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# how much does air weigh?

how much does 1 cubic foot of air at 1 atmosphere of pressure weigh?

(and for the lmgtfy.com answerers, a google search returned conflicting answers)

49374Views7Replies

how much does 1 cubic foot of air at 1 atmosphere of pressure weigh?

(and for the lmgtfy.com answerers, a google search returned conflicting answers)

Well, at standard temperature and pressure, 1 Atm @ 25 degrees celsius, 1 mole of air or any gas for that matter has a volume of 22.4 L.

So, using the ideal gas law, PV=nRT

Where: P = Pressure (1 atm)

V = 22.4 L

n = The moles of gas which needs to be determined to get the weight.

R = The ideal gas law constant 0.0821 (When working in Atm)

T = The temperature in Kelvin (25 + 273 = 298K)

So, we rearrange the equation to find the moles of gas: n = PV/RT.

n = (1 atm) x (22.4) / (0.0821) x (298K) = 1 Mole of Gas.

But, you want the weight for 1 cubic foot. Turns out that 1 cubic foot converts to 28.32 Liters. But the air we breathe is mainly composed of Nitrogen, Oxygen and Argon with trace amounts of other gases such as methane and helium. More specifically air is 78.03% Nitrogen, 20.99% Oxygen and 0.93% Argon.

Now, we go back to the Ideal Gas Law, but this time we use a volume of 28.32 L instead of 22.4 L. We should get a new value of 1.20 moles of gas.

So, no we take the percent composition of the 3 constituent gases and multiply them by the new value for "n", 1.20.

Nitrogen - .78 x 1.20 = 0.963 Moles

Oxygen - .21 x 1.20 = 0.252 Moles

Argon - .093 x 1.20 = 0.1116 Moles

Now, multiply the Moles of gas obtained by the molar masses of those gases:

1 Mole N2 = 28.02 g

1 Mole O2 = 32.0 g

1 Mole Ar = 39.95 g

0.963 x 28.02 = 26.98 g Nitrogen

0.252 x 32 = 8.064 g Oxygen

0.093 x 39.95 = 3.72 g Argon

Add up the weight and you get 38.76 g. Convert that to pounds and you have 0.0855 Lbs.

I hope this helped, rather than just Googling the answer.

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I actually have to pull S1L3N7 SWAT up there, you've mixed up your conditions.

Your maths was correct, you just mixed up SLC and STP

At Standard Temperature and Pressure, 1 mole does indeed occupy 22.4 L, however STP is 1 atm of pressure at 0 degrees Celsius.

Standard Laboratory Conditions is 1 atm at 25 degrees. And under these conditions the volume of air is 24.5L.

So the volume 22.4L and the temperature 25 degrees are actually incompatible.

Correct answer:

1 cubic foot of air at STP ( 1 atm, 0 degrees celsius) weighs

0.0807 lb

1 cubic foot of air at SLC ( 1 atm at 25 degrees celsius) weighs

0.0739 lb

for everyone outside the US, that's

1 cubic metre at STP

1.293 kg

1 cubic metre at SLC

1.184 kg

So the difference is small enough to be negligible, but I thought I'd just clear that up.

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Assuming the density of room temperature air, at 1 atm is about 1.2 kg/m^3

Multiply this times a volume of 1 ft^3, including an appropriate conversion factor.

1.2 kg /m^3 * (0.3048 m/ ft)^3 * 1 ft^3 = 0.033980 kg

That's the mass of the air in your 1 ft^3 volume. I'm guessing you want an answer for mass in the old American units, like:

0.033980 kg * (2.2 lb/kg) = 0.074756 lb

0.074756 lb * (16 oz/lb) = 1.2 oz

So, are you building a lighter-than-air-craft? A dirigible? Going to be the next balloon boy? No wait, forget I asked that. You're trying to keep things on the DL, strictly need-to-know. I understand completely.

;-)

See also:

http://en.wikipedia.org/wiki/Density_of_air

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nah, just something that popped in my head.

thanks to all.

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I can't help but notice you're going about this all wrong, enough with the mathematical B.S. just use common sense. Air = Weight of air + density of air + H2O and O2. Even though i used somewhat a math formula, mine is much easier and faster. thank you for your time.

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Well, if you're old fashioned, the answer is nothing. As the idea of air was nothingness...

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You need to specify a temperature; since PV=nRT, the density is a function of both pressure and temperature.

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