Thermodynamics II

[Meng 2132 ]
School of Mechanical & Industrial  
Engineering (AAiT)
March 2019
CHAPTER

Mixture of Ideal 
Gases
 Content
Chapter 1 Mixture of Ideal Gases
1.1.  Composition of gas mixtures
1.2.  P‐v‐T behavior of Gas Mixtures
1.3. Properties of Gas Mixtures
1.4.  Adiabatic Mixing of Perfect Gases
1.5. Mixing of Ideal Gases initially at different 
Pressure and Temperature
1.1. Composition of gas mixtures
 To determine the properties of a 
mixture, we need to know the 
composition of the mixture as well 
as the properties of the individual 
components.
 There are two ways to describe the 
composition of a mixture: either by
 specifying the number of moles of each 
component, called molar analysis, or
 By specifying the mass of each component, 
called gravimetric analysis
 The subscript m‐ denote the gas 
mixture and the subscript i will 
denote any single component of the 
mixture. 
 Consider a gas mixture composed of k components.
 The mass of the mixture mm is the sum of the masses of 
the individual components, and the mole number of the 
mixture Nm is the sum of the mole numbers of the 
individual components.

mm   mi and N m   N i
k k

i 1 i 1

The ratio of the mass of a component to the mass of the mixture 
is called the mass fraction (mf), and the ratio of the mole number 
of a component to the mole number of the mixture is called the 
mole fraction y:

mi Ni
mf i  and yi 
mm Nm
Dividing the 1st and 2nd equation by mm and Nm, the sum 
of the mass fraction or mole fraction

 mf y
k k

i
1 and i
1
i 1 i 1

The mass of a substance can be expressed in terms of the 
mole number N and molar mass M of the substance as

m  NM
The apparent (or average) molar mass of a mixture can 
be expressed as

mm  mi N M yM k

Mm    i i
i i
(kg/kmol)
Nm Nm N m
i 1
The average (or apparent) gas constant of the mixture
can be determined from 

Ru
Rm  [kJ/(kg.K)]
Mm

Ru  8.314 kJ/(kmol.K)]
 Example 1.1.
Consider a gas mixture that 
consists of 3 kg of O2 , 5 kg of 
N2 , and 12 kg of CH4 , as 
shown . Determine 
(a) the mass fraction of 
each component,
(b) the mole fraction of 
each component, and
(c) the average molar 
mass and gas constant 
of the mixture
Example 1.2.
The composition of moist air is given on a molar
basis to be 78 percent N2 , 20 percent O2 , and 2
percent water vapor. Determine the mass
fractions of the constituents of air.

Example 1.3.
A gas mixture consists of 8 kmol of H2 and
2 kmol of N2. Determine the mass of each gas
and the apparent gas constant of the mixture.
Example 1.4.
A gas mixture consists of 20 percent O2, 30
percent N2, and 50 percent CO2 on mass basis.
Determine the volumetric analysis of the mixture
and the apparent gas constant.

Example 1.5.
A gas mixture has the following composition on
a mole basis: 60 percent N2 and 40 percent CO2.
Determine the gravimetric analysis of the
mixture, its molar mass, and gas constant.
1.2. P‐v‐T Behavior of Gas Mixtures
 An ideal gas – whose molecules   When two or more 
are spaced far apart so that the  ideal gases are mixed, 
behavior of a molecule is not  the behavior of a 
influenced by the presence of  molecule normally is 
other molecules‐a situation  not influenced by the 
encountered at low densities. presence of other 
 The P‐v‐T behavior of an ideal  similar or dissimilar 
gas is expressed by the simple  molecules, and 
relatin Pv=RT, which is called  therefore a non-
the ideal-gas equation of state. reacting mixture of
 The P‐v‐T behavior of real gases  ideal gases also
is expressed by more complex  behaves as an ideal
equations of state or by Pv=ZRT,  gas.
where Z is the compressibility   Eg. Air
factor
The prediction of the P‐v‐T behavior of gas mixtures is 
usually based on two models: Dalton’s law of additive 
pressures and Amagat’s law of additive volumes.

Dalton’ law of additive pressures: The pressure of a gas

mixture is equal to the sum of the pressures each gas
would exert if it existed alone at the mixture temperature
and volume.

Amagat’s law of additive volumes: 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.
 Dalton’s and Amagat’s 
laws hold exactly for 
ideal‐gas mixtures, 
but only 
approximately for 
real‐gas mixtures. 
 This is due to 
intermolecular forces 
that may be 
significant for real 
gases at high 
densities. 
 For ideal gases, these 
two laws are identical 
and give identical 
results.
 Dalton’s and Amagat’s Law

Pm   Pi (Tm , Vm )
k

Dalton' s law :
i 1


Vm   Vi (Tm , Pm ) 
k

Amagat ' s law :

i 1

exact for ideal gases, approximate for real gases
 In these relations, Pi is called the component 
pressure and Vi is called the component volume.
 Note that Vi is the volume a component would 
occupy if it existed alone at Tm and Pm, not actual 
volume occupied by the component in the 
mixture.
 Ideal‐Gas Mixtures
 For ideal gases, Pi and Vi can be related to yi by using the 
ideal‐gas relation for both the components and the gas 
mixture:

Pi (Tm , Vm ) N i RuTm / Vm Ni
   yi
Pm N m RmTm / Vm N m
Vi (Tm , Pm ) N i RuTm / Pm Ni
   yi
Vm N m RuTm / Pm N m

Therefore,
Pi Vi Ni (*)
   yi
Pm Vm N m
 The above equation is strictly N.B.
valid for ideal-gas mixtures  The composition of an ideal-gas
since it is derived by assuming mixture (such as the exhaust gases
ideal-gas behavior for the gas leaving a combustion chamber) is
mixture and each of its frequently determined by a
components. volumetric analysis and Eq. (*).
 The quantity yiPm is called the  A sample gas at a known volume,
partial pressure (identical to pressure, and temperature is passed
the component pressure for into a vessel containing reagents
ideal gases), and that absorb one of the gases.
 The quantity yiVm is called the  The volume of the remaining gas is
partial volume (identical to the then measured at the original
component volume for ideal pressure and temperature.
gases).  The ratio of the reduction in
 Note that for an ideal gas volume to the original volume
mixture, the mole fraction , the (volume fraction) represents the
pressure fraction, and the mole fraction of that particular gas.
volume fraction of a
component are identical.
Example 1.6.
A rigid tank contains 8 kmol of O2 and 10 kmol
of CO2 gases at 290 K and 150 kPa. Estimate
the volume of the tank.(heated to 395k, find
final P)

Example 1.7.
A gas mixture at 350 K and 300 kPa has the
following volumetric analysis: 65 % N2, 20 %
O2, and 15% CO2. Determine the mass fraction
and partial pressure of each gas.
Example 1.8
A rigid tank that contains 1kg of N2 at 25 0C and
300 kPa is connected to another rigid tank that
contains 3kg of O2 at 25 0C and 500 kPa. The
valve connecting the two tanks is opened, and the
two gases are allowed to mix. If the final mixture
temperature is 25 0C, determine the
volume of each tank and the
final mixture pressure.
1.3. Properties of Gas Mixtures
 Consider a gas mixture 
that consists of 2 kg of N2
and 3 kg of CO2. 
 The mass (an extensive 
property) of this mixture 
is 5 kg.
 Extensive properties of a 
non‐reacting ideal‐ or 
real‐gas mixture: Just add
the contributions of each
component of the
mixture.
The total internal energy, enthalpy, and entropy of a gas 
mixture can be expressed, respectively, as

U m   U i   mi ui   N i ui
k k

k 1 k 1

H m   H i   mi hi   N i hi
k k

k 1 k 1

S m   S i   mi si   N i si
k k

k 1 k 1
By following a similar logic, the changes in internal 
energy, enthalpy, and entropy of a gas mixture during a 
process can be expressed, respectively, as

U m   U i   mi ui   N i ui
k k

k 1 k 1

H m   H i   mi hi   N i  hi
k k

k 1 k 1

S m   S i   mi si   N i  si
k k

k 1 k 1
The internal energy, enthalpy, and entropy of a gas mixture 
per unit mass or per unit mole of the mixture can be 
determined by dividing the equations above by the mass or 
the mole number of the mixture (mm or Nm).
um   mf i ui u m   yi u i
k k

and
k 1 k 1
Cv ,m   mf i Cv ,i and C v ,m   yi C v ,i
hm   mf i hi hm   yi hi
k k

and
C p ,m   mf i C p ,i C p ,m   yi C p ,i
k 1 k 1
and
sm   mf i si s m   y i si
k k

and
k 1 k 1

 The relations given above are generally valid and are applicable to
both ideal- and real-gas mixtures.
 The only major difficulty associated with these relations is the
determination of properties for each individual gas in the mixture.
 The analysis can be simplified greatly, by treating the individual
gases as an ideal gas.
 Ideal‐Gas Mixtures
 The gases that comprise a   Under the ideal‐gas 
mixture are often at a  approximation, the 
high temperature and low  properties of a gas are 
pressure relative to the  not influenced by the 
critical‐point values of  presence of other gases, 
individual gases. and each gas component 
 In such cases, the gas  in the mixture behaves as 
mixture and its  if it exists alone at the 
components can be  mixture temperature Tm, 
treated as ideal gases  and mixture volume Vm. 
with negligible error. This principle is known as 
the Gibbs-Dalton law.
Also, the h, u, Cv, and Cp of an ideal gas depend on 
temperature only and are independent of the 
pressure or the volume of the ideal‐gas mixture.
The partial pressure of a component in an ideal - gas
mixture is simply Pi  yi Pm , where Pm is the mixture pressure.

 Evaluation of change in u and h of the components of 

an ideal‐gas mixture during a process is relatively easy 
since it requires only a knowledge of the initial and 
final temperatures.
 In evaluating the changes in s of the components since 
the entropy of an ideal gas depends on the pressure 
or volume of the component as well as on its 
temperature.
The entropy change of individual gases in an ideal‐gas 
mixture during a process can be determined from 

Pi , 2 Ti , 2 Pi , 2
si  si , 2  si ,1  Ri ln
 
 C p ,i ln  Ri ln
Pi ,1 Ti ,1 Pi ,1
  Pi , 2 Ti , 2 Pi , 2
or  s i  s i , 2  s i ,1  Ri ln  C p ,i ln  Ri ln
Pi ,1 Ti ,1 Pi ,1
where Pi , 2  yi , 2 Pm , 2 and Pi ,1  yi ,1 Pm ,1
 Example 1.9.
A 0.90 m3 rigid tank is divided into two equal 
compartments by a partition. One compartment contains 
Ne at 20 0C and 100 kPa, and the other compartment 
contains Ar at 50 0C and 200 kPa. Now the partition is 
removed , and the two gases are allowed to mix. Heat is 
lost to the surrounding air during this process in the 
amount of 15 kJ. Determine,
(a) the final mixture temperature, and 
(b) the final mixture pressure
N.B.       Ru=8.314 kPa.m3/kmol.K ; MAr=39.95 kg/kmol ; 
MNe=20.18 kg/kmol ; Cv,Ar=0.3122 kJ/kg.K and 
Cv,Ne=0.6179 KJ/kg.K @ Room Temperature, 300 K.
1.4. Adiabatic Mixing of Perfect Gases
 Figure shows two gases A   The process can be simplified 
by the assumption that it is 
and B separated from each  adiabatic : this means that the 
other in a closed vessel by a  vessel is perfectly thermally 
thin diaphragm. insulated and there will 
therefore be an increase in
 If the diaphragm is  entropy of the system.
removed or punctured then 
the gases mix and each 
then occupies the total 
volume, having as if the 
other gas were not present.
 This process is equivalent 
to a free expansion of each 
gas, and is irreversible.
 In a free expansion process, the internal energy
initially is equal to the internal energy finally.

U 1  N ACv , ATA  N B Cv ,BTB

U 2  ( N ACv , A  N B Cv ,B )T
where C v is in kJ/kmole.K
If the result is extended to any number
of gases, we have

U 1   N i Cv ,iTi and U 2  T  N i Cv ,i
k k

i 1 i 1

Since U 1  U 2 , it follows that :

NC Ti  T  N i Cv ,i
k k

i v ,i
i 1 i 1

NC
k

i v ,i
Ti
T i 1

NC
k

i v ,i
i 1
When two streams of fluid meet to form a common
stream in steady flow, they give another form of
mixing

Applying steady‐flow energy equation to the mixing section
(neglecting changes in kinetic and potential energy ), we get

   

m A hA1  m B hB1  Q  m A hA2  m B hB2  W

 In case of adiabatic flow : Q  0, and W  0

   

m A hA1  m B hB1  m A hA2  m B hB2

Also h  c pT , hence, c p in [kJ / kgK ]

 

m A c pATA  mB c pBTB  m A c pAT  mB c pBT

 

m c T Tm c
k k

i i
pi i pi
i 1 i 1

m c T
k

i
pi i

T i 1


m c
k

i
pi
i 1
 Also, C p  Mc p and M  m / N
 NC p  mc p

NC T
k

i pi i

Hence, T i 1

NC
k

i pi
i 1

 The above equations represent one condition which

must be satisfied in an adiabatic mixing process of
perfect gas in steady flow.
 In a particluar problem some other information must 
be known (e.g., sp. Volume or the final pressure).
Example 1.10
A vessel of 2 m3 capacity contains O2 at 500 kPa
and 50 0C. The vessel is connected to another
vessel of 1.5 m3 capacity containing CO at 100
kPa and 20 0C. The connecting valve is opened
and the gases mixed adiabatically. Determine the
final temperature and pressure of the mixture.

Take : Cv = 21.1 kJ/kmol K for O2

Cv = 20.9 kJ/kmol K for CO
1.5. Mixing of Ideal Gases initially at 
different Pressure and Temperature
 Consider three ideal 
gases A,B and C initially 
at different pressures
and temperatures and 
separated by partitions.
 Let the gases be mixed 
by removing the 
partitions.
 The total volume 
occupied by the mixture 
and the mass of the 
mixture are given by :
V  VA  VB  VC and m  mA  mB  mC
In terms of the mixing masses, the internal energy of the 
gases after mixing can be related to the internal energy 
of the gases before mixing by the expression:
This image cannot currently be display ed.

mu  mAu A  mB u B  mC uC
Since u for a perfect gas is equal to cv T, this relation may 
be re‐written as; 
mcvT  mA cv , ATA  mB cv ,BTB  mC cv ,CTC

mA cv , ATA  mB cv ,BTB  mC cv ,CTC

T
mcv
with mcv equal to mA cv , A  mB cv ,B  mC cv ,C

mA cv , ATA  mB cv ,BTB  mC cv ,CTC

T
mA cv , A  mB cv ,B  mC cv ,C
Applying the perfect gas law
PAVA cv , A PBVB cv ,B PCVC cv ,C
 
RA RA RA
T
PAVA cv , A PBVB cv ,B PCVC cv ,C
 
TA RA TB RB TC RC
Multiplying each term in the numerator and
denominator by the respective molecular weights :
 M A PAVA cv , A M B PBVB cv ,B M C PCVC cv ,C 
 M R   
M B RB M C RC
T  A A

 M A PAVA cv , A  M B PBVB cv ,B  M C PCVC cv ,C 
 M A RATA M B RBTB M C RCTC 
The product Mcv is the same for ideal gases
and the product MR  R u is a constant for
all ideal gases.
PAVA  PBVB  PCVC
T
PAVA PBVB PCVC
 
TA TB TC
If the pressure are all equal before mixing, the
resulting mixing temperature will be obtained
from :
VA  VB  VC
T
VA VB VC
 
TA TB TC

For equal volumes before mixing,

PA  PB  PC
T
PA PB PC
 
TA TB TC