How much does 300 cc of air compressed to 230 bar at sea level at 24 degrees C weigh?
How much?
- Thread starter AnimalHitman
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Vapor density of 1, several correct answers. Like most things in life, not as simple as it may seem!!
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Endevour, obviously compressed air has more mass than the same volume of uncompressed air, ceteris paribus. That is high school physics class stuff pursuant to the ideal gas law. The issue is whether a filled gun bottle per the specifications I detailed adds significant weight to a gun vis-à-vis an unfilled bottle.
If I used the right units of measure and formula, I calculated the mass of the compressed air pursuant to my previous specifications as 80.942 grams. The mass of 300 cc of uncompressed air at sea level @ 75 degrees F is 36.605 grams. This means that a full bottle adds only 1.564 oz to the weight of a gun at sea level at 75 degrees F on Earth vis-à-vis an unfilled bottle, ceteris paribus. Does this sound right?
These are the kinds of things I think about when I'm up way beyond my bedtime watching old movies!
If I used the right units of measure and formula, I calculated the mass of the compressed air pursuant to my previous specifications as 80.942 grams. The mass of 300 cc of uncompressed air at sea level @ 75 degrees F is 36.605 grams. This means that a full bottle adds only 1.564 oz to the weight of a gun at sea level at 75 degrees F on Earth vis-à-vis an unfilled bottle, ceteris paribus. Does this sound right?
These are the kinds of things I think about when I'm up way beyond my bedtime watching old movies!
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Okay, here is the answer and why. A cubic meter of air at regular atmospheric pressure weighs roughly 1.2 kilograms, or 1200 grams.
A liter is 1/1000 of a the volume of a cubic meter. So a Liter of air (1000 CC) is 1200 grams divided by 1000, or, 1.2 grams of weight.
However, we are only talking about 300 CC, which is 300/1000 of the weight of a liter, so 300/1000 times 1.2 grams. Which is 0.36 Grams.
The 300 CC air tank is filled at 230 bars, which means 230 atmospheric volumes of air, or 230 times atmospheric pressure (roughly 14.7 PSI at sea level).
Thus, we take our 0.36 grams and multiply it by 230 to get a total weight of 82.8 grams of "MASS" of air. However, last calculation.....
As we are not in outer space, the air tank is not in a vacuum. It is on the ground where there is air all around it. The air around it offers buoyancy, which cancels out the "weight" of the one atmosphere of air in the container. This reduces the "measured weight on earth in an atmosphere by 0.36 grams".
Final answer 82.8 grams minus 0.36 grams...… equals 82.44 grams or roughly 2.9 ounces of weight.
Hope this clarified things.
Addertooth
A liter is 1/1000 of a the volume of a cubic meter. So a Liter of air (1000 CC) is 1200 grams divided by 1000, or, 1.2 grams of weight.
However, we are only talking about 300 CC, which is 300/1000 of the weight of a liter, so 300/1000 times 1.2 grams. Which is 0.36 Grams.
The 300 CC air tank is filled at 230 bars, which means 230 atmospheric volumes of air, or 230 times atmospheric pressure (roughly 14.7 PSI at sea level).
Thus, we take our 0.36 grams and multiply it by 230 to get a total weight of 82.8 grams of "MASS" of air. However, last calculation.....
As we are not in outer space, the air tank is not in a vacuum. It is on the ground where there is air all around it. The air around it offers buoyancy, which cancels out the "weight" of the one atmosphere of air in the container. This reduces the "measured weight on earth in an atmosphere by 0.36 grams".
Final answer 82.8 grams minus 0.36 grams...… equals 82.44 grams or roughly 2.9 ounces of weight.
Hope this clarified things.
Addertooth
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Thanks, adder; I got virtually the same result! I used the formula:
m = (pVM) / (RT)
where;
m = mass in grams
p = pressure in pascals = 2.3e+7
V = volume in cubic meters = 0.0003
M = molar mass of air at sea level in g/mol = 28.97
R = 8.314 (constant)
T = temperature in Kelvin = 297.039
Pluggun' 'n chuggin', I got:
(23,000,000 * 0.0003 * 28.97) / (8.314 * 297.039) = 80.942 g = 2.855 oz
It's interesting that in a 300 cc bottle, air compressed to 230 times its pressure at sea level weighs only approx. twice as much! Am I correct in assuming that as the capacity of the bottle increases, ceteris paribus, the rate that the relative weight increases is not linear?
The issue I had was getting all the inputs converted to the same system of measure (in this case, metric) to plug 'n chug. I guess I did it!
Your explained why the formula works. Thanks buddy!
m = (pVM) / (RT)
where;
m = mass in grams
p = pressure in pascals = 2.3e+7
V = volume in cubic meters = 0.0003
M = molar mass of air at sea level in g/mol = 28.97
R = 8.314 (constant)
T = temperature in Kelvin = 297.039
Pluggun' 'n chuggin', I got:
(23,000,000 * 0.0003 * 28.97) / (8.314 * 297.039) = 80.942 g = 2.855 oz
It's interesting that in a 300 cc bottle, air compressed to 230 times its pressure at sea level weighs only approx. twice as much! Am I correct in assuming that as the capacity of the bottle increases, ceteris paribus, the rate that the relative weight increases is not linear?
The issue I had was getting all the inputs converted to the same system of measure (in this case, metric) to plug 'n chug. I guess I did it!
Your explained why the formula works. Thanks buddy!
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Yep, you are there AnimalHitman. Small variations due to how far you carry it past the decimal point and the constants used makes up our very small differences.
Especially when you throw in variations in density due to the temperature, hotter air is less dense than cooler air at sea level.
Especially when you throw in variations in density due to the temperature, hotter air is less dense than cooler air at sea level.
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Thanks, adder; it's been a while since introductory physics! This is one of the things I appreciate about airguns; you don't have to dive too deep into advanced physics due to the low-velocity and short-range ballistics. The considerations start to become material only when you are dealing with supersonic velocities over long distances.
I just wanted to get an accurate weighing of my rifle with all the do-dads I added, and my bottle was only partially filled. I should have known the weight difference would be de minimis. In this case, only 0.594 oz. (My bottle is 62% full,) Christ, the rabbit holes I go down. All this because I was too lazy to break out the hand-pump and then weigh her with a full bottle. I expended twice the blood and treasure going the scientific route!
I just wanted to get an accurate weighing of my rifle with all the do-dads I added, and my bottle was only partially filled. I should have known the weight difference would be de minimis. In this case, only 0.594 oz. (My bottle is 62% full,) Christ, the rabbit holes I go down. All this because I was too lazy to break out the hand-pump and then weigh her with a full bottle. I expended twice the blood and treasure going the scientific route!
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I can tell, adder. I barely made it through physics in college. That's why I abandoned electrical engineering and went to law school!
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Hynzie, my man; I can't believe you chimed in! I wonder if Tommy Boy uses the ideal gas law when he fills footballs?
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