Chem_340_Key_Quiz_3_Spring_2012

Chem_340_Key_Quiz_3_Spring_2012 - CHEM 340: Physical...

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Unformatted text preview: CHEM 340: Physical Chemistry for Biochemists 1, Spring 2012 February 27, 2012 ' Quiz 3 Print Your Name: @ . ‘ 1. (2 points) Find AS when 1.00 mol of water vapor initially at 200°C and 1.00 bar undergoes a reversible cyclic process for which q = — 145 J. 5820 (67092 L M” LL98) 2. (5' points) One mol of an ideal gas at 300 K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because it is very large, the temperature of the water ' bath thermal reservoir in the surroundings remains essentially constant at 300 K during the process. Calculate AS, Assumundings and AStotal. AU '90 z - UU i’rw m h - WOW A V”; C 00/}; ziyflw'iu’Nsw/c r“ brew ’Ws “WT [7 = MILT 7 : Mam , K I/. 3,_ l 3" 2.- -- Z- Mx/D 'J be [7 a 1%: 4-2qu J 2 _ Hz 3U: 3W.ol<. 3 555w: ffii": “133* : Z'MW" ‘T z 7.51 we" 7 T 3W.b)<_ ‘54 7cm: 0 3. (2 points) You are told that AS = 0 for a process in which the system is coupled to its surroundings. Can you conclude that the process is reversible? Justify your answer. - No. The criterion for reversibility is AS + AS = 0. To decide if this criterion is satisfied, surroundings AS must be known. smoundin gs 3.21) _(2 points) Calculate AS if I mol of liquid water is heated from 0°Dto 100°C under constant pressure if CE... = 75.291 J K‘1 moi“. 1 .‘Jlfik'fifitfififélfl’mmfl‘r-Tremflwkhuaat. ,0 . cm. 1 1- ~_. - H20 (5,273.15 K)—> H20 (13,373.15 K) AS for the constant-pressure process is given by: AS = Cpm InE = (75.291 J mol“ K")x In (373 '15 K) — 23.49 J mol'] K71 Ti (273.15K)_ l3) (4 points) The meltin% point of water at the pressure of interest is 0°C and the enthalpy of fusion is 6.0095 kJ mol' . The boiling point is 100°C and the enthalpy of vaporization is 40.6563 kJ moi—1. Calculate AS for the transformation H2009, 0°C) —> H20(g,' 100°C). For the process: H20 (s, 273.15 K) —> H20 (a 373.15 K) the following path has to be considered: H20 (5, 273.15 K)——> H20 (12, 273.15 K)——> H20 (12, 373.15 K)—> H20 (g, 373.15 K) Therefore, ASfuS and ASvap have to be calculated: -1 AS _ AH.“ 2 (6.0095 kJmol )= 22.001116 “— T (273.15K)- -[ AS“ 3mm =[40.6563kJm01 LIOSBSJK1 P T (373.15K) —“— ASsys = ASfiJs + Cp,m (r, H20)ln%+ASvap 1 (373.15 K) = (22-00 J K'1)+ (75-3 1 KW In (273.15 K) +(108.95JK“) =154.4JK‘1 ...
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This note was uploaded on 04/07/2012 for the course CHEM 340 taught by Professor Staff during the Spring '08 term at Ill. Chicago.

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Chem_340_Key_Quiz_3_Spring_2012 - CHEM 340: Physical...

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