CHAPTER 5 PROCESSES OF IDEAL GAS

CHAPTER 5: PROCESSES OF IDEAL GAS

Any process of ideal gases can be well described by the equation relating the
relation between Pressure and Volume in P-V plane as:
𝑷𝑽𝒏 = 𝑪
where ‘P’ is the absolute pressure, ’V’ is the volume or specific volume of
certain working substance (ideal gas) and ‘n’ being the polytropic coefficient that
will define the process that the ideal gas will undergo.
In defining the processes of certain ideal gas such as heat, work etc., the use
of equation of state, properties of ideal gas and other applicable equations that will
govern the behavior of the working substance under consideration are still
inevitable. Proceeding discussions will show how the behavior of the ideal gas can
be mathematically defined.

INTENDED LEARNING OUTCOMES:

At the end of this chapter, the students should be able to:

1. Identify the 7-General equations that describe the process of ideal gas
2. Analyze the ideal gas processes and its mathematical relations

THE 7-GENERAL EQUATIONS

As a review of those mentioned equations in the previous topics, we
sort out relations that will define the behavior of ideal gas undergoing a
certain process. The 7-general equations will serve as a guide in deriving
equations necessary in calculating various ideal gas behavior. To summarize,
we have as follows:
1. Any Process of Ideal Gas
To define the behavior of ideal gas under any process, we have to
use the relation of Combined gas law as:
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CHAPTER 5 PROCESSES OF IDEAL GAS

𝑷𝑽
=𝑪
𝑻
From this equation, we can relate the three parameters (P-V-T)
between the ideal gas process state points.
2. Work non – flow

Non- flow work is the area under P-V plane which represents
the total work done of the ideal gas as it changes its volume and
pressure between two state points. Also it designates the work
done as the ideal gas exerts amount of energy to move a certain
thermodynamic boundary with infinitesimal change in volume
as dV. Therefore, work non-flow can be represented as the
integral of:
𝟐
𝑾𝒏 = ∫ 𝑷𝒅𝑽
𝟏

3. Internal Energy
As previously discussed, the change in internal energy is solely a
result of temperature change in a working substance inside the
thermodynamic system under constant volume but whether the

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CHAPTER 5 PROCESSES OF IDEAL GAS

volume remains constant or not, change in internal energy will

be represented by the equation as:
∆𝑼 = 𝒎𝑪𝒗 ∆𝑻

4. Heat Transferred
By neglecting any other energy interaction in the working
substance inside a thermodynamic system, heat added or
rejected can be quantified as the sum of internal energy and work
non-flow. Mathematically we have:
𝑸 = ∆𝑼 + 𝑾𝒏
5. Enthalpy
As previously discussed, the change in enthalpy is obtained by
considering the specific heat of ideal gas under constant pressure
but whether the pressure remains constant or not, change in
enthalpy will be represented by the equation as:
∆𝑯 = 𝒎𝑪𝒑 ∆𝑻
6. Change in Entropy
In the previous equation, Change in entropy can be expressed as:
∆𝑸
∆𝑺 =
𝑻
or it can be expressed as an integral of:
𝟐
𝒅𝑸
𝑺= ∫
𝟏 𝑻
where Q is the heat transferred as presented in part (4).
7. Work Steady-flow
To derive the relation of work steady-flow, we have to use the
formula,

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CHAPTER 5 PROCESSES OF IDEAL GAS

𝟐
𝑾𝒔 = − ∫ 𝑽𝒅𝑷
𝟏

Work steady flow is also the work done of the ideal gas as it
crosses the boundary of a certain thermodynamic system thus
maintaining a steady state flow conditions. Example of this is
when the air is compressed by the compressor. Certain mass of
air crosses the thermodynamic boundary of the compressor thus,
since it may be assumed that there is a constant flow, we use the
steady flow work equation to derive its work.
With the use of 7 general equations above, we can establish
equations in every processes of ideal gas to determine the
required parameters such as heat (Q), Work (W) and other
relations.

PROCESSES OF IDEAL GAS

As shown above, any process of ideal gas can be well presented by the
equation:
𝑷𝑽𝒏 = 𝑪
The polytropic coefficient ‘n’, depending on its values will define the
process that the ideal gas will undergo.
Value of
Process Process Definition
‘n’
0 Isobaric Constant Pressure Process
Isometric,
∞ Constant Volume Process
Isovolumic, Isochoric
1 Isothermal Constant Temperature Process
k Adiabatic, Isentropic Constant Entropy Process (Q = 0)
n≠k Polytropic Multiple Expansion and Compression (Q ≠ 0)

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CHAPTER 5 PROCESSES OF IDEAL GAS

1. Isobaric Process
As tabulated above, isobaric process is the process applied to
ideal gas by which the pressure is held constant and from this
process, the 7-general equations can be related to the process as
follow:
 Any Process relation:
𝑉
=𝐶
𝑇
𝑉1 𝑉2
=
𝑇1 𝑇2
𝑽𝟏 𝑻𝟏
=
𝑽𝟐 𝑻𝟐

 Work Non-flow:
2
𝑊𝑛 = ∫ 𝑃𝑑𝑉
1

from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 0 (𝑖𝑠𝑜𝑏𝑎𝑟𝑖𝑐)
𝑃=𝐶
substituting to the integral and evaluating:
𝑾𝒏 = 𝑷(𝑽𝟐 − 𝑽𝟏 ) = 𝑷𝟐 𝑽𝟐 − 𝑷𝟏 𝑽𝟏 = 𝒎𝑹(𝑻𝟐 − 𝑻𝟏 )

 Internal Energy:
Following the change in temperature, we have an internal
energy of:
∆𝑼 = 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 )
 Heat Transferred:
Following the relation of heat transfer, we have:
𝑄 = ∆𝑈 + 𝑊𝑛

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CHAPTER 5 PROCESSES OF IDEAL GAS

𝑄 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) + 𝑃(𝑉2 − 𝑉1 ) = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) + 𝑚𝑅(𝑇2 − 𝑇1 )

𝑄 = 𝑚(𝑇2 − 𝑇1 ) (𝐶𝑣 + 𝑅)
Simplifying we have:
𝑸 = 𝒎𝑪𝒑 (𝑻𝟐 − 𝑻𝟏 )

 Enthalpy:
Following the change in temperature, we have an enthalpy of:
∆𝑯 = 𝒎𝑪𝒑 (𝑻𝟐 − 𝑻𝟏 )
At this point, heat transferred is also equal with the change in
enthalpy of the working substance under isobaric process.
 Entropy:
Following the guiding equation as represented by the integral
above we will have:
2
𝑑𝑄
𝑆= ∫
1 𝑇
wherein we have:
𝑑𝑄 = 𝑚𝐶𝑝𝑑𝑇
2
𝑑𝑇
𝑆 = 𝑚𝐶𝑝 ∫
1 𝑇
Evaluating the integral, we will obtain:
𝑻𝟐 𝑽𝟐
𝜟𝑺 = 𝒎𝑪𝒑 𝐥𝐧 ( ) = 𝒎𝑪𝒑 𝐥𝐧 ( )
𝑻𝟏 𝑽𝟏

 Work Steady flow:

For work steady flow, since there is a constant pressure process,
we assume that P = C and therefore the differential of pressure,
dP, will be also zero. Thus;
𝑾𝒔 = 𝟎

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CHAPTER 5 PROCESSES OF IDEAL GAS

2. Isometric Process
Isometric process (also termed as Isovolumic Isochoric process)
is the process applied to ideal gas by which the volume is held
constant and from this process, the 7-general equations can be
related to the process as follow:
 Any Process relation:
𝑃
=𝐶
𝑇
𝑃1 𝑃2
=
𝑇1 𝑇2
𝑷𝟏 𝑻𝟏
=
𝑷𝟐 𝑻𝟐
 Work Non-flow:
As in process relation, by holding the volume constant under this
process, we can have V = C and that so, dV = 0. Thus,
𝑾𝒏 = 𝟎
 Internal Energy:
Following the change in temperature, we have an internal energy
of:
∆𝑼 = 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 )
 Heat Transferred:
Following the relation of heat transfer, we have:
𝑄 = ∆𝑈 + 𝑊𝑛
𝑄 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) + 0 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 )
And heat transferred under constant volume will be:
𝑸 = 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 )
 Enthalpy:
Following the change in temperature, we have an enthalpy of:
∆𝑯 = 𝒎𝑪𝒑 (𝑻𝟐 − 𝑻𝟏 )
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CHAPTER 5 PROCESSES OF IDEAL GAS

 Entropy:
Following the guiding equation as represented by the integral
above
we will have:
2
𝑑𝑄
𝑆= ∫
1 𝑇
wherein we have:
𝑑𝑄 = 𝑚𝐶𝑣𝑑𝑇
2
𝑑𝑇
𝑆 = 𝑚𝐶𝑣 ∫
1 𝑇
Evaluating the integral, we will obtain:
𝑻𝟐 𝑷𝟐
𝜟𝑺 = 𝒎𝑪𝒗 𝐥𝐧 ( ) = 𝒎𝑪𝒗 𝐥𝐧 ( )
𝑻𝟏 𝑷𝟏
 Work Steady flow:
For work steady flow, we follow the integral below:
2
𝑊𝑠 = − ∫ 𝑉𝑑𝑃
1

Also, from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = ∞ (𝑖𝑠𝑜𝑚𝑒𝑡𝑟𝑖𝑐, 𝑖𝑠𝑜𝑐ℎ𝑜𝑟𝑖𝑐)
𝑉=𝐶
Since the volume is constant, we can rewrite the integral above in
the
form,
2
𝑊𝑠 = −𝐶 ∫ 𝑑𝑃
1

Evaluating, we will have:

𝑊𝑠 = −𝐶(𝑃2 − 𝑃1 )
Substituting V = C,
𝑾𝒔 = −𝑽(𝑷𝟐 − 𝑷𝟏 ) = 𝑽(𝑷𝟏 − 𝑷𝟐 )
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CHAPTER 5 PROCESSES OF IDEAL GAS

SUPPLEMENTARY PROBLEMS
1. Carbon Dioxide undergoing an isobaric process has an initial condition of STP
and was heated until its temperature becomes 1200F. Calculate the heat applied in
the gas in Btu/lb and the change in entropy in Btu/lb-R. Use k of CO2 = 1.29

2. Two (2) kg of an ideal gas at 1atm and 300C is placed inside a piston cylinder
assembly. Heat was then applied allowing the piston to expand at constant pressure.
If the gas has Cp of 1.35kJ/kg-K and Cv of 1.09kJ/kg-K, determine the work done
by the gas (kJ) as it is heated to a temperature of 1000C. Determine also the final
volume of the gas after expansion (m3).

3. Five (5) pound of an ideal gas with R = 38 ft-lb/lb-R and k = 1.667 have 400 Btu of
heat added during reversible non-flow constant pressure process. The initial
temperature is 1100F. Determine the (a) Final temperature in 0F, (b) ΔH in Btu, (c) W
in Btu, (d) ΔU in lb-ft and (e) ΔS in Btu/R.

4. A perfect gas has a value of R = 319.2 J/kg.°K and k = 1.26. If 150 kJ are added to
3.35 kg of this gas at constant pressure when the initial temperature is 34.2°C. Find
(a) T2 (b) ΔH (c) ΔU (d) work non-flow (e) if the final temperature is 60.42°C, what
is ΔS.

5. A perfect gas has a value of R = 766.54 ft-lbf/lbm - °R and k = 1.26. If 30 Btu are
added to 5 lb of this gas at constant volume when the initial temperature is 95°F,
find (a) T2 in °F (b) ΔH in Btu (c) ΔS in Btu/°R (d) ΔU in Btu (e) Wn in Btu

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CHAPTER 5 PROCESSES OF IDEAL GAS

6. There are 2.33 kg of gas for which R = 377 J/kg-°K and k = 1.25, undergo a process
inside a rigid container from P1 = 657.8 kPa and T1 = 66°C to P2 = 1745 kPa. During
the process the gas is internally stirred and there are also added 108.25 kJ of heat.
Determine (a) T2 in °C (b) the work input in kJ and (c) the change of entropy in J/kg-
°K.

7. Twenty cu. ft. of air contained in a rigid vessel at 320 psia and 420°F is cooled to
150°F at constant volume. Cv = 0.1714 Btu/lb-°F and k = 1.4. What are (a) the final
pressure psig (b) the work in ft-lb (c) the change in internal energy in ft-lb (d) the
transferred heat in Btu (e) the change of enthalpy in ft-lb and (f) the change of
entropy in ft-lb/lb-°F

8. A 10-m3 vessel of Carbon Monoxide at a pressure of 300kPaa is vigorously stirred

by paddles until the pressure becomes 450kPaa. Determine (a) ΔU in kJ (b) Wn
assuming no heat transferred (kJ). Use k of CO = 1.399

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CHAPTER 5 PROCESSES OF IDEAL GAS

3. Isothermal Process
Isothermal process is the process applied to ideal gas by which
the temperature is held constant and from this process, the 7-
general equations can be related to the process as follow:
 Any Process relation:
𝑃𝑉 = 𝐶
𝑃1 𝑉1 = 𝑃2 𝑉2
𝑽𝟏 𝑷𝟐
=
𝑽𝟐 𝑷𝟏
 Work Non-flow:
For process relation, by holding the temperature constant under
this process, we can have T = C and that so employing again the
integral,
2
𝑊𝑛 = ∫ 𝑃𝑑𝑉
1

from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 1 (𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙)
𝑃𝑉 = 𝐶
𝐶
𝑃=
𝑉
substituting to the integral and evaluating:
2 2
𝐶 𝑑𝑉
𝑊𝑛 = ∫ 𝑑𝑉 = 𝐶 ∫
1 𝑉 1 𝑉
𝑽𝟐 𝑽𝟐 𝑷𝟏
𝑾𝒏 = 𝑷𝑽𝒍𝒏 ( ) = 𝒎𝑹𝑻𝒍𝒏 ( ) = 𝒎𝑹𝑻𝒍𝒏 ( )
𝑽𝟏 𝑽𝟏 𝑷𝟐
and where PV can be either obtain from state point 1 or 2.
 Internal Energy:
Since there is a constant temperature, we have an internal energy
of:

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CHAPTER 5 PROCESSES OF IDEAL GAS

∆𝑼 = 𝟎
 Heat Transferred:
Following the relation of heat transfer, we have:
𝑄 = 0 + 𝑊𝑛
𝑽𝟐 𝑽𝟐 𝑷𝟏
𝑸 = 𝑾𝒏 = 𝑷𝑽𝒍𝒏 ( ) = 𝒎𝑹𝑻𝒍𝒏 ( ) = 𝒎𝑹𝑻𝒍𝒏 ( )
𝑽𝟏 𝑽𝟏 𝑷𝟐
 Enthalpy:
Since there is a constant temperature, we have an enthalpy of:
∆𝑯 = 𝟎
 Entropy:
Following the guiding equation as represented by the to obtain
ΔS,
2
𝑑𝑄 𝑄
𝑆= ∫ =
1 𝑇 𝑇
wherein we have:
𝑉2 𝑃1
𝑄 = 𝑚𝑅𝑇𝑙𝑛 ( ) = 𝑚𝑅𝑇𝑙𝑛 ( )
𝑉1 𝑃2
Evaluating to obtain the change in entropy,
𝑽𝟐 𝑷𝟏
𝜟𝑺 = 𝒎𝑹𝒍𝒏 ( ) = 𝒎𝑹𝒍𝒏 ( )
𝑽𝟏 𝑷𝟐
 Work Steady flow:
For work steady flow, we follow the integral below:
2
𝑊𝑠 = − ∫ 𝑉𝑑𝑃
1

Also, from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 1 (𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙)
𝑃𝑉 = 𝐶
𝐶
𝑉=
𝑃
substituting to the integral and evaluating:
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CHAPTER 5 PROCESSES OF IDEAL GAS

2 2
𝐶 𝑑𝑃
𝑊𝑠 = − ∫ 𝑑𝑃 = − 𝐶 ∫
1 𝑃 1 𝑃

𝑷𝟏 𝑷𝟏 𝑽𝟐
𝑾𝒔 = 𝑷𝑽𝒍𝒏 ( ) = 𝒎𝑹𝑻𝒍𝒏 ( ) = 𝒎𝑹𝑻𝒍𝒏 ( )
𝑷𝟐 𝑷𝟐 𝑽𝟏
From the derived relation, we conclude that in isothermal process
non-flow work is equal to work steady flow.
𝑾𝒔 = 𝑾𝒏

4. Adiabatic Process
Adiabatic process is the process applied to ideal gas at which there
is neither heat rejected or added during the process. Thus we
assume that the heat transferred Q = 0. In the proceeding
derivation, we will prove also that Q = 0. In the two type of process
(reversible and irreversible), reversible adiabatic process will be
termed as Isentropic process or a constant entropy process.
Nevertheless, both of the two has a constant entropy between state
points of the process. And, as derived using the seven (7) general
equations, we have:
 Any Process relation:
𝑃𝑉
=𝐶
𝑇
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2
The relation between these three (3) parameters can be simplified
further to determine the relation between any of the two desired
parameters using:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 𝑘 (𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐, 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠)

𝑃𝑉 𝑘 = 𝐶

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CHAPTER 5 PROCESSES OF IDEAL GAS

Deriving the pressure – volume relation:

𝑃1 𝑉1 𝑘 = 𝑃2 𝑉2 𝑘
𝑷𝟐 𝑽𝟏 𝒌
=[ ]
𝑷𝟏 𝑽𝟐
Deriving the volume – temperature relation:
𝑃2 𝑉1 𝑇2 𝑃𝑉
= [ ] [ ] ; 𝑓𝑟𝑜𝑚 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝
𝑃1 𝑉2 𝑇1 𝑇
Substituting to the previously derived pressure - volume relation:
𝑉1 𝑘 𝑉1 𝑇2
[ ] = [ ][ ]
𝑉2 𝑉2 𝑇1
𝑻𝟐 𝑽𝟏 𝒌−𝟏
[ ]= [ ]
𝑻𝟏 𝑽𝟐
Deriving the pressure – temperature relation:
𝑉1 𝑃2 𝑇1 𝑃𝑉
= [ ] [ ] ; 𝑓𝑟𝑜𝑚 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝
𝑉2 𝑃1 𝑇2 𝑇
Substituting the previously derived volume - temperature
relation:
𝑘−1
𝑇2 𝑃2 𝑘−1 𝑇1
[ ]= [ ] [ ]
𝑇1 𝑃1 𝑇2
1−𝑘
𝑇2 𝑃2 𝑘−1 𝑇2
[ ]= [ ] [ ]
𝑇1 𝑃1 𝑇1
Simplifying, we will have:
𝒌−𝟏
𝑻𝟐 𝑷𝟐 𝒌
[ ]=[ ]
𝑻𝟏 𝑷𝟏

Also to derive ‘k’, we may use:

𝑃2 𝑉1
𝑙𝑛 [ ] = 𝑘 𝑙𝑛 [ ]
𝑃1 𝑉2

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CHAPTER 5 PROCESSES OF IDEAL GAS

𝑷
𝒍𝒏 [𝑷𝟐 ]
𝟏
𝒌=
𝑽
𝒍𝒏 [ 𝟏 ]
𝑽𝟐
 Work Non-flow:
Applying the integral to evaluate the work non-flow of the ideal
gas under adiabatic process, we will again employ the use of:
2
𝑊𝑛 = ∫ 𝑃𝑑𝑉
1

from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 𝑘 (𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐, 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐)
𝑃𝑉 𝑘 = 𝐶
𝑃 = 𝐶𝑉 −𝑘
substituting to the integral and evaluating:
2 2
2
−𝑘
𝑉 1−𝑘 𝑘
𝑉 1−𝑘 𝑃𝑉 2
𝑊𝑛 = 𝐶 ∫ 𝑉 𝑑𝑉 = 𝐶 [ ] = 𝑃𝑉 [ ] = |
1 1−𝑘 1 1−𝑘 1 1−𝑘 1
𝑷𝟐 𝑽𝟐 −𝑷𝟏 𝑽𝟏 𝒎𝑹(𝑻𝟐 − 𝑻𝟏 )
𝑾𝒏 = =
𝟏−𝒌 𝟏−𝒌
From the properties of ideal gas, the above relation can be
simplified further as:
𝑾𝒏 = −𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 )
Customarily, we apply that when the work is done by the
system, there is a positive value for Wn otherwise it is a work
being applied to the system.
 Internal Energy:
Following the relation in the change in internal energy, we have
an internal energy of:
∆𝑼 = 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 )
 Heat Transferred:
Following the relation of heat transfer, we have:
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CHAPTER 5 PROCESSES OF IDEAL GAS

𝑄 = ∆𝑈 + 𝑊𝑛
𝑸 = 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 ) − 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 ) = 𝟎
And from the relation above, we conclude by derivation that heat
transferred, Q, under adiabatic process is always equal to zero
(0).
 Enthalpy:
Following the relation in the change in enthalpy, we have an
enthalpy of:
∆𝑯 = 𝒎𝑪𝒑 (𝑻𝟐 − 𝑻𝟏 )

 Entropy:
Following the guiding equation as represented by the to obtain
ΔS,
2
𝑑𝑄 0
𝑆= ∫ = =0
1 𝑇 𝑇
Note that there is a constant entropy process and therefore:
𝜟𝑺 = 𝟎
 Work Steady flow:
For work steady flow, we follow the integral below:
2
𝑊𝑠 = − ∫ 𝑉𝑑𝑃
1

from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 𝑘 (𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐, 𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐)
𝑃𝑉 𝑘 = 𝐶
1 1
𝑉 = [𝐶 𝑘 ] [𝑃 −𝑘 ]

Substituting to the integral and evaluating:

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CHAPTER 5 PROCESSES OF IDEAL GAS

1 1
1
1 2 1 1 𝑃1−
𝑘 𝑘 [[𝑃𝑉 𝑘 ]𝑘 ] [𝑃1−𝑘 ]
𝑊𝑠 = − [𝐶 𝑘 ] ∫ [𝑃−𝑘 ] 𝑑𝑃 = [𝐶 𝑘 ] [ ]=
1 1−𝑘
1 −1
𝑘
Simplifying further we will arrive for the relation:
𝑃2 𝑉2 −𝑃1 𝑉1 𝑚𝑅(𝑇2 − 𝑇1 )
𝑊𝑠 = 𝑘 =𝑘
1−𝑘 1−𝑘
Therefore, at isentropic process, we have a work on steady flow
system as:
𝟐
𝑾𝒔 = 𝒌𝑾𝒏 = 𝒌 ∫ 𝑷𝒅𝑽
𝟏

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CHAPTER 5 PROCESSES OF IDEAL GAS

SUPPLEMENTARY PROBLEMS
1. A certain part of a thermodynamic cycle using 1kg air as a working substance
consist of an isothermal process where the pressure changes from STP to 135kPag.
For this condition, determine the heat transferred to the working substance (kJ),
Change in entropy (kJ/kg-K) and the work done by the piston as it changes its
volume between two state points (kJ)

2. Carbon Dioxide with specific heat ratio of 1.299 is used as a working fluid in a
piston-cylinder assembly which is designed handle fluid at constant temperature.
The heat was applied to CO2 at 3500F allowing an expansion in the piston cylinder
assembly. If the heat loss accounts to 20Btu/lb, determine the work done by piston
displacement (Btu/lb).

3. During an isothermal process at 880F, the pressure on 8lb of air drops from 80psia
to 5psig. For an internally reversible process, determine the following: (a) Wn
(Btu/lb), (b) Ws (Btu/lb) assuming ΔP and ΔK = 0 (c) Heat transferred (lb-ft/lb) (d)
ΔU and (e) ΔS (lb-ft/lb-R).

4. Air flows steadily through an engine at constant temperature, 400K. Find the work
per kilogram if the exit pressure is one-third of the inlet pressure and the inlet
pressure is 207kpag. Assume that the kinetic and potential energy variation is
negligible.

5. In a reversible adiabatic process, air is being compressed from 277K and 1atm
pressure to 700kPag. If the volume of the air in entrance condition has a rate of
2m3/s, determine the power required by the air compressor (kW). Calculate also
the change in enthalpy of the air between the two state points (kW).

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CHAPTER 5 PROCESSES OF IDEAL GAS

6. Air contained in a piston-cylinder assembly is being compressed in a reversible

adiabatic manner from a temperature of 300K and a pressure of 120kPag to a final
pressure of 480kPag. Determine (a) final pressure (0C) and the work done per kg of
air (kJ/kg)

7. Carbon Dioxide is being compressed isentropically in a piston cylinder assembly

with an initial condition at STP. If the resulting volume is half of the original volume
determine the final pressure (psia) final temperature ( 0F) and the work done in
compressing the gas (Btu/lb). Use k of CO2 = 1.299.

8. Hydrogen gas at a pressure of 100kPag has a volume of 0.32m3. The gas is

compressed in an adiabatic manner in a piston cylinder assembly until the
temperature is 1900C. If the reversible work is -63kJ. Determine the initial
temperature of the gas. Use k of H2 = 1.28.

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CHAPTER 5 PROCESSES OF IDEAL GAS

5. Polytropic Process
Polytropic process is an internally reversible process. The
previous four (4) processes discussed above is also a polytropic
process with various polytropic coefficient ‘n’. Now, we define a
special case of polytropic process wherein ‘n’ is any positive real
number and n ≠ k, 1, and ∞. Thus also, we may have concluded
that Q ≠ 0 also. And, as derived using the seven (7) general
equations, we have:
 Any Process relation:
𝑃𝑉
=𝐶
𝑇
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2
The relation between these three (3) parameters can be
simplified
further to determine the relation between any of the two
desired
parameters using:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑛 ≠ 𝑘, 1, 𝑎𝑛𝑑 ∞. )

𝑃𝑉 𝑛 = 𝐶
Deriving the pressure – volume relation:
𝑃1 𝑉1 𝑛 = 𝑃2 𝑉2 𝑛
𝑷𝟐 𝑽𝟏 𝒏
=[ ]
𝑷𝟏 𝑽𝟐
Also to derive ‘n’, we may use:
𝑃2 𝑉1
𝑙𝑛 [ ] = 𝑛 𝑙𝑛 [ ]
𝑃1 𝑉2

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CHAPTER 5 PROCESSES OF IDEAL GAS

𝑷
𝒍𝒏 [𝑷𝟐 ]
𝟏
𝒏=
𝑽
𝒍𝒏 [ 𝟏 ]
𝑽𝟐
Deriving the volume – temperature relation:
𝑃2 𝑉1 𝑇2 𝑃𝑉
= [ ] [ ] ; 𝑓𝑟𝑜𝑚 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝
𝑃1 𝑉2 𝑇1 𝑇
Substituting to the previously derived pressure - volume
relation:
𝑉1 𝑛 𝑉1 𝑇2
[ ] = [ ][ ]
𝑉2 𝑉2 𝑇1
𝑻𝟐 𝑽𝟏 𝒏−𝟏
[ ]= [ ]
𝑻𝟏 𝑽𝟐
Deriving the pressure – temperature relation:
𝑉1 𝑃2 𝑇1 𝑃𝑉
= [ ] [ ] ; 𝑓𝑟𝑜𝑚 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝
𝑉2 𝑃1 𝑇2 𝑇
Substituting the previously derived volume - temperature
relation:
𝑛−1
𝑇2 𝑃2 𝑛−1 𝑇1
[ ]= [ ] [ ]
𝑇1 𝑃1 𝑇2
1−𝑛
𝑇2 𝑃2 𝑛−1 𝑇2
[ ]= [ ] [ ]
𝑇1 𝑃1 𝑇1
Simplifying, we will have:
𝒏−𝟏
𝑻𝟐 𝑷𝟐 𝒏
[ ]=[ ]
𝑻𝟏 𝑷𝟏

 Work Non-flow:
Applying the integral to evaluate the work non-flow of the ideal
gas under polytropic process, we will again employ the use of:
2
𝑊𝑛 = ∫ 𝑃𝑑𝑉
1

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CHAPTER 5 PROCESSES OF IDEAL GAS

from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑛 ≠ 𝑘, 1, 𝑎𝑛𝑑 ∞. )

𝑃𝑉 𝑛 = 𝐶
𝑃 = 𝐶𝑉 −𝑛
substituting to the integral and evaluating:
2 2
2
−𝑛
𝑉 1−𝑛 𝑉 1−𝑛 𝑃𝑉 2
𝑊𝑛 = 𝐶 ∫ 𝑉 𝑑𝑉 = 𝐶 [ ] = 𝑃𝑉 𝑛 [ ] = |
1 1−𝑛 1 1−𝑛 1 1−𝑛 1
𝑷𝟐 𝑽𝟐 −𝑷𝟏 𝑽𝟏 𝒎𝑹(𝑻𝟐 − 𝑻𝟏 )
𝑾𝒏 = =
𝟏−𝒏 𝟏−𝒏

 Internal Energy:
Following the relation in the change in internal energy, we have
an internal energy of:
∆𝑼 = 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 )
 Heat Transferred:
Following the relation of heat transfer, we have:
𝑄 = ∆𝑈 + 𝑊𝑛
𝑚𝑅(𝑇2 − 𝑇1 )
𝑄 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) +
1−𝑛
From properties of ideal gas, we have:
𝑅
𝐶𝑣 = ; 𝑅 = 𝐶𝑣 (𝑘 − 1)
𝑘−1
Substituting from the above equation, we have:
𝑚𝐶𝑣 (𝑘 − 1)(𝑇2 − 𝑇1 )
𝑄 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) +
1−𝑛
(𝑘 − 1) 𝑘−𝑛
= 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) [1 + ] = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) [ ]
1−𝑛 1−𝑛

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CHAPTER 5 PROCESSES OF IDEAL GAS

Note that the relation of heat transferred under polytropic

process has a special value for specific heat. That is the specific
heat under polytropic process, Cn, with an equivalent below:
𝒌−𝒏
𝑪𝒏 = 𝑪𝒗 [ ] ;𝑛 ≠ 1
𝟏−𝒏
And heat transferred under polytropic process will be:
𝑸 = 𝒎𝑪𝒏 (𝑻𝟐 − 𝑻𝟏 )
 Enthalpy:
Following the relation in the change in enthalpy, we have an
enthalpy of:
∆𝑯 = 𝒎𝑪𝒑 (𝑻𝟐 − 𝑻𝟏 )
 Entropy:
Following the guiding equation as represented by the to obtain
ΔS,
2 2
𝑑𝑄 𝑚𝐶𝑛 𝑑𝑇
Δ𝑆 = ∫ = ∫
1 𝑇 1 𝑇
Evaluating the integral, we will have:
𝑻𝟐
𝚫𝑺 = 𝒎𝑪𝒏 𝒍𝒏 [ ]
𝑻𝟏
Note that the other equivalent of T2/T1 can be used also to
determine the value of ΔS. That is,
𝑇2 𝑉1 𝑛−1
[ ]= [ ]
𝑇1 𝑉2
𝑛−1
𝑇2 𝑃2 𝑛
[ ]=[ ]
𝑇1 𝑃1

 Work Steady flow:

For work steady flow, we follow the integral below:
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CHAPTER 5 PROCESSES OF IDEAL GAS

2
𝑊𝑠 = − ∫ 𝑉𝑑𝑃
1

from process definition:

𝑃𝑉 𝑛 = 𝐶 𝑤ℎ𝑒𝑟𝑒 𝑛 = 𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑛 ≠ 𝑘, 1, 𝑎𝑛𝑑 ∞. )
𝑃𝑉 𝑛 = 𝐶
1 1
𝑉 = [𝐶 𝑛 ] [𝑃 −𝑛 ]

substituting to the integral and evaluating in below:

1 1
1
1 2 1 1 𝑃1−𝑛 𝑛 [[𝑃𝑉 𝑛 ]𝑛 ] [𝑃1−𝑛 ]
𝑊𝑠 = − [𝐶 𝑛 ] ∫ [𝑃−𝑛 ] 𝑑𝑃 = [𝐶 𝑛 ] [ ]=
1 1−𝑛
𝑛−1
1

Simplifying further we will arrive for the relation:

𝑃2 𝑉2 −𝑃1 𝑉1 𝑚𝑅(𝑇2 − 𝑇1 )
𝑊𝑠 = 𝑛 =𝑛
1−𝑛 1−𝑛
Therefore, at polytropic process, we have a work on steady flow
system as:
𝟐
𝑾𝒔 = 𝒏𝑾𝒏 = 𝒏 ∫ 𝑷𝒅𝑽
𝟏

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CHAPTER 5 PROCESSES OF IDEAL GAS

SUPPLEMENTARY PROBLEMS
1. One kilogram per second of Nitrogen gas initially at 101kPaa and 300K is
compressed polytropically according to the process PV1.25 = C. Calculate the power
(kW) necessary to compress the Nitrogen to 1400kPaa. k of N2 = 1.40.

2. Air is compressed polytropically from 101kPag and 230C and delivered to an air
receiver at 1500kPag and 1750C. Determine the heat removed per kg of air (kJ/kg)
during compression.

3. Standard air is compressed according to the process PV2 = C. The work required
is 75Btu/lb. Determine the change in internal energy of the air (Btu/lb), the heat
transferred during compression (Btu/lb) and the work steady flow (Btu/lb)

4. Air is expanded from 1Mpaa, 3270C to 200Kpaa in a closed piston-cylinder device

executing PV1.2 = C process. Calculate the work produced during expansion of the
piston (kJ/kg).

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CHAPTER 5 PROCESSES OF IDEAL GAS

For ease of access, presented below is the summarized equations from the above derivations:

Any
Process Wn ΔU Q ΔH ΔS Ws
process
𝑉1 𝑉2 𝑇2
= = 𝑚𝐶𝑝 ln ( )
𝑇1 𝑇2 = 𝑃(𝑉2 − 𝑉1 ) 𝑇1
Isobaric 𝑉1 𝑇1 = 𝑚𝑅(𝑇2 − 𝑇1 )
= 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) 𝑉2 0
= = 𝑚𝐶𝑝 ln ( )
𝑉2 𝑇2 𝑉1
𝑃1 𝑃2 𝑇2
= 𝑚𝐶𝑣 ln ( )
𝑇1 𝑇2 𝑇1 = −𝑉(𝑃2 − 𝑃1 )
Isometric 𝑃1 𝑇1 0 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) 𝑃2 = 𝑉(𝑃1 − 𝑃2 )
= = 𝑚𝐶𝑣 ln ( )
𝑃2 𝑇2 𝑃1
𝑉2 𝑉2 𝑊𝑠 = 𝑊𝑛
= 𝑃𝑉𝑙𝑛 ( ) = 𝑃𝑉𝑙𝑛 ( ) 𝑃1
𝑉1 𝑉1 𝑉2 = 𝑃𝑉𝑙𝑛 ( )
𝑃1 𝑉1 = 𝑃2 𝑉2 𝑉2 𝑉2 = 𝑚𝑅𝑙𝑛 ( ) 𝑃2
𝑉1
Isothermal 𝑉1
=
𝑃2 = 𝑚𝑅𝑇𝑙𝑛 ( )
𝑉1
0 = 𝑚𝑅𝑇𝑙𝑛 ( )
𝑉1
0 𝑃1
𝑃1
= 𝑚𝑅𝑇𝑙𝑛 ( )
𝑉2 𝑃1 𝑃1 𝑃1 = 𝑚𝑅𝑙𝑛 ( ) 𝑃2
𝑃2 𝑉2
= 𝑚𝑅𝑇𝑙𝑛 ( ) = 𝑚𝑅𝑇𝑙𝑛 ( )
𝑃2 𝑃2 = 𝑚𝑅𝑇𝑙𝑛 ( )
𝑉1
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1 𝑇2 𝑊𝑠 = 𝑘𝑊𝑛
𝑃2 𝑉2 −𝑃1 𝑉1 𝑃2 𝑉2 −𝑃1 𝑉1
𝑃2 𝑉1 𝑘 =
=[ ] 1−𝑘 =𝑘
Adiabatic, 𝑃1 𝑉2 𝑚𝑅(𝑇2 − 𝑇1 ) 1−𝑘
𝑚𝑅(𝑇2 − 𝑇1 )
𝑇2 𝑉1 𝑘−1 = = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) 0 = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) 0
Isentropic [ ]= [ ] 1−𝑘 =𝑘
1−𝑘
𝑇1 𝑉2 = −𝑚𝐶𝑣 (𝑇2 2
𝑘−1 − 𝑇1 ) = 𝑘 ∫ 𝑃𝑑𝑉
𝑇2 𝑃2 𝑘 1
[ ]=[ ]
𝑇1 𝑃1
𝑇2
= 𝑚𝐶𝑛 𝑙𝑛 [ ] 𝑊𝑠 = 𝑛𝑊𝑛
𝑇1 𝑃2 𝑉2 −𝑃1 𝑉1
Same as 𝑃2 𝑉2 −𝑃1 𝑉1 = 𝑚𝐶𝑛 (𝑇2 − 𝑇1 )
also: =𝑛
= 1−𝑛
Isentropic 1−𝑛 𝑇2 𝑉1 𝑛−1
Polytropic (replace ‘k’ by 𝑚𝑅(𝑇2 − 𝑇1 ) = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) 𝑤ℎ𝑒𝑟𝑒 = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) [ ]= [ ] =𝑛
𝑚𝑅(𝑇2 − 𝑇1 )
= 𝑘−𝑛 𝑇1 𝑉2 1−𝑛
‘n’) 1−𝑛 𝐶𝑛 = 𝐶𝑣 [ ] 𝑛−1 2
1−𝑛 𝑇2 𝑃2 𝑛 = 𝑛 ∫ 𝑃𝑑𝑉
𝑛≠1 [ ]=[ ] 1
𝑇1 𝑃1

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CHAPTER 5 PROCESSES OF IDEAL GAS

CHAPTER PROBLEMS:

1. One pound of an ideal gas undergoes an isentropic process from 95.3psig and a
volume of 0.6ft3 to a final volume of 3.6 ft3. If cp = 0.124Btu/lb-R and cv = 0.093
Btu/lb-R, determine the final temperature of the gas (0F) and the work done by the
gas (Btu).

2. 1-kg of CO2 is compressed from 1MPag and 2000C to 3MPag in a piston-cylinder

device arranged to execute a polytropic process for which PV1.2 = constant.
Determine the final temperature of the gas (0C).

3. An ideal compressor compresses 12kg/min of air isothermally from 99kPaa and

a specific volume of 0.81m3/kg to a final pressure of 600kPaa. Determine the
required work of the compressor in kW.

4. A rigid tank contains an ideal gas with R = 3.07kJ/kg-K with k = 1.75. calculate
the change in entropy (kJ/kg-K) of the substance inside the tank if the heat added is
17kJ/kg and that the initial temperature of the gas is 270C.

5. A piston cylinder device which is designed to perform an isobaric process

contains 1-lb of hydrogen gas. Heat is being transferred to the gas until its
temperature increases by 700F. Determine the work done (Btu) by the gas due to
boundary expansion.

6. A piston–cylinder device initially contains 0.07 m3 of nitrogen gas at 130kPaa and

1800C. The nitrogen is now expanded to a pressure of 80kPaa polytropically with a
polytropic exponent whose value is equal to the specific heat ratio. Determine the
final temperature (0C) and the boundary work (kJ) done during this process.
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CHAPTER 5 PROCESSES OF IDEAL GAS

7. At a certain process, the specific heat at polytropic process of air is 0.3075kJ/kg-

k. Under the same process, air initially at STP is being compressed until the pressure
becomes 200kPaa. Determine the final temperature (0C) of air and the heat
transferred during compression (kJ/kg).

8. Nitrogen at an initial state of 300K, 150 kPaa, and 0.2m3 is compressed slowly in
an isothermal process to a final pressure of 800 kPa. Determine the work done
during this process (kJ).

9. A closed rigid container has a volume of 1m3 and holds Oxygen gas at 344.8kPaa
and 273K. Heat is added until the temperature is 600K. Determine the change in
internal energy (kJ) and the change in enthalpy (kJ). Use kO2 = 1.395.

10. Oxygen at 200kPAg, 270C is contained in a piston cylinder assembly arranged to

maintain a constant pressure process. How much work in kJ/kg is produced by the
system when it is heated to 3000C?

Page | 94