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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.
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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%: 42qu J 2 _ Hz 3U:
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‘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‘rTremﬂwkhuaat. ,0 . cm. 1 1 ~_.  H20 (5,273.15 K)—> H20 (13,373.15 K) AS for the constantpressure 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) = (2200 J K'1)+ (753 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.
 Spring '08
 STAFF

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