WOLDIA UNIVERSITY
INSTITUTE OF TECHNOLOGY
School of Mechanical and Chemical Engineering
Department of Mechanical Engineering
Thermodynamics II
Chapter 1
Mixture of Ideal Gases
By: Shumye G.
2
Introduction
• In the analysis and application of the basic laws of
thermodynamics, consideration is normally given to systems
involving constant and uniform chemical compositions. Many
Engineering applications, however, involve mixtures of gases
whose properties are determined in the same way as for a
single gas.
• An ideal gas is one which strictly obeys all the gas laws under
given conditions of temperature and pressure.
• The concept of the ideal gas represents a limiting state that can
be approached but not achieved by any real gas
• Analysis of many Mechanical Engineering processes are
conducted at pressures and temperatures that the ideal gas laws
can be used as simple and reasonably close approximations to
the behaviour of the real gas.
3
The characteristic Gas Equation
The Physical properties of a gas are controlled by:
• The pressure P exerted by the gas
• The Volume V occupied by the gas and
• The temperature T of the gas
The behaviour of an ideal gas, undergoing any change in the above variables, is
governed by such laws as Boyle’s and Charle’s law
1. Boyle’s law states that “the volume of a given mass of a gas varies inversely
as its absolute pressure provided the temperature remains constant
3. Avogadro’s Law
• Volume directly proportional to the number of gas molecules
 V = constant × n
 Constant P and T
p
v
1

2. Charles’ law states that “ the volume of a given mass varies directly as
its absolute temperature provided the pressure is kept constant”
T
v
4
• In Engineering practice, volume, pressure and
temperature of a system varies simultaneously.
When both temperature T and Pressure p vary
• The proportionality may be replaced by R
p
T
v
p
RT
v 
Ideal Gas Law
• The relationships that we have discussed so far can be
combined into a single law that encompasses all of them.
6
• The constant R is known as the characteristic gas constant . For the
mass m of a particular gas
mRT
pv 
mRT
pV 
P=absolute pressure of gas in N/m2
V= volume of gas in m3
T= absolute temperature of gas in K, m= mass of gas in kg
R=characteristic gas constant in Nm/kgK
The product of the characteristic gas constant R and the molecular weight
M of an ideal gas is constant known as the universal gas constant Ro.
K
mole
J
MR
Ro .
/
3
.
8314


Gas Formula Molar mass,
M kg/kmol
R
kJ/kgK
Air 0.2870
Steam H2O 18 0.4615
Hydrogen H2 2 4.124
7
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 T
u
 
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
8
Gravimetric analysis
A mixture of two or more gases of fixed chemical composition is
called a no reacting gas mixture.
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. 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 N
m i
i
k
m i
i
k
 
 
 
1 1
and
9
mf
m
m
y
N
N
i
i
m
i
i
m
 
and
mf y
i
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 M
i i i

To find the average molar mass for the mixture Mm , note
m m N M N M
m i
i
k
i i m m
i
k
  
 
 
1 1
Solving for the average or apparent molar mass Mm
M
m
N
N
N
M y M kg kmol
m
m
m
i
m
i
i
k
i i
i
k
  
 
 
1 1
( / )
10
The apparent (or average) gas constant of a mixture is expressed as
R
R
M
kJ kg K
m
u
m
 
( / )
Can you show that Rm is given as
R mf R
m i i
i
k



1
To change from a mole fraction analysis to a mass fraction
analysis, we can show that
mf
y M
y M
i
i i
i i
i
k



1
To change from a mass fraction analysis to a mole fraction
analysis, we can show that
y
mf M
mf M
i
i i
i i
i
k



/
/
1
11
Volume fraction (Amagat model)
Divide the container into k sub containers, such that each sub container has
only one of the gases in the mixture at the original mixture temperature
and pressure.
Amagat's law of additive vol­
umes 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 P
m i m m
i
k


 ( , )
1
The volume fraction of the vfi of any component is
vf
V T P
V
i
i m m
m

( , )
and
vfi
i
k


1
= 1
12
For an ideal gas mixture
V
N R T
P
and V
N R T
P
i
i u m
m
m
m u m
m
 
Taking the ratio of these two equations gives
vf
V
V
N
N
y
i
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 P
i i m

Dalton’s law: P P T V
m i m m
i
k


 ( , )
1
13
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' .
P
N R T
V
and P
N R T
V
i
i u m
m
m
m u m
m
' 
Note that the ratio of Pi' to Pm is
P
P
V
V
N
N
y
i
m
i
m
i
m
i
'
  
.
14
• Example 1: A mixture of ideal gases has the
following molar composition; Argon (yAr = 0.20),
helium (yHe = 0.54), and the balance is carbon
monoxide. Find mole fraction of carbon monoxide.
The mole fractions of the argon and helium are given.
Therefore, the mole fraction of carbon monoxide can be
found,
1
k
k
y
 
 Ar He CO 1
k
k
y y y y
   

CO Ar He
1 1 0.20 0.54 0.26
y y y
       
15
Example: 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.
(b) What is the mass of 1 m3
of this gas when P = 1.5 MPa and
T = 30o
C?
For ideal-gas mixtures, the percent by volume is the volume
fraction. Recall
y vf
i i

16
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 = 30o
C?
R
R
M
kJ kg K
kJ
kmol K
kg
kmol
kJ
kg K
m
u
m
 
  

( / )
.
.
.
8 314
34 4
0 242
m
P V
R T
MPa m
kJ kg K K
kJ
m MPa
kg
m
m m
m m


 

15 1
0 242 30 273
10
20 45
3 3
3
. ( )
( . / ( ))( )
.

CH-1 GASE MIXTURE therimo dainamics .pptx

  • 1.
    WOLDIA UNIVERSITY INSTITUTE OFTECHNOLOGY School of Mechanical and Chemical Engineering Department of Mechanical Engineering Thermodynamics II Chapter 1 Mixture of Ideal Gases By: Shumye G.
  • 2.
    2 Introduction • In theanalysis and application of the basic laws of thermodynamics, consideration is normally given to systems involving constant and uniform chemical compositions. Many Engineering applications, however, involve mixtures of gases whose properties are determined in the same way as for a single gas. • An ideal gas is one which strictly obeys all the gas laws under given conditions of temperature and pressure. • The concept of the ideal gas represents a limiting state that can be approached but not achieved by any real gas • Analysis of many Mechanical Engineering processes are conducted at pressures and temperatures that the ideal gas laws can be used as simple and reasonably close approximations to the behaviour of the real gas.
  • 3.
    3 The characteristic GasEquation The Physical properties of a gas are controlled by: • The pressure P exerted by the gas • The Volume V occupied by the gas and • The temperature T of the gas The behaviour of an ideal gas, undergoing any change in the above variables, is governed by such laws as Boyle’s and Charle’s law 1. Boyle’s law states that “the volume of a given mass of a gas varies inversely as its absolute pressure provided the temperature remains constant 3. Avogadro’s Law • Volume directly proportional to the number of gas molecules  V = constant × n  Constant P and T p v 1  2. Charles’ law states that “ the volume of a given mass varies directly as its absolute temperature provided the pressure is kept constant” T v
  • 4.
    4 • In Engineeringpractice, volume, pressure and temperature of a system varies simultaneously. When both temperature T and Pressure p vary • The proportionality may be replaced by R p T v p RT v 
  • 5.
    Ideal Gas Law •The relationships that we have discussed so far can be combined into a single law that encompasses all of them.
  • 6.
    6 • The constantR is known as the characteristic gas constant . For the mass m of a particular gas mRT pv  mRT pV  P=absolute pressure of gas in N/m2 V= volume of gas in m3 T= absolute temperature of gas in K, m= mass of gas in kg R=characteristic gas constant in Nm/kgK The product of the characteristic gas constant R and the molecular weight M of an ideal gas is constant known as the universal gas constant Ro. K mole J MR Ro . / 3 . 8314   Gas Formula Molar mass, M kg/kmol R kJ/kgK Air 0.2870 Steam H2O 18 0.4615 Hydrogen H2 2 4.124
  • 7.
    7 Many thermodynamic applicationsinvolve 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 T u   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
  • 8.
    8 Gravimetric analysis A mixtureof two or more gases of fixed chemical composition is called a no reacting gas mixture. 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. 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 N m i i k m i i k       1 1 and
  • 9.
    9 mf m m y N N i i m i i m   and mf y i 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 M i i i  To find the average molar mass for the mixture Mm , note m m N M N M m i i k i i m m i k        1 1 Solving for the average or apparent molar mass Mm M m N N N M y M kg kmol m m m i m i i k i i i k        1 1 ( / )
  • 10.
    10 The apparent (oraverage) gas constant of a mixture is expressed as R R M kJ kg K m u m   ( / ) Can you show that Rm is given as R mf R m i i i k    1 To change from a mole fraction analysis to a mass fraction analysis, we can show that mf y M y M i i i i i i k    1 To change from a mass fraction analysis to a mole fraction analysis, we can show that y mf M mf M i i i i i i k    / / 1
  • 11.
    11 Volume fraction (Amagatmodel) Divide the container into k sub containers, such that each sub container has only one of the gases in the mixture at the original mixture temperature and pressure. Amagat's law of additive vol­ umes 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 P m i m m i k    ( , ) 1 The volume fraction of the vfi of any component is vf V T P V i i m m m  ( , ) and vfi i k   1 = 1
  • 12.
    12 For an idealgas mixture V N R T P and V N R T P i i u m m m m u m m   Taking the ratio of these two equations gives vf V V N N y i 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 P i i m  Dalton’s law: P P T V m i m m i k    ( , ) 1
  • 13.
    13 Now, consider placingeach 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' . P N R T V and P N R T V i i u m m m m u m m '  Note that the ratio of Pi' to Pm is P P V V N N y i m i m i m i '    .
  • 14.
    14 • Example 1:A mixture of ideal gases has the following molar composition; Argon (yAr = 0.20), helium (yHe = 0.54), and the balance is carbon monoxide. Find mole fraction of carbon monoxide. The mole fractions of the argon and helium are given. Therefore, the mole fraction of carbon monoxide can be found, 1 k k y    Ar He CO 1 k k y y y y      CO Ar He 1 1 0.20 0.54 0.26 y y y        
  • 15.
    15 Example: 1 An ideal-gasmixture has the following volumetric analysis Component % by Volume N2 60 CO2 40 (a)Find the analysis on a mass basis. (b) What is the mass of 1 m3 of this gas when P = 1.5 MPa and T = 30o C? For ideal-gas mixtures, the percent by volume is the volume fraction. Recall y vf i i 
  • 16.
    16 Comp. yi MiyiMi 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 = 30o C? R R M kJ kg K kJ kmol K kg kmol kJ kg K m u m       ( / ) . . . 8 314 34 4 0 242 m P V R T MPa m kJ kg K K kJ m MPa kg m m m m m      15 1 0 242 30 273 10 20 45 3 3 3 . ( ) ( . / ( ))( ) .