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Vapor pressure of n-butane. From formula: log 10 P m m H g = 6.83029 − 945.90 240.0 + T {\displaystyle \scriptstyle \log _{10}P_{mmHg}=6.83029-{\frac {945.90}{240.0+T}}} obtained from Lange's Handbook of Chemistry , 10th ed.
log 10 of Diethyl Ether vapor pressure. Uses formula: log e P m m H g = {\displaystyle \scriptstyle \log _{e}P_{mmHg}=} log e ( 760 101.325 ) − 12.4379 log e ( T + 273.15 ) − 6340.514 T + 273.15 + 95.14704 + 1.412918 × 10 − 05 ( T + 273.15 ) 2 {\displaystyle \scriptstyle \log _{e}({\frac {760}{101.325}})-12.4379\log _{e}(T+ ...
Here is a similar formula from the 67th edition of the CRC handbook. Note that the form of this formula as given is a fit to the Clausius–Clapeyron equation, which is a good theoretical starting point for calculating saturation vapor pressures: log 10 (P) = −(0.05223)a/T + b, where P is in mmHg, T is in kelvins, a = 38324, and b = 8.8017.
Values are given in terms of temperature necessary to reach the specified pressure. Valid results within the quoted ranges from most equations are included in the table for comparison. A conversion factor is included into the original first coefficients of the equations to provide the pressure in pascals (CR2: 5.006, SMI: -0.875).
This is illustrated in the vapor pressure chart (see right) that shows graphs of the vapor pressures versus temperatures for a variety of liquids. [7] At the normal boiling point of a liquid, the vapor pressure is equal to the standard atmospheric pressure defined as 1 atmosphere, [1] 760 Torr, 101.325 kPa, or 14.69595 psi.
The table above gives properties of the vapor–liquid equilibrium of anhydrous ammonia at various temperatures. The second column is vapor pressure in kPa. The third column is the density of the liquid phase. The fourth column is the density of the vapor. The fifth column is the heat of vaporization needed to convert one gram of liquid to vapor.
(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 ...
Ethane vapor pressure vs. temperature. Uses formula ln P mm Hg = {\displaystyle \ln P_{\text{mm Hg}}=} ln 760 101.325 − 5.381564 ln T − 2626.728 T + 46.39131 + 1.601858 × 10 − 05 T 2 {\displaystyle \ln {\frac {760}{101.325}}-5.381564\ln T-{\frac {2626.728}{T}}+46.39131+1.601858\times 10^{-05}T^{2}} (with T in kelvins ...