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physical chemistry exam 2 practice problems, Exercises of Physical Chemistry

physical chemistry exam 2 practice problems

Typology: Exercises

2023/2024

Uploaded on 03/11/2025

K_Dirt87
K_Dirt87 🇺🇸

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{16.9+109+(-7.5)=118}
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Download physical chemistry exam 2 practice problems and more Exercises Physical Chemistry in PDF only on Docsity!

1. habits a pond. An organism i . In the course of its life, the organism transfers 100 kof heat to the pond water at O°C (273 K). i The resulting change in entropy of the water due to this transfer is: Grew co as — = — 1366) K The pond is large enough to ensure that the temperature of the water does not change as heat is transferred. The same transfer at results in +330) K-* ev _ 100 x 107 | as = Tee x 10°F Liquid water has @ high heat capacity | largely on account of its ability to absorb heat into the collective vibrational mades of the hydrogen-bonded clusters of molecules present in the liquid. When the temperature is raised a | to blood temperatur its molar entropy changes by: ' ! $3.0) K7! mol (75) K-! mot-* xn HO 298 K AS = Cln 3. The protein lysozyme, an enzyme that breaks down bacterial cell walls, unfolds at a transition temperature of ind the standard enthalpy of transition as determined by using DSC is It follows that the standard entropy of transition is: / Busting) C609 0K) J mol KU _Aeralt Ps) =)11.46 ky mol! KE | Tre” QIBaS 755) §usS Tos: 4 Suppose you are osked to calculate the entrony of waporization of water at @5E|but you know it only at the normal beiling point of 100 %E, which is First, caleulote the entropy change for heating liquld water trom 26 °C to 100 °C (using as = ln Ft and Cy y(H.0,1) AS, (heating liquid to 100°C) = emit f= (75.29 4 mol) x ae 16.9 J K~ 1 mt a 5 Now caleulate the change in entropy for cooling the vapor from 100°C to 25°C H,0(g, 100 °Cy (Cym(H2O,g) = 33.585 K™ mol): dd C qT; ASq(cool vapor to 25°C) — Cnt = cs. 58 J -t mol%) +109 20% te 54) R moe 18 228k asap emo x" K I The sum of the three entropy changes is the entropy of transition at 25 °C: H,0(), 100 *C xg AvapS(298 K) = AS, + AS; + ASS H,O01, 26 °C) 1118 pk ott {16.9+109+(-7.5)=118} 6. The loss of energy from a person through evaporation when! This energy is dissipated as heat into the surroundings hen the change in entropy of the surroundings is of water is perspired is / If the surroundings are af’ AH ~ = = +8.2 kJ K7*