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+// Given:-
+v = 2450.00 // volume of gaseous products in cm^3
+P = 7.00 // pressure of gaseous product in bar
+T = 867.00 // temperature of gaseous product in degree celcius
+T0 = 300.00 // in kelvin
+P0 = 1.013 // in bar
+
+// From table A-22
+u = 880.35 // in kj/kg
+u0 = 214.07 // in kj/kg
+s0T = 3.11883 // in kj/kg.k
+s0T0 = 1.70203 // in kj/kg.k
+
+// Calculations
+
+e = (u-u0) + (P0*(8.314/28.97)*(((T+273)/P)-(T0/P0))) - T0*(s0T-s0T0-(8.314/28.97)*log(P/P0)) // kj/kg
+
+// Results
+printf( ' The specific exergy of the gas is %.3f kJ/kg.',e)