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Unformatted text preview: CHEM 3615 Problem Set #10 – Ideal Mixtures & Colligative Properties Due November 13, 2007 Read Chapter 5. Note that we will start Chapter 7 after Thanksgiving break. The first six problems are due at the start of class and will be graded. Seven practice problems are provided below for concepts out of Chapter 5. These problems will not be graded. We may not cover them by the time the problem set is due, however, they are useful for understanding the material in Chapter 5, and depending on where we end lecture on November 15, they may be covered on the third exam. Graded Problems 1. Given the normal boiling point of water is 100 °C and the normal freezing point is 0 °C, construct a phase diagram for water. Assume T c = 647.4 K, the density of ice is 0.917 g/cm 3 , the density of water is 1.000 g/cm 3 , the molar enthalpy of fusion is 6.008 kJ/mol and the molar enthalpy of vaporization is 40.656 kJ/mol. (Extrapolate for the region 0 < p < 2 atm and 0 < T < 700 K, assuming the enthalpies of fusion, vaporization and sublimation, along with the volumes of the solid and gas are independent of temperature). Compare the slopes [Pa/K] of the solidliquid coexistencecurve for benzene (previous problem set) and water and comment on the most significant feature? Solution: For the boiling point, p b = 101325 Pa and T b = 373.15 K For the melting point, p m = 101325 Pa and T m = 273.15 K At the critical point, T c = 647.5 K ∆ . / . / ∆ . . / ∆ ∆ ∆ . / . . ,, . . . , , . . ∆ . . (where p is in atm) Unlike the previous problem set, we cannot start with the triple point. First we need to find where the vaporliquid and liquidsolid coexistence lines intersect. For liquidsolid coexistence, we can use the normal melting point as a reference point, whereby ∆ ∆ . / . . For liquidgas coexistence, we can use the normal boiling point as a reference point, whereby ∆ . / . . Unfortunately, these two equations cannot be solved analytically so we need to find T from the expression . / . . . / . . A numerical solution using Solver in Excel is: T TP = 273.157 K with a corresponding p TP of 836.2 Pa or p TP = 0.008253 atm. The value for T TP is exactly the same as the experimental value (0.01 °C). The value for p TP is approximately 30% larger than the experimental value, p TP = 0.00603 atm. This error arises from the assumption that the molar enthalpy of vaporization is independent of T. We can also find this result by graphing the liquidsolid line from p = 0 to p = 2 atm graphing the liquidgas line from T ~ 0 to T = T c = 647.4 K: ffd8ffe000104a4649460001020100c800c80000ffe20c584943435f50524f46494c4500010100000c484c6 96e6f021000006d6e74725247422058595a2007ce00020009000600310000616373704d53465400000000 49454320735247420000000000000000000000000000f6d6000100000000d32d485020200000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000116 3707274000001500000003364657363000001840000006c77747074000001f000000014626b7074000002...
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 Fall '07
 AREsker
 Physical chemistry, pH, Vapor pressure, molar enthalpy, normal boiling point

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