initial commit / add all books

19 files changed, 508 insertions, 0 deletions

diff --git a/527/CH8/EX8.1/8_1exam.sce b/527/CH8/EX8.1/8_1exam.sce
new file mode 100755
index 000000000..43a95902c
--- /dev/null
+++ b/527/CH8/EX8.1/8_1exam.sce
@@ -0,0 +1,7 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.1 

+//Page 369

+

+clear ; clc ;

+disp("  Example: 8.1   Page no : 369") ;

+disp("           The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file
diff --git a/527/CH8/EX8.10/8_10exam.sce b/527/CH8/EX8.10/8_10exam.sce
new file mode 100755
index 000000000..0c4ad6cc0
--- /dev/null
+++ b/527/CH8/EX8.10/8_10exam.sce
@@ -0,0 +1,66 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.10

+//Page No:390

+

+clear ; clc ;

+//Given

+R = 8.314 ;

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

+A_B = 9.2806 ; // From Appendix A , Table A1.1

+B_B = 2788.5 ; // From Appendix A , Table A1.1

+C_B = -52.36 ; // From Appendix A , Table A1.1

+A_C = 9.1325 ; // From Appendix A , Table A1.1

+B_C = 2766.63 ; // From Appendix A , Table A1.1

+C_C = -50.50 ; // From Appendix A , Table A1.1

+

+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] ; //From table E8.9A

+

+P_1_sat = 6073 ; //[Pa]

+P_2_sat = 6344 ; //[Pa] 

+A = 1390:1410 ;

+B = 60:80 ;

+w = 1 / (R * T) ;

+for k = 1:21

+    y = A(k) ;

+    for i = 1:21 

+        z = B(i) ;

+        for j = 1:12

+        P(1,j) = x1(1,j) * exp((y + 3 * z) * (1 - (x1(1,j)))^2 *w-4*z*(1-x1(1,j))^3* w )* P_1_sat + (1-x1(1,j))*exp((y -3*z)*(x1(1,j))^2 * w + 4 * z * (x1(1,j)^3) * w )*P_2_sat ;

+        R(1,j) =(P(1,j) - P_exp(1,j))^2 ;

+        end

+        

+        m = 0 ;

+        for l = 1:12

+            m = m + R(1,l) ;

+        end

+        S(k,i) = m ;

+     end  

+end

+for i = 1:21

+    k = S(i,1) ;

+    for l = 2:21

+        if S(i,l) < k then

+           k = S(i,l) ;

+        end

+    end

+        D(1,i) = k ;

+end

+

+a = D(1,1) ;

+for i = 2:21

+    if D(1,i) < a then

+        a = D(1,i) ;

+    end

+end

+disp(" Example: 8.10   Page no : 390") ;

+for i = 1:21

+    if D(1,i) == a then

+        for l = 1:21

+            if S(i,l) == a then

+                printf("\n     A = %g J/mol",A(1,i));

+                printf("\n     B = %g J/mol",B(1,l)) ;

+            end

+        end

+    end

+end

diff --git a/527/CH8/EX8.11/8_11exam.sce b/527/CH8/EX8.11/8_11exam.sce
new file mode 100755
index 000000000..2dbea636d
--- /dev/null
+++ b/527/CH8/EX8.11/8_11exam.sce
@@ -0,0 +1,45 @@
+//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 .
\ No newline at end of file
diff --git a/527/CH8/EX8.12/8_12exam.sce b/527/CH8/EX8.12/8_12exam.sce
new file mode 100755
index 000000000..f2a0851d7
--- /dev/null
+++ b/527/CH8/EX8.12/8_12exam.sce
@@ -0,0 +1,15 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.12

+//Page no :395

+

+clear ; clc ;

+//Given 

+H_O2 = 44253.9 ;//[bar] , From table 8.1

+p_O2 = 0.21 ; //[bar]

+

+x_O2 = p_O2 / H_O2 ;

+v_H2O = 1/(1/0.001 * 1/0.018 * 0.001 );

+_O2_ = x_O2 / v_H2O ; //[M]

+disp(" Example: 8.12   Page no : 395") ;

+printf("\n     Mole fraction of O2 = %g",x_O2 ) ;

+printf("\n     Concentration of O2 = %g M ",_O2_) ;

diff --git a/527/CH8/EX8.13/8_13exam.sce b/527/CH8/EX8.13/8_13exam.sce
new file mode 100755
index 000000000..bb7443fb6
--- /dev/null
+++ b/527/CH8/EX8.13/8_13exam.sce
@@ -0,0 +1,25 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.13

+//Page no :396

+

+clear ; clc ;

+//Given 

+P = 300 ; //[bar]

+V_bar_inf_N2 = 3.3 * 10^-5 ;

+R = 8.314 ;

+T = 298 ; //[K]

+y_N2 = 1 ; // At 25*C vapour pressure of water is small

+H_N2_1 = 87365 ; //[bar]

+P_c = 33.8 ; //[bar]

+T_c = 126.2 ;// [K]

+w = 0.039 ; // From Appendix A.1 

+log_w_0 = 0.013 ;

+log_w_1 = 0.210 ;

+H_N2_300 = H_N2_1 * exp((V_bar_inf_N2 * (P -1) * 10^5 )/ (R * T)) ;

+

+k = log_w_0 + w * log_w_1 ;

+sai_N2 = 10^k ;

+x_N2 = y_N2 * sai_N2 * P / H_N2_300 ;

+

+disp(" Example: 8.13   Page no : 396") 

+printf("\n     Solubility of N2 in water = %.5f",x_N2) ;
\ No newline at end of file
diff --git a/527/CH8/EX8.14/8_14exam.sce b/527/CH8/EX8.14/8_14exam.sce
new file mode 100755
index 000000000..4a1db0861
--- /dev/null
+++ b/527/CH8/EX8.14/8_14exam.sce
@@ -0,0 +1,7 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.14

+//Page no :400

+

+clear  ; clc;

+disp("  Example: 8.14   Page no : 400") ;

+disp("           The problem does not contain any numerical calculation . The readers can go through the text book to get the required answer .")
\ No newline at end of file
diff --git a/527/CH8/EX8.15/8_15exam.sce b/527/CH8/EX8.15/8_15exam.sce
new file mode 100755
index 000000000..85c6dfb8b
--- /dev/null
+++ b/527/CH8/EX8.15/8_15exam.sce
@@ -0,0 +1,47 @@
+//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)
\ No newline at end of file
diff --git a/527/CH8/EX8.16/8_16exam.sce b/527/CH8/EX8.16/8_16exam.sce
new file mode 100755
index 000000000..697bc3782
--- /dev/null
+++ b/527/CH8/EX8.16/8_16exam.sce
@@ -0,0 +1,19 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.16

+//Page no :403

+

+clear  ; clc;

+//Given

+A = 6349 ; //[J/mol]

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

+R = 8.314 ;

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

+

+k = 0.000001 ;

+disp(" Example: 8.16   Page no : 403") ;

+function y816 = f816(x_a) , y816 = R * T * (1/x_a + 1/(1 - x_a)) - 2 * A +6 * B * (1 - 2 * x_a) + k

+endfunction

+ans1 = fsolve([0.1],f816) ;

+ans2 = fsolve([0.5],f816) ;

+

+printf("\n    %.3f < x_a < %.3f ",ans1,ans2)

diff --git a/527/CH8/EX8.17/8_17exam.sce b/527/CH8/EX8.17/8_17exam.sce
new file mode 100755
index 000000000..dbfe93906
--- /dev/null
+++ b/527/CH8/EX8.17/8_17exam.sce
@@ -0,0 +1,27 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.17

+//Page no :406

+clear  ; clc;

+//Given

+

+T = 300 ; //[K]

+A = 6235 ; //[J/mol] 

+P_a_sat = 100 * 10^3 ; //[Pa]

+P_b_sat = 50 * 10^3 ; //{Pa}

+R = 8.314 ;

+w = 1/(R * T) ;

+function Z817 = f817(R)

+    x_a_a = R(1) ;

+    x_a_b = R(2) ;

+    Z817(1) = x_a_b * exp(A * (1 - x_a_b) ^ 2 * w) - x_a_a * exp(A * (1 - x_a_a) ^ 2 * w) ; // E8.17A

+    Z817(2) = (1 - x_a_b) * exp(A * ( x_a_b) ^ 2 * w) - (1 - x_a_a) * exp(A * (x_a_a) ^ 2 * w ) ; // E8.17B

+endfunction

+x0 = [0.75 ; 0.1] ;

+[z,fxs,m] = fsolve(x0,f817) ;

+disp(" Example: 8.17  Page no : 406") ;

+printf("\n        The compositions are : x_a_a = %.3f and x_a_b = %.3f",z(1,1), z(2,1)) ;

+

+P = z(1,1) * exp(A * z(2,1) ^ 2 * w) * P_a_sat + z(2,1) * exp(A * z(1,1) ^ 2 * w) * P_b_sat ;

+printf("\n        Total pressure = %d kPa",P * 10^-3) ;

+y_a = z(1,1) * exp(A * z(2,1) ^ 2 * w) * P_a_sat / P ;

+printf("\n            y_a = %.3f" , y_a ) ;
\ No newline at end of file
diff --git a/527/CH8/EX8.18/8_18exam.sce b/527/CH8/EX8.18/8_18exam.sce
new file mode 100755
index 000000000..7d7d4d281
--- /dev/null
+++ b/527/CH8/EX8.18/8_18exam.sce
@@ -0,0 +1,20 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.18

+//Page no :418

+

+clear ; clc ;

+//Given 

+T_b = 373.15 ; //[K]

+del_h_vap = 2257 ; //[J/g]

+MW_salt = 58.5 ; //[g/mol]

+MW_water = 18 ; //[g/mol]

+w_salt = 3.5 ;

+w_water = 100 - w_salt ;

+R = 8.314 ;

+

+x_salt = (w_salt / MW_salt) / (w_salt / MW_salt + w_water / MW_water) ;

+x_b = 2 * x_salt ;// We assume NaCl completely dissociates into Na+ & Cl- ions

+

+del_T = R * T_b^2 / (del_h_vap * MW_water)* x_b ;

+disp(" Example: 8.18   Page no : 418") 

+printf("\n    The temperature that sea water boils is = %.2f degreeC",100 + del_T);
\ No newline at end of file
diff --git a/527/CH8/EX8.19/8_19exam.sce b/527/CH8/EX8.19/8_19exam.sce
new file mode 100755
index 000000000..b2abe96e0
--- /dev/null
+++ b/527/CH8/EX8.19/8_19exam.sce
@@ -0,0 +1,20 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.19

+//Page no :418

+

+clear ; clc ;

+//Given

+rho_w = 1000 ; // [kg/m^3]

+g = 9.8 ; // [m/s^2]

+h = 0.0071 ;//[m]

+m_b = 1.93 * 10^-3 ; // [kg]

+V = 520 * 10^-6 ; //[m^3]

+R = 8.314 ;

+T = 298 ;

+

+PI = rho_w * g * h ;

+C_b = m_b / V ;

+MW_b = R * T * C_b / PI ;

+

+disp(" Example: 8.19   Page no : 418") 

+printf("\n      The molecular weight of the protein = %d kg/mol", MW_b );

diff --git a/527/CH8/EX8.2/8_2exam.sce b/527/CH8/EX8.2/8_2exam.sce
new file mode 100755
index 000000000..12cba0773
--- /dev/null
+++ b/527/CH8/EX8.2/8_2exam.sce
@@ -0,0 +1,29 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.2

+//Page no :369

+

+clear ; clc ;

+//Given

+A_C5H12 = 9.2131 ; //From table E8.2A

+B_C5H12 = 2477.07 ; //From table E8.2A

+C_C5H12 = -39.94 ; //From table E8.2A

+A_C6H12 = 9.1325 ; //From table E8.2A

+B_C6H12 = 2766.63 ; //From table E8.2A

+C_C6H12 = -50.50 ; //From table E8.2A

+A_C6H14 = 9.2164 ; //From table E8.2A

+B_C6H14 = 2697.55 ; //From table E8.2A

+C_C6H14 = -48.78 ; //From table E8.2A

+A_C7H16 = 9.2535 ; //From table E8.2A

+B_C7H16 = 2911.32 ; //From table E8.2A

+C_C7H16 = -56.51 ; //From table E8.2A

+

+x_C5H12 = 0.3 ;

+x_C6H12 = 0.3 ;

+x_C6H14 = 0.2 ;

+x_C7H16 = 0.2 ;

+

+function y82 = f82(T), y82 = -1 + (x_C5H12 * exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) + x_C6H12 * exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) + x_C6H14 * exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) + x_C5H12 * exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) + x_C7H16 * exp(A_C7H16 - B_C7H16 / (T + C_C7H16)));

+endfunction ;

+y =fsolve([300],f82) ;

+disp(" Example: 8.2   Page no : 369") ;

+printf("\n    The temperature at which the liquid develops the first bubble of vapour = %d K",y); 
\ No newline at end of file
diff --git a/527/CH8/EX8.3/8_3exam.sce b/527/CH8/EX8.3/8_3exam.sce
new file mode 100755
index 000000000..fe13850aa
--- /dev/null
+++ b/527/CH8/EX8.3/8_3exam.sce
@@ -0,0 +1,43 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.3

+//Page no :370

+

+clear  ; clc ;

+//Given

+A_C5H12 = 9.2131 ; //From table E8.2A

+B_C5H12 = 2477.07 ; //From table E8.2A

+C_C5H12 = -39.94 ; //From table E8.2A

+A_C6H12 = 9.1325 ; //From table E8.2A

+B_C6H12 = 2766.63 ; //From table E8.2A

+C_C6H12 = -50.50 ; //From table E8.2A

+A_C6H14 = 9.2164 ; //From table E8.2A

+B_C6H14 = 2697.55 ; //From table E8.2A

+C_C6H14 = -48.78 ; //From table E8.2A

+A_C7H16 = 9.2535 ; //From table E8.2A

+B_C7H16 = 2911.32 ; //From table E8.2A

+C_C7H16 = -56.51 ; //From table E8.2A

+

+y_C5H12 = 0.3 ;

+y_C6H12 = 0.3 ;

+y_C6H14 = 0.2 ;

+y_C7H16 = 0.2 ;

+P = 1 ; //[bar]

+

+function y83 = f83(T), y83 = -1 + P * ( y_C5H12 / exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) + y_C6H12 / exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) + y_C6H14 / exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) + y_C7H16 / exp(A_C7H16 - B_C7H16 / (T + C_C7H16)));

+endfunction ;

+y =fsolve([300],f83) ;

+disp(" Example: 8.3   Page no : 370") ;

+printf("\n\n           The temperature at which vapour develops the first drop of liquid = %.2f K",y) ;

+

+T = y ;

+P_sat_C5H12 = exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) ;

+p_sat_C6H12 = exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) ;

+P_sat_C6H14 = exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) ;

+P_sat_C7H16 = exp(A_C7H16 - B_C7H16 / (T + C_C7H16)) ;

+

+x_C5H12 = y_C5H12 * P / P_sat_C5H12 ;

+x_C6H12 = y_C6H12 * P / p_sat_C6H12 ;

+x_C6H14 = y_C6H14 * P / P_sat_C6H14 ;

+x_C7H16 = y_C7H16 * P / P_sat_C7H16 ;

+

+printf("\n\n         x_C5H12 = %f     x_C6H12 = %f\n\n         x_C6H14 = %f     x_C7H16 = %f",x_C5H12,x_C6H12 ,x_C6H14,x_C7H16) ;

diff --git a/527/CH8/EX8.4/8_4exam.sce b/527/CH8/EX8.4/8_4exam.sce
new file mode 100755
index 000000000..1b607c2ae
--- /dev/null
+++ b/527/CH8/EX8.4/8_4exam.sce
@@ -0,0 +1,30 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.4

+//Page no :371

+

+clear ; clc ;

+//Given 

+P_a_sat = 0.53 ;  //[bar]

+P_b_sat = 0.16 ; //[bar]

+X = 1/3 ;

+Y = 1- X ;

+x_a_feed = 0.5 ;

+x_b_feed = 0.5 ;

+a = Y * -(x_a_feed + x_b_feed) + Y^2 ;

+b = X * Y *(P_a_sat + P_b_sat) - (x_a_feed * P_b_sat + x_b_feed * P_a_sat)*X ;

+c =  P_a_sat * P_b_sat * X^2;

+

+k=poly(0,'k');

+P = c + b*k^1 + a*k^2 ;

+M = roots(P);

+

+disp(" Example: 8.4   Page no : 371") ;

+for i = 1:2

+    sign(M(i,1)) ;

+        if ans == 1 then

+          printf("\n\n     Pressure  = %.2f bar",M(i,1)) ;  

+          Xa = x_a_feed / (P_a_sat / M(i,1) * X + Y) ;//....E8.4D

+          Ya = Xa * P_a_sat / M(i,1)  ;//.....E8.4B

+          printf("\n\n       Xa = %.2f \n       Ya = %.2f\n",Xa,Ya);

+        end

+end

diff --git a/527/CH8/EX8.5/8_5exam.sce b/527/CH8/EX8.5/8_5exam.sce
new file mode 100755
index 000000000..9dc1facd6
--- /dev/null
+++ b/527/CH8/EX8.5/8_5exam.sce
@@ -0,0 +1,20 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.5

+//Page no :378

+

+clear  ; clc;

+//Given

+P1_sat = 0.72 ; //[bar]

+P2_sat = 0.31 ; //[bar]

+A = 3590 ;

+B = -1180 ;

+R = 8.314 ;

+T = 70 + 273 ;//[K]

+function y85 = f85(x1) , y85 = -.48 + ( x1 * exp((A + 3*B) * (1 - x1)^2 / (R * T) - 4 * B * (1 - x1)^3  / (R * T)) * P1_sat) / ( x1 * exp((A + 3*B) * (1 - x1)^2 / (R * T) -4 * B * (1 - x1)^3  / (R * T)) * P1_sat +(1 - x1) * exp((A - 3*B) * x1^2 / (R * T) -4 * B * x1^3  / (R * T)) * P2_sat ) ; 

+endfunction

+y = fsolve([0.1],f85);

+x1 = y ;

+P = ( x1 * exp((A + 3*B) * (1 - x1)^2 / (R * T) - 4 * B * (1 - x1)^3  / (R * T)) * P1_sat) + (1 - x1) * exp((A - 3*B) * x1^2 / (R * T) -4 * B * x1^3  / (R * T)) * P2_sat ;

+disp(" Example: 8.5   Page no : 378") ;

+printf("\n     The value of x1 = %.3f\n\n",y) ;

+printf("      Pressure = %.2f bar",P) ;
\ No newline at end of file
diff --git a/527/CH8/EX8.6/8_6exam.sce b/527/CH8/EX8.6/8_6exam.sce
new file mode 100755
index 000000000..396b31aeb
--- /dev/null
+++ b/527/CH8/EX8.6/8_6exam.sce
@@ -0,0 +1,7 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.6 

+//Page no :378

+

+clear ; clc ;

+disp("  Example: 8.6   Page no : 378") ;

+disp("           The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file
diff --git a/527/CH8/EX8.7/8_7exam.sce b/527/CH8/EX8.7/8_7exam.sce
new file mode 100755
index 000000000..a1774c4c8
--- /dev/null
+++ b/527/CH8/EX8.7/8_7exam.sce
@@ -0,0 +1,22 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.7

+//Page no :385

+

+clear ; clc;

+//Given 

+P = 0.223 ; //[bar]

+P_a_sat = 0.156 ; // [bar]

+P_b_sat = 0.124 ; //[bar]

+R = 8.314 ;

+T = 50 + 273 ;

+Xa = 0.554 ;

+Xb = 1 - Xa ;

+

+gama_a = P / P_a_sat ;

+A1 = R * T * log(gama_a) / (Xb^2) * 10^-3 ;

+gama_b = P / P_b_sat ;

+A2 = R * T * log(gama_b) / (Xa^2) * 10^-3 ;   

+

+A = ceil((A1 + A2) / 2) ;

+disp(" Example: 8.7   Page no : 385") ;

+printf("\n    Value of two suffix Marguels parameter = %.1f kJ/mol",A);
\ No newline at end of file
diff --git a/527/CH8/EX8.8/8_8exam.sce b/527/CH8/EX8.8/8_8exam.sce
new file mode 100755
index 000000000..b2983472b
--- /dev/null
+++ b/527/CH8/EX8.8/8_8exam.sce
@@ -0,0 +1,7 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.8 

+//Page no :385

+

+clear ; clc ;

+disp("  Example: 8.8   Page no : 385") ;

+disp("           The problem contains only theory and different substitutions.There is no numerical part involved. Users can go through the book to obtain the required expression .") ;
\ No newline at end of file
diff --git a/527/CH8/EX8.9/8_9exam.sce b/527/CH8/EX8.9/8_9exam.sce
new file mode 100755
index 000000000..4deed432f
--- /dev/null
+++ b/527/CH8/EX8.9/8_9exam.sce
@@ -0,0 +1,52 @@
+//Engineering and Chemical Thermodynamics

+//Example 8.9

+//Page No:388

+

+clear  ; clc ;

+//Given

+R = 8.314 ;

+T = 10 + 273 ; //[K]

+A_B = 9.2806 ; // From Appendix A , Table A1.1

+B_B = 2788.5 ; // From Appendix A , Table A1.1

+C_B = -52.36 ; // From Appendix A , Table A1.1

+A_C = 9.1325 ; // From Appendix A , Table A1.1

+B_C = 2766.63 ; // From Appendix A , Table A1.1

+C_C = -50.50 ; // From Appendix A , Table A1.1

+

+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] ; //From table E8.9A

+

+P_1_sat = 6072.15 ; //[Pa]

+P_2_sat = 6344 ; //[Pa]  

+

+A = [1390 ,1391 ,1392 ,1393 ,1394 ,1395 ,1396 ,1397 ,1398 ,1399 ,1400 ,1401 ,1402 ,1403 ,1404 ,1405 ,1406 ,1407 ,1408 ,1409 ,1410 ] ;

+

+for k = 1:21

+    y = A(1,k) ;

+   for i = 1:12

+        P(1,i) = x1(1,i) * exp( y / (R * T ) * (1 - x1(1,i))^2) * P_1_sat+(1 - x1(1,i)) * exp(y / (R * T ) * x1(1,i)^2) * P_2_sat ;

+        C(k,i) = (P(1,i) - P_exp(1,i))^2 ;

+    end

+end 

+

+for k = 1:21 

+   y = 0 ;

+   for i = 1:12

+     y = y + C(k,i) ;       

+   end

+    R(1,k) = y ;

+end

+

+k = 100000 ;

+for i = 1:21

+    K = R(1,i) ; 

+    if K < k then

+        k = K ;

+    end

+end

+disp(" Example: 8.9   Page no : 388") ;

+for i = 1:21

+    if R(1,i) == k then

+      printf("\n      The two suffix Margules co-efficient is = %g J/mol" ,A(1,i)) ; 

+    end

+end