# Weight of a container of gas and container shape [duplicate]

Consider a cube shaped container with one mole of gas inside.

Acceleration due to gravity would give the difference in force between the top of the container and the bottom to be mg, with net force mg on the bottom, where m is the mass of the molecule.

I think I understand this idea, but what I am confused about is that at any one time, only a subset of the molecules collide with the bottom of the container at any instance in time; therefore, only a subset of all the molecules are contributing to the net difference in force - some, for example are colliding with the sides, some occupy the space within the container. So is the mass of the gas, determined by weighing, less than the actual mass of the gas?

What if I changed the shape of the container, so that the surface area in contact with the balance is greater or smaller? The pressure per unit area is the same for all sides, but the overall force on that surface is different.

I hope that makes sense, I would appreciate any help with this.

## marked as duplicate by user191954, Sebastian Riese, glS, Qmechanic♦Sep 26 '18 at 11:59

• Related and possibly a duplicate: Why do gases have weight? – John Rennie Aug 24 '18 at 12:46
• Hi John, I agree it is related, but perhaps not a duplicate: more of an extension. I am interested to know what happens with a collection of molecules. I may have misunderstood that other post, but it seems limited to one molecule. My main point of confusion regards what happens with a collection, as only a subset of the molecules collide with the bottom of the container at any one instance. – user204786 Aug 24 '18 at 12:49
• Possible duplicate of Does the weight of fluid in a conical container act entirely on the base? – sammy gerbil Aug 24 '18 at 17:39

only a subset of the molecules collide with the bottom of the container at any instance in time; therefore, only a subset of all the molecules are contributing to the net difference in force

This part isn't correct. For the moment, let's pare this down to a much simpler example.

Let's have a person standing on a scale holding a very heavy ball. When they're standing still, the scale shows the sum of the weights of both the person and the ball.

Now the person begins to juggle the ball. If the scale is very fast (responsive to changes), then we'll see it bounce up and down a lot. It will show less weight when the ball is in the air, and more weight when the person is juggling it. In fact, it will show a lot more weight because the person isn't just holding the ball at that point, they're accelerating it upward.

To keep from getting dizzy with the scale indicator bouncing around, we can dampen the scale. The indicator no longer moves instantaneously, but slowly in response to the actual weight. If we dampen it sufficiently, the pointer will converge to combined weight of the person and the ball. The fact that the ball is often in the air (for an unknown amount of time and to an unknown height) doesn't matter. It also doesn't matter that the ball isn't hitting the scale directly.

The longer it is in the air not contributing to the weight, the harder the person has to accelerate it, causing more force on the scale.

The same thing happens in the box. The particles can be moving fast or slow, be large or small. But over time, all of them have to be accelerated by the bottom of the container (either directly, or by collisions with other particles), or gravity would pull them down. The combined effect on the bottom of the container is equal to the weight of all the mass inside, plus any pressure from the top of the container.

What if I changed the shape of the container, so that the surface area in contact with the balance is greater or smaller? The pressure per unit area is the same for all sides, but the overall force on that surface is different.

So let's simplify - consider a 2D container and our cardinal directions are U(p), D(own), L(eft) and R(right). Good?

Consider the container in the shape of a square. In this case you'll want to think of the weight as being the contribution of the molecules at the D end of the interior hitting the D side of the container... as in your example.

Now let's set that concept on its head.

Consider a container in the shape of a funnel with the top and bottom closed. You say "well, the bottom is smaller, so isn't the force smaller"?

Consider a molecule in that container that is near the L side, and travelling LD. Do you see that when it hits the L side it does so at an angle? And that when you apply simple vector decomposition, it has a D-directed force? So even though it's hitting the "side" of the container, its still causing a downward force we call weight.

Now, what do you think the integral of every interaction at all of the container walls might be? Might it ultimately be exactly the same in every case?

Yes!

You're confusing yourself because you're doing the one thing you should never do with any fluid, considering "this part" of the fluid! That way lies dragons!