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