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A normal boiling point of Zn(l0) is 1181K. At this temperature AHvap is 114,440 (J/mol) and molar volume of Zn(l) is 9.42 (cm3/mol). Calculate the followings at 850K. (R = 0.082 atm-L/mol-K, 1J = 9.87cm³-atm)e (a) Calculate the pressure of Zn(g) in equilibrium With Zn(l).

A normal boiling point of Zn(l0) is 1181K. At this temperature AHvap is 114,440 (J/mol) and molar volume of Zn(l) is 9.42 (cm3/mol). Calculate the followings at 850K. (R = 0.082 atm-L/mol-K, 1J = 9.87cm³-atm)e (a) Calculate the pressure of Zn(g) in equilibrium With Zn(l).

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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A normal boiling point of Zn(1) is 1181K. At this temperature AHvap is 114,440 (J/mol) and molar volume of Zn(l) is 9.42 (cm3/mol). Calculate the followings at 850K. (R = 0.082 atm-L/mol-K, 1J = 9.87cm³-atm)- (a) Calculate the pressure of Zn(g) in equilibrium With Zn(l).
Transcribed Image Text:A normal boiling point of Zn(1) is 1181K. At this temperature AHvap is 114,440 (J/mol) and molar volume of Zn(l) is 9.42 (cm3/mol). Calculate the followings at 850K. (R = 0.082 atm-L/mol-K, 1J = 9.87cm³-atm)- (a) Calculate the pressure of Zn(g) in equilibrium With Zn(l).
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