// Given:- nN2 = 0.79 // initial moles of nitrogen in kmol pN2 = 2.0 // initial pressure of nitrogen in bars TN2 = 250.0 // initial temperature of nitrogen in kelvin nO2 = 0.21 // initial moles of oxygen in kmol pO2 = 1.0 // initial pressure of oxygen in bars TO2 = 300.0 // initial temperature of oxygen in kelvin // Part(a) MN2 = 28.01 // molar mass of nitrogen in kg/kmol MO2 = 32.0 // molar mass of oxygen in kg/kmol // Calculations // With the help of table A-20 cvbarN2 = MN2*0.743 // in kj/kmol.K cvbarO2 = MO2*0.656 // in kj/kmol.K T2 = (nN2*cvbarN2*TN2+nO2*cvbarO2*TO2)/(nN2*cvbarN2+nO2*cvbarO2) // Result printf( 'The final temperature of the mixture in kelvin is: %f',T2); // Part(b) // Calculation p2 = ((nN2+nO2)*T2)/(nN2*TN2/pN2 + nO2*TO2/pO2) // Result printf( 'The final pressure of the mixture in bar is: %f',p2); // Part(c) Rbar = 8.314 // universal gas constant // Calculations cpbarN2 = cvbarN2 + Rbar cpbarO2 = cvbarO2 + Rbar yN2 = nN2/(nN2+nO2) // mole fraction of N2 yO2 = nO2/(nN2+nO2) // mole fraction of O2 sigma = nN2*(cpbarN2*log(T2/TN2)-Rbar*log(yN2*p2/pN2)) + nO2*(cpbarO2*log(T2/TO2)-Rbar*log(yO2*p2/pO2)) // Result printf( 'The amount of entropy produced in the mixing process, in kJ/K is: %f',sigma);