Weight of Compressed Air

[TRIAD]

It is unclear to me how a gas with a higher molecular weight would benefit the propulsion of a paintball. At any given temperature, all substances have the same average KE (temperature being a relative measure of the average kinetic energy of a substance). Thus, while one gas molecule might be more massive (let's compare a molecule of Ar to a molecule of N2), the one that is more massive will also be traveling at a lower velocity. (KE=(1/2)M Vsquared).
So, a mole of Ar gas and a mole of N2 gas will have (essentially) the same kinetic energy (considering they have the same average kinetic energy, and you have a specific amount of each). When you fill a tank wwith Ar, since Ar has a molecular mass of 18, the molecules have, on average, a higher velocity than molecules of molecular nitrogen (molecular mass of 28.0134). Molecular weight is therefore a moot point when dealing with the propulsion of a paintball (as far as I can see), because the average KE of each molecule of gas is still the same.

oh yeah, and for temperature to pressure comparisons, you can use the formula:

P/T=P/T

T is in K
P units must be the same on both sides

*Finally, since this is Deep Blue and we're getting all technical here, K is an absolute measurement, and thus has no units. It is not degrees K, it is just K. There is no unit used with K. Also, you want to use the term mass and not weight.

[athomas]

Its not molecular weight, its molecular density that is key to which gases can produce more propulsion.

The density dictates how many of the molecules actually affect an area at a given instant.

That's why helium does not work well as a propellant. It displaces a lot of area and produces a high pressure with a lot less molecules.

[TRIAD]

I do not believe there is a such term as "molecular density". Perhaps you meant "particle density", which I do not know if that's a real term either, but we'll use it.

The "particle density" of all gases is essentially the same. This means that no matter what gas you're dealing with, given the same size container and the same temperature and pressure, you will have the same number of particles of gas. This is why a conversion factor of 22.4 liters of gas to one mole of gas (assuming room temperature and atmospheric pressure) can be employed. The distance between the molecules of gas is so great relative to the size of the individual particles that particle size becomes negligible. So, whether I have a liter of N2 or a liter of F or a liter of Ar, the number of particles in the same size container at the same temperature and pressure will ALWAYS be the same.

Helium should, in theory, work just as well as any other propellant. However, I am ignorant of the physics involved in paintball propulsion, and am analyzing this from a strictly chemistry-oriented view. Helium is a relatively small molecule, so that may be why it does not work very well as a propellant.

[athomas]

Originally posted by TRIAD
I do not believe there is a such term as "molecular density". Perhaps you meant "particle density", which I do not know if that's a real term either, but we'll use it.

I did indeed explain this wrong.

Yes, I believe I used the wrong term here. The correct terms I believe are gas density and molar mass.

I'm gonna stop because I'm just messing myself up here. Once I get it totally straight I'll try again.

[bjjb99]

Originally posted by TRIAD
It is unclear to me how a gas with a higher molecular weight would benefit the propulsion of a paintball. At any given temperature, all substances have the same average KE (temperature being a relative measure of the average kinetic energy of a substance). Thus, while one gas molecule might be more massive (let's compare a molecule of Ar to a molecule of N2), the one that is more massive will also be traveling at a lower velocity. (KE=(1/2)M Vsquared).

The mean kinetic energy of a gas molecule is related to its temperature by the following formula:

KE = (3/2) * k * T, where
k is Boltzmann's Constant (1.38x10^-23 J/K), and
T is temperature.

The mean kinetic energy of a gas molecule is also related to its velocity by the following formula:

KE = (1/2) * m * v^2, where
m is the mass of the molecule, and
v is the molecule's velocity.

We can combine these two equations and solve for velocity as a function of temperature and molecular mass:

v = sqrt(3 * k * T / m)

At room temperature (300K), the average velocity for a molecule of N2 is right around 517 meters per second. For CO2, it is around 412 meters per second.

We cannot talk solely in terms of energy transfer when considering the acceleration of a paintball. Since the molecule-paintball collisions are not elastic, kinetic energy is not necessarily conserved. Momentum, however, is conserved. If you look at things from a momentum standpoint, you can see how a heavier gas could accelerate a paintball more easily than a lighter gas at the same temperature.

Momentum is simply the product of an object's mass and velocity.

P = m * v

A molecule of N2 at 300K has a momentum of approximately 2.4x10^-23 kg-m/s.
A molecule of CO2 at the same temperature has a momentum of approximately 3.0x10^-23 kg-m/s.

The heavier CO2 molecule has more momentum to transfer when colliding with the paintball. Since we're apparently dealing with the same number of molecules in each case, a paintball fired using a heavier gas should accelerate faster than a paintball fired using the same number of molecules of a lighter gas.

Just my guess on how things work.

BJJB

[athomas]

Well said.

[TRIAD]

a=f/m

where a is acceleration
f is force
and m is mass (of the thing being accelerated, in this case the paintball)
m will not change, because the mass of the paintball is a constant

P=F/A

A is area (of the thing being accelerated, in this case the paintball)
A will not change because the area of the paintball is constant

Therefore, the force applied to paintball is directly proportion to the pressure of the propellant. The type of propellant is a moot point.

[bjjb99]

Originally posted by TRIAD
a=f/m

where a is acceleration
f is force
and m is mass (of the thing being accelerated, in this case the paintball)
m will not change, because the mass of the paintball is a constant

P=F/A

A is area (of the thing being accelerated, in this case the paintball)
A will not change because the area of the paintball is constant

Therefore, the force applied to paintball is directly proportion to the pressure of the propellant. The type of propellant is a moot point.

It's moot until you start considering that no gas is truly ideal, which is now where I think the CO2-vs-N2 effects will be seen. For example, a cubic meter of CO2 at 300K and 1 atmosphere (101325 Pa) contains 40.796 moles of gas. The same volume, pressure, and temperature for N2 contains 40.651 moles. The rates at which pressure changes as a function of expanding volume are different for these two gases once Van der Waals effects are taken into account.

Start with a fixed initial volume V of gas at a temperature of 300K and a pressure P. Compute the number of moles of CO2 and N2 that would be contained in that volume under those conditions, taking Van der Waals effects into account. Then use the same Van der Waals equation to compute P as a function of V at 300K for the CO2 and N2 mole amounts you obtained. The P-vs-V curves are not identical.

While this is not necessarily where my original post was headed, your reply made me think a bit more and I stumbled onto the possibility that these gases were not as ideal as I originally presumed them to be.

BJJB

[TRIAD]

Most gases behave close to ideal. The only time they do not behave in an ideal fashion is at extremely low temperatures (the average velocity of the gases is such that they exhibit a lower pressure than would normally be seen. The reason for this is that the IMFA of the gases is effective enough that when they interact with each other, the molecules of gas travel in slightly curved paths, and thus they do not hit the sides of the container as often as they would at higher temperatures, or if the gases were ideal.) They also don't behave in an ideal fashion at extremely high pressures. The reason for this is that the molecules of gas actually take up space, as opposed to ideal behavior. However, we are not dealing with extremely low temperatures (near O K). I will find the equation for calculating the effect of 4500 psi on the volume of space taken up by the gas, and I should be able to calculate it.

However, our subject has changed to not whether a certain gas provides better propulsion (after talking to my physics teacher, I now know that the type of gas does not matter) but whether there is MORE of a gas in a certain volume. I do not believe that 4500 psi is as high enough pressure to affect the volume, though I am not sure, and will check. (I also believe my physics teacher would've mentioned that, but I'll talk to my chem teacher to get the final word).

"It's moot until you start considering that no gas is truly ideal, which is now where I think the CO2-vs-N2 effects will be seen. For example, a cubic meter of CO2 at 300K and 1 atmosphere (101325 Pa) contains 40.796 moles of gas. The same volume, pressure, and temperature for N2 contains 40.651 moles. The rates at which pressure changes as a function of expanding volume are different for these two gases once Van der Waals effects are taken into account."

I do not know where you are getting these mole amounts. I'll have to check the deviations as I said.

[bjjb99]

Yes, most gases behave close to ideal at normal temperatures and pressures. However, we're looking at small differences in paintball velocities, and the devil's in those details. A first-order ideal gas treatment may not be sufficient to cover the phenomenon in question.

And yes, we are looking at whether there is more of a given gas contained in a given volume at a given pressure and temperature... when that different amount of gas is released, it's going to provide a different pressure curve behind the paintball and thus different paintball muzzle velocities.

Let's take a look at some gas behavior if we include Van der Waals effects.

The Van der Waals equation:

[P + a * (n/V)^2 ] * (V - n * b) = n * R * T

P in pascals, V in m^3, T in K, n in moles, R in J/mol-K, a and b in appropriate units (listed further down).

I'm going to use a hypothetical paintgun based (loosely) off an automag.

The dump chamber volume is 0.55 cubic inches (9.01x10^-6 cubic meters).

The dump chamber pressure is 400 psig (2.76x10^6 Pascals)

I'm gonig to assume a constant temperature of 300K throughout the decompression process to keep things simple.

A bit of algebra transforms the Van der Waals equation into the following form:

(a * b / V^2) * n^3 - (a / V) n^2 + (P * b - R * T) * n - (P * V) = 0.

The constants a and b are as follows for CO2 and N2:

CO2: a = 0.3643 J-m^3/mole
CO2: b = 4.27x10^-5 m^3/mole

N2: a = 0.1361 J-m^3/mole
N2: b = 3.85x10^-5 m^3/mole

[ constants obtained from http://www.ac.wwu.edu/~vawter/Physic...atOfState.html ]

Plugging in these values along with P, V, and T yields the following results for n:

CO2: n = 0.01142 moles
N2: n = 0.01013 moles

At the same 400 psig pressure, 0.55 ci volume, and 300K temperature, there are almost 13% more molecules of CO2 than there are of N2. An ideal gas at these same conditions should amount to 0.00996 moles.

Now let's start with those values for n and see how the pressure varies as a function of increasing volume. Starting at 0.55 ci and increasing the volume to encompass a 12 inch long, 0.68 inch diameter barrel.

V = 0.55 ci --> 4.91 ci

Solving the Van der Waals equation for P gives us:

P = [ (n*R*T) / (V - n*b) - (a * (n/V)^2) ]

Evaluating this equation at V = 0.55, 1, 2, 3, 4, and 4.91 ci for both CO2 and N2 resulted in the following:

CO2:
V = 0.55 ci, P = 400 psig
V = 1.00 ci, P = 234 psig
V = 2.00 ci, P = 122 psig
V = 3.00 ci, P = 82 psig
V = 4.00 ci, P = 62 psig
V = 5.00 ci, P = 51 psig

N2:
V = 0.55 ci, P = 400 psig
V = 1.00 ci, P = 222 psig
V = 2.00 ci, P = 111 psig
V = 3.00 ci, P = 74 psig
V = 4.00 ci, P = 56 psig
V = 5.00 ci, P = 45 psig

The pressure of N2 drops off faster. This shouldn't be surprising since there were fewer N2 molecules in the dump chamber to begin with, even though both dump chambers had the same initial 400 psig pressure.

If we assume constant temperature, the CO2 maintains a higher pressure as the gas expands. This would make a difference in how the paintball is accelerated.

Of course, temperature changes will also factor into the rapidly expanding gas as it propels the paintball out of the barrel. A full numerical treatment of P, V, and T for CO2 and N2 as they expand to push a paintball along is a bit more than I want to do.

To first-order, gas type does not matter. Once you start to factor in second order effects, you begin to see non-neglegible differences in gas behavior (and subsequently, paintball behavior).

BJJB

[RobAGD]

I was noodling around after a customer asked whould be get more shots from a 4500 91 or a 88 5000.

I didnt quite know off hand, did some quick math and said that the 5k would be the better deal for him.

So I started screwing around and made this.



the argon numbers are not 100% and are not indicitive of a good comparsion between Air and Argon.

-Robert