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HW Sol#4 - Chemistry 131 Fall 2007 Solutions to Homework No...

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Chemistry 131, Fall 2007 Solutions to Homework No. 4 Problem 1. See Levine, problem 4.39. (a) False. The chemical potential of each substance depends on temperature, pressure, and composition of the system (Levine p. 128). At a given temperature and pressure, the chemical potentials of toluene and benzene will depend on the relative amounts of each substance in the solution, so those chemical potentials will in general be different. (b) False. This statement is not true in general. The state- ment would be true if the solution was saturated with su- crose, in which case solid sucrose would be in equilibrium with dissolved sucrose and the chemical potentials of the two phases would be the same (Levine p. 131) and would be equal to the molar Gibbs energy of solid sucrose, because solid sucrose is a one-phase pure substance (Levine eq. 4.86). (c) True. See explanation for part (b). (d) False. The statement does not specify that α and β are phases of the same substance (Levine p. 131). Chemical potentials of phases of different substances need not be the same at phase equilibrium. Problem 3. See Levine, problem 4.53. (a) The condition for phase equilibrium is (Levine eq. 4.88) μ β j = μ δ j , for all phases β and δ of each substance j present in the system. (b) The condition for reaction equilibrium is (Levine eq. 4.98) i ν i μ i = 0 . (c) The two conditions above are special cases of the general condition for material equilibrium (Levine eq. 4.85) α i μ α i dn α i = 0 . The condition dG = 0 applies to material equilibrium only when the temperature and pressure are constant, in which case G is minimized (Levine p. 128). Problem 5. See Levine, problem 5.9. (a) True. The system does no work because the volume of the calorimeter is constant, so w = 0. Also, assuming that the “contents of the calorimeter” include the bomb walls and the surrouding water bath, no heat flows out of the system because the calorimeter is adiabatic, so q = 0. Therefore Δ U = q + w = 0 + 0 = 0.
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