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+clear;

+clc;

+printf("\t\t\tProblem Number 6.18\n\n\n");

+// Chapter 6: The Ideal Gas

+// Problem 6.18 (page no. 261) 

+// Solution

+

+//data,

+cp=0.9093; //Specific heat at constant pressure //kJ/kg*R

+p2=150; //kPa //final pressure

+p1=500; //kPa //initial pressure

+T2=273+0;  //final temperature //Celsius converted to kelvin

+T1=273+100; //initial temperature //Celsius converted to kelvin

+//J=778; //conversion factor

+R=8.314/32; //moleculer weight of oxygen=32 //Unit:ft*lbf/lbm*R //constant of proportionality

+//Using equation, and dropping J gives,

+deltas=(cp*(log(T2/T1)))-((R)*(log(p2/p1))); //change in entropy //kJ/kg*K

+//For 2 kg,

+deltaS=2*deltas; //The change in enthalpy in kJ/K

+printf("For 2 kg oxygen,The change in enthalpy is %f kJ/K\n",deltaS);