi pressure of air against a vacuum, it would float up a bit faster than a helium balloon. (the difference being the weight of that volume of helium).

The only real differences between the balloon and the box of the same volume are that

1) the box's BASE weight is heavier, and

2) the box is able to withstand the net pressures of containing a vacuum.

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The weight is the same if its on your desk. It is dependent on the medium the box is in. Weight is determined by gravity and gravity itself is simply the effect of a pull on denser vs less dense molecules. If the box is sitting in air - the air has no effect on its weight because in relation to the air the box sits in there is zero weight, the same as a vaccuum. If the box is in water (more dense) the box with air weighs less. If the box with air is in a vaccuum, it weighs more. Its mass is different no matter the medium.

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Short answer: No. The air in the box has a mass, and a weight -- but that weight is decreased by buoyancy -- i.e. the mass of the air that is displaced by the box.. (i.e. also the mass of the air in the box). This seems to make the air weigh nothing.

So let's presume that we have a box of 10kg mass, that holds about 1KG worth of air (about 1m3). The weight of the box would be 10KG (box) + 1KG (air) - 1KG (buoyancy) == 10KG. ... I.e. The air seems to have no weight.

If filled with a vacuum, however, the weight of the box would be: 10KG(box) + 0KG (vacuum) - 1KG (buoyancy) = 9KG In other words the vacuum seems to weigh about -1KG.

The same thing happens in water -- but because the water is heavier, the cubic meter of water has a buoyancy of about 1 tonne vs 1KG for air -- so the buoyancy effect is much more noticeable.

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  • $\begingroup$ The buoyancy is due to the air present outside or inside the box? $\endgroup$
    – Yashas
    Apr 13, 2017 at 8:54
  • $\begingroup$ It's due to the air outside the box $\endgroup$
    – Stephen Samuel
    Apr 14, 2017 at 11:41
  • $\begingroup$ It's actually the weight (density) of the air outside the box times the space occupied by the box. $\endgroup$
    – Stephen Samuel
    Apr 14, 2017 at 11:46
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The physical principle that applies here is called the Archimedes' principle.

Emilio Pisanty's and other answers explains well what is going on. This answer is solely to provide the name of the underlying theory.

The Archimedes' principle states basically that for a submerged object (in your case the box once with air in it, once without) the apparent weight, once submerged, is equal to the weight of the object (as measured in vacuum) minus the weight of the displaced fluid (air in this case). Or simpler:

$$\text{apparent immersed weight} = \text{weight of object} - \text{weight of displaced fluid}.$$

So in your case, you change the weight (in vacuum) of the object by filling it up with air. You do not change the volume and thus the weight of the displaced fluid stays the same.

From here on it should be easy to figure out which one is heavier.

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Unless you want the box containing vacuum float up in the air like a balloon because of something called buoyancy, you have to tie it to the scale with a rope. If you do tie the box to the scale, it will pull the scale upwards, making the scale show a negative weight.

So no, the two boxes would not weigh the same as measured by a scale.

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