2013-3-6 Describing Mixture Composition Example:The molar analysis of a gas mixture is 50%N2,35%CO2,and 15%02.Determine (a)the apparent molecular weight of the mixture and (b)the analysis in terms of mass fractions. Solution: (a)The apparent molecular weight of the mixture is found using Eq.12.9 and molecular weights (rounded)from Table A-1 M=0.50(28)+0.35(44)+0.1532)=34.2kg/kmol Describing Mixture Composition ent,in kmol,is equal to its mole fraction,as shown in column (ii). Column(i)lists the respective molecular weights. Column (iv)gives the mass of each component,in kg per kmole of mixture,obtained using=nM,(Eq.12.1). hemasactionsnlsedaspercenagesncohecnatioial e values in column(iv)by the column 0.50 14 40.94 CO, 025 15.4 45.03 0.15 1404 42013-3-6 4 Describing Mixture Composition Example: The molar analysis of a gas mixture is 50% N2, 35% CO2, and 15% O2. Determine (a) the apparent molecular weight of the mixture and (b) the analysis in terms of mass fractions. (a) The apparent molecular weight of the mixture is found using Eq. 12.9 and molecular weights (rounded) from Table A-1 M = 0.50( ) 28 + 0.35(44)+ 0.15(32) = 34.2 kg/kmol Solution: Describing Mixture Composition ►Then, the amount ni of each component, in kmol, is equal to its mole fraction, as shown in column (ii). ►Column (iii) lists the respective molecular weights. ►Column (iv) gives the mass mi of each component, in kg per kmole of mixture, obtained using mi = ni Mi (Eq. 12.1). ►The mass fractions, listed as percentages in column (v), are obtained by dividing the values in column (iv) by the column total and multiplying by 100. (i) Component (ii) ni × (iii) Mi = (iv) mi (v) mfi % N2 0.50 × 28 = 14 40.94 CO2 0.35 × 44 = 15.4 45.03 O2 0.15 × 32 = 4.8 14.04 1.00 34.2 100 (b) Although the actual amount of mixture is not known, the calculations can be based on any convenient amount. We use 1 kmol of mixture