7.55. A liquid mixture of benzene and toluene is to be separated in a continuous single-stage equilibriumflash tank.Vapor productny(mol)YB(mol C,H/mol)Liquid feedI mol (basis)T(°C)P(mm Hg)ZB (mol C,H/mol)(1 – zg)(mol C,Hg/mol)130°CLiquid productn (mol)Xg(mol CgH/mol)HeatThe pressure in the unit may be adjusted to any desired value, and the heat input may similarly beadjusted to vary the temperature at which the separation is conducted. The vapor and liquid productstreams both emerge at the temperature T(°C) and pressure P(mm Hg) maintained in the vessel.Assume that the vapor pressures of benzene and toluene are given by the Antoine equation,Table B.4 or APEX; that Raoult's law–Equation 6.4-1¬applies; and that the enthalpies of benzeneand toluene liquid and vapor are linear functions of temperature. Specific enthalpies at two tempera-tures are given here for each substance in each phase.C,H6(1) (T = 0°C, Ĥ = 0 kJ/mol)C,H6(v) (T = 80°C, Ĥ = 41.61 kJ/mol) (T = 120°C, Ĥ = 45.79 kJ/mol)(T = 80°C, Ħ = 10.85 kJ/mol)C,Hg(1)(T = 0°C, Ĥ = 0 kJ/mol)(T = 111°C, Ĥ = 18.58 kJ/mol)C;Hg(v) (T = 89°C, Ĥ = 49.18 kJ/mol) (T = 111°C, Ĥ = 52.05 kJ/mol)(a) Suppose the feed is equimolar in benzene and toluene (ZB = 0.500). Take a basis of 1 mol of feedand do the degree-of-freedom analysis on the unit to show that if T and P are specified, you cancalculate the molar compositions of each phase (xB and yß), the moles of the liquid and vaporproducts (nį and ny), and the required heat input (Q). Don’t do any numerical calculations in thispart.(b) Do the calculations of Part (a) for T = 90°C and P = 652 mm Hg. (Suggestion: First derive anequation for xB that can be solved by trial and error from known values of T and P.)(c) For zB = 0.5 and T = 90°C, there is a range of feasible operating pressures for the evaporator,P< Pmax- If the evaporator pressure P fell outside this range, no separation of benzene andPmintoluerte would be achieved. Why not? What would emerge from the unit if P < Pmin? What wouldemerge if P > Pmax? [Hint: Look at your solution to Part (b) and think about how it would changeif you lowered P.]APTER 7 Energy and Energy Balances(d) Set up a spreadsheet to perform the calculation of Part (b) and then use it to determine Pmax andPmin- The spreadsheet should appear as follows (some solutions are shown):Problem 7.55-Flash vaporization of benzene and toluene sucais vour emerge at uie temperature 1TC) and pIessure r(1IIIII ng) mamameu II Uie vesseI,Assume that the vapor pressures of benzene and toluene are given by the Antoine equation,Table B.4 or APEX; that Raoult's law-Equation 6.4-1-applies; and that the enthalpies of benzeneand toluene liquid and vapor are linear functions of temperature. Specific enthalpies at two tempera-tures are given here for each substance in each phase.C6H6(1) (T = 0°C, H = 0 kJ/mol)C6H6(v) (T = 80°C, Ĥ = 41.61 kJ/mol) (T = 120°C, Ĥ = 45.79 kJ/mol)(T = 80°C, H = 10.85 kJ/mol)C,Hg(1)(T = 0°C, Ĥ = 0 kJ/mol)(T = 111°C, ÂĤ = 18.58 kJ/mol)C,H3(v) (T = 89°C, Ĥ = 49.18 kJ/mol) (T = 111°C, Ĥ = 52.05 kJ/mol)(a) Suppose the feed is equimolar in benzene and toluene (zB = 0.500). Take a basis of 1 mol of feedand do the degree-of-freedom analysis on the unit to show that if T and P are specified, you cancalculate the molar compositions of each phase (xB and yg), the moles of the liquid and vaporproducts (nį and ny), and the required heat input (Q). Don't do any numerical calculations in thispart.(b) Do the calculations of Part (a) for T = 90°C and P = 652 mm Hg. (Suggestion: First derive anequation for XB that can be solved by trial and error from known values of T and P.)(c) For zB = 0.5 and T = 90°C, there is a range of feasible operating pressures for the evaporator,Pmin < P< Pmax- If the evaporator pressure P fell outside this range, no separation of benzene andtoluene would be achieved. Why not? What would emerge from the unit if P < Pmin? What wouldemerge if P > Pmax? [Hint: Look at your solution to Part (b) and think about how it would changeif you lowered P.]7 Energy and Energy Balances(d) Set up a spreadsheet to perform the calculation of Part (b) and then use it to determine Pmax andPmin. The spreadsheet should appear as follows (some solutions are shown):Problem 7.55–Flash vaporization of benzene and toluenezBpB pT*xB yBnLnV0.500 90.0 652 10210.55438.1440.500 90.0 | 714-6,0930.500 90.0Additional columns may be used to store other calculated variables (e.g., specific enthalpies).Briefly explain why Q is positive when P = 652 mm Hg and negative when P = 714 mm Hg.(e) In successive rows, repeat the calculation for the same zg and T at several pressures between Pminand Pmax. Generate a plot (using the spreadsheet program itself, if possible) of ny versus P. Atapproximately what pressure is half of the feed stream vaporized?

Vapor product ny(mol) YB(mol C,H/mol) Liquid feed I mol (basis) T(°C) P(mm Hg) ZB(mol C,H,/mol) (1– 2g)(mol C,Hg/mol) 130°C Liquid product (mol) xg(mol C,H/mol) Heat The pressure in the unit may be adjusted to any desired value, and the heat input may similarly be adjusted to vary the temperature at which the separation is conducted. The vapor and liquid product streams both emerge at the temperature T(°C) and pressure P(mm Hg) maintained in the vessel. Assume that the vapor pressures of benzene and toluene are given by the Antoine equation, Table B.4 or APEX; that Raoult's law–Equation 6.4-1¬applies; and that the enthalpies of benzene and toluene liquid and vapor are linear functions of temperature. Specific enthalpies at two tempera- tures are given here for each substance in each phase. C,H6(1) (T = 0°C, Ĥ =0 kJ/mol) C6H6(v) (T = 80°C, Ĥ = 41.61 kJ/mol) (T = 120°C, Ĥ = 45.79 kJ/mol) C,Hg(1) (T = 0°C, Ĥ =0 kJ/mol) C;Hg(v) (T = 89°C, Ĥ = 49.18 kJ/mol) (T = 111°C, Ĥ = 52.05 kJ/mol) (T = 80°C, H = 10.85 kJ/mol) (T = 111°C, Ĥ = 18.58 kJ/mol) (a) Suppose the feed is equimolar in benzene and toluene (zB = 0.500). Take a basis of 1 mol of feed and do the degree-of-freedom analysis on the unit to show that if T and P are specified, you can calculate the molar compositions of each phase (xâ and yp), the moles of the liquid and vapor products (nį and ny), and the required heat input (Q). Don't do any numerical calculations in this part. (b) Do the calculations of Part (a) for T = 90°C and P = 652 mm Hg. (Suggestion: First derive an equation for xg that can be solved by trial and error from known values of T and P.) (c) For zB = 0.5 and T = 90°C, there is a range of feasible operating pressures for the evaporator, Pmin P< Pmax- If the evaporator pressure P fell outside this range, no separation of benzene and toluerte would be achieved. Why not? What would emerge from the unit if P < Pmin ? What would emerge if P > Pmax? [Hint: Look at your solution to Part (b) and think about how it would change if you lowered P.] y and Energy Balances (d) Set up a spreadsheet to perform the calculation of Part (b) and then use it to determine Pmax and Pmin. The spreadsheet should appear as follows (some solutions are shown): Problem 7.55-Flash vaporization of benzene and toluene