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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 ...
One way to write the van der Waals equation is: [8] [9] [10] = where is pressure, is temperature, and = / is molar volume. In addition is the Avogadro constant, is the volume, and is the number of molecules (the ratio / is a physical quantity with base unit mole (symbol mol) in the SI).
The boiling point of water is the temperature at which the saturated vapor pressure equals the ambient pressure. Water supercooled below its normal freezing point has a higher vapor pressure than that of ice at the same temperature and is, thus, unstable. Calculations of the (saturation) vapor pressure of water are commonly used in meteorology.
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]
where the pressure, p, is the atmospheric pressure, V is the measured volume of the vessel, T is the absolute temperature of the hot bath, and R is the gas constant. The molecular weight of the chemical is then simply the mass in grams of the vapor within the vessel divided by the calculated number of mole.
Sometimes specific volume is expressed in terms of the number of cubic centimeters occupied by one gram of a substance. In this case, the unit is the centimeter cubed per gram (cm 3 /g or cm 3 ·g −1). To convert m 3 /kg to cm 3 /g, multiply by 1000; conversely, multiply by 0.001. Specific volume is inversely proportional to density.
M is the molar mass of the solvent. T b is boiling point of the pure solvent in kelvin. ΔH vap is the molar enthalpy of vaporization of the solvent. Through the procedure called ebullioscopy, a known constant can be used to calculate an unknown molar mass. The term ebullioscopy means "boiling measurement" in Latin.
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 /mol 2. 1 dm 3 /mol = 1 L/mol = 1 m 3 /kmol = 0.001 m 3 /mol (where kmol is kilomoles = 1000 moles)