## Air Pressure Question

Peeno
Posts: 2
Joined: Fri May 08, 2009 3:49 am

### Air Pressure Question

How much air pressure (psi) is created when you put two liters of air in a one liter container? And how much is created when you put three in the same one liter container? Is the resulting difference in air pressure linear or exponential?

Silence
Posts: 3825
Joined: Sun Apr 09, 2006 9:01 pm

### Re: Air Pressure Question

Welcome to the forums! The ideal gas law gives us a couple formulas relating pressure and amount (or moles, represented by n) of gas, or pressure and the container's volume, given a constant R.

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``````  P * V = n * R * T
P * V = k       (Boyle's law;
rolling the constants n, R, and T into the constant k)
P1 * V1 = k = P2 * V2
P1 * V1 = P2 * V2``````
One thing to note is that at ground level, air is 15 PSI. However, what we can measure is gauge pressure, the difference between the pressure in a container and the outside pressure. Generally when we say a water gun holds "50 PSI," we're talking gauge pressure, so that's really 50 PSIg or 50+15 = 65 PSI.

Also, when I say "doubling the volume of air," I'm doubling n (the moles of gas) and not V (the volume of the container).

An "empty" tank is equalized with the atmosphere to 15 PSI but only 15−15 = 0 PSIg, so it won't fire at all. Twice the volume of air gives it 30 PSI and 30−15 = 15 PSIg. Three times the volume gives 45 PSI and 45−15 = 30 PSIg.

The correlation between actual pressure and amount of gas is linear. Let's go back to the ideal gas law, but set the container volume and temperature constant.

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``P = n * k         (rolling the constants V, R, and T into the constant k)``
But remember you have to subtract 15 PSI to get the gauge pressure, which is what we care about.
Last edited by Silence on Sat May 09, 2009 12:37 am, edited 1 time in total.

SSCBen
Posts: 6449
Joined: Sat Mar 22, 2003 1:00 pm

### Re: Air Pressure Question

Well put Silence.

It'd be best to use the mass based ideal gas law because it's easier to work with in my experience. This can be ignored if you don't care, but I think it's worth mentioning.

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``P * V = n * R_bar * T	molar ideal gas law typically taught in high school``
We work in terms of mass and not moles so let's convert this to work with mass.

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``````n = m / M	The number of moles equals the mass divided by the molar mass

R = R_bar / M

--> P * V = m * R_bar * T / M

--> P * V = m * R * T``````
Note that R_bar is the constant here. This is the same constant as what Silence used as it's the same equation, however, the notation is changed. Typically R is the specific gas constant, which is R_bar / M where M is the molar mass of the gas. This form I find is much more convenient.

R for air is about 640.12 in / R where R is degrees Rankine (Fahrenheit plus 459.67).

Peeno
Posts: 2
Joined: Fri May 08, 2009 3:49 am

### Re: Air Pressure Question

Its very early in the morning, I just got off work. So, my brain isn't functioning at 100% right now. But I'll really read into this later. But from what it seems, this will go a long way to helping my design.

Thank you both.
Last edited by Peeno on Sat May 09, 2009 10:33 am, edited 1 time in total.

cantab
Posts: 1492
Joined: Fri Oct 19, 2007 1:35 pm

### Re: Air Pressure Question

Well - for instantaneous pressurisation you should use isentropic formulae. T will increase when you compress the gas.

Then the system comes to thermal equilibrium, and P tends to the value expected from isothermal compression. How long that takes depends on the thermal properties of the container.

According to Wikipedia,

P2/P1 = (V1/V2)^y

T2/T2 = (V1/V2)^(y-1)

Where P2, V2, T2 are final values and P1, V1, T1 are initial values. Pressures and temperatures should be absolute.
y is the ratio of specific heat capacities, Cp/Cv where Cp is at constant pressure and Cv at constant volume. For air, y = 1.40

Thus, in the original question:

V2 = 1
V1 = 2

P2/P1 = 2.64
P2 = 264 kPa
That's the absolute pressure. For gauge pressure, we subtract 100 kPa.
P2g = 164 kPa

Note that this is 64% more than the isothermal value.

For the temperature increase,
T1 = 298K
V1/V2 = 2
then T2 = 393 K

?!

That seems wrong...I've never had a water gun get that hot. Maybe the thermal time constant is very short. That will depend on the geometry of the container and the thickness and material of the walls.
Last edited by cantab on Sat May 09, 2009 11:51 am, edited 1 time in total.
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dutchguy
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Joined: Fri May 01, 2009 7:20 pm

### Re: Air Pressure Question

WOW
I thought i was good at Math (i am the best of class in it) but..... cantab
Posts: 1492
Joined: Fri Oct 19, 2007 1:35 pm

### Re: Air Pressure Question

Ah. I know what I did wrong. The above applies for air in a well-insulated container. But in a water gun we don't have that - we have air and water in a container. So heat can transfer from the air into the water. The specific heat capacity per unit volume for water is a few orders of magnitude higher than that for air (simply because water is a thousand times as dense as air). Thus, when the air and water thermally equilibrate, the water will have warmed by a fraction of a degree.

Therefore, isothermal compression, as described by Silence and Ben, is what matters for water guns.

Isentropic compression could be relevant for WBLs though.
I work on Windows. My toolbox is Linux.
Arsenal:
Super Soaker: XP215, 2xXP220, Liquidator, Aquashock Secret Strike M(odded), Arctic Blast M, CPS1200, CPS2100, SC Power Pak, 3l aquapack, 1.5l aquapack
Water Warriors: Jet, Sting Ray M, Shark, Argon M, Tiger Shark, PulseMaster
Others: Waterbolt, The Blaster, Storm 500, Shield Blaster 2000, generic PR gun, generic backpack piston pumper (broken), 3l garden sprayer M, 10l water carrier:

SSCBen
Posts: 6449
Joined: Sat Mar 22, 2003 1:00 pm

### Re: Air Pressure Question

You're right cantab. Isothermal assumptions make sense for water guns unless you want to take into account weird problems with very high water flow (and I don't).

I neglect heat transfer in my pneumatic gun simulation, but that's because the time's on the order of 10^-2 s. So the adiabatic (not isentropic as friction is involved) assumption is good for WBLs.

Silence
Posts: 3825
Joined: Sun Apr 09, 2006 9:01 pm

### Re: Air Pressure Question

@dutchguy: Then you'll probably get here at some point. By the time you take calculus and college chemistry/thermochemistry/physics this should be cake.

Ben, I'm got tot he ideal gas law from a chemistry background...it seems like mass-based would work well for kinetics, so I'll go with your equation.