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thermo_ism_ch09

# thermo_ism_ch09 - Chapter 9 Ideal and Real Solutions...

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Chapter 9: Ideal and Real Solutions Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions Q9.1) Using the differential form of G , dG = VdP SdT , show that if ln mixing i i i G nRT x x = , then 0. mixing mixing H V = ∆ = 1 2 , , ln ln and 0 ln ln 0 mixing i i i mixing mixing i i i P mixing mixing T n n mixing mixing mixing i i i i i i G nRT x x G S nR x x T G V P H G T S nRT x x T nR x x = ∂∆ = - = - ∂∆ = = = ∆ + = - = Q9.2) For a pure substance, the liquid and gaseous phases can only coexist for a single value of the pressure at a given temperature. Is this also the case for an ideal solution of two volatile liquids? No, an ideal solution of two volatile liquids can exist over a range of pressures that are limited by the pressure for which only a trace of liquid remains, and the pressure for which only a trace of gas remains. Q9.3) Fractional distillation of a particular binary liquid mixture leaves behind a liquid consisting of both components in which the composition does not change as the liquid is boiled off. Is this behavior characteristic of a maximum or a minimum boiling point azeotrope? This behavior is characteristic of a maximum-boiling azeotrope. After initially giving off the more volatile component, the liquid remaining tends to the composition of the maximum boiling point at intermediate composition. After the more volatile component has boiled away, the azeotrope evaporates at constant composition. 9-1

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Chapter 9/Ideal and Real Solutions Q9.4) Why is the magnitude of the boiling point elevation less than that of the freezing point depression? The boiling point elevation is less than the freezing point depression because the chemical potential of the vapor is a much more steeply decreasing function of temperature than the solid, as seen in Figure 9.11a. This is due to the relation d μ = – SdT at constant P and the fact that the molar entropy of a vapor is much larger than that of a solid. When the μ liquid curve is displaced down by the addition of the solute (see Figure 9.11a), the intersection of the μ liquid curve with the μ solid curve and the μ gas curve dertermine the shift in the freezing and boiling temperatures. Because the magnitude of the slope of the μ gas curve is greater than that of the μ solid curve, T b moves up less than the T m moves down. Q9.5) Why is the preferred standard state for the solvent in an ideal dilute solution the Roualt's law standard state set? Why is the preferred standard state for the solute in an ideal dilute solution the Henry's law standard state? Is there a preferred standard state for the solution in which x solvent = x solute = 0.5?
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thermo_ism_ch09 - Chapter 9 Ideal and Real Solutions...

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