![]() ![]() ![]() Your second method is the exact same process, except you deal in enthalpy. This is your first method, and it is the correct answer. ![]() subtract the change in entropy as the vapor is cooled from 100 deg C to 85 deg C. sum the change in entropy from 85 degC to 100 deg Cģ. If someone could tell me what I did wrong, I would really appreciate it! I thought to find the :delta: H vaporization at 100☌ by doing 373.15K *109.0 J/K*mol = 40673.35 J/mol. However, I initially did it a different way and wanted to know what was wrong with what I did: The answer obtained from this method is 111 J/K*mol. I realize that it is possible to do this problem by finding the entropy change to heat liquid water from 85☌ to 100☌, finding the entropy change in vaporizing this liquid water (which would just be 109.0J/K*mol), and finding the entropy change to lower the temperature of the water vapor back to 85☌. Calculate the standard entropy of vaporization of water at 85☌, given that its standard entropy of vaporization at 100.☌ is 109.0 J/K*mol and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J/K*mol and 33.6 J/K*mol, respectively, in this range. ![]()
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