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+//ques7
+//Evaluation of the du of an Ideal Gas
+clear
+clc
+//(a) One way of determining the change in internal energy of air is to read the values at T1 and T2 from Table A–17 and take the difference
+u1=214.07;//internal energy in kJ @ 300K
+u2=434.78;//Internal energy in kJ @ 600K
+du=u2-u1;//Change in in internal energy in kJ
+printf('(a) Change in Internal Energy(from air data table ) = %.2f kJ \n',du);
+  
+//(b) the functional form of the specific heat (Table A–2c)
+//constant
+a=28.11;
+b=0.1967*10^-2;
+c=0.4802*10^-5;
+d=-1.966*10^-9;
+Ru=8.314;//Universal gas constant
+//Cp=a+b*T+c*T^2+d*T^3;
+//Cv=Cp-Ru
+T1=300;//Initial Temp in K
+T2=600;//Final  temp in K
+U=integrate('a-Ru+b*T+c*T^2+d*T^3','T',T1,T2);
+M=28.97;//molicular mass
+u=U/M;//specific internal energy in KJ/Kg
+printf(' (b) Change in Internal Energy using functional form of the specific heat =  %.2f kJ\n',u);
+
+//(c)the average specific heat value (Table A–2b)
+Tavg=(T1+T2)/2;//avg temp in K
+Cv=0.733;//heat capacity at constant volume in kJ/K @Tavg from Table-2A
+u=Cv*(T2-T1);//average change in internal energy in kJ/kg
+printf(' (c) Change in Internal Energy using avg specific heat value = %.2f kJ/kg ',u);
+\ No newline at end of file