then the mass fraction of each component becomeschapter 12 gas mixtures then the mass fraction of...
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Chapter 12 Gas Mixtures
Then the mass fraction of each component becomes
5 kg
23 kg
8kg .
23kg
10 kg
= 0.217
--
--
=~=
mm
mN2
mm
mCO2-=-=0.435mj( -
CO2 -mm kj 1<g
(b) To find the mole fractions, we need to determine the mole numbers of each component first,
I
Thus,
N m = No2 + NN2 + NCO2 = 0.156 kmol + 0.286 kmol + 0.227 kmol = 0.669 kmo1
and
f!1
M m = !!!.!!!.- = 23 kgN () f;f;a lr nl = 34.4 kg / kmol
and
Rm = ~ -Mm -8.314 kJ/kmol.K
= 0.242 kJ/kg-K34.4 kg/kmol
12-4
Analysis (a) The total mass of the mixture is
mm = mo2 +mN2 +mco2 = 5 kg+8 kg+l0kg = 23 kg
Chapter 12 Gas Mixtures
12-33 The masses of the constituents of a gas mixture at a specified pressure and temperature are given. Thepartial 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 asan ideal gas mixture.
Properties The molar masses ofCO2 and C~ are 44.0 and 16.0 kg/kmol, respectively (Table A-I)
Analysis The mole numbers of the constituents are
mCO2 = 1 kg 1 kg CO2
3 kg CH4~
300K200 kPa
mCH4 = 3 kg --+
Nm =NCo2 +NCH4 =O.O227kmo]+0.]875kmo]=0.2]O2kmo]
= 0.108
= 0.892
Then the partial pressures become
Pco = Yco Pm = (0. I 08X200 kPa)= 21.6 kPa2 2
PCH4 =YCH4Pm =(0.892X200kPa)=178.4 kPa
The apparent molar mass of the mixture is
~
12-9
Chapter 12 Gas Mixtures
12-36 The volumetric fractions of the constituents of a gas mixture at a specified pressure and temperatureare given. The mass fraction and partial pressure of each gas are to be determined.
Assumptions Under specified conditions all N2, O2 and C02 can be treated as ideal gases, and the mixtureas an ideal gas mixture.
Properties The molar masses ofN2, O2 and C02 are 28.0, 32.0, and 44.0 kg/kmol, respectively {Table A-l)
Analysis For convenience, consider 100 kmol of mixture. Then the mass of each component and the totalmass are
65% N2
20% O2
15% CO2
350K
300 kPa
NN2 = 65 krno1 ~ mN2 = NN2M N2 = (65 krno1X28 kg/krno1)= 1820 kg
No2 = 20 krno1 ~ m02 = N02M O2 = (20 krno1X32 kg/krno1)= 640 kg
VC02 =15krno1~mC02 =NCO2Mc02 =(15krn01X44kg/krn01)=660kg
mm = mN2 +mO2 +mCO2 = 1,820 kg +640 kg +660 kg = 3,120 kg
Then the mass fraction of each component (gravimetric analysis) becomes
For ideal gases, the partial pressure is proportional to the mole fraction, and is determined from
PN2 =YN2Pm =(0.65X300kPa)=195 kPa
PO2 = Yo2 p m = (0.20X300 kPa)= 60 kPa
PCO2 =YCO2Pm =(0.15X300 kPa)=45 kPa
12-11
:.-
Chapter 12 Gas Mixtures
12-49 The temperatures and pressures of two gases forming a mixture are given. The final mixturetemperature and pressure are to be determined.
Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture asan ideal gas mixture. 2 There are no other forms of work involved.
Properties The molar masses and specific heats of Ne and Ar are 20.18 kg/kmol, 39.95 kg/kmol, 0.6179kJ/kg.oC, and 0.3122 kJ/kg.oC, respectively. (Tables A-l and A-2b).
Analysis The mole number of each gas is
/ ) 3 !:1YJ- = ( 100kPa)(0.45m ) ~ ~ ~ -~
Ru TI Ne
~VI
= O.O185kmoJN = l -
Ne Ne
100 kPa
20°C
Ar
200 kPa
50°C
(8.314kPa .m3 /kmol. K)(293K)
(200kPa)(0.45m3 )= O.O335kmolNAr = =
\115kJ
(8.314kPa .m 3 /kmol. K)(323K)RuT1 /' Ar
Thus,
N m = NNe + N AT = 0.0185 kmol + 0.0335 kmol = 0.0520 kmol
(a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is aclosed system with W = 0. Then the conservation of energy equation for this closed system reduces to
Ein -Eoul = Msyslem
-QOUI = t!U = t!U Ne + t!U Ar -+ -QOUI = [mC,.(T m -TI )]Ne + [mC,,(T m -~ )JAr
Using Cv values at room temperature and noting that m = NM, the final temperature of the mixture is
determined to be
-15kJ = {0.0185 x 20. 18kgXO.6179kJlkg .oCXT m -20°C)
+ {0.0335 x 39.95kgXo.3122kJlkg .oCXT m -50°C)
T m = 16.2°C {289.2K)
(b) The final pressure in the tank is determined from
Pm = NmRuTm
Vm
~
12-17