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+clear;
+//clc();
+
+// Example 4.2
+// Page: 77
+printf("Example-4.2  Page no.-77\n\n");
+
+//***Data***//
+
+// let we denote graphite by 'g' and diamond by 'd' 
+// Gibb's free energies of graphite and diamond are given by
+g_g = 0.00;//[kJ/mol] 
+g_d = 2.90;//[kJ/mol]
+
+// Specific volumes of graphite and diamond are given by
+v_g = 5.31*10^(-1);//[kJ/(mol*kbar)]
+v_d = 3.42*10^(-1);//[kJ/(mol*kbar)]
+
+// Now from the equation 4.32 ( page 74) given in the book, we have
+// (dg/dP) = v , at constant temperature
+// where 'v' is specific volume
+// let us denote (dg/dP) by 'D' ,so
+D_g = v_g;//[J/(mol*Pa)] For graphite
+D_d = v_d;//[J/(mol*Pa)] For diamond
+
+// Now we can take our plot from P = 0( =1 ), however, total pressure is 1 atm. 
+// If we consider specific volumes of the given species to be constant with changing the pressure then g-P curve will be a straight line
+// So the equation of the line for graphite is 
+// g = D_g*P + g_g
+// and that for diamond
+// g = D_d*P + g_d
+
+P = [0:1:30]';
+
+plot2d(P,[ D_d*P+g_d D_g*P+g_g ],style=[color("darkgreen"),color("red")]);
+
+xlabel("Pressure, P, kbar");
+ylabel("Gibb''s free energy per mol, g, kJ/mol");
+
+printf(" Gibb''s free energy-pressure diagram for graphite-diamond system at 25 degC is as shown in the graphic window. ");
+hl=legend(['Diamond, slope = 0.342 (kJ/mol)/kbar';'Graphite, slope = 0.532 (kJ/mol)/kbar']);
+
+