Tangent intercepts

The curve on this plot shows the molar volume (volume per mol) as a function of mol fraction of species a, ethanol. For Xq = 0 the molar volume is that of pure species b, water, vl K 0.018 L/mol, while for = 1 it is the molar volume of pure species a, ethanol, v° 0.058 L/mol. The curve appears to be straight, but careful examination shows that is it slightly S-shaped. 

FIGURE 6.5 Molar volume as a function of mol fraction of ethanol, Xea on for ethanol and water at 20°C. Data from [1]. 

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The evaluation involves drawing a tangent to the initial process response curve at the point of maximum slope and noting the point where this tangent intercepts the final steady-state value. In this case, the process time constant is the time between the tangent intercepting the original and... 

Fig. 9.10. Integral heat of solution as a function of mole fraction. The tangent intercepts are the partial molar heats of solution.

A plot of AG versus X may be constructed in which case the tangent intercepts are (/r a — and (/xb — respectively. These quantities, the partial molar free... 

It is important to realize that all the features we have discussed - the tangent intercepts, activity coefficients, etc. - operate just the same no matter how complex the mixing curve becomes. They are just easier to discuss with a simple model. 

Values of cb for various molalities were calculated from density data of Chemical Engineers Han ook (McGraw-Hill Book Company, Inc., New York, 1950). Partial molal volumes Vb were calculated from these density data by the method of tangent intercepts described by Lewis and Randall (loc, cit.f p. 39). Relative viscosities m/mb are available for NaCl solutions at 18°C. in International Critical Tables (Vol. V, p. 15). The accompanying tabulation summarizes the results of the various calculations, and Fig. 5.6 compares the calculated results with the experimental data as reported in International Critical Tables (Vol. V, p. 67). 

Figure 8.8 To create the Legendre transform, a function y(x) is expressed as a tangent slope function c(x), and a tangent intercept function b(x). The tangent slopes and intercepts of points x and x" are shown here.

The partial enthalpies of Eq. (8.76) can be obtained in a number of ways, e.g., from activity-coefficient data [12] or from integral heats of solution by the method of tangent intercepts [22, 40]. Since, however, the equilibrium partial pressures of components A and C from these solutions are required over a range of temperatures in any event, the simplest procedure is that described earlier in connection with Fig. 8.2 and Eq. (8.3). For ideal solutions,, is the enthalpy of pure liquid J at /, for all Xj... 

FIGURE 6.6 Redraft of Figure 6.5, with curvature exaggerated, to be used in demonstrating the idea of tangent intercepts. 

If we have experimental data, or an equation that is believed to represent such experimental data for some extensive property as a function of concentration, we can compute the partial molar values by the method of tangent slopes or the method of tangent intercepts. In either case we could plot the data and make the geometric constructions. However, the mathematical procedure, which our computers can do for us is much more useful. 

This equation has no common name a good name for it would be the tangent intercepts calculating equation. It is shown here for volume, but equally applicable for any property whose molar values can be expressed as a function of mol fraction. It is not restricted to binary mixtures it can be applied to species a (or b or c) in mixtures of any number of species. Example 6.4 shows the application of this equation. 

Example 6.5 is clearly a longer and messier way to do by tangent slopes what we did more easily by tangent intercepts in Example 6.4. So why bother Because some functions are easier to do by tangent slopes, and this method appears in the historical literature, so you must understand it to understand that literature. Most often we use tangent intercepts and Eqs. 6.4 and 6.5... 

If we do that (Problem 6.9) we find analogs to most of the equations and procedures shown in this chapter, with partial molar properties replaced by those same properties divided by the molecular weight. Such properties have no common name perhaps they are best called partial mass properties. The most commonly seen application is with plots of enthalpy per unit mass vs. mass fraction, as shown for water and sulfuric acid in Figure 6.8. As shown in Problem 6.9, if we apply the method of tangent intercepts to this figure, the intercept values are the partial mass enthalpies. 

Example 6.7 Using Figure 6.8 and the method of tangent intercepts, estimate the partial molar enthalpies of both water... 

The method of tangent intercepts is shown in the main text, based on the partial molar equation, Eq. 6.5. It may also be shown purely geometrically as follows. Figure 6.13 is the same as Figure 6.6, but the point of tangency has been moved to move points b and d further apart, points c, d, and e have been added, and unnecessary text has been deleted. 

Figure 6.14 is the same as Figure 6.6, but the mol fraction has been replaced by the mass fraction, and the molar volume has been replaced by the specific volume (volume per unit mass). Show that, for the construction shown, the two tangent intercepts, a and b, are the partial mass volumes. Start with Eq. 6.5 and observe that each mol fraction is the number of mols that species divided by the total number of mols, tij. Cancel the rij-s in the denominators. Then replace each of the values with the mass of that species divided by its molecular weight. Then divide both sides by the total mass of the system. 

FIGURE 7.9 Finding the partial molar volume residual amethane by the method of tangent intercepts. The line is tangent to the curve at Jmethane = 0.784. Its intercept on the right hand axis is 0.6 ft / Ibmol. 

Eigure 9.7 shows the form of Eq. 9.18. From this figure (and tangent intercepts, Chapter 6) it is clear that if 5 = 0,... 

In principle, we should be able to begin with Eq. F.6, take the partial derivatives with respect to yi, and use the method of tangent intercepts to find fi/ytP. In practice, this is very difficult, because it is very difficult to write out the derivatives of z from a pressure-explicit EOS. Instead, we begin with Eq. 7.15, rewritten as... 

The tangent-intercept method can also be applied to the partial molar property change of mixing if is plotted vs. X2, the intercept gives (AV ,j,)i = Vi — Oi. 

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