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At IUPAC standard temperature and pressure (0 °C and 100 kPa), dry air has a density of approximately 1.2754 kg/m 3. At 20 °C and 101.325 kPa, dry air has a density of 1.2041 kg/m 3. At 70 °F and 14.696 psi, dry air has a density of 0.074887 lb/ft 3.
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
The standard unit is the meter cubed per kilogram (m 3 /kg or m 3 ·kg −1). Sometimes specific volume is expressed in terms of the number of cubic centimeters occupied by one gram of a substance. In this case, the unit is the centimeter cubed per gram (cm 3 /g or cm 3 ·g −1). To convert m 3 /kg to cm 3 /g, multiply by 1000; conversely ...
The official SI symbols are g/cm 3, g·cm −3, or g cm −3. It is equivalent to the units gram per millilitre (g/mL) and kilogram per litre (kg/L). The density of water is about 1 g/cm 3, since the gram was originally defined as the mass of one cubic centimetre of water at its maximum density at 4 °C (39 °F). [1]
At standard mean sea level it specifies a temperature of 15 °C (59 °F), pressure of 101,325 pascals (14.6959 psi) (1 atm), and a density of 1.2250 kilograms per cubic meter (0.07647 lb/cu ft). It also specifies a temperature lapse rate of −6.5 °C (−11.7 °F) per km (approximately −2 °C (−3.6 °F) per 1,000 ft).
Since the density of dry air at 101.325 kPa at 20 °C is [10] 0.001205 g/cm 3 and that of water is 0.998203 g/cm 3 we see that the difference between true and apparent relative densities for a substance with relative density (20 °C/20 °C) of about 1.100 would be 0.000120. Where the relative density of the sample is close to that of water (for ...
In case of air, using the perfect gas law and the standard sea-level conditions (SSL) (air density ρ 0 = 1.225 kg/m 3, temperature T 0 = 288.15 K and pressure p 0 = 101 325 Pa), we have that R air = P 0 /(ρ 0 T 0) = 287.052 874 247 J·kg −1 ·K −1. Then the molar mass of air is computed by M 0 = R/R air = 28.964 917 g/mol. [11]
An amagat (denoted amg or Am [1]) is a practical unit of volumetric number density.Although it can be applied to any substance at any conditions, it is defined as the number of ideal gas molecules per unit volume at 1 atm (101.325 kPa) and 0 °C (273.15 K). [2]