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+
+// Theory and Problems of Thermodynamics
+// Chapter 5
+//Second Law of Thermodynamcis
+// Example 16
+
+clear ;clc;
+
+//Given data
+N = 1 //number of moles of air to be seperated
+P = 0.1*1e6 //pressure of of air before seperation in Pa
+T = 300 //temperature of air before seperation in K
+R = 8.314 //gas constant
+
+// Calculations for determining the minimum work required
+// From first and second law of thermodynamics
+// dU = T dS - P dV
+// initial and final temperatures are identical => dU = 0
+S_mix = -R*(0.21*log(0.21) + 0.79*log(0.79)) // entorphy for mixing
+S_sep = - S_mix // entrophy change for seperation
+W = T * S_sep // minimum work to be done
+
+// Output Results
+mprintf('Minimum work to be done for seperation = %6.1f kJ', -W)
+