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is the molecular mass of dry air, approximately 4.81 × 10 −26 in kg. [note 1], the specific gas constant for dry air, which using the values presented above would be approximately 287.050 0676 in J⋅kg −1 ⋅K −1. [note 1] Therefore:
Air is given a vapour density of one. For this use, air has a molecular weight of 28.97 atomic mass units, and all other gas and vapour molecular weights are divided by this number to derive their vapour density. [2] For example, acetone has a vapour density of 2 [3] in relation to air. That means acetone vapour is twice as heavy as air.
1 Nm 3 of any gas (measured at 0 °C and 1 atmosphere of absolute pressure) equals 37.326 scf of that gas (measured at 60 °F and 1 atmosphere of absolute pressure). 1 kmol of any ideal gas equals 22.414 Nm 3 of that gas at 0 °C and 1 atmosphere of absolute pressure ... and 1 lbmol of any ideal gas equals 379.482 scf of that gas at 60 °F and ...
ISO TR 29922-2017 provides a definition for standard dry air which specifies an air molar mass of 28,965 46 ± 0,000 17 kg·kmol-1. [2] GPA 2145:2009 is published by the Gas Processors Association. It provides a molar mass for air of 28.9625 g/mol, and provides a composition for standard dry air as a footnote. [3]
The gaseous state of water is lighter than air (density 0.804 g/L at STP, average molecular mass 18.015 g/mol) due to water's low molar mass when compared with typical atmospheric gases such as nitrogen gas (N 2). It is non-flammable and much cheaper than helium.
Molecular weight (M.W.) (for molecular compounds) and formula weight (F.W.) (for non-molecular compounds), are older terms for what is now more correctly called the relative molar mass (M r). [8] This is a dimensionless quantity (i.e., a pure number, without units) equal to the molar mass divided by the molar mass constant .
Using the same example of a 2,325 road trip requiring 97 gallons of gas, take an eyeballed rough average of the gas prices the Gas Buddy or Gas Guru apps or Google Maps shows you’ll be paying ...
where R is the ideal gas constant, T is temperature, M is average molecular weight, and g 0 is the gravitational acceleration at the planet's surface. Using the values T =273 K and M =29 g/mol as characteristic of the Earth's atmosphere, H = RT / Mg = (8.315*273)/(29*9.8) = 7.99, or about 8 km, which coincidentally is approximate height of Mt ...