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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.
At about 891 kJ/mol, methane's heat of combustion is lower than that of any other hydrocarbon, but the ratio of the heat of combustion (891 kJ/mol) to the molecular mass (16.0 g/mol, of which 12.0 g/mol is carbon) shows that methane, being the simplest hydrocarbon, produces more heat per mass unit (55.7 kJ/g) than other complex hydrocarbons.
= air pollutant concentration, in parts per million by volume mg/m 3 = milligrams of pollutant per cubic meter of air = atmospheric temperature in kelvins = 273.15 + °C 0.08205 = Universal Gas Law constant in atm·l/(mol·K) = molecular weight of the air pollutant (dimensionless)
1 dm 3 /mol = 1 L/mol = 1 m 3 /kmol = 0.001 m 3 /mol (where kmol is kilomoles = 1000 moles) References This page was last ...
For example, Paraffin has very large molecules and thus a high heat capacity per mole, but as a substance it does not have remarkable heat capacity in terms of volume, mass, or atom-mol (which is just 1.41 R per mole of atoms, or less than half of most solids, in terms of heat capacity per atom).
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 ...
? kJ/mol Standard molar entropy, S o liquid? J/(mol K) Heat capacity, c p? J/(mol K) Gas properties Std enthalpy change of formation, Δ f H o gas: −74.6 kJ/mol [8] Standard molar entropy, S o gas: 186.3 J/(mol K) [8] Enthalpy of combustion Δ c H o: −802 kJ/mol [9] Heat capacity, c p: 35.7 J/(mol K) [8] van der Waals' constants [10] a ...
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]