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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 ...
As an example, given a concentration of 260 mg/m 3 at sea level, calculate the equivalent concentration at an altitude of 1,800 meters: C a = 260 × 0.9877 18 = 208 mg/m 3 at 1,800 meters altitude Standard conditions for gas volumes
The sum of molar concentrations gives the total molar concentration, namely the density of the mixture divided by the molar mass of the mixture or by another name the reciprocal of the molar volume of the mixture. In an ionic solution, ionic strength is proportional to the sum of the molar concentration of salts.
The ideal gas law follows from the van der Waals equation whenever the molar volume is sufficiently large (when , so ), or correspondingly whenever the molar density, = / , is sufficiently small (when (/) / , so + / ).
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 ...
As an example, a measured NO x concentration of 45 ppmv in a dry gas having 5 volume % O 2 is: 45 × ( 20.9 - 3 ) ÷ ( 20.9 - 5 ) = 50.7 ppmv of NO x. when corrected to a dry gas having a specified reference O 2 content of 3 volume %. Note: The measured gas concentration C m must first be corrected to a dry basis before using the above equation.
Volume percent is the concentration of a certain solute, measured by volume, in a solution.It has as a denominator the volume of the mixture itself, as usual for expressions of concentration, [2] rather than the total of all the individual components’ volumes prior to mixing:
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).