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The molar volume of gases around STP and at atmospheric pressure can be calculated with an accuracy that is usually sufficient by using the ideal gas law. The molar volume of any ideal gas may be calculated at various standard reference conditions as shown below: V m = 8.3145 × 273.15 / 101.325 = 22.414 dm 3 /mol at 0 °C and 101.325 kPa
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 ...
Until 1982, STP was defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 101.325 kPa (1 atm). Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (1 bar). Conversions between each volume flow metric are calculated using the following formulas: Prior to 1982,
Avogadro's law states that the volume occupied by an ideal gas is proportional to the amount of substance in the volume. This gives rise to the molar volume of a gas, which at STP is 22.4 dm 3 /mol (liters per mole).
The van der Waals equation of state may be written as (+) =where is the absolute temperature, is the pressure, is the molar volume and is the universal gas constant.Note that = /, where is the volume, and = /, where is the number of moles, is the number of particles, and is the Avogadro constant.
A Assuming an altitude of 194 metres above mean sea level (the worldwide median altitude of human habitation), an indoor temperature of 23 °C, a dewpoint of 9 °C (40.85% relative humidity), and 760 mmHg sea level–corrected barometric pressure (molar water vapor content = 1.16%). B Calculated values *Derived data by calculation.
Using the number density as a function of spatial coordinates, the total number of objects N in the entire volume V can be calculated as = (,,), where dV = dx dy dz is a volume element. If each object possesses the same mass m 0 , the total mass m of all the objects in the volume V can be expressed as m = ∭ V m 0 n ( x , y , z ) d V ...
For a substance X with a specific volume of 0.657 cm 3 /g and a substance Y with a specific volume 0.374 cm 3 /g, the density of each substance can be found by taking the inverse of the specific volume; therefore, substance X has a density of 1.522 g/cm 3 and substance Y has a density of 2.673 g/cm 3. With this information, the specific ...