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+//Engineering and Chemical Thermodynamics

+//Example 8.11

+//Page No:390

+

+clear ; clc ;

+//Given

+R = 8.314 ;

+T = 10 + 273.15 ; //[K]

+x1 = [0 ,0.0610 ,0.2149 ,0.3187 ,0.4320 ,0.5246 ,0.6117 ,0.7265 ,0.8040 ,0.8830 ,0.8999 ,1] ; //From table E8.9A

+P_exp = [6344 ,6590 ,6980 ,7140 ,7171 ,7216 ,7140 ,6974 ,6845 ,6617 ,6557 ,6073 ,6073] ; //From table E8.9A

+y1 = [ 1 ,0.0953 ,0.2710 ,0.3600 ,0.4453,0.5106 ,0.5735 ,0.6626 ,0.7312 ,0.8200 ,0.8382, 0 ] ;//From table E8.9A

+P_1_sat = 6073 ; //[Pa]

+P_2_sat = 6344 ; //[Pa]

+

+n = 0 ;

+for i = 2:11

+    x2(1,i) = 1 - x1(1,i) ;

+    y2(1,i) = 1 - y1(1,i) ;

+    g_E(1,i) = R * T *( x1(1,i) * log (( y1(1,i) * P_exp(1,i)) / (x1(1,i)* P_1_sat)) + x2(1,i) * log((y2(1,i) * P_exp(1,i)) / (x2(1,i) * P_2_sat)) ) ;

+    n = n + g_E(1,i) / ((x1(1,i) * x2(1,i)) * 10) ;

+    ydata(1,i-1) = (g_E(1,i)/(x1(1,i)*x2(1,i)));

+    xdata(1,i-1) = x1(1,i) - x2(1,i) ;

+end

+m= 0 ; n=0 ; o = 0 ; p= 0 ;N = 10 ;

+for i = 2:11

+    m = m + g_E(1,i) * (2 * x1(1,i) - 1) / ( x1(1,i) *  x2(1,i)) ;

+    n = n + g_E(1,i) / ( x1(1,i) *  x2(1,i)) ;

+    o = o + (2 * x1(1,i) - 1)  ;

+    p = p + (2 * x1(1,i) - 1)^2 ;

+end

+x_bar = o / N ;

+y_bar = n / N ;

+a1 = (N * m - n * o)/(N * p - o^2) ;

+a0 = y_bar - a1 * x_bar ;

+

+for i = 1:10

+      ydata2(1,i) = a0 + a1*xdata(1,i) ;

+end

+plot(xdata,ydata,"+") ;

+plot(xdata,ydata2) ;

+xtitle("Figure E8.11","x1-x2","g_E/x1*x2") ;

+disp(" Example: 8.11   Page no : 390") ;

+printf("\n      From average ,  the value of A = %d J/mol\n",n/10) ;

+printf("\n      From linear regression best fit line the values of A and B are     %.1f J/mol    &     %.1f J/mol   respectively .",a0 , a1) ;

+//Readers can refer figure E8.11 .
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