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+clear;
+clc;
+disp('Example 5.22');
+
+//  aim : To determine the
+//  (a) new pressure of the air in the receiver
+//  (b) specific enthalpy of air at 15 C
+
+//  Given values
+V1 = .85;//  [m^3]
+T1 = 15+273;//  [K]
+P1 = 275;// pressure,[kN/m^2]
+m = 1.7;//  [kg]
+cp = 1.005;//  [kJ/kg*K]
+cv = .715;//  [kJ/kg*K]
+
+//  solution
+
+//  (a)
+
+R = cp-cv;//  [kJ/kg*K]
+//  assuming m1 is original mass of the air, using P*V=m*R*T
+m1 = P1*V1/(R*T1);//  [kg]
+m2 = m1+m;//  [kg]
+//  again using P*V=m*R*T
+//  P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so
+P2 = P1*m2/m1;//  [kN/m^2]
+mprintf('\n (a) The new pressure of the air in the receiver is = %f kN/m^2\n',P2);
+
+//  (b)
+//  for 1 kg of air, h2-h1=cp*(T1-T0)
+//  and if 0 is chosen as the zero enthalpy, then
+h = cp*(T1-273);//  [kJ/kg]
+mprintf('\n (b) The specific  enthalpy of the air at 15 C is =  %f kJ/kg\n',h);
+
+//  End