A normal boiling point of Zn(l) is 1181K. At this temperature AHvap is 114,440 (J/mol) and mola volume of Zn(1) is 9.42 (cm3/mol). Calculate the followings at 850K. (R = 0.082 atm-L/mol-K, 1J = 9.87cm³-atm)<

Introduction to Chemical Engineering Thermodynamics
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Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
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A normal boiling point of Zn(1) is 1181K. At this temperature AHyap 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).-
(b) Calculate dP/dT, the slope of the liquid-vapor curve.
Transcribed Image Text:A normal boiling point of Zn(1) is 1181K. At this temperature AHyap 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).- (b) Calculate dP/dT, the slope of the liquid-vapor curve.
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