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Calculate AS for the transformation n. Ice at 273K and 1 atm 423K y 1.1atm Data: C (H₂O) = 75.291 2 (') Р ΔΗ C (H₂O) = 33,577- P 2 (g) ΔΗ fits vap kJ mol = 6.0095- → steam at J molk J. molk = 40.6563- kJ mol

Calculate AS for the transformation n. Ice at 273K and 1 atm 423K y 1.1atm Data: C (H₂O) = 75.291 2 (') Р ΔΗ C (H₂O) = 33,577- P 2 (g) ΔΗ fits vap kJ mol = 6.0095- → steam at J molk J. molk = 40.6563- kJ mol

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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Calculate AS for the transformation n. Ice at 273K and 1 atm → 423K y 1.1atm Data: С (НО C₁ (H₂00) Р = 75.291 C (H₂O) = 33,577- Р 2 (g) ΔΗ ΔΗ fits vap kJ mol = 6.0095- J molk J. molk kJ mol = 40.6563- steam at
Transcribed Image Text:Calculate AS for the transformation n. Ice at 273K and 1 atm → 423K y 1.1atm Data: С (НО C₁ (H₂00) Р = 75.291 C (H₂O) = 33,577- Р 2 (g) ΔΗ ΔΗ fits vap kJ mol = 6.0095- J molk J. molk kJ mol = 40.6563- steam at
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