Chapter 13: Gas Mixtures

Gas Mixtures

MATTER

is

solid liquid gas

melts to

freezes to

evaporates to

condenses to

anything that has mass and takes

up space

can be

can be

can be

MATTER FACT

Water can be found naturally on earth as a solid, liquid, and a gas.

MIXTURES

Two or more kinds of matter

put together

are

solutions

separated

can be

can be

by

filtration

using sieves

using magnets

floating vs. sinking

evaporation

sorting

distillation

chromatography

one kind of material dissolves

in another

made when

cannot be separated by

filtration

which

4

Many thermodynamic applications involve mixtures of ideal

gases. That is, each of the gases in the mixture individually

behaves as an ideal gas. In this section, we assume that the

gases in the mixture do not react with one another to any

significant degree.

We restrict ourselves to a study of only ideal-gas mixtures. An

ideal gas is one in which the equation of state is given by

PV mRT or PV NR Tu

Air is an example of an ideal gas mixture and has the following approximate

composition.

Component % by Volume

N2 78.10

O2 20.95

Argon 0.92

CO2 + trace elements 0.03

5

Definitions

Consider a container having a volume V that is filled with a mixture of k different

gases at a pressure P and a temperature T.

A mixture of two or more gases of fixed chemical composition is called a nonreacting

gas mixture. Consider k gases in a rigid container as shown here. The properties of

the mixture may be based on the mass of each component, called gravimetric

analysis, or on the moles of each component, called molar analysis.

k gases

T = Tm V = Vm

P = Pm m = mm

The total mass of the mixture mm and the total moles of mixture

Nm are defined as

m m N Nm i

i

k

m i

i

k

1 1

and

6

mfm

my

N

Ni

i

m

ii

m

and

mf yi

i

k

i

i

k

1 1

1 = 1 and Note that

The composition of a gas mixture is described by specifying either the mass fraction

mfi or the mole fraction yi of each component i.

The mass and mole number for a given component are related through the molar

mass (or molecular weight).

m N Mi i i

To find the average molar mass for the mixture Mm , note

m m N M N Mm i

i

k

i i m m

i

k

1 1

Solving for the average or apparent molar mass Mm

Mm

N

N

NM y M kg kmolm

m

m

i

m

i

i

k

i i

i

k

1 1

( / )

7

The apparent (or average) gas constant of a mixture is expressed as

RR

MkJ kg Km

u

m

( / )

Can you show that Rm is given as

R mf Rm i i

i

k

1

To change from a mole fraction analysis to a mass fraction analysis, we can show

that

mfy M

y Mi

i i

i i

i

k

1

To change from a mass fraction analysis to a mole fraction analysis, we can show that

ymf M

mf Mi

i i

i i

i

k

/

/1

8

Volume fraction (Amagat model)

Divide the container into k subcontainers, such that each subcontainer has only one

of the gases in the mixture at the original mixture temperature and pressure.

Amagat's law of additive volumes states that the volume of a gas mixture is equal to

the sum of the volumes each gas would occupy if it existed alone at the mixture

temperature and pressure.

Amagat's law: V V T Pm i m m

i

k

( , )1

The volume fraction of the vfi of any component is

vfV T P

Vi

i m m

m

( , )

and

vf i

i

k

1

= 1

9

For an ideal gas mixture

VN R T

Pand V

N R T

Pi

i u m

m

mm u m

m

Taking the ratio of these two equations gives

vfV

V

N

Nyi

i

m

i

m

i

The volume fraction and the mole fraction of a component in an ideal gas mixture are

the same.

Partial pressure (Dalton model)

The partial pressure of component i is defined as the product of the mole fraction and

the mixture pressure according to Dalton’s law. For the component i

P y Pi i m

Dalton’s law: P P T Vm i m m

i

k

( , )1

10

Now, consider placing each of the k gases in a separate container having the volume

of the mixture at the temperature of the mixture. The pressure that results is called

the component pressure, Pi' .

PN R T

Vand P

N R T

Vi

i u m

m

mm u m

m

'

Note that the ratio of Pi' to Pm is

P

P

V

V

N

Nyi

m

i

m

i

m

i

'

For ideal-gas mixtures, the partial pressure and the component pressure are the

same and are equal to the product of the mole fraction and the mixture pressure.

11

Other properties of ideal-gas mixtures

The extensive properties of a gas mixture, in general, can be determined by summing

the contributions of each component of the mixture. The evaluation of intensive

properties of a gas mixture, however, involves averaging in terms of mass or mole

fractions:

U U m u N u

H H m h N h

S S m s N s

m i

i

k

i i

i

k

i i

i

k

m i

i

k

i i

i

k

i i

i

k

m i

i

k

i i

i

k

i i

i

k

1 1 1

1 1 1

1 1 1

(kJ)

(kJ)

(kJ / K)

and u mf u u y u

h mf h h y h

s mf s s y s

m i i

i

k

m i i

i

k

m i i

i

k

m i i

i

k

m i i

i

k

m i i

i

k

1 1

1 1

1 1

and (kJ / kg or kJ / kmol)

and (kJ / kg or kJ / kmol)

and (kJ / kg K or kJ / kmol K)

C mf C C y C

C mf C C y C

v m i v i

i

k

v m i v i

i

k

p m i p i

i

k

p m i p i

i

k

, , , ,

, , , ,

1 1

1 1

and

and

12

These relations are applicable to both ideal- and real-gas mixtures. The properties or

property changes of individual components can be determined by using ideal-gas or

real-gas relations developed in earlier chapters.

Ratio of specific heats k is given as

kC

C

C

Cm

p m

v m

p m

v m

,

,

,

,

The entropy of a mixture of ideal gases is equal to the sum of the entropies of the

component gases as they exist in the mixture. We employ the Gibbs-Dalton law that

says each gas behaves as if it alone occupies the volume of the system at the

mixture temperature. That is, the pressure of each component is the partial pressure.

For constant specific heats, the entropy change of any component is

13

The entropy change of the mixture per mass of mixture is

The entropy change of the mixture per mole of mixture is

14

In these last two equations, recall that

P y P

P y P

i i m

i i m

, , ,

, , ,

1 1 1

2 2 2

Example 13-1

An ideal-gas mixture has the following volumetric analysis

Component % by Volume

N2 60

CO2 40

(a)Find the analysis on a mass basis.

For ideal-gas mixtures, the percent by volume is the volume fraction. Recall

y vfi i

15

Comp. yi Mi yiMi mfi = yiMi /Mm

kg/kmol kg/kmol kgi/kgm

N2 0.60 28 16.8 0.488

CO2 0.40 44 17.6 0.512

Mm = yiMi = 34.4

(b) What is the mass of 1 m3 of this gas when P = 1.5 MPa and T = 30oC?

RR

MkJ kg K

kJ

kmol Kkg

kmol

kJ

kg K

mu

m

( / )

.

.

.

8 314

34 4

0 242

mP V

R T

MPa m

kJ kg K K

kJ

m MPa

kg

mm m

m m

15 1

0 242 30 273

10

20 45

3 3

3

. ( )

( . / ( ))( )

.

16

(c) Find the specific heats at 300 K.

Using Table A-2, Cp N2 = 1.039 kJ/kgK and Cp CO2 = 0.846 kJ/kgK

C mf C

kJ

kg K

p m i p i

m

, , ( . )( . ) ( . )( . )

.

1

2

0 488 1039 0512 0846

0 940

C C RkJ

kg K

kJ

kg K

v m p m m

m

m

, , ( . . )

.

0 940 0 242

0 698

17

(d) This gas is heated in a steady-flow process such that the temperature is

increased by 120oC. Find the required heat transfer. The conservation of mass and

energy for steady-flow are

( )

( ),

m m m

m h Q m h

Q m h h

mC T T

in

in

p m

1 2

1 1 2 2

2 1

2 1

The heat transfer per unit mass flow is

qQ

mC T T

kJ

kg KK

kJ

kg

inin

p m

m

m

( )

. ( )

.

, 2 1

0 940 120

112 8

18

(e) This mixture undergoes an isentropic process from 0.1 MPa, 30oC, to 0.2 MPa.

Find T2.

The ratio of specific heats for the mixture is

kC

C

p m

v m

,

,

.

..

0 940

0 6981347

Assuming constant properties for the isentropic process

(f) Find Sm per kg of mixture when the mixture is compressed isothermally from 0.1

MPa to 0.2 MPa.

19

But, the compression process is isothermal, T2 = T1. The partial pressures are given

by

P y Pi i m

The entropy change becomes

For this problem the components are already mixed before the compression process.

So, y yi i, ,2 1

Then,

20

s mf s

kg

kg

kJ

kg K

kg

kg

kJ

kg K

kJ

kg K

m i i

i

N

m N

CO

m CO

m

1

2

0 488 0 206 0512 0131

0167

2

2

2

2

( . )( . ) ( . )( . )

.

Why is sm negative for this problem? Find the entropy change using the average

specific heats of the mixture. Is your result the same as that above? Should it be?

(g) Both the N2 and CO2 are supplied in separate lines at 0.2 MPa and 300 K to a

mixing chamber and are mixed adiabatically. The resulting mixture has the

composition as given in part (a). Determine the entropy change due to the mixing

process per unit mass of mixture.

21

Take the time to apply the steady-flow conservation of energy and mass to show that

the temperature of the mixture at state 3 is 300 K.

But the mixing process is isothermal, T3 = T2 = T1. The partial pressures are given by

P y Pi i m

The entropy change becomes

22

But here the components are not mixed initially. So,

y

y

N

CO

2

2

1

2

1

1

,

,

and in the mixture state 3,

y

y

N

CO

2

2

3

3

0 6

0 4

,

,

.

.

Then,

23

Then,

s mf s

kg

kg

kJ

kg K

kg

kg

kJ

kg K

kJ

kg K

m i i

i

N

m N

CO

m CO

m

1

2

0 488 0152 0512 0173

0163

2

2

2

2

( . )( . ) ( . )( . )

.

If the process is adiabatic, why did the entropy increase?

Extra Assignment

Nitrogen and carbon dioxide are to be mixed and allowed to flow through a

convergent nozzle. The exit velocity to the nozzle is to be the speed of sound for the

mixture and have a value of 500 m/s when the nozzle exit temperature of the mixture

is 500°C. Determine the required mole fractions of the nitrogen and carbon dioxide to

produce this mixture. From Chapter 17, the speed of sound is given by

C kRTMixture

N2 and CO2

C = 500 m/s

T = 500oC

NOZZLE

Answer: yN2 = 0.589, yCO2 = 0.411

Gas-Vapor Mixtures

Atmospheric air

The air in the atmosphere normally contains some water vapor and is referred to atmospheric air.

Air that contains no water is called dry air.

Although the amount of water vapor in the air is small, it plays a major role in human comfort. Therefore it is an important consideration in air conditioning applications

The dry air and vapor of atmospheric air in air conditioning application range (temperature changes from -10 ℃ to 50 ℃) can be treated as ideal gas

The pressure of atmospheric air

va ppp

The total pressure of atmospheric air

The partial pressure of dry air

The partial pressure of vapor, increases with the amount of vapor in air

s

p2

T

p1

T1

pmax

Dew-point Temperature

T

p1

T1

1

As to vapor state 1, decrease the temperature , then pressure of vapor won’t change

Dew- point

The dew-point Tdp is defined as the temperature at which condensation begins if the air is cooled at constant pressure

The properties of air

Specific humidity of air

The ratio of the mass of water vapor present in a

unit mass of dry air, denoted by ω

a

v

m

m

aa

aa

vv

vv

TRVp

TRVp

Since Vv=Va and Tv=Ta

a

v

v

a

P

P

R

R

a

v

P

P622.0

v

v

PP

P

622.0

Ra= 287.1 J/kg.K

Rv= 461.4 J/kg.K

Relative humidity of air

The comfort level depends more on the amount of moisture the air holds(mv) relative to the maximum amount of moisture the air can hold at the same temperature(mg).

The ratio of these two quantities is called the relative humidity

g

v

m

m

TR

Vp

TRVp

v

g

v

v

g

v

p

p

The saturated pressure under the temperature of atmospheric air

v

v

PP

P

622.0

g

g

PP

P

622.0

Enthalpy of air

va HHH vvaa hmhm

Specific enthalpy:

v

a

va h

m

mhh

va hh

In the field of air-conditioning, the reference temperature is 0℃. Then:

ha= 1.005t kJ/kg.℃

hv= 2501 + 1.863t kJ/kg.℃

h= 1.005t + ω( 2501 + 1.863t) kJ/kg.℃

2501kJ/kg means

the enthalpy at 0 ℃ 1.863kJ/kg. ℃ is the

average specific heat

of vapor

1.863kJ/kg. ℃ is the

average specific heat

of dry air

Adiabatic Saturation and Wet-Bulb Temperature

Adiabatic saturation

t1, ω1, 1 t2, ω2, 2100%

Unsaturated air Saturated air

hf

Conservation of mass

aaa mmm 21

21 vfv mmm

Conservation of energy

22221111 vvaaffvvaa hmhmhmhmhm

12 vvf mmm

2222121111 )( vvaafvvvvaa hmhmhmmhmhm

2222121111 )( vvaafvvvvaa hmhmhmmhmhm

Dividing by gives: 21 aaa mmm

22212111 )( vafva hhhhh