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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 ...
At about 891 kJ/mol, methane's heat of combustion is lower than that of any other hydrocarbon, but the ratio of the heat of combustion (891 kJ/mol) to the molecular mass (16.0 g/mol, of which 12.0 g/mol is carbon) shows that methane, being the simplest hydrocarbon, produces more heat per mass unit (55.7 kJ/g) than other complex hydrocarbons.
? kJ/mol Standard molar entropy, S o liquid? J/(mol K) Heat capacity, c p? J/(mol K) Gas properties Std enthalpy change of formation, Δ f H o gas: −74.6 kJ/mol [8] Standard molar entropy, S o gas: 186.3 J/(mol K) [8] Enthalpy of combustion Δ c H o: −802 kJ/mol [9] Heat capacity, c p: 35.7 J/(mol K) [8] van der Waals' constants [10] a ...
1 dm 3 /mol = 1 L/mol = 1 m 3 /kmol = 0.001 m 3 /mol (where kmol is kilomoles = 1000 moles) References This page was last ...
In that case, the specific volume would equal 0.4672 in 3 /lb. However, if the temperature is changed to 1160 °R, the specific volume of the super heated steam would have changed to 0.2765 in 3 /lb, which is a 59% overall change. Knowing the specific volumes of two or more substances allows one to find useful information for certain applications.
1 Nm 3 of any gas (measured at 0 °C and 1 atmosphere of absolute pressure) equals 37.326 scf of that gas (measured at 60 °F and 1 atmosphere of absolute pressure). 1 kmol of any ideal gas equals 22.414 Nm 3 of that gas at 0 °C and 1 atmosphere of absolute pressure ... and 1 lbmol of any ideal gas equals 379.482 scf of that gas at 60 °F and ...
Note that the especially high molar values, as for paraffin, gasoline, water and ammonia, result from calculating specific heats in terms of moles of molecules. If specific heat is expressed per mole of atoms for these substances, none of the constant-volume values exceed, to any large extent, the theoretical Dulong–Petit limit of 25 J⋅mol ...
The buoyancy depends upon the difference of the densities (ρ air) − (ρ gas) rather than upon their ratios. The lifting force for a volume of gas is given by the equation: F B = (ρ air - ρ gas) × g × V. Where F B = Buoyant force (in newton); g = gravitational acceleration = 9.8066 m/s 2 = 9.8066 N/kg; V = volume (in m 3).