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+clear;
+clc;
+printf("\t\t\tProblem Number 6.17\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.17 (page no. 261) 
+// Solution
+
+//data of problem6.16
+cp=0.24; //Specific heat at constant pressure //Btu/lbm*R
+p2=15; //psia //final pressure
+p1=100; //psia //initial pressure
+T2=460+0;  //absolute  final temperature //unit:R
+T1=460+100; //absolute initial temperature //unit:R
+J=778; //conversion factor
+R=1545/29; //moleculer weight=29 //Unit:ft*lbf/lbm*R //constant of proportionality
+//Because cp and R are given,let us first solve for cv,
+//cp=(R*k)/(J*(k-1))
+k=(cp*J)/((cp*J)-R); //k=cp/cv //ratio of specific heats
+printf("Ratio of specific heats k is %f\n",k);
+//k=cp/cv
+cv=cp/k; //Specific heat at constant volume //Btu/lbm*R
+printf("Specific heat at constant volume is %f Btu/lbm*R\n",cv);
+//Now, deltas=(cv*log(p2/p1))+(cp*log(v2/v1));
+//But, v2/v1=(T2*p1)/(T1*p2)
+v2byv1=(T2*p1)/(T1*p2); // v2/v1 //unitless
+deltas=(cv*log(p2/p1))+(cp*log(v2byv1)); //The change in enthalpy  //unit:Btu/lbm*R
+printf("The change in enthalpy is %f Btu/lbm*R\n",deltas);
+//The agreement is very good.