Pressure mole fractions and

Here Xgq may be a mixed expression involving pressures, mole fractions, and molar concentrations. 

Combination of Eqs. (6.73) and (6.74) gives the full general relation between vapor pressure, mole fraction, and temperature of coexistence of a two-component liquid -gas system, the solute being not volatile at all ... 

When a gas or gaseous mixture remains in contact with a liquid surface, it will acquire vapor from the liquid until the partial pressure of the vapor in the gas mixture equals the vapor pressure of the liquid at the existing temperature. In drying applications, the gas frequently used is air and the liquid used is water. Although common concentration units (partial pressure, mole fraction, and others) based on total quantity of gas and vapor are useful, for operations that involve changes in vapor content of a vapor-gas mixture without changes in the amount of gas, it is more convenient to use a unit based on the unchanging amount of gas. 

Unfortunately, many commonly used methods for parameter estimation give only estimates for the parameters and no measures of their uncertainty. This is usually accomplished by calculation of the dependent variable at each experimental point, summation of the squared differences between the calculated and measured values, and adjustment of parameters to minimize this sum. Such methods routinely ignore errors in the measured independent variables. For example, in vapor-liquid equilibrium data reduction, errors in the liquid-phase mole fraction and temperature measurements are often assumed to be absent. The total pressure is calculated as a function of the estimated parameters, the measured temperature, and the measured liquid-phase mole fraction. 

Osmotic pressure is one of four closely related properties of solutions that are collectively known as colligative properties. In all four, a difference in the behavior of the solution and the pure solvent is related to the thermodynamic activity of the solvent in the solution. In ideal solutions the activity equals the mole fraction, and the mole fractions of the solvent (subscript 1) and the solute (subscript 2) add up to unity in two-component systems. Therefore the colligative properties can easily be related to the mole fraction of the solute in an ideal solution. The following review of the other three colligative properties indicates the similarity which underlies the analysis of all the colligative properties ... 

Feed analyses in terms of component concentrations are usually not available for complex hydrocarbon mixtures with a final normal boihng point above about 38°C (100°F) (/i-pentane). One method of haudhug such a feed is to break it down into pseudo components (narrow-boihng fractions) and then estimate the mole fraction and value for each such component. Edmister [2nd. Eng. Chem., 47,1685 (1955)] and Maxwell (Data Book on Hydrocarbons, Van Nostrand, Princeton, N.J., 1958) give charts that are useful for this estimation. Once values are available, the calculation proceeds as described above for multicomponent mixtures. Another approach to complex mixtures is to obtain an American Society for Testing and Materials (ASTM) or true-boihng point (TBP) cui ve for the mixture and then use empirical correlations to con-strucl the atmospheric-pressure eqiiihbrium-flash cui ve (EF 0, which can then be corrected to the desired operating pressure. A discussion of this method and the necessary charts are presented in a later subsection entitled Tetroleum and Complex-Mixture Distillation. ... 

Assuming an ideal gas mixture at amiospheric pressure, calculate llie mole fraction and ppm of a component if its partial pressure is 19 mniHg. 

Strategy First, calculate the number of moles of each gas (remember that oxygen is in excess). Then determine the mole fractions and finally the partial pressures. 

There are several ways to describe the chemical composition of a mixture of gases. The simplest method is merely to list each component with its partial pressure or number of moles. Two other descriptions, mole fractions and parts per million, also are used frequently. 

For a differential reactor, the change in composition across the reactor will be very small, and the bulk fluid composition may be estimated from the inlet molal flow rates. Assuming that the inlet air is 79% nitrogen and 21% oxygen, the calculations below indicate the bulk fluid mole fractions and partial pressures of the various components of the reaction mixture. 

Fig. 4. Vapor pressure of the AlCl3-NaCl molten salt as a function of the AICI3 mole fraction and the temperature. The data used to construct this plot were taken from Viola et al. [41]...

The first equation simply states the balance in chemical potentials inside and outside of the cell. The expression for the chemical potential inside the protocell separates into a term involving the mole fraction and the chemical potential associated with the pressure difference. The work done by the cell in opposing the pressure change, assuming that the cell remains at constant volume, is given below, where the change in pressure is from p to p + tv. 

Table 1.1 Equilibrium composition of gas mixtures at 2300 K and 0.1 atmos pressure (mole fractions of the major gases only)...

With the equations entered as listed above, press F5 or solve/sweep under the solutions menu to solve the equations. The software indicates that x = 0.333. From this, the following mole fractions and partial pressures are obtained ... 

This is an important relationship between the mole fraction and the ratio of partial and mixture pressures. 

At this point, you re usually given the temperature versus mole fraction diagram for two miscible liquids (Fig. 140), and you re told it s a consequence of Raoult s Law. Well, yes. But not directly. Raoult s Law is a relationship of pressure, not temperature, versus mole fraction and Raoult s Law is pretty much a straight line. You don t need all your orbitals filled to see that you ve been presented with a temperature versus mole fraction diagram, there are two lines (not one), and neither of them are very straight. 

The least squares data correlation was carried out on HC1 vapor mole fraction and total pressure with 8 ],... 

For solvents, 1, is equal to V because the standard state is the pure solvent, if we neglect the small effect of the difference between the vapor pressure of pure solvent and 1 bar. As the standard state for the solute is the hypothetical unit mole fraction state (Fig. 16.2) or the hypothetical 1-molal solution (Fig. 16.4), the chemical potential of the solute that follows Henry s law is given either by Equation (15.5) or Equation (15.11). In either case, because mole fraction and molality are not pressure dependent. 

A water-cooled stainless-steel probe (4.1-millimeter internal diameter) with four inlet holes (0.50-millimeter diameter) was used to continuously sample combustion products 2 cm above the burner. The samples were drawn through an ice-bath-cooled water trap, a drying column, and a 5-micron filter to reduce the water mole-fraction and to remove particles. Temperature and static pressure in the absorption cell were monitored using a type-S thermocouple and a pressure gauge. The flow entered the cell on the same end as the optical beam and exited on the opposite end through 0.5-inch windows before... 

Cowan Fickett (Addnl Ref E) added the 6 to T to prevent pressure from tending to infinity as the temperature tends to zero and to keep (dp/e T)v positive over the range of volumes of interest. They found that the values a =0.5 and /3 = 0.09 were satisfactory for reporducing experimental detonation velocity-density curve and C-J pressure of Comp B. The value of 6 they used was 400 k was defined as /tSx -k(, where K was 11,85, x was the mole fraction and kz- were the individual geometrical covolumes. The final version of Cowan-Fickett equation is given here under Kistiakowsky-Wilson Equation of State... 

Using nitrogen as the adsorbate at a concentration of 0.3 mole fraction and assuming Pq is 15 torrs above ambient pressure, equation (15.14) can be expressed as... 

Vapor-liquid equilibrium data at atmospheric pressure (690-700 mmHg) for the systems consisting of ethyl alcohol-water saturated with copper(II) chloride, strontium chloride, and nickel(II) chloride are presented. Also provided are the solubilities of each of these salts in the liquid binary mixture at the boiling point. Copper(II) chloride and nickel(II) chloride completely break the azeotrope, while strontium chloride moves the azeotrope up to richer compositions in ethyl alcohol. The equilibrium data are correlated by two separate methods, one based on modified mole fractions, and the other on deviations from Raoult s Law. 

Assume that we wish to design a high-pressure combustion chamber where complete oxidation of CO to C02 is an important design consideration. For this purpose we extrapolate our global rate expression for CO oxidation to higher pressure. The right-hand side of Eq. 13.6 can be rewritten in terms of mole fractions and the total molar concentration [M],... 

X moles each of H2 anci I2. partial pressure of each gas is given by the product of its mole fraction and the total pressure (see p 163). 

STANDARD STATE. The stable form of a substance at unit activity. The stable state for each substance of a gaseous system is the ideal gas at 1 atmosphere pressure for a solution it is taken al unit mole fraction and for a solid or liquid element it is taken at 1 atmosphere pressure and ordinary temperature. 

These units for diA and CA will eventually lead to the units m3 for the volume of the reactor when we come to use equation 1.35. When we substitute for 0iA in equation 1.35, however, to integrate, we must express CA in terms of aA, where the reactant A is CjHt. To do this we first note that CA-yAC where yA is the mole fraction and C is the molar density of the gas mixture (kmol/m3). Assuming ideal gas behaviour, C is the same for any gas mixture, being dependent only on pressure and temperature in accordance with the ideal gas laws. Thus 1 kmol of gas occupies 22.41m3 at 1 bar (= 1.013 x 105 N/m2) and 273 K. Therefore at 1.4 bar = 1.4 x 10s N/m3 and 1173 K it will occupy ... 

PI5.1 Show that in reaction (15.1), assuming ideal behavior of the gases, the maximum (equilibrium) conversion of nitrogen and hydrogen into ammonia at a given temperature and total pressure, is obtained when the reacting gases are in the proportion of 1 to 3. (To do so, suppose that the N2 and H2 molecules are present in the ratio of 1 to r, with x as the mole fraction of NH3 present at equilibrium. Then express Kp in terms of mole fractions and the total pressure, and find the condition that makes dx/dr equal to zero.)... 

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