The vapor pressure of benzene fits the expression: log₁0 [P(mmHg)] = 7.960 1780 T(K) Calculate the enthalpy of vaporization and normal boiling point of benzene.
The vapor pressure of benzene fits the expression: log₁0 [P(mmHg)] = 7.960 1780 T(K) Calculate the enthalpy of vaporization and normal boiling point of benzene.
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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![The vapor pressure of benzene fits the expression:
log₁0 [P(mmHg)] = 7.960
1780
T(K)
Calculate the enthalpy of vaporization and normal boiling point
of benzene.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90d9b958-0bcc-49a1-a89f-df62fee032d1%2F98db992b-426b-4944-bc3c-2ba022be8fda%2Fofw95d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The vapor pressure of benzene fits the expression:
log₁0 [P(mmHg)] = 7.960
1780
T(K)
Calculate the enthalpy of vaporization and normal boiling point
of benzene.
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