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+//Engineering and Chemical Thermodynamics

+//Example 8.15

+//Page no :402

+

+clear ; clc;

+//Given 

+R = 8.314 ;

+T = 20 + 273 ;//[K]

+A = 6000 ; //[J/mol]

+B = -384 ; //[J/mol]

+x_a = [0.001 ,0.03 ,0.05 ,0.06 ,0.075 ,0.1 ,0.12 , 0.13 ,0.15 ,0.2 ,0.25 ,0.3 ,0.35 ,0.4 ,0.45,0.475 ,0.5 ,0.55 ,0.6 ,0.65  ,0.7 ,0.75 ,0.8 ,0.8475 ,0.85  ,0.9 ,0.925 ,0.95 ,0.975 ,0.999] ;

+

+for i = 1:30

+    y_data(1,i) = R * T * ( x_a(1,i) * log(x_a(1,i)) + (1 - x_a(1,i)) * log(1- x_a(1,i))) + x_a(1,i) * (1 - x_a(1,i)) * (A + B * (2*x_a(1,i) - 1 )) ;

+    y_data2(1,i) = - 82 * x_a(1,i)- 185.6 ;

+end

+

+m = min(y_data) ;

+for i = 1:30

+    if y_data(1,i) == m then

+       a = x_a(1,i) ;

+    end    

+end

+

+

+for i = 1: 30 

+     y_data2(1,i) = -(R * T *( log(a)  - log(1 - a)) + A * (1 - 2*a) + B * (6 * a - 1 - 6 * a^2)) * (x_a(1,i) - a) + m ;

+end

+

+for i = 1:20

+    y_data3(1,i) = y_data(1,i) - y_data2(1,i) ;

+end

+n = min(y_data3) ;

+

+for i = 1:20

+    if y_data3(1,i) == n then

+       b = x_a(1,i) ;

+    end    

+end

+

+

+disp(" Example: 8.15   Page no : 402") ;

+plot(x_a ,y_data) ;

+plot(x_a ,y_data2) ;

+xtitle(" Figure E8.15","x_a","g - x_a * g_a - x_b * g_b") ;

+

+printf("\n\n             The equilibrium composition can be found by drawing a line tangent to the minima .\n\n                     In this case the answer is %.2f and %.1f   ." , a ,b)
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