A vapor mixture of n-butane and n-hexane contains 50.0 mol % butane at 120'C and 1.0 atm. A stream of this mixture flowing at a rate of 150.0 L/s is cooled and compressed, causing some but not all of the vapor to condense. Liquid and vapor product streams emerge rom the process in equilibrium at T and 1100 mm Hg. The vapor product contains 60.0 mol % butane. Determine the required final emperature, and composition of the liquid product (in mole fractions. Jse the following: log P |mmHg|- A- T|°C]+ C For Butane: A = 6.52485, B = 943.453, C = 239.711 For Hexane: A = 6.88555, B = 1175.817, C = 224.867
A vapor mixture of n-butane and n-hexane contains 50.0 mol % butane at 120'C and 1.0 atm. A stream of this mixture flowing at a rate of 150.0 L/s is cooled and compressed, causing some but not all of the vapor to condense. Liquid and vapor product streams emerge rom the process in equilibrium at T and 1100 mm Hg. The vapor product contains 60.0 mol % butane. Determine the required final emperature, and composition of the liquid product (in mole fractions. Jse the following: log P |mmHg|- A- T|°C]+ C For Butane: A = 6.52485, B = 943.453, C = 239.711 For Hexane: A = 6.88555, B = 1175.817, C = 224.867
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 vapor mixture of n-butane and n-hexane contains 50.0 mol % butane at 120°C and 1.0 atm. A stream of this mixture flowing at a rate
of 150.0 L/s is cooled and compressed, causing some but not all of the vapor to condense. Liquid and vapor product streams emerge
from the process in equilibrium at T and 1100 mm Hg. The vapor product contains 60.0 mol % butane. Determine the required final
temperature, and composition of the liquid product (in mole fractions.
Use the following: log P |mmHg |- A-
T°C]+ C
For Butane: A = 6.52485, B = 943.453, C = 239.711
For Hexane: A = 6.88555, B = 1175.817, C = 224.867](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F724a14e1-400d-42e0-bda8-cc31820c5b64%2F8a5f7acc-1197-4e62-af97-18f2ebbff7bd%2F3affivi_processed.png&w=3840&q=75)
Transcribed Image Text:A vapor mixture of n-butane and n-hexane contains 50.0 mol % butane at 120°C and 1.0 atm. A stream of this mixture flowing at a rate
of 150.0 L/s is cooled and compressed, causing some but not all of the vapor to condense. Liquid and vapor product streams emerge
from the process in equilibrium at T and 1100 mm Hg. The vapor product contains 60.0 mol % butane. Determine the required final
temperature, and composition of the liquid product (in mole fractions.
Use the following: log P |mmHg |- A-
T°C]+ C
For Butane: A = 6.52485, B = 943.453, C = 239.711
For Hexane: A = 6.88555, B = 1175.817, C = 224.867
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