Thermodynamics I is a course on the fundamentals of thermodynamics including common terms and concepts, fluid properties, processes like isothermal and adiabatic, the laws of thermodynamics, heat engines and refrigeration. The course will cover these topics through lectures, recommended textbooks, videos and assessments including assignments, a test and exam. Students will learn about systems, surroundings, energy, work, heat, entropy and more. Properties of gases can be determined using equations of state from ideal gas laws, tables and diagrams.

1. Thermodynamics I: CHEP 121
Department of Chemical and Processing
Engineering
Lecturer: Mr Rashama 0773251373; 0714329030
c.rashama@mail.com

2. Course Overview
1. Understand Common Thermodynamics Terms and Concepts
System, sorroundings, closed, open,
2. Fluid properties (Zero and 1st Law of Thermodynamics)
Potential energy, kinetic energy, enthalpy, internal energy, Work, Heat,
Entropy, Gibbs free Energy.
3. Where to get these Properties
Ideal gas eqtn. Non-ideality. Steam Tables. Mollier Diagrams
4. Processes PVT diagrams
- isothermal, isobaric, isentropic, adiabatic, isochoric. Reversible. Irreversible,
equilibrium

3. Overview Cont’d
5. 2nd, and 3rd Thermodynamic Laws
6. Heat Engines and Refridgeration
Recommended Textbooks
1.Vaness, Smith 8th Edition (Chapters 1-9)
2.Tarik-Al Shemeri, Engineering Thermodynamics
3.Engineering Thermo Fundamentals
Youtube videos
Assessment – 1 Test, 1 Assignments. 1 Exam.
•Assignment 1 Due on: 15 June 2022
•Test 1 on: 23 June 2022 (Chapters 1-3)
3

4. 4
Thermodynamics Terms & Concepts
Example systems
• Gas in a container
• Charging & discharging a
battery
• Chemical reactions
System
Sorroundings
Universe
Thermodynamics is the study of the inter-relation between
heat, work and internal energy of a system and its interaction
with its environment..

5. 5
The system contains many molecular The most important
step at the start of solving a problem in thermodynamics is to
carefully define the system boundaries.
Closed System:
Thermal transfer but no mass transfer, say an ice cube melts
into a puddle and the ice cube is the system.
Open System:
Mass and thermal transfer occurs, a system is a section of a
river.
Isolated System:
No heat or mass transfer. A perfectly insulated box in which
a match is lit.
Thermodynamic Terms & Concepts

6. 6
Thermodynamics States
Examples of state variables:
• P = pressure (Pa or N/m2),
• T = temperature (K),
• V = volume (m3),
• n = number of moles, and
• U = internal energy (J).
A state variable describes the state of a system at time
t, but it does not reveal how the system was put into
that state.

7. 7
Intensive Properties:
Pressure, Temperature, Free Energy, Internal Energy, Specific Volume
Things that do not depend on system size.
Extensive Properties
Volume, Mass, Total Energy
Things that are determined by the system size.
Thermodynamic Properties

8. Where to get Properties
PVT Equations of State – Ideal & Non-Ideal
-Compressibility factor and reduced P, T.
- Virial equations
-Van der Waals
-Redlich Kwong
-Soave
Standard Tables – Steam/Ammonia/Air 8

9. 9
“Quality, q”
When a mixture of two phases (vapor/liquid) exist the fraction
vapor is called the “quality”. The intrinsic properties (M) such as V,
U, H, S can be calculated for a two phase single component system
using the “quality” and the values for the saturated liquid and vapor
phases:
M = (1-q) ML + q MV ;or
M = ML + q (ΔM) = ML + q (MV-ML)

10. Phase Behavior for Single Component, C = 1
Water for example.
10
F = C – P + 2

11. Gibbs Phase Rule
F free parameters
C components
P phases
So for saturated water vapor we have one component, two
phases and one free parameter. That is if T is known we
know the vapor pressure. If we know the pressure we know
the temperature.
For supersaturated steam we have one component, one
phase and we can vary P and T and these will determine the
specific volume or density, internal energy, enthalpy, etc.
11
F = C – P + 2

12. Gibbs Phase Rule
F free parameters
C components
P phases
12
F = C – P + 2

13. Zeroth Law of Thermodynamics
“Systems that are in thermal equilibrium exist at
the same temperature”.
Zeroth law of thermodynamics takes into
account that temperature is something worth
measuring because it predicts whether the heat
will transfer between objects or not.
13

14. 14
First Law of Thermodynamics
ΔU = Q + W (Conservation of Energy)
Work Done
Heat Energy
Internal Energy

15. 15
The First Law of Thermodynamics
The first law of thermodynamics says the change in
internal energy of a system is equal to the heat flow
into the system plus the work done on the system
(conservation of energy).
W
Q
U 



16. 16
Sign Conventions

17. 17
Thermodynamic Processes
A thermodynamic process is represented by a change in
one or more of the thermodynamic variables describing
the system.
Each point on the curve
represents an equilibrium state of
the system.
Our equation of state, the ideal
gas law (PV = nRT), only
describes the system when it is
in a state of thermal equilibrium.

18. MFMcGraw Chap15d-Thermo-Revised 5/5/10 18
A PV diagram can be used to represent the state changes of a
system, provided the system is always near equilibrium.
The area under a PV curve
gives the magnitude of the
work done on a system. W>0
for compression and W<0 for
expansion.
Thermodynamic Processes

19. 19
The work done on a system depends on the path taken in the PV
diagram. The work done on a system during a closed cycle can
be nonzero.
To go from the state (Vi, Pi) by the path (a) to the state (Vf, Pf)
requires a different amount of work then by path (b). To return to
the initial point (1) requires the work to be nonzero.

20. 20
An isothermal process
implies that both P and V
of the gas change
(PVT).

21. 21
Summary of Thermodynamic Processes

22. 22
Thermodynamic Processes for an Ideal Gas
No work is done on a system when
its volume remains constant
(isochoric process). For an ideal
gas (provided the number of moles
remains constant), the change in
internal energy is
.
T
nC
U
Q V 




23. 23
For a constant pressure (isobaric) process, the change in internal
energy is
W
Q
U 


.
T
nC
Q P

CP is the molar specific heat at constant
pressure. For an ideal gas CP = CV + R.
T
nR
V
P
W 





where and

24. 24
For a constant temperature (isothermal) process, U = 0 and the
work done on an ideal gas is
.
ln
f
i









V
V
nRT
W

25. 25
Isothermal Process
W = -Q

26. 26
Isobaric Process
W = -p(V2 - V1)

27. 27
Isometric Process

28. MFMcGraw Chap15d-Thermo-Revised 5/5/10 28
Adiabatic Process
W = ΔU

29. 29
Summary of Thermal Processes
f i
W = -P(V -V)
W
Q
U 


The First Law of Thermodynamics
 
 
 
 
i
f
V
W = nRT ln
V
 
 
 
 
i
f
V
+ nRT ln
V
f i
3
+ nR( T - T )
2

30. 30
Check this in
Thermo
Fundamentals
Text
Summary of
Thermo
Processes
Formulae

31. 31
Example: An ideal monatomic gas is taken through a cycle in the
PV diagram.
(a) If there are 0.0200 mol of this gas, what are the temperature
and pressure at point C?
From the graph: Pc
= 98.0 kPa
Using the ideal gas law
K.
1180
c
c
c 

nR
V
P
T

32. 32
Example continued:
(b) What is the change in internal energy of the gas as it is
taken from point A to B?
This is an isochoric process so W = 0 and U = Q.
 
  J
200
2
3
2
3
2
3
























A
B
A
A
B
B
A
A
B
B
V
P
P
V
V
P
V
P
nR
V
P
nR
V
P
R
n
T
nC
Q
U

33. 33
(c) How much work is done by this gas per cycle?
(d) What is the total change in internal energy of this gas in
one cycle?
Example continued:
The work done per cycle is the area between the curves on the
PV diagram. Here W=½VP = 66 J.
  0
2
3
2
3
i
i
f
f
i
i
f
f




















V
P
V
P
nR
V
P
nR
V
P
R
n
T
nC
U V
The cycle ends where it
began (T = 0).

34. 34
Example:
An ideal gas is in contact with a heat reservoir so that it remains at
constant temperature of 300.0 K. The gas is compressed from a
volume of 24.0 L to a volume of 14.0 L. During the process, the
mechanical device pushing the piston to compress the gas is found
to expend 5.00 kJ of energy.
How much heat flows between the heat reservoir and the gas, and
in what direction does the heat flow occur?
This is an isothermal process, so U = Q + W = 0 (for an
ideal gas) and W = Q = 5.00 kJ. Heat flows from the gas
to the reservoir.