# Them peskie PSI's

Discussion in 'Question of the Day' started by ship, Dec 29, 2005.

1. ### shipSenior Team EmeritusPremium Member

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So in an air tank it's normally at about 120 or more PSI per square inch worth of air pressure.

In lifting my portable compressor and it's storage tank from my car's trunk, I got to thinking. Would they be any more heavy if they were fully pressurized with 15 gallons of 120PSI air? While air for the most part is weightless, under pressure and under sufficifient capacity, would it not add up to something?

2. ### sound_nerdActive Member

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Off the top of my head, I'd say no it doesn't add weight at all. But now you've got me thinking...

3. ### JahJahwarriorActive Member

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you could always weight it with and weigh it without and see...I know there are ways to calculate, but I really don't enjoy doing math at all.

4. ### shipSenior Team EmeritusPremium Member

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could... but than where would the challenge be?

5. ### MayhemSenior Team EmeritusPremium Member

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Well then you would have to take into consideration the small amount of water that is produced when compressing air!

6. ### ricc0lukeActive Member

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yes, it would be heavier full of compressed air. what i don't know is if the air would be heavy enough to notice. Air is gas, and gas has mass... Think of scuba tanks... definatly heavier when full of compressed air (well... its a very carefully controlled mixture but still basicly air). I just don't know if 15gal of 120psi is enough to feel a huge difference.... try it!

7. ### ScooterMember

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on my farm we have a portable air tank for taking out into the field to air up tires and such.

it might just be a mental thing but i think it feels a little heavier when it is full but i dont have any scientific evidence to prove it.

8. ### MayhemSenior Team EmeritusPremium Member

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Well you are correct as an empty tank will still have air in it, when it is full, it will have a lot more air in it.

Given that the temperature remains the same, Boyles law states that P<sub>1</sub>V<sub>1</sub> = P<sub>2</sub>V<sub>2</sub>

You can use this to figure out the volume of gas at 120 PSI and at 0 PSI (lets assume that this is the pressure in the empty tank).

Given the volumes, you can then figure out the composition of air (just the main 3 should do), which you can then look up the atomic mass of each.

But hang on a minute! How do you know how many atoms of each are present in the volumes provided?

This is where Avogadro's number N(a) = 6.02 x 10<sup>23</sup> comes into play. You can calculate the number of moles of each gas present in an empty tank (yes there is gas in there still) and when the tank is full, as we know that the molar volume of any gas is 22.4L at S.T.P.

From here you can then figure out the number of atoms of each element and thus figure out the mass of each.

All this to tell ship that he is getting on in years and everything feels heavier at the end of a long day!

Anyone care to take up the chemistry challenge?

9. ### ScooterMember

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ok, gimme a minute, i'm gonna figure this out

10. ### ScooterMember

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well i work on it reall hard and for now im kinda stumped, but i'll come back to it later, maybe i'll get extra credit in chemistry to?

11. ### kingfisher1Active Member

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this is the only chemistry that I think is interesting....

http://www.uky.edu/Projects/Chemcomics/

(actually that's a lie, i like chem. and physics and stuff, but in the intrest of comedy....)

12. ### shipSenior Team EmeritusPremium Member

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So in getting old, should I wish to lessen my lifting load, I should provide a negative pressure within the tank correct?

13. ### MayhemSenior Team EmeritusPremium Member

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In theory!

Over the next few days I might see if I have the time to check that my initial thinking is correct and see if I can give you a total mass for 15 gal of air at 120 PSI. Just for the hell of it as I have not really done anything like this for some years now. I may even be way off the mark with my initial thinking.

14. ### shipSenior Team EmeritusPremium Member

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Sometimes the most stupid of momentary of thoughts....

15. ### jwl868Active Member

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A simple approach is with the ideal gas law.

PV = nRT

P is pressure, n absolute terms (that, a perfect vacuum is 0 psi)
V is volume
n is number of moles of gas (a mole is a fixed number of atoms or molecules of a substance, moles therefore represent mass)
R is the ideal gas constant
T is temperature, absolute

Because the volume of the container is constant, V is a constant. For all practical purposes, the temperature of the low pressure tank is the same as the high pressure tank and can be considered a constant.

In the problem, (Is a pressurized tank heavier than a low pressure tank?), pressure goes up compared to the low pressure tank.

Now re-arrange the equation ideal gas law equation so the constants are on one side:

V/RT = n/P

Label pressurized case “1” and low pressure case “2”,

V/RT = n1/P1 = n2/P2

n1/P1 = n2/P2

P1 is greater than P2, so to balance the equation, n1 must be greater than n2.

(Or re-arrange again:

P1/P2 = n1/n2

P1/P2 is greater than 1, therefore, n1/n2 is greater than 1.)

More mass is added in the pressurized case, so the pressurized case is heavier. (How much heavier is a function of the tank volume.)

The weight proportion of gas in a 150 psig (psi gauge) to the empty tank (0 psig)? Convert the pressure to absolute by adding atmospheric pressure, about 14.7 psia. (“a” for absolute) (A “gauge” pressure is usually relative to atmospheric pressure. 0 psig then on a tank means nothing will leak out or in.) I’m going to cheat a little and round 14.7 psia up to 15 psia, so the ratio is:

(150 + 15)/(0+15) = 165/15 = 11

So there is 11 times the mass of gas in the pressurized tank than in the empty tank.

Now say the tank volume is 1 cubic foot, and typical air density at atmospheric conditions and typical temperature is about 0.075 pound per cubic feet,

The weight of air in the low pressure (0 psig) tank is 0.075 pounds, and the weight of air in the pressurized tank weighs about 0.075 x 11 = 0.825 pounds.

[That's not quite ship's example, but it can be seen that it's not much added weight. ship's tank at the example pressure would be about twice the weight - 1.6 pounds.]

Joe

16. ### ScooterMember

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well there you go, now i dont have to pick my brain to figure it out, you already did it for me

17. ### MayhemSenior Team EmeritusPremium Member

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Well done Joe.

I didn't know the typical air density at atmospheric conditions and typical temperature is about 0.075 pound per cubic feet. This is why I was venturing down the mole mass path.

18. ### shipSenior Team EmeritusPremium Member

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Very interesting. Also taking this, were there a negative pressure, of 120# it would be about 1.6# lighter.

19. ### jwl868Active Member

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There is no negative 120 psi (at least not using any reference points of I know of.) Pressure is typically measured relative to atmospheric pressure (usually with the word "gauge" included in the measurement) or to a perfect vacuum (with the word "absolute" in the measurement).

A typical pressure gauge shows the tank/line pressure relative to the atmospheric pressure. Thus, at 0 psig, the tank/line pressure is the same as atmospheric.

For pressures less than atmospheric (that is vacuum), you need either a separate gauge to read it, or a special gauge that reads both positive and negative pressures. Because atmospheric pressure is typically 14.7 psi over a perfect vacuum, the lowest you can go into negative, relative to atmospheric is –14.7 psi. Or the term 14.7 psi vacuum. Typically, vacuum pressures are given in inches of mercury, but I don’t know how this convention came about. (I don’t know the conversion from psi to inches of mercury off the top of my head.)

For many engineering and chemical applications, pressure is referenced to perfect vacuum of 0 psi. I believe a pressure gauge that references to absolute will have “psia” or “absolute” printed on the face.

[scooter – I apologize for answering the question so quickly...]

[Mayhem – the density I cited is for certain “standard” conditions, which in this case I believe are 68 degrees F, 14.7 psia, and 60 % relative humidity. There are a couple of different “standard” conditions, depending on the type of science and technology involved.]

Joe

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