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1 mole of Bi changed 820K -> 298K. Find AH.“ SBis) = 14 (cal/mol-K) (T = 298K) %3D CPBIG) + 4 +5x10-3T (cal/mol-K) (T = 298~544K)- CPBI) = 5 +1x10-3T (cal/mol-K) (T = 544~820K)- AHm.Bi = 2600(cal/mol) (T = 544K)<

1 mole of Bi changed 820K -> 298K. Find AH.“ SBis) = 14 (cal/mol-K) (T = 298K) %3D CPBIG) + 4 +5x10-3T (cal/mol-K) (T = 298~544K)- CPBI) = 5 +1x10-3T (cal/mol-K) (T = 544~820K)- AHm.Bi = 2600(cal/mol) (T = 544K)<

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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1 mole of Bi changed 820K -> 298K. Find AH- SBirs) = 14 (cal/mol-K) (T = 298K)- CPB(s) + 4 +5x10-3T (cal/mol-K) (T = 298~544K)- CPBI() = 5 +1x10-3T (cal/mol-K) (T = 544~820K)< %3D AHm,Bi = 2600(cal/mol) (T = 544K)- %3D %3D
Transcribed Image Text:1 mole of Bi changed 820K -> 298K. Find AH- SBirs) = 14 (cal/mol-K) (T = 298K)- CPB(s) + 4 +5x10-3T (cal/mol-K) (T = 298~544K)- CPBI() = 5 +1x10-3T (cal/mol-K) (T = 544~820K)< %3D AHm,Bi = 2600(cal/mol) (T = 544K)- %3D %3D
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