At 27.85°C, which is the normal boiling point of 2-methylbutane, the enthalpy of vaporization is 24.69 kJ mol1. The molar heat capacity Cp of the liquid can be assumed to be temperature independent in this problem and has the value 165.5 J K' mol1. The molar heat capacity of the vapor Cp is given as 11.72 + 0.4302 T-0.0001537 T² J K1 mol1 (a) Calculate AH for the following change for 5.000 mol of 2-methylbutane at constant pressure of 101.32 kPa :
At 27.85°C, which is the normal boiling point of 2-methylbutane, the enthalpy of vaporization is 24.69 kJ mol1. The molar heat capacity Cp of the liquid can be assumed to be temperature independent in this problem and has the value 165.5 J K' mol1. The molar heat capacity of the vapor Cp is given as 11.72 + 0.4302 T-0.0001537 T² J K1 mol1 (a) Calculate AH for the following change for 5.000 mol of 2-methylbutane at constant pressure of 101.32 kPa :
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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![At 27.85°C, which is the normal boiling point of 2-methylbutane, the enthalpy of vaporization is
24.69 kJ mol1. The molar heat capacity Cp of the liquid can be assumed to be temperature
independent in this problem and has the value 165.5 J K1 mol4. The molar heat capacity of the
vapor Cp is given as 11.72 + 0.4302 T-0.0001537 T² J K1 mol1
(a) Calculate AH for the following change for 5.000 mol of 2-methylbutane at constant pressure
of 101.32 kPa :
5 CSH12(liquid, 25.00°C) ·
→ 5 CSH12 (gas, 100.00°C)
(b) Calculate q, w, AU, and AH for the vaporization of 5.000 mol of 2-methylbutane at 101.32
kPa and 27.85°C. At
pressure, 2-methylbutane behaves as an ideal gas. The density of
the liquid is 0.6117 g cm³ at 27.85°C.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc068a02a-a2fd-4579-8966-c452da369d78%2Ff16f55fe-19e7-4e32-925d-6b25d071951c%2F3ifvvvb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:At 27.85°C, which is the normal boiling point of 2-methylbutane, the enthalpy of vaporization is
24.69 kJ mol1. The molar heat capacity Cp of the liquid can be assumed to be temperature
independent in this problem and has the value 165.5 J K1 mol4. The molar heat capacity of the
vapor Cp is given as 11.72 + 0.4302 T-0.0001537 T² J K1 mol1
(a) Calculate AH for the following change for 5.000 mol of 2-methylbutane at constant pressure
of 101.32 kPa :
5 CSH12(liquid, 25.00°C) ·
→ 5 CSH12 (gas, 100.00°C)
(b) Calculate q, w, AU, and AH for the vaporization of 5.000 mol of 2-methylbutane at 101.32
kPa and 27.85°C. At
pressure, 2-methylbutane behaves as an ideal gas. The density of
the liquid is 0.6117 g cm³ at 27.85°C.
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