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+clear;
+clc;
+disp('Example 5.16');
+
+//  aim : To determine the 
+//  (a) initial partial pressure of the steam and air
+//  (b) final partial pressure of the steam and air
+//  (c) total pressure in the container after heating
+
+//  Given values
+T1 = 273+39;//  initial temperature,[K]
+P1 = 100;//  pressure, [MN/m^2]
+T2 = 273+120.2;//  final temperature,[K]
+
+//  solution
+
+//  (a)
+//  from the steam tables, the pressure of wet steam at 39 C is
+Pw1 = 7;//  partial pressure of wet steam,[kN/m^2]
+//  and by Dalton's law
+Pa1 = P1-Pw1;//  initial pressure of air, [kN/m^2]
+
+mprintf('\n (a) The initial partial pressure of the steam is  =  %f  kN/m^2',Pw1);
+mprintf('\n      The initial partial pressure of the air is  =  %f kN/m^2\n',Pa1);
+
+//  (b)
+//  again from steam table, at 120.2 C the pressure of wet steam is
+Pw2 = 200;//  [kN/m^2]
+
+//  now since volume is constant so assuming air to be ideal gas so for air  P/T=contant, hence
+Pa2 = Pa1*T2/T1 ;//  [kN/m^2]
+
+mprintf(' \n(b) The final partial pressure of the steam is  =  %f kN/m^2',Pw2);
+mprintf('\n     The final partial pressure of the air is  =  %f  kN/m^2\n',Pa2);
+
+//  (c)
+Pt = Pa2+Pw2;//  using dalton's law, total pressure,[kN/m^2]
+mprintf('\n (c) The total pressure after heating is  =  %f  kN/m^2\n',Pt);
+
+//  End