Volume moles and

In this equation, P, V, n, and T represent pressure, volume, moles, and temperature, just as in the ideal gas law. R is still the universal gas constant. 

Mass, volume, mole and pressure (note that pressure is an intensive quantity) are linked in the general gas equation (which, however, is only valid for diluted mixtures, which is true when we consider traces in air), as seen in Eq. (4.14). We also can apply the gas law based on the partial quantities m = 2 m/, V = J, Vi, p = Ipi and n = Itii in two forms ... 

PURE calculates pure liquid standard-state fugacities at zero pressure, pure-component saturated liquid molar volume (cm /mole), and pure-component liquid standard-state fugacities at system pressure. Pure-component hypothetical liquid reference fugacities are calculated for noncondensable components. Liquid molar volumes for noncondensable components are taken as zero. 

For a titration to be accurate we must add a stoichiometrically equivalent amount of titrant to a solution containing the analyte. We call this stoichiometric mixture the equivalence point. Unlike precipitation gravimetry, where the precipitant is added in excess, determining the exact volume of titrant needed to reach the equivalence point is essential. The product of the equivalence point volume, Veq> and the titrant s concentration, Cq, gives the moles of titrant reacting with the analyte. 

Substituting the molarity and volume of titrant for moles, and rearranging gives... 

Heat of vaporization in kilocalories per mole and molecule volume in cubic angstroms at — 89 C [56]. 

Consider the entropy change in forming one mole of solution at constant T from 4>m moles of 1 (of molar volume V ) and ni2 moles of 2 (of molar volume V2). In the pure state, the 0m 1 moles of 1 occupy (or have available) the volume 0mIF], and the 0m2 moles of 2 have available the volume 0m2 2- However, when the solution is formed, both the 0m 1 moles of 1 and the 0m2 moles of 2 have available to them the entire volume of the solution 0mi V +0m2 2- The entropy change experienced by component 1 due to this available volume change is... 

These ealeulations show that the volume dependeney of a gas-phase reaetion is a funetion not only of the stoiehiometry, but also of the inerts eontent of the reaeting mixture. The sensitivity of volume to eonversion is lowered as the inerts inerease. The expansion faetor, , is positive for reaetions produeing a net inerease in moles, negative for a deerease in moles, and 8 = 0 for reaetions produeing no net ehanges and at eonstant volume. 

A three-necked round-bottom flask is fitted with a dropping funnel, a thermometer, and a magnetic stirrer and is heated in a water bath to 30°. Tetralin (1.32 g, 0.01 mole) and 50 ml of 3.5 Anitric acid solution are placed in the flask and brought to temperature. Ceric ammonium nitrate (21.9 g, 0.04 mole) is dissolved in 100 ml of 3.5 N nitric acid, and the solution is added dropwise to the reaction mixture at a rate such that the temperature does not rise and only a pale yellow color is evident in the reaction mixture. At the completion of the reaction (1 to 2 hours), the mixture should be colorless. The solution is cooled to room temperature, diluted with an equal volume of water, and extracted twice with ether. The ether solution is dried with anhydrous sodium sulfate, filtered, and the ether is evaporated. The residue may be distilled to yield a-tetralone (bp 113-11676 mm or 170749 mm) or may be converted directly to the oxime, which is recrystallized from methanol, mp 88-89°. 

The volume fractions and mole fractions become identical in ideal gas mixtures at fixed conditions of pressure and temperature. In an isolated, nonreactive system, the molar composition does not vary with temperature. 

Start with a basis of 1 lb-mole of the natural gas at T = 80°F = 540°R and P = 40 psig = 54.7 psia. The volume percent and mole percent compositions are identical for a perfect-gas mixture. 

In this very useful form, Rg is known as the universal gas constant, has a value of 1545 and is the same for all gases. The specific gas constant (i i) for any gas can be obtained by dividing 1545 by the molecular weight. Rg is only equal to 1545 when gas pressure (p) is in PSIA volume (y) is expressed as cubic feet per pound mole and temperature (T) is in Rankine or absolute, i.e. °F + 460. 

The molecular weight of H2O being 18.0, 1 mole of water at room temperature occupies 18.0 cm. If several water molecules are added to a quantity of water, the increment in volume per molecule added will be 18.0 cm3 divided by Avogadro s constant. In other words, omitting Avogadro s constant, the increment will be 18.0 cms/mole. As we shall be interested in ion pairs, we may remark that the increment in volume per pair of H2O molecules will be 3G cma/mole and we may use this value as a basis of comparison for a pair of atomic ions. 

Notice that R has the units of atmospheres, liters, moles, and K. These units must be used for pressure, volume, amount, and temperature in any problem in which this value of R is used. 

The pragmatic consideration is that if a student were to undertake this reaction, then it would be important to react corresponding amounts of the two reactants. Amount here implies the number of moles, and the unbalanced version of the equation would imply that equal volumes of reactant solutions (if the same concentration) were needed, when actually twice as much alkali solution would be needed as acid solution because the acid is dibasic. The principled point is that the equation represents a chemical process, which is subject to the constraints of conservation rules matter (as energy) is conserved. In a chemical change, the elements present (whether as elements or in compounds), must be conserved. A balanced equation has the same elements in the quantities represented on both sides ... 

Because moles are the currency of chemistry, all stoichiometric computations require amounts in moles. In the real world, we measure mass, volume, temperature, and pressure. With the ideal gas equation, our catalog of relationships for mole conversion is complete. Table lists three equations, each of which applies to a particular category of chemical substances. 

Consider the binary batch distillation column, represented in Fig. 3.58, and based on that of Luyben (1973, 1990). The still contains Mb moles with liquid mole fraction composition xg. The liquid holdup on each plate n of the column is M with liquid composition x and a corresponding vapour phase composition y,. The liquid flow from plate to plate varies along the column with consequent variations in M . Overhead vapours are condensed in a total condenser and the condensate collected in a reflux drum with a liquid holdup volume Mg and liquid composition xq. From here part of the condensate is returned to the top plate of the column as reflux at the rate Lq and composition xq. Product is removed from the reflux drum at a composition xd and rate D which is controlled by a simple proportional controller acting on the reflux drum level and is proportional to Md-... 

Ans. The volume, temperature, and pressure given allow us to calculate the number of moles of oxygen. 

Arts. Since the volumes, temperatures, and pressures arc the same, the numbers of moles of the two gases are the same. Therefore, there are equal numbers of molecules of the two gases. 

Arts. From the pressure, volume, and temperature data, we can calculate the number of moles of gas present. From the number of moles and the mass, we can calculate the molecular weight. 

One way to make Example 19.13 harder is by giving numbers of moles and a volume instead of concentrations at equilibrium. Since the equilibrium constant is defined in terms of concentrations, we must first convert the numbers of moles and volume to concentrations. Note especially that the volume of all the reactants is the same, since they are all in the same system. 

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