For the solid, liquid, and vapor phases of phenol, follow the hypothetical process path shown. Calculate the each enthalpy. Use the formula DeltaH = VdeltaP. Also use Table B.1 for the phenol properties in the Elementary Principles of Chemical Processes. The problem begins where it is underlined in blue. ||!PDF *Elementary Principles of Chemic X +60°FClearFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDraw2 T Read aloud406 CHAPTER 8 Balances on Nonreactive ProcessesTest Yourself426 of 695ODHowever, we do not have such a table. Our task is then to construct a hypothetical process pathfrom the solid at 25°C and 1 atm to the vapor at 300°C and 3 atm. To do so, we will look ahead a bitand note that Table B.1 gives enthalpy changes for the melting of phenol at 1 atm and 42.5°C (thenormal melting point of phenol) and for the vaporization of phenol at 1 atm and 181.4°C (thenormal boiling point of phenol). We therefore choose the following hypothetical process path:Ph (s, 25°C, 1 atm)Q SearchPh (s, 42.5°C, 1 atm)Ph (1, 42.5°C, 1 atm)Ph (1, 181.4°C, 1 atm)Ph=phenol(C,H,OH)(True path)AĤAĤ₁ΔΗ͂,AĤ3ΔΗ͂ΑPh (v. 300.0°C, 3 atm)AĤ6Ph (v, 300.0°C, 1 atm)ΔῆςPh (v, 181.4°C, 1 atm)Notice that in this path, the first, third, and fifth steps are Type 2 (change in T at constant P), thesecond and fourth steps are Type 3 (change in phase at constant T and P), and the sixth step isType 1 (change in P at constant T). Also notice that the phase changes were made to occur at theconditions for which tabulated enthalpy changes are available.The next step in the calculation would be to determine the values of AĤ for Steps 1, 3, 5, and 6using methods to be given in Section 8.2; read the values of AH₂ and AĤ4 from Table B.1; andthen use the fact that enthalpy is a state property to calculate the desired AĤ (AĦ for the upperdashed line in the figure) asAĤ = AĤ₁ + AĤ₂ + AĤ3 + AĤ4 + AĤ5 + AĤ6Construct a process nath for each of the following processes consisting of sequential stens of the{"J6350^ ENGSign inI003:40 AM5/8/2023+☐{0} X +PDF *Elementary Principles of Chemic X b My Questions | bartlebyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDraw(T)Read aloudof 695 3its mole fraction in the mixture). A state property does not depend on how the species reached itsstate. Consequently, when a species passes from one state to another, both ▲Û and ▲H for theprocess are independent of the path taken from the first state to the second one.←→425In most of this chapten and in Chapter 9, you will learn how to calculate internal energy andenthalpy changes associated with certain processes: specifically,Changes in P at constant T and state of aggregation (Section 8.2).2. Changes in T at constant P and state of aggregation (Section 8.3).3. Phase changes at constant T and P-melting, solidifying, vaporizing, condensing, sublimating(Section 8.4).4. Mixing of two liquids or dissolving of a gas or a solid in a liquid at constant T and P (Section 8.5).5. Chemical reaction at constant T and P (Chapter 9).Q SearchFor example, compressing hydrogen gas from 1 atm to 300 atm at 25°C is a Type 1 process;melting ice at 0°C and then heating the liquid water to 30°C, all at 1 atm, is a Type 3 processfollowed by a Type 2 process; mixing sulfuric acid and water at a constant temperature of 20°Cand a constant pressure of 1 atm is a Type 4 process.Once we know how to calculate AU and AH for these five types of processes, we cancalculate these quantities for any process by taking advantage of the fact that Û and Ĥ are stateproperties. The procedure is to construct a hypothetical process path from the initial state to thefinal state consisting of a series of steps of the given five types. Having done this, we calculate AĤfor each of the steps, and then add the AA's for the steps to calculate AĤ for the total process.Since Ĥ is a state property, AĤ calculated for the hypothetical process path-which weconstructed for convenience is the same as AĤ for the path actually followed by the process.The same procedure can be followed to calculate AU for any process.Suppose, for example, that we wish to calculate AĤ for a process in which solid phenol at25°C and I atm is converted to phenol vapor at 300°C and 3 atm. If we had a table of enthalpies forphenol, we could simply subtract Ĥ at the initial state from Ĥ at the final state, orAĤ = Ĥ(vapor, 300°C, 3 atm) - Ĥ(solid, 25°C, 1 atm)However, we do not have such a table. Our task is then to construct a hypothetical process pathfrom the solid at 25°C and 1 atm to the vapor at 300°C and 3 atm. To do so, we will look ahead a bitand note that Table B.1 gives enthalpy changes for the melting of phenol at 1 atm and 42.5°C (thenormal melting point of phenol) and for the vaporization of phenol at 1 atm and 181.4°C (thenormal boiling point of phenol). We therefore choose the following hypothetical process path:Phnhenol(C₂H-OH).{"J6350ENGSign inI●5:53 AM5/8/2023+O{0}

Question

For the solid, liquid, and vapor phases of phenol, follow the hypothetical process path shown.  Calculate the each enthalpy.  Use the formula DeltaH = VdeltaP.  Also use Table B.1 for the phenol properties in the Elementary Principles of Chemical Processes.  The problem begins where it is underlined in blue. ||!PDF *Elementary Principles of Chemic X +60°FClearFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDraw2 T Read aloud406 CHAPTER 8 Balances on Nonreactive ProcessesTest Yourself426 of 695ODHowever, we do not have such a table. Our task is then to construct a hypothetical process pathfrom the solid at 25°C and 1 atm to the vapor at 300°C and 3 atm. To do so, we will look ahead a bitand note that Table B.1 gives enthalpy changes for the melting of phenol at 1 atm and 42.5°C (thenormal melting point of phenol) and for the vaporization of phenol at 1 atm and 181.4°C (thenormal boiling point of phenol). We therefore choose the following hypothetical process path:Ph (s, 25°C, 1 atm)Q SearchPh (s, 42.5°C, 1 atm)Ph (1, 42.5°C, 1 atm)Ph (1, 181.4°C, 1 atm)Ph=phenol(C,H,OH)(True path)AĤAĤ₁ΔΗ͂,AĤ3ΔΗ͂ΑPh (v. 300.0°C, 3 atm)AĤ6Ph (v, 300.0°C, 1 atm)ΔῆςPh (v, 181.4°C, 1 atm)Notice that in this path, the first, third, and fifth steps are Type 2 (change in T at constant P), thesecond and fourth steps are Type 3 (change in phase at constant T and P), and the sixth step isType 1 (change in P at constant T). Also notice that the phase changes were made to occur at theconditions for which tabulated enthalpy changes are available.The next step in the calculation would be to determine the values of AĤ for Steps 1, 3, 5, and 6using methods to be given in Section 8.2; read the values of AH₂ and AĤ4 from Table B.1; andthen use the fact that enthalpy is a state property to calculate the desired AĤ (AĦ for the upperdashed line in the figure) asAĤ = AĤ₁ + AĤ₂ + AĤ3 + AĤ4 + AĤ5 + AĤ6Construct a process nath for each of the following processes consisting of sequential stens of the{&#34;J6350^ ENGSign inI003:40 AM5/8/2023+☐{0} X +PDF *Elementary Principles of Chemic X b My Questions | bartlebyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDF Drive%20).pdfDraw(T)Read aloudof 695 3its mole fraction in the mixture). A state property does not depend on how the species reached itsstate. Consequently, when a species passes from one state to another, both ▲Û and ▲H for theprocess are independent of the path taken from the first state to the second one.←→425In most of this chapten and in Chapter 9, you will learn how to calculate internal energy andenthalpy changes associated with certain processes: specifically,Changes in P at constant T and state of aggregation (Section 8.2).2. Changes in T at constant P and state of aggregation (Section 8.3).3. Phase changes at constant T and P-melting, solidifying, vaporizing, condensing, sublimating(Section 8.4).4. Mixing of two liquids or dissolving of a gas or a solid in a liquid at constant T and P (Section 8.5).5. Chemical reaction at constant T and P (Chapter 9).Q SearchFor example, compressing hydrogen gas from 1 atm to 300 atm at 25°C is a Type 1 process;melting ice at 0°C and then heating the liquid water to 30°C, all at 1 atm, is a Type 3 processfollowed by a Type 2 process; mixing sulfuric acid and water at a constant temperature of 20°Cand a constant pressure of 1 atm is a Type 4 process.Once we know how to calculate AU and AH for these five types of processes, we cancalculate these quantities for any process by taking advantage of the fact that Û and Ĥ are stateproperties. The procedure is to construct a hypothetical process path from the initial state to thefinal state consisting of a series of steps of the given five types. Having done this, we calculate AĤfor each of the steps, and then add the AA's for the steps to calculate AĤ for the total process.Since Ĥ is a state property, AĤ calculated for the hypothetical process path-which weconstructed for convenience is the same as AĤ for the path actually followed by the process.The same procedure can be followed to calculate AU for any process.Suppose, for example, that we wish to calculate AĤ for a process in which solid phenol at25°C and I atm is converted to phenol vapor at 300°C and 3 atm. If we had a table of enthalpies forphenol, we could simply subtract Ĥ at the initial state from Ĥ at the final state, orAĤ = Ĥ(vapor, 300°C, 3 atm) - Ĥ(solid, 25°C, 1 atm)However, we do not have such a table. Our task is then to construct a hypothetical process pathfrom the solid at 25°C and 1 atm to the vapor at 300°C and 3 atm. To do so, we will look ahead a bitand note that Table B.1 gives enthalpy changes for the melting of phenol at 1 atm and 42.5°C (thenormal melting point of phenol) and for the vaporization of phenol at 1 atm and 181.4°C (thenormal boiling point of phenol). We therefore choose the following hypothetical process path:Phnhenol(C₂H-OH).{&#34;J6350ENGSign inI●5:53 AM5/8/2023+O{0}

Accepted Answer

Answered: Image /qna-images/answer/1a5ec9de-3f43-4cf7-acd1-2be4543f76ec.jpg

Ph (s. 25°C, 1 atm) Ph (s. 42.5°C, 1 atm) Ph (1, 42.5°C, 1 atm) Ph (1,181.4°C, 1 atm) Ph=phenol(C,H,OH) (True path) AR, AR₂ AR, ارد AR₂ Ph (v. 300.0°C, 3 atm) AR Ph (v. 300.0°C, 1 atm) AR₂ Ph (v. 181.4°C, 1 atm)