Question

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Chapter 12 Ideal Gas Mixtures and Psychrometric Applications Converting Between Mass Fraction and Mole Fraction Mass Fraction Mole Fraction m/M y, M M mf M cy, MM m/M Example 1: Determine the mf CO2 0.04 MW mix mixture molecular weight mf_N2 0.7 m mix (kg) (kg/kmol), specific volume mf_02 0.2 0.06 (m®/kg), and mole fraction for mf_H20 T(C) 40 a gas mixture given the mass P (bar) 1 fraction, temperature, pressure and volume. V(m3) Example 2: Determine the mixture 0.08 MW mix molecular weight, specific volume v (m^3/kg) (m²/kg), and mass fraction for a gas 0.27 mf CO2 mixture given the mole fraction, 0.11 mf_N2 temperature, and pressure. TC) 60 mf_02 P (bar) mf H2O Y_CO2 Y N2 y 02 Y_H20 2.5 0.54 y CO2 Y N2 y O2 Y H20 1.2

Answer 1

M= molecular weight

n= mole

Me= equivalent molecular weight of the mixture

Req= Equivalent R of the mixture

Hope it will help you.

solution 0-000909 kmole 44 kg mcor mah moz MIRO = 0:2 = 0.06 = 0.06 = 0.04 Mi = 44 mi = 0.04 = 0.7 M₂ = 29 th = 0.7 = 0.2 Mz = 32 ng M4 = 18 n4 Molecular weight of the mixture - Me Me Minit M2M + M373 + M4n4 nit me + N₂ + N4 14*0.04 + ke xo:7) 32x2) 0.04 + (97) 0?) + ( 02) + 44 + (18 XQ:06 + 0:06 ~ 28.175 kg Ikmal 0.0355 Me = 28:175 lg 1 kmole Now PV MRT Protal V total = mtotal Regt Ptotal o total = Rea T & total ReqT Ptotal • Specific volume of mixture - Req = m, Ri+ M2 R2 + Mz R3 + mere mit mt m3 + m4 + 102 xBe)+(0.06X) + (0.7%) 10:04x E) 05299 0.295 kilkgok

mde fraction (HO) = (0.06/18) D 0.295X 313 T = 40°C = 313 K 100 0.9236 Rem3/kg Mole fraction mole of a component total mole mole fraction (CO2) 0.04 47 0.04 0.06 0.2 + 32 18 mole fraction (12) 0.0256 (07/28) (0.04144) + (0:7/28) + (0-2132) + (0-06/18) 0.7044 (0-2/32) = 0.1761 0.03549 mole fraction (O2) = 0.0939 0-03549