4. The molar enthalpy of vaporization of pentane is AvapH = 25.79 kJ/mol at its normal boiling point. Its molar entropy of vaporization at that temperature is AvapS = 83.41 J/K. %3D (a) Compute AvapG in kJ/mol at 305 K, 315 K, and 325 K, and determine if vaporization would be expected to occur spontaneously at any of these temperatures. Assume that AvapH and AvapS have the same values at all of these temperatures. (b) Determine the temperature (in K) of vaporization, Tvap, the temperature at which AvapG = 0. Is this temperature consistent with the results you obtained in part (a)? (c) The values of Cp for liquid and for gaseous pentane are 168.0 and 120.0 J/mol·K, respectively. Use this information to estimate AvapH (in kJ/mol) at T= 298 K, assuming that the heat capacities of the liquid and gaseous states are constant over this temperature range. Hint: Construct a multi-step process that includes vaporization at 298 K, but ends at the same state as vaporization at Tvap. Is this enthalpy of vaporization larger or smaller in magnitude from the standard value? %3D
4. The molar enthalpy of vaporization of pentane is AvapH = 25.79 kJ/mol at its normal boiling point. Its molar entropy of vaporization at that temperature is AvapS = 83.41 J/K. %3D (a) Compute AvapG in kJ/mol at 305 K, 315 K, and 325 K, and determine if vaporization would be expected to occur spontaneously at any of these temperatures. Assume that AvapH and AvapS have the same values at all of these temperatures. (b) Determine the temperature (in K) of vaporization, Tvap, the temperature at which AvapG = 0. Is this temperature consistent with the results you obtained in part (a)? (c) The values of Cp for liquid and for gaseous pentane are 168.0 and 120.0 J/mol·K, respectively. Use this information to estimate AvapH (in kJ/mol) at T= 298 K, assuming that the heat capacities of the liquid and gaseous states are constant over this temperature range. Hint: Construct a multi-step process that includes vaporization at 298 K, but ends at the same state as vaporization at Tvap. Is this enthalpy of vaporization larger or smaller in magnitude from the standard value? %3D
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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Transcribed Image Text:4. The molar enthalpy of vaporization of pentane is AvapH = 25.79 kJ/mol at its normal boiling
point. Its molar entropy of vaporization at that temperature is AvapS = 83.41 J/K.
%3D
(a) Compute AvapG in kJ/mol at 305 K, 315 K, and 325 K, and determine if vaporization would
be expected to occur spontaneously at any of these temperatures. Assume that AvapH
and AvapS have the same values at all of these temperatures.
(b) Determine the temperature (in K) of vaporization, Tvap, the temperature at which AvapG =
0. Is this temperature consistent with the results you obtained in part (a)?
(c) The values of Cp for liquid and for gaseous pentane are 168.0 and 120.0 J/mol·K,
respectively. Use this information to estimate AvapH (in kJ/mol) at T= 298 K, assuming
that the heat capacities of the liquid and gaseous states are constant over this
temperature range. Hint: Construct a multi-step process that includes vaporization at 298
K, but ends at the same state as vaporization at Tvap. Is this enthalpy of vaporization
larger or smaller in magnitude from the standard value?
%3D
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