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Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." [1] For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.
[1] To convert from L 2 b a r / m o l 2 {\displaystyle \mathrm {L^{2}bar/mol^{2}} } to L 2 k P a / m o l 2 {\displaystyle \mathrm {L^{2}kPa/mol^{2}} } , multiply by 100. To convert from L 2 b a r / m o l 2 {\displaystyle \mathrm {L^{2}bar/mol^{2}} } to m 6 P a / m o l 2 {\displaystyle \mathrm {m^{6}Pa/mol^{2}} } , divide by 10.
R ∗ = 8.314 32 × 10 3 N⋅m⋅kmol −1 ⋅K −1 = 8.314 32 J⋅K −1 ⋅mol −1. Note the use of the kilomole, with the resulting factor of 1000 in the constant. The USSA1976 acknowledges that this value is not consistent with the cited values for the Avogadro constant and the Boltzmann constant. [ 13 ]
For convenience in avoiding conversions in the imperial (or US customary units), some engineers adopted the pound-mole (notation lb-mol or lbmol), which is defined as the number of entities in 12 lb of 12 C. One lb-mol is equal to 453.592 37 g‑mol, [6] which is the same numerical value as the number of grams in an international avoirdupois pound.
For example, Paraffin has very large molecules and thus a high heat capacity per mole, but as a substance it does not have remarkable heat capacity in terms of volume, mass, or atom-mol (which is just 1.41 R per mole of atoms, or less than half of most solids, in terms of heat capacity per atom).
The relative activity of a species i, denoted a i, is defined [4] [5] as: = where μ i is the (molar) chemical potential of the species i under the conditions of interest, μ o i is the (molar) chemical potential of that species under some defined set of standard conditions, R is the gas constant, T is the thermodynamic temperature and e is the exponential constant.
The molar mass of atoms of an element is given by the relative atomic mass of the element multiplied by the molar mass constant, M u ≈ 1.000 000 × 10 −3 kg/mol ≈ 1 g/mol. For normal samples from Earth with typical isotope composition, the atomic weight can be approximated by the standard atomic weight [ 2 ] or the conventional atomic weight.
m(NaCl) = 2 mol/L × 0.1 L × 58 g/mol = 11.6 g. To create the solution, 11.6 g NaCl is placed in a volumetric flask, dissolved in some water, then followed by the addition of more water until the total volume reaches 100 mL. The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02 = 0.055 mol/g). Therefore ...