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Liquid equilibrium with vapor

The outline of Teller [70, 133] suggests using the differential form above. Vapor is assumed to be in equilibrium with liquid. [Pg.46]

The vapor pressure of water, which is 24 mm Hg at 25°C, becomes 92 mm Hg at 50°C and 1 atm (760 mm Hg) at 100°C. The data for water are plotted at the top of Figure 9.2. As you can see, the graph of vapor pressure versus temperature is not a straight line, as it would be if pressure were plotted versus temperature for an ideal gas. Instead, the slope increases steadily as temperature rises, reflecting the fact that more molecules vaporize at higher temperatures. At 100°C, the concentration of H20 molecules in the vapor in equilibrium with liquid is 25 times as great as at 25°C. [Pg.229]

Curve AB is a portion of the vapor pressure-temperature curve of liquid water. At any temperature and pressure along this line, liquid water is in equilibrium with water vapor. At point A on the curve, these two phases are in equilibrium at 0°C and about 5 mm Hg (more exactly, 0.01°C and 4.56 mm Hg). At B, corresponding to 100°C, the pressure exerted by the vapor in equilibrium with liquid water is 1 atm this is the normal boiling point of water. The extension of line AB beyond point B gives the equilibrium vapor pressure of the liquid above the normal boiling point. The line ends at 374°C, the critical temperature of water, where the pressure is 218 atm. [Pg.233]

The bulbs contain tleft to right) gaseous chlorine and the vapors in equilibrium with liquid bromine and solid iodine. [Pg.556]

A triple point is a point where three phase boundaries meet on a phase diagram. For water, the triple point for the solid, liquid, and vapor phases lies at 4.6 Torr and 0.01°C (see Fig. 8.6). At this triple point, all three phases (ice, liquid, and vapor) coexist in mutual dynamic equilibrium solid is in equilibrium with liquid, liquid with vapor, and vapor with solid. The location of a triple point of a substance is a fixed property of that substance and cannot be changed by changing the conditions. The triple point of water is used to define the size of the kelvin by definition, there are exactly 273.16 kelvins between absolute zero and the triple point of water. Because the normal freezing point of water is found to lie 0.01 K below the triple point, 0°C corresponds to 273.15 K. [Pg.438]

The vapor-liquid equilibrium for noncondensable gases in equilibrium with liquids can often be approximated by Henry s Law1-3 ... [Pg.61]

The standard partial molal properties of neutral organic aqueous molecules behave similarly to their inorganic counterparts (i.e., at pressures corresponding to liquid-vapor equilibrium, with increasing T they reach a minimum at intermediate T values and then approach +00 as P approaches the critical point of H2O see figure 8.27B). [Pg.565]

Figures 16-18 and 16-19 give the dew-point water-vapor content of a nitrogen-free. natural gas in equilibrium with liquid water. Figures 16-18 and 16-19 give the dew-point water-vapor content of a nitrogen-free. natural gas in equilibrium with liquid water.
Line Q2C is parallel to and slightly above the vapor-pressure line for the pure hydrocarbon. The dashed line to the left of point Q2 is simply an extension of the vapor-pressure line of the hydrocarbon. Point C is the three-phase critical point at which the properties of the hydrocarbon gas and liquid merge to form a single hydrocarbon phase in equilibrium with liquid water. [Pg.476]

Isotopic fractionation resulting from evaporation from standing water bodies can be described in terms of equilibrium and nonequilibrium fractionation effects. Equilibrium fractionation occurs when the isotopic composition of the evaporated water or lake evaporate is in thermodynamic equilibrium with the lake water (23). Equilibrium fractionation, however, can occur only when the water vapor in the air mass above the lake is 100% saturated. The process of equilibrium isotopic fractionation is described by Raleigh fractionation. The isotopic composition of water vapor in equilibrium with liquid water at any time is given by... [Pg.79]

In this chapter we get to know the second essential equation of surface science — the Kelvin5 equation. Like the Young-Laplace equation it is based on thermodynamic principles and does not refer to a special material or special conditions. The subject of the Kelvin equation is the vapor pressure of a liquid. Tables of vapor pressures for various liquids and different temperatures can be found in common textbooks or handbooks of physical chemistry. These vapor pressures are reported for vapors which are in thermodynamic equilibrium with liquids having planar surfaces. When the liquid surface is curved, the vapor pressure changes. The vapor pressure of a drop is higher than that of a flat, planar surface. In a bubble the vapor pressure is reduced. The Kelvin equation tells us how the vapor pressure depends on the curvature of the liquid. [Pg.15]

Reaction (6) as an allowed transition means that absorption of the 1849 A line by mercury vapor is extraordinarily high, and with mercury vapor in equilibrium with liquid mercury at room temperature nearly all of the radiation will be absorbed in a distance of less than one millimeter from the window through which radiation enters the vessel. [Pg.6]

Vapors in equilibrium with liquid in fine capillaries or pores will have depressed vapor pressure as a result of the Kelvin effect. In fact, if the pores are adequately small, the vapor will condense at pressures far below normal. By measuring the volume of nitrogen adsorbed at a relative pressure, i.e., p/po, of 0.99 and with prior knowledge of the surface area, the average pore diameter can be calculated. [Pg.44]

Saturated water vapor is in equilibrium with liquid water. If the vapor is not saturated and any liquid water is present, enough of this evaporates to make the vapor saturated. Compressing a saturated vapor would tend to increase its concentration, which would increase its pressure. The observed fact that there is no such increase in pressure shows that the vapor does not become more concentrated, that is, that it does not become supersaturated but condenses to liquid water to maintain the exact state of saturation as the piston is pressed down. When the piston is drawn up the liquid must evaporate to maintain a saturated condition of the vapor. [Pg.43]

Once the air is saturated with water vapor, the water (if the water vapor is in equilibrium with liquid water) exerts a pressure equal to its vapor pressure at 15°C. [Pg.91]

Correction for the Vapor Pressure of Water. When a sample of gas is collected over water (Fig. 8-2), the pressure of the gas is due in part to the water vapor in it. The pressure due to the water vapor in the gas in equilibrium with liquid water is equal to the vapor pressure of... [Pg.155]

If water vapor is in equilibrium with liquid water, must the vapor be saturated Can a vapor be saturated if there is no liquid present in the system ... [Pg.253]

At 100°C cyclohexane has a partial pressure of433 mm and toluene a partial pressure of 327 mm the sum of the partial pressures is 760 mm and so the liquid boils. If some of the liquid in equilibrium with this boiling mixture were condensed and analyzed, it would be found to be 433/760 or 57 mole percent cyclohexane (pointB, Fig. 2). This is the best separation that can be achieved on simple distillation of this mixture. As the simple distillation proceeds, the boiling point of the mixture moves toward 110°C along the line from A, and the vapor composition becomes richer in toluene as it moves from B to 110°C. In order to obtain pure cyclohexane, it would be necessary to condense the liquid at B and redistill it. When this is done it is found that the liquid boils at 90°C (point C) and the vapor equilibrium with this liquid is about 85 mole percent cyclohexane (point D). So to separate a mixture of cyclohexane and toluene, a series of fractions would be collected and each of these partially redistilled. If this fractional distillation were done enough times the two components could be separated. [Pg.61]

Schoonmaker and Porter ( ) analyzed the vapors in equilibrium with liquid CsOH and mixed KOH-CsOH condensed phases with a mass spectrometer and reported the presence of appreciable concentrations of dimer in the temperature range 650-700 K. By applying the method of relative equilibrium constants, these workers calculated the difference in the free energies of dimerization for KOH-CsOH. A 3rd law analysis of their free energy data for this pair leads to a difference in the enthalpies of dimerization of 4.9 kcal mol at 298.15 K for CsOH and KOH. Based upon the adopted value for KOH(g), AjH (dimerization, 298. 15 K) = -45.3 3.0 kcal mol" (2), we derive A H (dimerization, 298.15 K) -40.4 4.0 kcal mol" for 2 CsOH(g) = Cs2(0H)2(g). Combining this result with the enthalpy of formation for the gaseous monomer (3), that for the dimer is AjH (CS2(0H)2, g. 298.15 K) = -164.4+10.0 kcal mol" (-687.8+41.8 kJ mol" ). [Pg.962]

From (71), we see that mx +b is just the vapor concentration in equilibrium with liquid at concentration x, which we denote at y. After minor rearrangement, the above equation becomes... [Pg.86]

Composition of vapor in equilibrium with liquid on the tray... [Pg.500]

Gaseous CHCl was obtained from the vapor phase in equilibrium with liquid. Several freeze-pump-thaw cycles were used before vapor was admitted to the chamber. Vapor was admitted with all electron guns turned off and the ion pump isolated from the chamber. Exposure levels were monitored with an ionization gauge. [Pg.63]

We can derive the simplest form of the Young-Laplace equation for a spherical vapor bubble in equilibrium with liquid in a one-component system (or a liquid drop in air) from Newton mechanics. In the absence of any external field such as gravitational, magnetic or electrical fields, the bubble will assume a spherical shape, and the force acting towards the boundary of the bubble (or liquid drop) from the interior of the bubble is given as... [Pg.127]

EXAMPLE 15 (fl) A mixture of oxygen and water vapor at a total pressure of 107 kPa is in equilibrium with liquid water at 25°C, at which temperature the water vapor pressure is 3.2 kPa. Calculate the pressure of the oxygen, (b) Oxygen is collected over water at 25°C under a barometric pressure of 107 kPa. (PH2O = 3.2 kPa) Calculate the pressure of the oxygen. [Pg.95]

The humidity H of HjO-HjS vapor in equilibrium with liquid mixtures, defined as... [Pg.781]


See other pages where Liquid equilibrium with vapor is mentioned: [Pg.1381]    [Pg.42]    [Pg.42]    [Pg.730]    [Pg.70]    [Pg.565]    [Pg.38]    [Pg.896]    [Pg.207]    [Pg.253]    [Pg.676]    [Pg.1204]    [Pg.295]    [Pg.42]    [Pg.42]    [Pg.289]    [Pg.765]    [Pg.143]    [Pg.434]    [Pg.95]    [Pg.232]   
See also in sourсe #XX -- [ Pg.3 ]




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