fisherdvm
Contributor
- Messages
- 3,577
- Reaction score
- 52
- # of dives
- 200 - 499
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...
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...