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+clear;
+clc;
+disp('Example 8.17');
+
+// aim : To determine
+// (a) the partial pressure of each gas in the vessel
+// (b) the volume of the vessel
+// (c)  the total pressure in the gas when temperature is raised to228 C
+
+// given values
+MO2 = 8;//  mass of O2, [kg]
+MN2 = 7;//  mass of N2, [kg]
+MCO2 = 22;//  mass of CO2, [kg]
+
+P = 416;// total pressure in the vessel, [kN/m^2]
+T = 273+60;// vessel temperature, [K]
+R = 8.3143;// gas constant, [kJ/kmol K]
+
+mO2 = 32;// molculer mass of O2 
+mN2 = 28;// molculer mass of N2
+mCO2 = 44;// molculer mass of CO2
+
+// solution
+// (a)
+n1 = MO2/mO2;// moles of O2, [kmol]
+n2 = MN2/mN2;// moles of N2, [kmol]
+n3 = MCO2/mCO2;// moles of CO2, [kmol]
+
+n = n1+n2+n3;// total moles in the vessel, [kmol]
+// since,Partial pressure is proportinal, so
+P1 = n1*P/n;// partial pressure of O2, [kN/m^2]
+P2 = n2*P/n;// partial pressure of N2, [kN/m^2]
+P3 = n3*P/n;// partial pressure of CO2, [kN/m^2]
+
+mprintf('\n (a)The partial pressure of O2 is  =  %f  kN/m^2,\n,   The partial pressure of N2 is  =  %f  kN/m^2,\n    The partial pressure of CO2 is  =  %f  kN/m^2,\n',P1,P2,P3);
+
+// (b)
+// assuming ideal gas 
+V = n*R*T/P;// volume of the container, [m^3]
+mprintf('\n (b) The volume of the container is  =  %f  m^3\n',V);
+
+// (c)
+T2 = 273+228;// raised vessel temperature, [K]
+// so volume of vessel  will constant , P/T=constant
+P2 = P*T2/T;// new pressure in the vessel , [kn/m62]
+mprintf('\n (c) The new total pressure in the vessel is  =  %f  kN/m^2\n',P2);
+
+//  End