![]() ![]() ![]() A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1989. Here's a link to the official CODATA site showing the values:Ĭox, J. Not quite an answer, but too long for a comment: What is the precise definition (preferably with reference) of standard molar entropy? Most of the following issues will be covered by the exact definition, but to be more elaborate, these immediately come to my mind: what exactly is the standard state for each element? Do all elements form a perfect crystal at 0 K, and how can we prove this? Should I think of the third law (zero entropy at zero kelvin) as being a definition that sets an otherwise arbitrary baseline, or can it be proven that the quantum mechanical von Neumann entropy really is zero for pure elements at 0 K?Īn internet search turned up plenty of information about the values of standard molar entropy for various compounds and how to use them, but I wasn't able to find a resource with a detailed description of its definition. However, I'm not even sure that's exactly correct, and there are a lot of details that need to be filled in if it is right. Then, for the compound of interest, calculate and/or measure the entropy difference between the compound and its constituent elements at 0 K. My recollection from a long time ago is that it goes roughly like this:įor each element, assume that its entropy goes to zero at 0 kelvin when it's in some standard state Once we get all the values of entropy change, just add them up to get the standard molar entropy of vaporization of water.I understand what the standard molar entropy is, and how to use it in calculations, but I'm interested in understanding exactly how it's defined and measured. This was the best explanation for this problem! I just followed these steps and finally got the right answer :) ![]() Once we get all the values of entropy change, just add them up to get the standard molar entropy of vaporization of water. Just plug and solve to get that entropy change Only this time, we use the Cpm 33.6 J⋅K−1⋅mol−1 for water vapor and the temperature change is from 100 ☌ to 59 ☌. To solve the third part, we use again the equation ∆S = Cpm ln (T2/T1). Since the Cpm is 75.3 J⋅K−1⋅mol−1 for liquid water and the temperature change is from 59.0 ☌ to 100 ☌, just plug and solve to get the entropy changeĪs for the second part, we already are given the entropy of vaporization of water at 100 ☌ which is 109.0 J⋅K−1⋅mol−1. ![]() To solve the first part, we should use the equation ∆S = Cpm ln (T2/T1). To solve this problem, we need to know three steps that involve finding the standard molar entropy of vaporization: the entropy change heating water from 59.0 ☌ to 100 ☌, the entropy of vaporization of water at 100.0 ☌, and the entropy change cooling water from 100 ☌ to 59 ☌. ![]()
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