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Triple point: 85.47 K (−187.68 °C), 0.0001 Pa ... Propane is highly temperature dependent. [3] ... log of propane vapor pressure.
The density of propane gas at 25 °C (77 °F) is 1.808 kg/m 3, about 1.5× the density of air at the same temperature. The density of liquid propane at 25 °C (77 °F) is 0.493 g/cm 3, which is equivalent to 4.11 pounds per U.S. liquid gallon or 493 g/L. Propane expands at 1.5% per 10 °F. Thus, liquid propane has a density of approximately 4.2 ...
The boiling point corresponds to the temperature at which the vapor pressure of the liquid equals the surrounding environmental pressure. Thus, the boiling point is dependent on the pressure. Boiling points may be published with respect to the NIST, USA standard pressure of 101.325 kPa (1 atm), or the IUPAC standard pressure of 100.000 kPa (1 ...
The pressure and temperature of the gas are directly proportional: As temperature increases, the pressure of the propane gas increases by the same factor. A simple consequence of this proportionality is that on a hot summer day, the propane tank pressure will be elevated, and thus propane tanks must be rated to withstand such increases in pressure.
The pressure at which LPG becomes liquid, called its vapour pressure, likewise varies depending on composition and temperature; for example, it is approximately 220 kilopascals (32 psi) for pure butane at 20 °C (68 °F), and approximately 2,200 kilopascals (320 psi) for pure propane at 55 °C (131 °F).
Boiling point (°C) K b (°C⋅kg/mol) Freezing point (°C) K f (°C⋅kg/mol) Data source; Aniline: 184.3 3.69 –5.96 –5.87 K b & K f [1] Lauric acid: 298.9 44 –3.9 Acetic acid: 1.04 117.9 3.14 16.6 –3.90 K b [1] K f [2] Acetone: 0.78 56.2 1.67 –94.8 K b [3] Benzene: 0.87 80.1 2.65 5.5 –5.12 K b & K f [2] Bromobenzene: 1.49 156.0 6. ...
Ref. SMI uses temperature scale ITS-48. No conversion was done, which should be of little consequence however. The temperature at standard pressure should be equal to the normal boiling point, but due to the considerable spread does not necessarily have to match values reported elsewhere. log refers to log base 10
(760 mmHg = 101.325 kPa = 1.000 atm = normal pressure) This example shows a severe problem caused by using two different sets of coefficients. The described vapor pressure is not continuous—at the normal boiling point the two sets give different results. This causes severe problems for computational techniques which rely on a continuous vapor ...