Mole Fractions, Partial P and V

13-12 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, the average molar mass, and gas constant are to be determined.

Properties The molar masses of O2, N2, and CO2 are 32.0, 28.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis (a) The total mass of the mixture is m = m + m + m = 5 kg + 8 kg +10 kg = 23 kg m O2 N2 CO2 Then the mass fraction of each component becomes

m O 5 kg mf = 2 = = 0.217 5 kg O O2 m 23 kg 2 m 8 kg N 2 m N 8 kg mf = 2 = = 0.348 N2 10 kg CO m m 23 kg 2

m CO 10 kg mf = 2 = = 0.435 CO2 m m 23 kg (b) To find the mole fractions, we need to determine the mole numbers of each component first,

m O 5 kg N = 2 = = 0.156 kmol O2 M 32 kg/kmol O2

m N 8 kg N = 2 = = 0.286 kmol N2 M 28 kg/kmol N2

m CO 10 kg N = 2 = = 0.227 kmol CO2 M 44 kg/kmol CO2 Thus, N = N + N + N = 0.156 kmol + 0.286 kmol + 0.227 kmol = 0.669 kmol m O2 N2 CO2 and

NO 0.156 kmol y = 2 = = 0.233 O2 Nm 0.699 kmol

NN 0.286 kmol y = 2 = = 0.428 N2 Nm 0.669 kmol

NCO 0.227 kmol y = 2 = = 0.339 CO2 Nm 0.669 kmol (c) The average molar mass and gas constant of the mixture are determined from their definitions:

m m 23 kg M m = = = 34.4 kg/kmol Nm 0.669 kmol and Ru 8.314 kJ/kmolЧK R m = = = 0.242 kJ/kgЧK M m 34.4 kg/kmol 13-15E The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined.

Properties The molar masses of H2, and N2 are 2.0 and 28.0 lbm/lbmol, respectively (Table A-1E). Analysis The mass of each component is determined from

N H = 5 lbmol ҫҫ ® mH = N H M H = (5 lbmol)(2.0 lbm/lbmol)= 10 lbm 2 2 2 2 N N = 4 lbmol ҫҫ ® mN = N N M N = (4 lbmol)(28 lbm/lbmol)= 112 lbm 5 lbmol H 2 2 2 2 2 4 lbmol N The total mass and the total number of moles are 2 m = m + m = 10 lbm +112 lbm = 122 lbm m H2 N2 N = N + N = 5 lbmol + 4 lbmol = 9 lbmol m H2 N2 The molar mass and the gas constant of the mixture are determined from their definitions,

m m 122 lbm M m = = = 13.56 lbm/lbmol Nm 9 lbmol and

Ru 1.986 Btu/lbmolЧR Rm = = = 0.1465 Btu/lbm ЧR Mm 13.56 lbm/lbmol

13-31 The masses of the constituents of a gas mixture at a specified pressure and temperature are given. The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined.

Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture.

Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol, respectively (Table A-1) Analysis The mole numbers of the constituents are m CO2 1 kg mCO = 1 kg ҫҫ ® NCO = = = 0.0227 kmol 2 2 M 44 kg / kmol CO2 1 kg CO 2 m 3 kg CH 4 3 kg CH mCH = 3 kg ҫҫ ® NCH = = = 0.1875 kmol 4 4 4 M 16 kg / kmol CH 4 N = N + N = 0.0227 km ol+ 0.1875 km ol= 0.2102 km ol m CO2 CH 4 300 K 200 kPa NCO 0.0227 kmol y = 2 = = 0.108 CO2 Nm 0.2102 kmol N 0.1875 kmol y = CH4 = = 0.892 CH 4 Nm 0.2102 kmol Then the partial pressures become P = y P = 0.108 200 kPa = 21.6 kPa CO2 CO2 m ( )( ) P = y P = 0.892 200 kPa = 178.4 kPa CH4 CH4 m ( )( ) The apparent molar mass of the mixture is mm 4 kg Mm = = = 19.03 kg / km ol Nm 0.2102 km ol

13-31 The masses of the constituents of a gas mixture at a specified pressure and temperature are given. The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined.

Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture.

Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol, respectively (Table A-1) Analysis The mole numbers of the constituents are m CO2 1 kg mCO = 1 kg ҫҫ ® NCO = = = 0.0227 kmol 2 2 M 44 kg / kmol CO2 1 kg CO 2 mCH 3 kg 4 3 kg CH mCH = 3 kg ҫҫ ® NCH = = = 0.1875 kmol 4 4 4 M 16 kg / kmol CH 4 N = N + N = 0.0227 km ol+ 0.1875 km ol= 0.2102 km ol m CO2 CH 4 300 K 200 kPa NCO 0.0227 kmol y = 2 = = 0.108 CO2 Nm 0.2102 kmol N 0.1875 kmol y = CH4 = = 0.892 CH 4 Nm 0.2102 kmol Then the partial pressures become P = y P = 0.108 200 kPa = 21.6 kPa CO2 CO2 m ( )( ) P = y P = 0.892 200 kPa = 178.4 kPa CH4 CH4 m ( )( ) The apparent molar mass of the mixture is

mm 4 kg Mm = = = 19.03 kg / km ol Nm 0.2102 km ol

13-33 The masses, temperatures, and pressures of two gases contained in two tanks connected to each other are given. The valve connecting the tanks is opened and the final temperature is measured. The volume of each tank and the final pressure are to be determined.

Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture

Properties The molar masses of N2 and O2 are 28.0 and 32.0 kg/kmol, respectively (Table A-1) Analysis The volumes of the tanks are mRT (1 kg)(0.2968 kPam 3 /kg K)(298 K) V = ж ц = Ч Ч = 0.295 m 3 N2 з ч 1 kg N 3 kg O и P шN 300 kPa 2 2 2 3 25C 25C жmRT ц (3 kg)(0.2598 kPaЧm /kg ЧK)(298 K) 3 VO = з ч = = 0.465 m 2 P 500 kPa 300 kPa 500 kPa и шO2 V =V +V = 0.295 m 3 + 0.465 m 3 = 0.76 m 3 total N2 O2 Also, m N2 1 kg m N = 1 kg ҫҫ ® NN = = = 0.03571 kmol 2 2 M 28 kg/kmol N2 m O2 3 kg m O = 3 kg ҫҫ ® NO = = = 0.09375 kmol 2 2 M 32 kg/kmol O2 N = N + N = 0.03571 kmol + 0.09375 kmol = 0.1295 kmol m N2 O2 Thus, 3 жNRuT ц (0.1295 kmol)(8.314 kPaЧm /kmolЧK)(298 K) Pm = з ч = = 422.2 kPa з V ч 3 и шm 0.76 m

13-35E A mixture is obtained by mixing two gases at constant pressure and temperature. The volume and specific volume of the mixture are to be determined. Properties The densities of two gases are given in the problem statement. Analysis The volume of constituent gas A is m 1 lbm V = A = = 1000 ft 3 A 3 r A 0.001 lbm/ft and the volume of constituent gas B is 1 lbm gas A m 2 lbm V = B = = 1000 ft 3 B 3 r B 0.002 lbm/ft 2 lbm gas B Hence, the volume of the mixture is 3 V =V A +VB = 1000 +1000 = 2000 ft The specific volume of the mixture will then be V 2000 ft 3 v = = = 666.7 ft 3 /lbm m (1+ 2) lbm