how much does air weigh?

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The ideal gas law is PV = nRT

Lets assume g is the mass of the air we are calculating.
M is the molecular weight of air = (0.8 x 28) + (0.2x32) = 28.8 moles / liter
where 0.8 is the amount of nitrogen, and 0.2 is the amount of oxygen.
n is the number of moles of air.
Where R = 0.082, the gas constant.
Substituting: Pressure = P = 1 atm
Volume in liter = V
Temperature in Kelvin (Temp in Celcius plus 273) = 298 (at 25 deg celcius)
Mass of air per liter = M = 28.8 per liter at 25 degree celcius


P x V = n x R x T = g / M x (R x T) or g = (M x P x V) / (R x T)

Plugging them in, g will be 1.2 gram of air per liter at 1 atm, at 25 degree celcius, without any co2 or water vapor.

Now if you want to add water vapor, carbon dioxide, etc... You have to calculate the molecular weight of these molecules, and fraction....

Then you can adjust the pressure, to 3000 psi, by converting psi to ATM.

Then you have to convert Fahrenheit to Celcius, etc...
 
Based on the fact that 1 cu ft of air has 28.3168466 liters , my calculation for the weight of a cu ft of air is 33.98 grams.
 
fisherdvm:
No, it is 0.074756 lbs.


at what temperature? and altitude?
 
At 25 degree celcius and 1 atm or sea level ...

And at a bad *** attitude....
 
ah, see my room was 23 degrees celcius
 
fisherdvm:
... (0.8 x 28) + (0.2x32) = 28.8 moles / liter ...
Excuse me????
I do believe you meant 28.8 grams/mole.
At STP (physics) 1 mole of gas occupies 22.4 liters, or about .045 moles/liter.
What's amazing is that in the end you managed to arrange the numbers to come up with close to the right answer!
The "real" math using moles follows:
.79X28 + .21X32 = 28.84 g/mole
28.84g/mole / 22.4 liters/mole = 1.2875 g/liter
1.2875 g/liter X 28.32 liters/CF = 36.462 g/CF
36.462 g/CF / 454g/lb = .0803 lb/CF
:D
Rick
 
https://www.shearwater.com/products/teric/

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