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+// Implementation of example 7.5
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+// T for temperature,dS is change in entropy,p for pressure,V is volume
+n=1.3
+p1=500 // kPa
+V1=0.2 // m^3
+V2=0.05 // m^3
+// the fluid is undergoing reversible adiabatic compression according to the law p*(V^1.3)=constant
+p2=p1*(V1/V2)^1.3;
+dH=[n*(p2*V2-p1*V1)]/(n-1);
+dU=dH-(p2*V2-p1*V1);
+dS=0;
+Q12=0;
+W12=-dU;
+printf("change in enthalpy = %.2f kJ \n change in entropy = %.2f  \n change in internal energy = %.2f kJ \n heat transfer = %.2f \n work transfer = %.2f kJ",dH,dS,dU,Q12,W12);
+// end
+\ No newline at end of file