# Compression of air, and its mass density

1. Nov 8, 2014

### Low-Q

If you have a tank with compressed air. Say this tank is a cube of 1x1x1m. One of the walls are "mobile" but sealed so no air can excape. A spring on the outside is pushing on this wall so it will remain in its position at for example 10 bar pressure inside the tank.
Now, atmospheric air has a density of approx 1.3 kg pr. 1000 litre. If we compress this air to a density of 13 kg pr 1000 litre (Say 10 bar pressure), would the spring push on that wall with 100 000 kg + the extra 13 kg. of air? Or how can we determine the correct pressure the spring must counterforce?

( No, this is not a school project - I'm just curious)

Vidar.

2. Nov 13, 2014

### Bystander

Are your walls and spring massless? What magnitude is your gravitational field? What is the orientation of the moving wall relative to field direction? How are you turning 1.3 kg into 13 kg?

3. Nov 13, 2014

### Simon Bridge

You can work that out with a free body diagram.

4. Nov 13, 2014

### Low-Q

OOPS! Ofcourse it is still 1.3 kg air - blush... I meant 100 000 kg + 1.3kg of air.
Assume we are dealing with massless spring and walls. I'm just interesting in the forces, and the content in the tank. The tank could be made of 10km thick steel walls for that matter.
Gravitational field? I assume the Earth would be the best place to do this experiment. The moving wall moves sideways - if it moves.

Last edited: Nov 13, 2014
5. Nov 13, 2014

### Bystander

Forces you want. Enter the unforgivably sloppy world of "pressure" units in common use, specifically, "kg." Pressure is defined as force per unit area; in SI, Newtons per square meter, or the Pascal. Atmospheric pressure is 1.01x105 Pa, 10 atm. is 1.01 MPa. You're playing with a one square meter piston; that area times pressure gives you a force --- not a mass. That abominable "kg" that shows up on pressure gauges is a "kilogram force" (kg times earth surface gravity, 9.8 m/s2) per cm2(?). Is there a pressure gradient inside the tank? Only if it's in a gravitational field --- pressure at the bottom is going to be greater than pressure at the top.

As SB has suggested, time to play with the free body diagram.

6. Nov 14, 2014

### Low-Q

I know all this. Your reply appears to be more like a cavil one, than trying to understand what I mean. Maybe you are so good at physics that you cant "see" what people mean if they don't use the correct academic language, hihi ;-)

The pressure on the bottom is greater than the top. When I think of this again, the tank is still 1x1x1m after compressing the air to 10 bar and will increase the air mass inside the tank to about 10 times greater - 13 kg of air pr. 1x1x1m, I would assume. That is how I turn 1.3kg air into 13 kg. of air - by taking air from outside the tank with a compressor.

7. Nov 14, 2014

### Bystander

You start with a movable wall, 1m3 volume, 1 atm. pressure, then increase pressure to 10 atm. without moving the wall, and want to know what force or forces are acting on the wall? If it doesn't move, there is zero net force. At 1 atm., 1.01x105 Pa x 1m2 pushing the wall out of the box and 1.01x105 N pushing the wall into the box; at 10 atm., 1.01 Pa x 1m2 out and 1.01 x 106 N pushing in. If you want to apply the inward force with a spring and prevent movement of the wall, the spring has an infinite Hooke's Law constant; i.e., you have a "fixed" wall rather than movable, and "a rose by any other name is still a rose."

8. Nov 14, 2014

### Low-Q

I talk about a wall that CAN move freely if the spring and pressure wasn't there at all. I know that the wall stands still if the counterforce from the spring is equal to the force from the pressurized air inside. The total force acting on the wall is therfor zero (Force equals counterforce), but the force from the spring alone is what I am asking for.
Let us boil this down to a chamber that already has compressed air inside, where the (possible) movable wall is held in place by a spring (So the wall doesn't fly away like a projectile).
Anyways, if I place a scale between the spring and the wall. What will the scale display in kg? (Yes, it is a very very solid weighing scale)
Will it read if I have 10 bar pressurized air while the "movable" wall is one square meter in area? The wall is vertical, the spring is pushing back on the wall horizontally. The scale is placed between the wall and the spring.

Vidar

9. Nov 14, 2014

### Fredrik

Staff Emeritus
If the movable wall is not moving, the forces on it add up to zero. If the pressure is 10 bar = 105 Pa = 105 N/m2 and the area is 1 m2, the force that the air exerts on the wall is 105 N. So the force that the spring exerts on the wall is 105 N, in the opposite direction.

If you insert a scale between the wall and the spring (calibrated to display zero when it's turned sideways), it will display the mass m (in kilograms) that solves the equation F=mg, because a scale is the same type of device that measures force; it's just designed so that the number displayed is the force in newtons divided by g in meters per second squared. So the force displayed is (105 N)/(9.81 m/s2) =105/9.81 ((kg m)/s2) / (m/s2) = 105/9.81 kg ≈ 1.02×104 kg.

However, if you now pump air into the box while holding the wall fixed by other means until you reach ten times the pressure and therefore ten times the force, and then release the wall so that the spring is only pushing it back, the total force on the wall will be 106-105 N=9×105 N. So it will have an acceleration of that number divided by the mass of the wall in kilograms. This is assuming that there's no friction and that the mass of the spring (and the scale) can be neglected.

Unfortunately I'm having a bit of a brain malfunction at the moment, so I don't see how to handle a scale that's being squeezed by two forces with different magnitudes. I'll take another look at this later, to see if I can figure it out.

Last edited: Nov 14, 2014
10. Nov 15, 2014

### Low-Q

Thanks! I don't need to squeeze more air into that tank. 10 bar will do, and thanks for the explanation. I might seem to overcomplicate things here, while it should be easy (after some thoughts) to puzzle this out by myself :-))

Vidar