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+//(12.3) A mixture of 0.3 kg of carbon dioxide and 0.2 kg of nitrogen is compressed from p1 = 1 bar, T1 = 300 K to p2 = 3 bars in a polytropic process for which n=  1.25. Determine (a) the final temperature, in K, (b) the work, in kJ, (c) the heat transfer, in kJ, (d) the change in entropy of the mixture, in kJ/K.
+
+//solution
+
+//variable initialization
+m1 = .3 //mass of CO2 in kg
+m2 = .2 //mass of N2 in kg
+p1 = 1 //in bar
+T1 = 300 //in kelvin
+p2 = 3 //in bar
+n = 1.25
+
+//part(a)
+T2 = T1*(p2/p1)^[(n-1)/n] //in kelvin
+printf('the final temperature in Kelvin is: %f',T2)
+
+//part(b)
+Rbar = 8.314 //universal gas constant in SI units
+M = (m1+m2)/(m1/44 + m2/28) //molar mass of mixture in kg/kmol
+
+W = [(m1+m2)*(Rbar/M)*(T2-T1)]/(1-n) //in kj
+printf('\nthe work in kj is: %f',W)
+
+//part(c)
+//from table A-23
+uCO2T1 = 6939 //internal energy of CO2 on molar mass basis at temperature T1
+uCO2T2 = 9198 //internal energy of CO2 on molar mass basis at temperature T2
+uN2T1 = 6229 //internal energy of N2 on molar mass basis at temperature T1
+uN2T2 = 7770 //internal energy of N2 on molar mass basis at temperature T2
+deltaU = (m1/44)*[uCO2T2-uCO2T1] + (m2/28)*[uN2T2-uN2T1] //internal energy change of the mixture in KJ
+
+//with assumption, The changes in kinetic and potential energy between the initial and final states can be ignored
+Q = deltaU + W
+printf('\nthe heat transfer in kj is: %f',Q)
+
+//part(d)
+//from table A-23
+sbarT2CO2 = 222.475
+sbarT1CO2 = 213.915
+sbarT2N2 = 198.105
+sbarT1N2 = 191.682
+Rbar = 8.314 //universal gas constant
+
+deltaS = (m1/44)*[sbarT2CO2-sbarT1CO2-Rbar*log(p2/p1)] + (m2/28)*[sbarT2N2-sbarT1N2-Rbar*log(p2/p1)]
+printf('\nthe change in entropy of the mixture in kj/k is: %f',deltaS)
+
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