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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 ...
A normal cubic meter (Nm 3) is the metric expression of gas volume at standard conditions and it is usually (but not always) defined as being measured at 0 °C and 1 atmosphere of pressure. A standard cubic foot (scf) is the USA expression of gas volume at standard conditions and it is often ( but not always ) defined as being measured at 60 ...
Effect of temperature and relative humidity on air density. The addition of water vapor to air (making the air humid) reduces the density of the air, which may at first appear counter-intuitive. This occurs because the molar mass of water vapor (18 g/mol) is less than the molar mass of dry air [note 2] (around 29 g/mol).
Water: 5.536 0.03049 Xenon: 4.250 0.05105 Units. 1 J·m 3 /mol 2 = 1 m 6 ·Pa/mol 2 = 10 L 2 ·bar/mol 2. 1 L 2 atm/mol 2 = 0.101325 J·m 3 /mol 2 = 0.101325 Pa·m 6 ...
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.
Note that the especially high molar values, as for paraffin, gasoline, water and ammonia, result from calculating specific heats in terms of moles of molecules. If specific heat is expressed per mole of atoms for these substances, none of the constant-volume values exceed, to any large extent, the theoretical Dulong–Petit limit of 25 J⋅mol ...
If the size of the chamber remains constant and some atoms are removed, the density decreases and the specific volume increases. 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).
Note that the form of this formula as given is a fit to the Clausius–Clapeyron equation, which is a good theoretical starting point for calculating saturation vapor pressures: log 10 (P) = −(0.05223) a / T + b , where P is in mmHg, T is in kelvins, a = 38324, and b = 8.8017.