Problem set #4SolutionCHE 425 __________________ LAST NAME, FIRSTProblem set #41-8 Copy the program UnitOp4.exe from the CHE 425 class distribution folder to your flashor H drive. You can also download the program from the website: https://www.csupomona.edu/~tknguyen/che425/homework.htm. Run the program and choose problem 1-8 in any order. Solve the problems with the dataprovided by the program, copy the problem statement to Word. The program will check youranswer and provide an answer code when you click on “Check”. Copy the answer code andpaste them after the problem statement. You need to present all your work with a diagram indetails to get full credit. Your work should look like this:CHE425 Problem set #1 NGUYEN, THUAN 1/11. A vapor at the dew point and 200 kPa containing a mole fraction of 0.40 benzene (1) and0.60 toluene (2) and 100 kmol total is brought into contact with 110 kmol of a liquid at theboiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streamsare contacted in a single stage, and the outlet streams leave in equilibrium with each other.Assume constant molar overflow, calculate the amounts and compositions of the exit streams.Data: Vapor pressure, Psat, data: ln Psat = A  B/(T + C), where Psat is in kPa and T is in K. Compound A B CBenzene (1) 14.1603 2948.78  44.5633Toluene (2) 14.2515 3242.38  47.1806Problem 1: Correct, Code =4313336481Solutionx0 = 0.3, L0 = 110 kmol, x and y are mole fraction of benzene in the liquid and vapor phase, respectively.Equilibrium stageV2V1L0L1y2 = 0.4, V2 = 100 kmolFor CMO, L1 = L0 = 110 kmol, V2 = V1 = 100 kmol. Making a balance on benzene givesL0x0 + V2y2 = L1x1 + V1y1110(0.30) + 100(0.40) = 110x1 + 100y111x1 + 10y1 = 7.3  y1 = 0.73  1.1x1(E-1)Since the two streams V1 and L1 are in equilibrium, we havey1x1= 1200satP  200y1 = x1exp(14.1603  2948.78/(T  44.5633)) (E-2)1− y11− x1= 2200satP  200(1  y1) = (1  x1)exp(14.2515  3242.38/(T  47.1806)) (E-3)The three equations (E-1,2,3) can be solved for T, x1, and y1 either by graphical or numerical method.4.2 Construct a Txy and a xy phase equilibrium diagram for n-hexane (1) and n-octane (2) at14.7 psia. The equilibrium data for these two species are given:T(R) = 615.67, x = 1.000, y = 1.000 T(R) = 619.51, x = 0.934, y = 0.989 T(R) = 623.35, x = 0.872, y = 0.978 T(R) = 627.19, x = 0.814, y = 0.965 T(R) = 631.03, x = 0.759, y = 0.951 T(R) = 634.87, x = 0.708, y = 0.937 T(R) = 638.71, x = 0.660, y = 0.921 T(R) = 642.55, x = 0.615, y = 0.903 T(R) = 646.38, x = 0.573, y = 0.885 T(R) = 650.22, x = 0.532, y = 0.865 T(R) = 654.06, x = 0.494, y = 0.843 T(R) = 657.90, x = 0.458, y = 0.820 T(R) = 661.74, x = 0.424, y = 0.796 T(R) = 665.58, x = 0.391, y = 0.769 T(R) = 669.42, x = 0.359, y = 0.741 T(R) = 673.25, x = 0.329, y = 0.710 T(R) = 677.09, x = 0.301, y = 0.678 T(R) = 680.93, x = 0.273, y = 0.643 T(R) = 684.77, x = 0.247, y = 0.606 T(R) = 688.61, x = 0.221, y = 0.567 T(R) = 692.45, x = 0.196, y = 0.524 T(R) = 696.29, x = 0.172, y = 0.479 T(R) = 700.13, x = 0.149, y = 0.432 T(R) = 703.96, x = 0.126, y = 0.381T(R) = 707.80, x = 0.104, y = 0.326 T(R) = 711.64, x = 0.083, y = 0.269 T(R) = 715.48, x = 0.061, y = 0.208 T(R) = 719.32, x = 0.041, y = 0.143 T(R) = 723.16, x = 0.020, y = 0.073 T(R) = 727.00, x = 0.000, y = 0.000 Note: Use Matlab to plot the diagrams with tie lines and label the figures with your nameusing the Title command. (See example 4.1-1).7. Read the problem statement in the program UnitOp4.exe. You can follow the following procedure assuming 100 kmol feed.- Assume a temperature- Calculate equilibrium ratios- Calculate V/F- Calculate y1- Calculate the amount of toluene in the vapor. If the calculate value is (0.1)(0.6)(100) = 6 kmol, then temperature guess is correct. Otherwise, guess another T, and repeat the above