chap5 - Physical Chemistry II Chem 402 Spring 2010 Chapter...

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Physical Chemistry II Chem 402 Spring 2010 Chapter 5 ( 12 , 19, 21c, 42, 44 ) P5.12) 1.85 moles of N 2 at 15.2°C and 5.75 bar undergoes a transformation to the state described by 195°C and 2.10 bar. Calculate S if 2 , 35 11 2 3 8 3 30.81 11.87 10 2.3968 10 Jmol K K K 1.0176 10 K Pm C TT T    , -1 -1 23 8 468.15 -1 288.35 -1 ln 2.10 bar 5.75 mol 8.314 Jmol K ln 5.75 bar 30.81 11.87 10 2.3968 10 1.0176 10 KK K JK K 15.5 J K 30.81ln K f i T f i T P C Sn R n d T PT T T d T T       468.15 3 2 58 - 1 2 288.35 - 111 1 0.01187 1.1984 10 0.3392 10 K 15.5 JK 27.6 JK 3.94 JK 3.02 JK 0.49 JK 41.7 JK T        P5.19) At the transition temperature of 95.4°C, the enthalpy of transition from rhombic to monoclinic sulfur is 0.38 kJ mol –1 . a. Calculate the entropy of transition under these conditions. b. At its melting point, 119°C, the enthalpy of fusion of monoclinic sulfur is 1.23 kJ mol –1 . Calculate the entropy of fusion. c. The values given in parts (a) and (b) are for 1 mol of sulfur; however, in crystalline and liquid sulfur, the molecule is present as S 8 . Convert the values of the enthalpy and entropy of fusion in parts (a) and (b) to those appropriate for S 8 . a)  -1 -1 -1 0.38 kJ mol 1.0 J K mol 273.15 + 95.4 K transition transition transition H S T
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b)  -1 -1 -1 1.23 kJ mol 3.14 J K mol 273.15 119 K fusion fusion fusion H S T  c) Each of the S in parts (a) and (b) should be multiplied by 8.
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This note was uploaded on 05/05/2010 for the course CHEM 402 taught by Professor Mueller,karltodd during the Spring '10 term at Pennsylvania State University, University Park.

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chap5 - Physical Chemistry II Chem 402 Spring 2010 Chapter...

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