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+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.4\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.4  (page no. 325) 
+// Solution
+
+//five moles of oxygen and 10 moles of hydrogen are mixed
+//The total number of moles is 10+5=15.Therefore,mole fraction of each constituent is
+xO2=5/15; //The mole fraction of oxygen 
+xH2=10/15; //The mole fraction of hydrogen
+printf("The mole fraction of oxygen is %f and of hydrogen is %f\n",xO2,xH2);
+//the molecular weight of mixture=sum of products of mole fraction of each gas component(MW of O2=32 and MW of H2=2.016)
+printf("The molecular weight of the final mixture is %f\n",((5/15)*32)+((10/15)*2.016));
+R=1545/32; //the gas constant of oxygen
+T=460+70; //absolute temperature //Unit:R
+p=14.7; //pressure //psia
+//The partial volume of the oxygen can be found as follows:per pound of oxygen,
+//p*vO2=R*T;
+vO2=(R*T)/(p*144); //ft^3/lbm //1 in^2=144 ft^2  
+//Because there are 5 moles of oxygen,each containing 32 lbm,
+VO2=vO2*5*32; //ft^3 //partial volume of oxygen
+printf("The partial volume of oxygen is %f ft^3\n",VO2);
+//For the hydrogen,we can simplify the procedure by noting that the fraction of the total volume occupied by the oxygen is the same as its mole fraction.Therefore,
+Vm=3*VO2; //total volume occupied //ft^3
+printf("The mixture volume is %f ft^3\n",Vm);
+//and the hydrogen volume
+VH2=Vm-VO2; //Ft^2 //partial volume of hydrogen
+printf("From simplified procedure,The partial volume of hydrogen is %f ft^3\n",VH2);
+
+//We could obtain the partial volume of hydrogen by proceeding as we did for the oxygen.Thus,
+//p*vH2=R*T;
+R=1545/2.016; //the gas constant of hydrogen
+vH2=(R*T)/(p*144); //ft^3/lbm //1 in^2=144 ft^2  
+//Because there are 10 moles of hydrogen,each containing 2.016 lbm,
+VH2=vH2*10*2.016; //ft^3 //partial volume of hydrogen
+printf("The partial volume of hydrogen is %f ft^3\n\n",VH2);
+//Which checks our previous values.
+
+
+printf("From another method,\n");
+//As an alternative to the foregoing,we could also use the fact that at 14.7 psia and 32F a mole of any gas occupies a volume of 358 ft^3.
+printf("At 70F and 14.7 psia,a mole occupies %f ft^3\n",358*((460+70)/(460+32))); 
+//Therefore, 5 moles of oxygen occupies 
+VO2=5*358*((460+70)/(460+32)); //The partial volume of oxygen //ft^3
+printf("The partial volume of oxygen is %f ft^3\n",VO2);
+//and 10 moles of hydrogen occupies
+VH2=10*358*((460+70)/(460+32)); //The partial volume of hydrogen //ft^3
+printf("The partial volume of hydrogen is %f ft^3\n",VH2);
+//Both values are in good agreement with the previous calculations.
+