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240 mol of an liquid mixture of 30 mol% species M and 70 mol% species E is fed to a flash drum. The mixture comes to equilibrium at 90°C and 1225 mmHg. A liquid stream and vapor stream leaves the process unit. Data for species M: py(T = 90°C) = 500 mmHg Data for species E: p:(T = 90°C) = 1950 mmHg A flowchart of the process is shown below. %3D Vapor (2) n2 =? YM2 =? YE2 = 1 - yM2 Feed Flash Drum п, 3D 240 тol = 0.300 (1) T=90°C XM1 P=1225 mmHg Liquid (3) XE1 = 0.700 n3 =? XM3 =? ХЕз — 1 — Хмз a. Which species is more volatile, species M or species E? Explain using the data provided. b. Use Raoult's law to determine the mole fraction of species M in streams 2 and 3, yM2 and XM3.

240 mol of an liquid mixture of 30 mol% species M and 70 mol% species E is fed to a flash drum. The mixture comes to equilibrium at 90°C and 1225 mmHg. A liquid stream and vapor stream leaves the process unit. Data for species M: py(T = 90°C) = 500 mmHg Data for species E: p:(T = 90°C) = 1950 mmHg A flowchart of the process is shown below. %3D Vapor (2) n2 =? YM2 =? YE2 = 1 - yM2 Feed Flash Drum п, 3D 240 тol = 0.300 (1) T=90°C XM1 P=1225 mmHg Liquid (3) XE1 = 0.700 n3 =? XM3 =? ХЕз — 1 — Хмз a. Which species is more volatile, species M or species E? Explain using the data provided. b. Use Raoult's law to determine the mole fraction of species M in streams 2 and 3, yM2 and XM3.

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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240 mol of an liquid mixture of 30 mol% species M and 70 mol% species E is fed to a flash drum. The mixture comes to equilibrium at 90°C and 1225 mmHg. A liquid stream and vapor stream leaves the process unit. Data for species M: py(T = 90°C) = 500 mmHg Data for species E: p:(T = 90°C) = 1950 mmHg A flowchart of the process is shown below. Vapor (2) n2 =? YM2 =? Ув2 — 1 — ум2 Feed Flash Drum (1) n, = 240 mol T=90°C Хм1 — 0.300 XE1 = 0.700 P=1225 mmHg Liquid n3 =? XM3 =? ХЕЗ 3 1 — хмз a. Which species is more volatile, species M or species E? Explain using the data provided. b. Use Raoult's law to determine the mole fraction of species M in streams 2 and 3, yM2 and XM3.
Transcribed Image Text:240 mol of an liquid mixture of 30 mol% species M and 70 mol% species E is fed to a flash drum. The mixture comes to equilibrium at 90°C and 1225 mmHg. A liquid stream and vapor stream leaves the process unit. Data for species M: py(T = 90°C) = 500 mmHg Data for species E: p:(T = 90°C) = 1950 mmHg A flowchart of the process is shown below. Vapor (2) n2 =? YM2 =? Ув2 — 1 — ум2 Feed Flash Drum (1) n, = 240 mol T=90°C Хм1 — 0.300 XE1 = 0.700 P=1225 mmHg Liquid n3 =? XM3 =? ХЕЗ 3 1 — хмз a. Which species is more volatile, species M or species E? Explain using the data provided. b. Use Raoult's law to determine the mole fraction of species M in streams 2 and 3, yM2 and XM3.
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