How heavy is air?

This isn’t for FRC but is there a way to calculate how much air weighs when compressed? For example, if I have 1 cubic feet of compressed air at 100 psi, how much heavier would that be versus an empty tank? If possible please give me the formula on how to calculate this. Thanks

2 Likes

Now I might be totally wrong, but some chemistry equations can be helpful on this one, such as the Ideal Gas Law. The formula is :

PV=nrt
Where:
P = pressure in atm
V = volume in L
n = moles of gas
r = ideal gas constant 0.08206
T = temperature in K

So you might be wondering, where is the mass? Well, we need to separate n into 2 separate components that we will divide, m and M. m is the mass (g) of the gas and M is the molar mass of the gas (g/mol). This works because when you divide m/M you are left will moles. So now we need to know the molar mass of air, which after a quick google search is 28.97 g/mol.

PV = (m/M)RT

Rearrange to solve for m.

m = PVM/RT

So that’s the equation (I think). m will be in grams. If you wanted weight and not mass, just multiply m by 9.81. Make sure you convert all your numbers to the correct units that I specified above (google can easily convert them).

(Disclaimer: I might be wrong on this one, I’m just a student. Also, this is in no way 100% correct, it’s called the ideal gas law for a reason.)
Learn more about the ideal gas law here: 11.9: The Ideal Gas Law: Pressure, Volume, Temperature, and Moles - Chemistry LibreTexts

3 Likes

You can use the Ideal Gas Law formula: PV = nRT

Note that I’m assuming by weight that you actually meant mass, ie the number of grams

Here’s each variable:
P: Pressure (most common units are atm, Pa, and torr)
V: Volume
n: moles of gas
R: gas Constant (The value differs depending on the units for pressure)
T: Temperature (Usually in Kelvin)

Now the units are extremely important, so make sure you keep track of the units for each variable. Now depending on the units you’re using, here’s a handy table to determine which value of R you should use. If you don’t find an exact match with the units you are using, then convert your units to ones easier to work with.

Personally I would convert psi to atmospheric pressure (atm), volume to meters^3 (m^3), and temperature to Kelvin (K). This will make R = 0.08206

So with the values you gave, let’s do some conversions (I’m just using Google’s conversion tool):

P = 100 psi = 6.805 atm
V = 1 cubic foot = 0.0283 m^3
R = 0.08206 (This has some crazy units, don’t worry about it)
T (I’m going to assume room temperature) = 68 F = 293 K

Plugging things into PV = nRT, lets solve for n (the number of moles):
(6.805)(0.0283) = n(0.08206)(293)
n = 0.00801 moles of air

Now we multiply that number with the molar mass of air. A quick Google search tells us that the molar mass of dry air is about 28.97 g/mol.

Mass of air = 0.00801 * 28.97 = 0.232 kg of air or 232 grams

And there you have it!

2 Likes

For the units you’re using,

R = 8.2e-5 atm•m3/(K•mol)

So mass of air will be 232 grams. To be a little nitpicky, OP probably means ‘guage pressure’ of 100 psi, or 7.8 atm actual pressure.

That 232 grams is about 1/2 lb… but don’t panic about your robot weight! one of the typical AndyMark or Clippard tanks are in the 500-600ml range. That’s about 0.02 cubic feet! Even a 1 gallon air tank is only 0.13 cubic feet. So for the vast majority of robots, pressurized air will account for less than 1/10th of a pound.

Am I mistaken in that robots can and should go to weighin depressurized?

They should, yes. Unless otherwise necessary (like putting a mechanism inside your frame perimeter), inspection should happen at the lowest energy state the robot has. And even when necessary, we’ll check everything else first, then have you demonstrate as needed with a fully pressurized system.

But you would be surprised how many teams come up to inspection with a pressurized system!

1 Like

Yes you’re right, thank you for the corrections. I went and changed the original post, but I didn’t add the pressure gauge caveat because I don’t want to redo the calculations

Thank you guys for the info. It’s for a side project not FRC. Thanks so much again!

Can you think of a way to easily increase the mass of a robot and easily remove it as well? Like a submarine where it can add water and remove water to get the desired amount of mass?

I recognize that this isn’t for FRC, but if you could describe what you’re trying to accomplish, it would be much easier for us to provide you with information. Submarines can easily gather mass from their environment by pumping in water. However, air is much less dense than water, and a similar process isn’t really possible with it (as far as I know). Why are you trying to increase the mass of a robot?
The only reason I can think to do this is to increase normal force/traction/friction, there are other ways of doing it such as manipulating air pressure above and below the robot.

1 Like

25 Likes

The world was not ready for dirtman’s out of the box and revolutionary thinking.

9 Likes

The question is: Does the robot need to do this live and on its own, or would humans be able to safely interface with it?

Y’see, if humans can interface with the robot to drop weight, you can’t go wrong with the old steel plate trick. (NOT lead, please… unless you know how to handle it safely.) That’s as easy as a wingnut on a bolt through the frame, potentially.

If not… we’d need to know about the environment you’re operating in. Water, you’ve got a pretty good shot. Sand/loose dirt, you can probably scoop some up (though maybe you want to be careful about how) and dump it again later. Air? See previous calculations.

You can reduce the weight of your robot significantly if you pull a vacuum on the frame… of course I’m assuming that you started with the frame full of water.

2 Likes

If you need something REALLY heavy, go with a chunk of tungsten. Even a cubic centimeter of tungsten weighs over 19 grams, which is over 19 times heavier than water. You should be able to get it in a form (like a small cube) that can easily be picked up by the robot if it needs to be automatic, and if a human needs to add it, that’s pretty easy too.
Just be aware, tungsten is ridiculously heavy and tough so it’s not too easy to machine.

If you’re willing to use exotic metals to shoot for density, why think small? U-238 is where it’s at.

4 Likes

Clearly upward facing solid rocket engines is the right answer.

10 Likes

Manipulate the Higgs field or go home.

5 Likes

Uh… like concrete or something lol