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How was 146 degrees Celcius obtained?  I know how the 5250 mmhg was obtained.

How was 146 degrees Celcius obtained?  I know how the 5250 mmhg was obtained.

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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How was 146 degrees Celcius obtained?  I know how the 5250 mmhg was obtained.  

||! PDF *Elementary Principles of Chemic X + 69°F Mostly cloudy File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw T Read aloud Example 8.4-2 Single Substance + 446 of 695 Phase changes ofter-occur at temperatures other than the temperature for which the latent heat is tabulated. When faced with this situation, you must select hypothetical process path that permits the available data to be used. Solution An energy balance yields Suppose, for example, that a substance is to be vaporized isothermally at 130°C but the only available value of the heat of vaporization is at 80°C. A process path from the liquid al 130°C to the yaper at the same temperature must then be chosen that includes an isothermal vaporization step at 80°C: specifically, cool the liquid from 130°C to 80°C, vaporize the liquid at 80°C, and then heat the vapor back to 130°C. Summing the changes in enthalpy for each of these steps yields the change in enthalpy for the given process. (By definition, the calculated value is the latent heat of vaporization at 130°C.) Vaporization and Heating One hundred g-moles per hour of liquid n-hexane at 25°C and 7 bar is vaporized and heated to 300°C at constant pressure. Neglecting the effect of pressure on enthalpy, estimate the rate at which heat must be supplied. ETA Q Search ῥ =ΔΗ (W, = ΔÉp = 0, ΔΕk = 0) Therefore an evaluation of AH will yield the desired value of Q. From the Antoine equation (Table B.4) the APEX function Antoine hexame, 5250), the temperature at which the vapor pressure of n-hexafe is 7 bar (5250 mm Hg) is 146°C, and this is therefore the temperature at which the vaporization actually occurs. However, Table B. lists a value of AH, at the normal boiling point of n-hexane, AH, = 28.85 kJ/mol at 69°C We must therefore find a path that takes hexane from a liquid to a vapor at 69°C, rather than at the true vaporization temperature of 146°C. As noted previously, the change in enthalpy associated with a process may be determined from any convenient path as long as the initial and final points of the chosen path correspond to those of the process. The diagram shown on the following page illustrates several possible paths from liquid hexane at 25°C to hexane vapor at 300°C. ■ H {" Ⓡ J 63 60 ENG Sign in (0) 10:48 PM 5/16/2023 + 0
Transcribed Image Text:||! PDF *Elementary Principles of Chemic X + 69°F Mostly cloudy File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw T Read aloud Example 8.4-2 Single Substance + 446 of 695 Phase changes ofter-occur at temperatures other than the temperature for which the latent heat is tabulated. When faced with this situation, you must select hypothetical process path that permits the available data to be used. Solution An energy balance yields Suppose, for example, that a substance is to be vaporized isothermally at 130°C but the only available value of the heat of vaporization is at 80°C. A process path from the liquid al 130°C to the yaper at the same temperature must then be chosen that includes an isothermal vaporization step at 80°C: specifically, cool the liquid from 130°C to 80°C, vaporize the liquid at 80°C, and then heat the vapor back to 130°C. Summing the changes in enthalpy for each of these steps yields the change in enthalpy for the given process. (By definition, the calculated value is the latent heat of vaporization at 130°C.) Vaporization and Heating One hundred g-moles per hour of liquid n-hexane at 25°C and 7 bar is vaporized and heated to 300°C at constant pressure. Neglecting the effect of pressure on enthalpy, estimate the rate at which heat must be supplied. ETA Q Search ῥ =ΔΗ (W, = ΔÉp = 0, ΔΕk = 0) Therefore an evaluation of AH will yield the desired value of Q. From the Antoine equation (Table B.4) the APEX function Antoine hexame, 5250), the temperature at which the vapor pressure of n-hexafe is 7 bar (5250 mm Hg) is 146°C, and this is therefore the temperature at which the vaporization actually occurs. However, Table B. lists a value of AH, at the normal boiling point of n-hexane, AH, = 28.85 kJ/mol at 69°C We must therefore find a path that takes hexane from a liquid to a vapor at 69°C, rather than at the true vaporization temperature of 146°C. As noted previously, the change in enthalpy associated with a process may be determined from any convenient path as long as the initial and final points of the chosen path correspond to those of the process. The diagram shown on the following page illustrates several possible paths from liquid hexane at 25°C to hexane vapor at 300°C. ■ H {" Ⓡ J 63 60 ENG Sign in (0) 10:48 PM 5/16/2023 + 0
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