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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.
New York: The International Nickel Company, Inc., 1941: 16. — "Values ranging from 21.3 to 21.5 gm/cm 3 at 20 °C have been reported for the density of annealed platinum; the best value being about 21.45 gm/cm 3 at 20 °C." 21.46 g/cm 3 — Rose, T. Kirke. The Precious Metals, Comprising Gold, Silver and Platinum.
Specific volume is a property of materials, defined as the number of cubic meters occupied by one kilogram of a particular substance. 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.
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 ...
Nitrogen is the most common pure element in the earth, making up 78.1% of the volume of the atmosphere [9] (75.5% by mass), around 3.89 million gigatonnes (3.89 × 10 18 kg). Despite this, it is not very abundant in Earth's crust, making up somewhere around 19 parts per million of this, on par with niobium , gallium , and lithium .
If the relative density is exactly 1 then the densities are equal; that is, equal volumes of the two substances have the same mass. If the reference material is water, then a substance with a relative density (or specific gravity) less than 1 will float in water. For example, an ice cube, with a relative density of about 0.91, will float.
Nitrogen frost has a density of 0.85 g cm −3. [49] As a bulk material the crystals are pressed together and density is near that of water. It is temperature dependent and given by ρ = 0.0134T 2 − 0.6981T + 1038.1 kg/m 3. [48] The volume coefficient of expansion is given by 2×10 −6 T 2 − 0.0002T + 0.006 K −1. [48]
R ∗ = 8.314 32 × 10 3 N⋅m⋅kmol −1 ⋅K −1 = 8.314 32 J⋅K −1 ⋅mol −1. Note the use of the kilomole, with the resulting factor of 1000 in the constant. The USSA1976 acknowledges that this value is not consistent with the cited values for the Avogadro constant and the Boltzmann constant. [ 13 ]