Liquefying of Nitrogen & Oxygen Gaseous to Liquid Phase. Thermodynamics. How to plot solution? Solved [closed]

Question

Process of Liquefying of air composed of 79% Nitrogen & 21% Oxygen from Gaseous to Liquid Phases as Follows:

1) When Temperature is higher than T>83 k they both are gas.

2) at T=83 k and less slightly the phase transformation take place to right O2 and to Left N2. as expected N2 Liquefy faster than O2, Where mixture are presents of both substances.

3)Finally at T=73 k both Substances are liquid with slight percentage of O2 gas.

Solved! How do i plot last part of the code my solution on same plot.

Here's the question to calcify things

Here is my full code!

t1[x_] := 77.4 + 25.27*x - 33.66*x^2 + 51.52*x^3 - 42.65*x^4 + 12.32*x^5; 

t2[x_] := 77.4 + 8.372*x - 6.162*x^2 + 14.62*x^3 + 2.201*x^4 - 6.235*x^5; 


    p1 = Plot[{t1[x], t2[x]}, {x, 0, 1}, Frame -> True, PlotRange -> {{0, 1}, {76, 92}}, 
   PlotLegends -> "Expressions", ImageSize -> Large]

no = 40*0.21; 
nn = 40*0.79; 

x1 = FindRoot[t1[x] == 78, {x, 1}]
x2 = FindRoot[t2[x] == 83, {x, 1}]
xg = x /. x1
xl = x /. x2


sol = Flatten[Thread[Solve[{xg == ngo/(ngn + ngo), xl == nlo/(nln + nlo), 
       no == ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}, Reals]]] /. Rule -> Equal

Plot must be similar to Figure

Answer 1

t1[x_] := 77.4 + 25.27*x - 33.66*x^2 + 51.52*x^3 - 42.65*x^4 + 12.32*x^5; 
t2[x_] := 77.4 + 8.372*x - 6.162*x^2 + 14.62*x^3 + 2.201*x^4 - 6.235*x^5; 

pp1 = Plot[{t1[x], t2[x]}, {x, 0, 1}, Frame -> True, PlotRange -> {{0, 1}, {76, 92}}, 
    PlotLegends -> "Expressions", ImageSize -> Large];


"Number of moles 21% nOxygen 79% nNitrogen"
no = 40*0.21; 
nn = 40*0.79; 



"Part Liquefying of Oxygen"


####PURE*GAS####
xg1 = x /. FindRoot[t1[x] == 83, {x, 1}]

sol1 = Flatten[Thread[Solve[{xg1 == ngo/(ngn + ngo), xl1 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g1 = Graphics[{Arrow[{{xg1, 90}, {xg1, 83}}]}]; 

p1 = Graphics[{Blue, PointSize[0.01], Point[{xg1, 83}]}];



####MIX####

xm1 = x /. FindRoot[t2[x] == 83, {x, 0.2, 0.6}]

sol2 = Flatten[Thread[Solve[{xg1 == ngo/(ngn + ngo), xl1 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g2 = Graphics[{Arrow[{{xg1, 83}, {xm1, 83}}]}];

p2 = Graphics[{Green, PointSize[0.01], Point[{xm1, 83}]}]; 



####PURE*LIQUID####

xl1 = x /. FindRoot[t2[x] == 78, {x, 0.6, 0}]

sol3 = Flatten[Thread[Solve[{xg1 == ngo/(ngn + ngo), xl1 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g3 = Graphics[{Arrow[{{xm1, 83}, {xl1, 78}}]}];

p3 = Graphics[{Red, PointSize[0.01], Point[{xl1, 78}]}];




"Part Liquefying of Nitrogen"


####PURE*GAS####
xg2 = x /. FindRoot[t1[x] == 83, {x, 1}]

sol4 = Flatten[Thread[Solve[{xg2 == ngo/(ngn + ngo), xl2 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g4 = Graphics[{Arrow[{{xg2, 90}, {xg2, 83}}]}];

p4 = Graphics[{Blue, PointSize[0.01], Point[{xg2, 83}]}]; 


####MIX####

xm2 = x /. FindRoot[t1[x] == 78, {x, 0.2, 0.6}]

sol5 = Flatten[Thread[Solve[{xg2 == ngo/(ngn + ngo), xl2 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g5 = Graphics[{Arrow[{{xg2, 83}, {xm2, 78}}]}]; 

p5 = Graphics[{Green, PointSize[0.01], Point[{xm2, 78}]}];


####PURE*LIQUID####
xl2 = x /. FindRoot[t2[x] == 78, {x, 0.6, 0}]
sol6 = Flatten[Thread[Solve[{xg2 == ngo/(ngn + ngo), xl2 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g6 = Graphics[{Arrow[{{xm2, 78}, {xl2, 78}}]}]; 

p6 = Graphics[{Red, PointSize[0.01], Point[{xl2, 78}]}]; 



"Graph"
    Show[{pp1, g1, g2, g3, p1, p2, p3, g4, g5, g6, p4, p5, p6}, 
      {FrameLabel -> {"X", "T(k)"}}]