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diff --git a/647/CH3/EX3.6/Example3_6.sce b/647/CH3/EX3.6/Example3_6.sce
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+clear;
+clc;
+
+// Example: 3.6
+// Page: 96
+
+printf("Example: 3.6 - Page: 96\n\n");
+
+// Solution
+
+//*****Data*****//
+P = 15*10^5;// [Pa]
+T = 773;// [K]
+R = 8.314;// [J/mol K]
+//**************//
+
+// Solution (a)
+printf("Ideal Equation of State\n")
+// Applying ideal Eqn. of State:
+Vm = R*T/P;// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (b)
+printf("Van der Wall Equation of State\n");
+a = 0.2303;// [Nm^4/square mol]
+b = 4.3073*10^(-5);// [cubic m/mol]
+deff('[y] = f1(Vm)','y = P - (R*T/(Vm-b)) + (a/Vm^2)');
+Vm = fsolve(Vm,f1);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+//Solution (c)
+printf("Virial Equation of State\n");
+// Z = 1 + B/V
+// (P*V/(R*T)) = (1 + B/V)
+// V^2 - V*R*T/P - B*R*T/P = 0
+B = 1.3697*10^(-5);// [cubic m/mol]
+deff('[y] = f2(Vm)','y = Vm^2 - (Vm*R*T/P) - (B*R*T/P)');
+Vm = fsolve(7,f2);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (d)
+printf("Redlich Kwong Equation of State\n");
+Tc = 190.6;// [K]
+Pc = 45.99*10^5;// [Pa]
+a = 0.4278*R^2*Tc^2.5/Pc;// [N/m^4 square mol]
+b = 0.0867*R*Tc/Pc;// [cubic m/mol]
+deff('[y] = f3(Vm)','y = P - (R*T/(Vm - b)) + (a/((T^0.5)*Vm*(Vm+b)))');
+Vm = fsolve(Vm,f3);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+\ No newline at end of file