PERFECT GAS
(GAS UNGGUL)
1.
2.
3.
4.
Did you know, one important type of fluid that has many
applications in thermodynamics is the type in which the working
temperature of the fluid remains well above the critical
temperature of the fluid?
In this case, the fluid cannot be liquefied by an isothermal
compression, i.e. if it is required to condense the fluid, then
cooling of the fluid must first be carried out.
In the simple treatment of such fluids, their behavior is likened to
that a perfect gas. Although, strictly speaking, a perfect gas is an
ideal which can never be realized in practice.
The behavior of many ‘permanent’ gases, e.g. hydrogen, oxygen,
air etc is very similar to the behavior of a perfect gas to a first
approximation.
A perfect gas is a collection of particles
that:
1. Are in constant, random motion,
2. Have no intermolecular attractions
(which leads to elastic collisions in
which no energy is exchanged or
lost),
3. Are considered to be volume-less
points.
The principle properties used to define the
state of a gaseous system and its SI units
(Systems International are as follow
respectively:
1. Pressure (P), Pascal (Pa).
2. Volume (V) , m3 for volume (although
liters and cm3 are often substituted),
3. Temperature (T). and the absolute scale
of temperature or Kelvin (K).
Two of the laws describing the behaviour of a perfect
gas are Boyle’s Law and Charles’ Law.
The Boyle’s Law may be stated as follows:
Provided the temperature T of a perfect gas remains constant,
then volume, V of a given mass of gas is inversely proportional to
the pressure P of the gas, i.e. P  1/V (as shown in Fig. 3.1-1), or P
x V = constant if temperature remains constant.
If the process is represented on a graph having axes
of pressure P and volume V, the results will be as
shown in Fig. 3.1-2. The curve is known as a
rectangular hyperbola, having the mathematical
equation xy = constant.
P
P  1/V
1/V
Figure 3.1-1 Graph P  1/V
If a gas changes from state 1 to state 2 during an
isothermal process, then
P1 V1 = P2 V2 = constant
Figure 3.1-2 P-V graph for constant temperature
The Charles’s Law may be stated as follows:
Provided the pressure P of a given mass of gas remains constant, then
the volume V of the gas will be directly proportional to the absolute
temperature T of the gas, i.e.
V  T, or V = constant x T. Therefore V/T = constant, for constant
pressure P.
If gas changes from state 1 to state 2 during a constant pressure process, then
V1 V2

 constant
T1 T2
If the process is represented on a P – V diagram as before, the result will be as shown in Fig. 3.2.
P
1
2
0
V1
V2
V
Figure 3.2 P-V graph for constant pressure process
P
V1
V
P 2
T1
T2
Charles’ Law gives us the change in volume of a gas with
temperature when the pressure remains constant. Boyle’s
Law gives us the change in volume of a gas with pressure if
 constant
the temperature remains constant
(P1 V1)/T = ( P2 V2)/T = constant
PV
 constant  R
T
characteristic equation of state of a
perfect gas
The relation which gives the volume of
a gas when both temperature and the
pressure are changed is stated in the
following equation
No gases in practice obey this law
rigidly, but many gases tend
towards it. An imaginary ideal that
obeys the law is called a perfect gas
P1V1 P2V2

T1
T2
When the state is changing from 1 to
state 2
PV
 constant  R
T
1. The constant, R, is called the gas constant.
2. The unit of R is Nm/kg K or J/kg K. Each
perfect gas has a different gas constant.
3. The characteristic equation is usually written
as PV = RT
4. Or for m kg, occupying V m3, PV = mRT
1.
Another form of the characteristic equation can be derived using
the kilogram-mole as a unit.
2. The kilogram-mole is defined as a quantity of a gas equivalent to
m kg of the gas, where M is the molecular weight of the gas (e.g.
since the molecular weight of oxygen is 32, then 1 kg mole of
oxygen is equivalent to 32 kg of oxygen).
3.
From the definition of the kilogram-mole, for m kg of a gas we
have, m = nM, (where n is the number of moles).
4. Note: Since the standard of mass is the kg, kilogram-mole will
be written simply as mole.
Substituting for m from equation m = nM in PV = mRT, we
have
PV
PV = nMRT or
MR 
nT
1.
Now Avogadro’s hypothesis states that the volume of 1 mole of any
gas is the same as the volume of 1 mole of any other gas, when the
gases are at the same temperature and pressure.
2.
Therefore V/n is the same for all gases at the same value of P and T.
That is the quantity PV/nT is constant for all gases.
3.
This constant is called the universal gas constant, and is given the
symbol Ro.
universal gas constant, Ro
PV
MR  Ro 
or PV  nRoT
nT
or since MR = Ro
then,
Ro
R
M
Experiment has shown that the volume of 1 mole of any perfect
gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from
equation
Untuk
semua
jenis
gas
?
PV 1 x 105 x 22.71
R0 

 8314.4 J/mole K
nT
1 x 273.15
Experiment has shown that the volume of 1 mole of any perfect
gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from
equation
Untuk
semua
jenis
gas
PV 1 x 105 x 22.71
R0 

 8314.4 J/mole K
nT
1 x 273.15
Experiment has shown that the volume of 1 mole of any perfect
gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from
equation
Untuk
semua
jenis
gas
PV 1 x 105 x 22.71
R0 

 8314.4 J/mole K
nT
1 x 273.15
Note: Please refer to last page of Thermodynamics and transport
Properties of Fluid (STEAM TABLE)
Experiment has shown that the volume of 1 mole of any perfect
gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from
equation
PV 1 x 105 x 22.71
R0 

 8314.4 J/mole K
nT
1 x 273.15
Untuk
semua
jenis
gas
Note: Please refer to last page of Thermodynamics and transport
Properties of Fluid (STEAM TABLE)
From equation below, the gas constant for any gas can be found when the
molecular weight is known, e.g. for oxygen of molecular weight 32, the gas
constant is
Untuk
jenis gas
tertentu,
Oksigen
Ro 8314.4
R

 259.8 J/kg K
M
32
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300
kN/m2 and a temperature of 45 oC. The gas is compressed until the
pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is
assumed to be a perfect gas, determine:
a)
b)
the mass of gas (kg)
the final volume of gas (m3)
Given: R = 0.29 kJ/kg K
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300
kN/m2 and a temperature of 45 oC. The gas is compressed until the
pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is
assumed to be a perfect gas, determine:
a)
b)
the mass of gas (kg)
the final volume of gas (m3)
Given: R = 0.29 kJ/kg K
Solution to Example 3.3
From the question
V1 = 0.046 m3
P1 = 300 kN/m2
T1 = 45 + 273 K = 318 K
P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2
T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300
kN/m2 and a temperature of 45 oC. The gas is compressed until the
pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is
assumed to be a perfect gas, determine:
a)
b)
the mass of gas (kg)
the final volume of gas (m3)
Given: R = 0.29 kJ/kg K
Solution to Example 3.3
From the question
V1 = 0.046 m3
P1 = 300 kN/m2
T1 = 45 + 273 K = 318 K
P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2
T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K
From equation PV = mRT
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300
kN/m2 and a temperature of 45 oC. The gas is compressed until the
pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is
assumed to be a perfect gas, determine:
a)
b)
the mass of gas (kg)
the final volume of gas (m3)
Given: R = 0.29 kJ/kg K
Solution to Example 3.3
From the question
V1 = 0.046 m3
P1 = 300 kN/m2
T1 = 45 + 273 K = 318 K
P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2
T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K
a) From equation PV = mRT
m
P1V1 300 x 0.046

 0.1496 kg
RT1
0.29 x 318
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300
kN/m2 and a temperature of 45 oC. The gas is compressed until the
pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is
assumed to be a perfect gas, determine:
a)
b)
the mass of gas (kg)
the final volume of gas (m3)
Given: R = 0.29 kJ/kg K
Solution to Example 3.3
From the question
V1 = 0.046 m3
P1 = 300 kN/m2
T1 = 45 + 273 K = 318 K
P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2
T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K
b) From equation
a) From equation PV = mRT
m
P1V1 P2V2

, the constant volume process i.e. V1 = V2
T1
T2
P1V1 300 x 0.046

 0.1496 kg
RT1
0.29 x 318
0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300
kN/m2 and a temperature of 45 oC. The gas is compressed until the
pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is
assumed to be a perfect gas, determine:
a)
b)
the mass of gas (kg)
the final volume of gas (m3)
Given: R = 0.29 kJ/kg K
Solution to Example 3.3
From the question
V1 = 0.046 m3
P1 = 300 kN/m2
T1 = 45 + 273 K = 318 K
P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2
T2 = 83 + 273 K = 356 K
R = 0.29 kJ/kg K
b) From equation
a) From equation PV = mRT
m
P1V1 P2V2

, the constant volume process i.e. V1 = V2
T1
T2
P1V1 300 x 0.046

 0.1496 kg
RT1
0.29 x 318
P1 P2

T1 T2
 P2 
 1.27 x 103 
  1346 K
T2  T1    318
 300 
 P1 
The specific heat capacities of any substance is
defined as the amount of heat energy required to
raise the unit mass through one degree
temperature raise.
Mass of
substance
Q=?
If 1 kg of a gas is supplied with an amount of heat
energy sufficient to raise the temperature of the
gas by 1 degree whilst the volume of the gas
remains constant, then the amount of heat energy
supplied is known as the specific heat capacity at
constant volume, and is denoted by Cv. The unit of
Cv is J/kg K or kJ/kg K.
For a reversible non-flow process at constant volume, we have:
dQ = mCvdT-------equation (1)
For a perfect gas the values of Cv are constant for any one gas at all
pressures and temperatures. Equations (1) can then be expanded as
follows :
Heat flow in a constant volume process, Q12 = mCv(T2 – T1)
Also, from the non-flow energy equation:
Q – W = (U2 – U1)
mcv(T2 – T1) – 0 = (U2 – U1)
(U2 – U1) = mCv(T2 – T1)
i.e.
dU = Q
P
P2
2
P1
1
V1 = V 2
Note:
In a reversible constant volume process, no work energy transfer can take place
since the piston will be unable to move i.e. W = 0.
V
3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the
temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine:
a)
b)
c)
The heat flow during the process
The beginning pressure of gas
The final pressure of gas
Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the
temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine:
a)
b)
c)
The heat flow during the process
The beginning pressure of gas
The final pressure of gas
Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
Solution to Example 3.4
From the question, m = 3.4 kg
V1 = V2 = 0.92 m3
T1 = 17 + 273 K = 290 K
T2 = 147 + 273 K = 420 K
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the
temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine:
a)
b)
c)
The heat flow during the process
The beginning pressure of gas
The final pressure of gas
Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
Solution to Example 3.4
From the question, m = 3.4 kg
V1 = V2 = 0.92 m3
T1 = 17 + 273 K = 290 K
T2 = 147 + 273 K = 420 K
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
a) From equation: Q12 = mCv(T2 – T1)
= 3.4 x 0.72(420 – 290)
= 318.24 kJ
3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the
temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine:
a)
b)
c)
The heat flow during the process
The beginning pressure of gas
The final pressure of gas
Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
Solution to Example 3.4
From the question, m = 3.4 kg
V1 = V2 = 0.92 m3
T1 = 17 + 273 K = 290 K
T2 = 147 + 273 K = 420 K
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
b)
From equation : PV = mRT
Hence for state 1,
P1V1 = mRT1
mRT1 3.4 kg x 0.287 kJ/kgK x 290 K
P1 

V1
0.92 m 3
 307.6 kN/m 2
a) From equation: Q12 = mCv(T2 – T1)
= 3.4 x 0.72(420 – 290)
= 318.24 kJ
3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17 oC until the
temperature rise to 147 oC. If the gas is assumed to be a perfect gas, determine:
a)
b)
c)
The heat flow during the process
The beginning pressure of gas
The final pressure of gas
Given
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
Solution to Example 3.4
From the question, m = 3.4 kg
V1 = V2 = 0.92 m3
T1 = 17 + 273 K = 290 K
T2 = 147 + 273 K = 420 K
Cv = 0.72 kJ/kg K
R = 0.287 kJ/kg K
b)
From equation : PV = mRT
Hence for state 1,
P1V1 = mRT1
mRT1 3.4 kg x 0.287 kJ/kgK x 290 K
P1 

V1
0.92 m 3
 307.6 kN/m 2
c)
a) From equation: Q12 = mCv(T2 – T1)
= 3.4 x 0.72(420 – 290)
= 318.24 kJ
P2 
For state 2, P2V2 = mRT2
mRT2 3.4 kg x 0.287 kJ/kgK x 420 K

V2
0.92 m 3
 445.5 kN/m 2
If 1 kg of a gas is supplied with an amount of heat
energy sufficient to raise the temperature of the gas by 1
degree whilst the pressure of the gas remains constant,
then the amount of heat energy supplied is known as
the specific heat capacity at constant pressure, and is
denoted by Cp. The unit of Cp is J/kg K or kJ/kg K.
For a reversible non-flow process at constant
pressure, we have: dQ = mCpdT--------------(2)
For a perfect gas the values of Cp are constant for any
one gas at all pressures and temperatures. Equation (2)
can then be expanded as follows:
Heat flow in a reversible constant pressure process
Q = mCp(T2 – T1)
1. Let a perfect gas be heated at constant pressure from T1 to T2.
2. With reference to the non-flow equation: Q = U2 – U1 + W,
3. The equation for a perfect gas: U2 – U1 = mCv(T2 – T1), (refer to
4.
5.
6.
7.
8.
heat flow@ constant volume)
Hence, Q = mCv(T2 – T1) + W
In a constant pressure process, the work done by the fluid is
given by the pressure times the change in volume, i.e. W =
P(V2 – V1). (refer to heat flow@ constant pressure)
Then using equation PV = mRT, we have: W = mR(T2 – T1)
Therefore substituting, Q = mCv(T2 – T1) + mR(T2 – T1) =
m(Cv + R)(T2 – T1)
But for a constant pressure process from equation: Q =
mCp(T2 – T1)
9. Hence, by equating the two expressions for the heat flow Q,
we have: mCp(T2 – T1) = m(Cv + R)(T2 – T1)
10. Cp = Cv + R
11. Alternatively, it is usually written as:
R = Cp - Cv
i.e.
=
The ratio of the specific heat at constant pressure to the specific heat at
constant volume is given the symbol  (gamma),
i.e.

Cp
Cv
1. Note that since Cp - Cv= R, it is clear that Cp must be greater than Cv
for any perfect gas.
2. It follows therefore that the ratio Cp/Cv =  , is always greater than
unity
 (gamma)
1.4
1.6
1.3
1.11-1.22
Type of gases
Diatomic
Monatomic
Triatomic
Hydro-carbons
argon (A), and helium
(He),
carbon dioxide (CO2),
and sulphur dioxide
(SO2),
ethane (C2H6),  =
1.22, and for isobutane (C4H10),  = 1.11.
Example
carbon monoxide
(CO), hydrogen (H2),
nitrogen (N2), and
oxygen (O2).
The Derivations
1
7
Cp - Cv= R
2 Dividing through by Cv
3
5
equation at
Cp
R
1 
Cv
Cv
4 Therefore using equation
Also from equation
Cp = Cv hence
substituting in
6
,
become as the
following equation:

R
 1 
Cv
6
Cp
Cp  Cv 
then,
Cv
R
(  1)
OR
R
Cv 
(  1)
8
R
Cp 
(  1)
A certain perfect gas has specific heat as follows: Cp = 0.846
kJ/kg K and Cv = 0.657 kJ/kg K
Find the gas constant and the molecular weight of the gas.
Solution to Example 3.5
From equation R = Cp – Cv
i.e. R = 0.846 – 0.657 = 0.189 kJ/kg K
or R = 189 Nm/kg K
R0
From equation M =
R
J/kg K
8314
 44
i.e. M =
189
J/kg K
Be careful with
the unit
conversion……..
What is the unit
of Ro stated in
Steam Table
PV = mRT Where:
P-Pressure(N/m2)
V –volume (m3)
1
m-Mass (kg)
R-Gas constant (Nm/kg K or J/kg K)
T- Temperature (Kelvin)
Ro
R
M
2
1. Ro-Universal Gas Constant,
which refer to Last page of
Steam table
2. M-molecular weight which refer
to periodic table
Cp - Cv= R
3
Relationship between Cp and Cv

Cp
Cv
4
Ratio between Cp and Cv
Thank you