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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
Molar concentration or molarity is most commonly expressed in units of moles of solute per litre of solution. [1] For use in broader applications, it is defined as amount of substance of solute per unit volume of solution, or per unit volume available to the species, represented by lowercase : [2]
Isotherms of an ideal gas for different temperatures. The curved lines are rectangular hyperbolae of the form y = a/x. They represent the relationship between pressure (on the vertical axis) and volume (on the horizontal axis) for an ideal gas at different temperatures: lines that are farther away from the origin (that is, lines that are nearer to the top right-hand corner of the diagram ...
Specific volume is inversely proportional to density. If the density of a substance doubles, its specific volume, as expressed in the same base units, is cut in half. If the density drops to 1/10 its former value, the specific volume, as expressed in the same base units, increases by a factor of 10.
Relative density with respect to air can be obtained by =, where is the molar mass and the approximately equal sign is used because equality pertains only if 1 mol of the gas and 1 mol of air occupy the same volume at a given temperature and pressure, i.e., they are both ideal gases. Ideal behaviour is usually only seen at very low pressure.
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.
Air is given a vapour density of one. For this use, air has a molecular weight of 28.97 atomic mass units, and all other gas and vapour molecular weights are divided by this number to derive their vapour density. [2] For example, acetone has a vapour density of 2 [3] in relation to air. That means acetone vapour is twice as heavy as air.