How was 146 degrees Celcius obtained? I know how the 5250 mmhg was obtained. ||!PDF *Elementary Principles of Chemic X +69°FMostly cloudyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloudExample 8.4-2SingleSubstance+446 of 695Phase changes ofter-occur at temperatures other than the temperature for which the latent heatis tabulated. When faced with this situation, you must select hypothetical process path thatpermits the available data to be used.Solution An energy balance yieldsSuppose, for example, that a substance is to be vaporized isothermally at 130°C but the onlyavailable value of the heat of vaporization is at 80°C. A process path from the liquid al 130°C to theyaper at the same temperature must then be chosen that includes an isothermal vaporization step at80°C: specifically, cool the liquid from 130°C to 80°C, vaporize the liquid at 80°C, and then heatthe vapor back to 130°C. Summing the changes in enthalpy for each of these steps yields thechange in enthalpy for the given process. (By definition, the calculated value is the latent heat ofvaporization at 130°C.)Vaporization and HeatingOne hundred g-moles per hour of liquid n-hexane at 25°C and 7 bar is vaporized and heated to 300°C atconstant pressure. Neglecting the effect of pressure on enthalpy, estimate the rate at which heat must besupplied.ETAQ Searchῥ =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), thetemperature at which the vapor pressure of n-hexafe is 7 bar (5250 mm Hg) is 146°C, and this is therefore thetemperature at which the vaporization actually occurs. However, Table B. lists a value of AH, at the normalboiling point of n-hexane,AH, = 28.85 kJ/mol at 69°CWe must therefore find a path that takes hexane from a liquid to a vapor at 69°C, rather than at the truevaporization temperature of 146°C.As noted previously, the change in enthalpy associated with a process may be determined from anyconvenient 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 tohexane vapor at 300°C.■H{"ⓇJ6360ENGSign in(0)10:48 PM5/16/2023+0

Question

How was 146 degrees Celcius obtained?  I know how the 5250 mmhg was obtained. ||!PDF *Elementary Principles of Chemic X +69°FMostly cloudyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloudExample 8.4-2SingleSubstance+446 of 695Phase changes ofter-occur at temperatures other than the temperature for which the latent heatis tabulated. When faced with this situation, you must select hypothetical process path thatpermits the available data to be used.Solution An energy balance yieldsSuppose, for example, that a substance is to be vaporized isothermally at 130°C but the onlyavailable value of the heat of vaporization is at 80°C. A process path from the liquid al 130°C to theyaper at the same temperature must then be chosen that includes an isothermal vaporization step at80°C: specifically, cool the liquid from 130°C to 80°C, vaporize the liquid at 80°C, and then heatthe vapor back to 130°C. Summing the changes in enthalpy for each of these steps yields thechange in enthalpy for the given process. (By definition, the calculated value is the latent heat ofvaporization at 130°C.)Vaporization and HeatingOne hundred g-moles per hour of liquid n-hexane at 25°C and 7 bar is vaporized and heated to 300°C atconstant pressure. Neglecting the effect of pressure on enthalpy, estimate the rate at which heat must besupplied.ETAQ Searchῥ =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), thetemperature at which the vapor pressure of n-hexafe is 7 bar (5250 mm Hg) is 146°C, and this is therefore thetemperature at which the vaporization actually occurs. However, Table B. lists a value of AH, at the normalboiling point of n-hexane,AH, = 28.85 kJ/mol at 69°CWe must therefore find a path that takes hexane from a liquid to a vapor at 69°C, rather than at the truevaporization temperature of 146°C.As noted previously, the change in enthalpy associated with a process may be determined from anyconvenient 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 tohexane vapor at 300°C.■H{&#34;ⓇJ6360ENGSign in(0)10:48 PM5/16/2023+0

Accepted Answer

Temperature corresponds to the vapor pressure of the n-hexane is to be calculated.