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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.
where temperature T is in degrees Celsius (°C) and saturation vapor pressure P is in kilopascals (kPa). According to Monteith and Unsworth, "Values of saturation vapour pressure from Tetens' formula are within 1 Pa of exact values up to 35 °C." Murray (1967) provides Tetens' equation for temperatures below 0 °C: [3]
The correct result would be P = 101.325 kPa, the normal (atmospheric) pressure. The deviation is −1.63 kPa or −1.61 %. The deviation is −1.63 kPa or −1.61 %. It is important to use the same absolute units for T and T c as well as for P and P c .
The vapour pressure above the curved interface is then higher than that for the planar interface. This picture provides a simple conceptual basis for the Kelvin equation. The change in vapor pressure can be attributed to changes in the Laplace pressure. When the Laplace pressure rises in a droplet, the droplet tends to evaporate more easily.
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.
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