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+// Example 2_7
+clc;funcprot(0);
+// Given data
+T=500+273;// K
+rho=24;// The density in kg/m^3
+R=0.462;// kJ/kg.K
+v=1/rho;// m^3/kg
+
+// Calculation
+// (a)
+P=rho*R*T;// kPa
+printf("\n(a)Using the ideal gas equation,The pressure of steam(P)=%4.0f kPa.",P);
+// (b)
+// Using values for a and b from Table B-8,the vander Waals equation provides
+a=1.703;
+b=0.00169;
+P=((R*T)/(v-b))-(a/v^2);// kPa
+printf("\n(b)Using the vander Waals equation,the pressure of steam(P)=%4.0f kPa.",P);
+// (c)
+// Using values for a and b from Table B-8,the Redlich-Kwong equation provides
+a=43.9;
+b=0.00117;
+P=((R*T)/(v-b))-(a/(v*(v+b)*sqrt(T)));// kPa
+printf("\n(c)Using the Redlich-Kwong equation,the pressure of steam(P)=%4.0f kPa.",P);
+// (d)
+T_c=647.4;// The critical temperature in K
+T_R=T/T_c;// The reduced temperature
+P_c=8000;// The critical pressure in kPa
+P_R=P/P_c;// The reduced pressure 
+// By using the reduced temperature and the reduced pressure
+Z=0.93;// The compressibilty factor
+P=(Z*R*T)/v;// kPa
+printf("\n(d)By using the compressibilty factor,the pressure of steam(P)=%4.0f kPa.",P);
+// (e)
+// By using the steam tables,
+P=8000;// kPa
+printf("\n(e)By using the steam tables,the pressure of steam(P)=%4.0f kPa.",P);