This document discusses thermodynamic properties of fluids and covers several key topics: 1) It outlines the objectives of studying thermodynamic properties, including developing property relations from the first and second laws of thermodynamics. 2) Fundamental thermodynamic properties like pressure, volume, temperature, and entropy are defined. 3) The first and second laws of thermodynamics are summarized, including equations for closed systems and ideal gases. 4) Methods for determining thermodynamic properties from tables or correlations are introduced.

1. ChE 103
Instructor:
Dr. Md. Easir Arafat Khan
Associate Professor
Department of Chemical Engineering, BUET, Dhaka-1000
January, 2023
Chemical Engineering Thermodynamics
1

2. Thermodynamic properties of fluids
Objective
• Develop property relations from 1st and 2nd law
• Derive equations to calculate ΔH and ΔS from Cp and PVT
data
• Use Ready PVT data from property tables or develop
generalized correlations to estimate property values
2

3. Thermodynamic properties
A quantity which is either an attribute of an entire system or is a
function of position which is continuous and does not vary rapidly
over microscopic distances, except possibly for abrupt changes at
boundaries between phases of the system; examples are
temperature, pressure, volume, concentration, surface tension,
and viscosity. Also known as macroscopic property.
3

4. # First Law of Thermodynamic:
Although energy assumes many forms, the total quantity of energy is
constant, and when energy disappears in one form it appears
simultaneously in other forms.
ΔUt = Q+ W → Δ(nU) = Q+ W
dUt = dQ+ dW→ d(nU) = dQ+ dW
There exists a form of energy,
known as internal energy U.
Δ(Energy of the system) + Δ(Energy of surroundings) = 0
Ideal Gas Equation
First law for a closed system
Cp= Cv+R
RdT=PdV+VdP 4

5. # 2nd Law of Thermodynamics
It is impossible for any device that operates on a cycle to receive
heat from a single reservoir and produce a net amount of work.
The Second Law of Thermodynamics: Kelvin–Planck
Statement
The Second Law of Thermodynamics: Clausius
Statement
related to heat engines
It is impossible to construct a device that operates in a cycle and
produces no effect other than the transfer of heat from a lower-
temperature body to a higher-temperature body.
related to refrigerators or heat pumps
5

6. Thermodynamic Properties of Fluids
• Application of thermodynamics to practical problems requires
numerical values of thermodynamic properties.
• A very simple example is calculation of the work required for a
steady-state gas compressor. If designed to operate adiabatically
with the purpose of raising the pressure of a gas from P1 to P2, this
work can be determined by an energy balance [Eq. (2.32)], wherein
the small kinetic and potential energy changes of the gas are
neglected:
Ws = ΔH = H2 − H1
6

7. Primary Thermodynamic Properties
These have traditionally been called
Helmholtz free energy and the Gibbs
free energy. The word free originally had
the connotation of energy available to
perform useful work, under appropriate
conditions.
 All of the primitive thermodynamic properties P, V, T, U, and S are
included in this equation.
 It is a fundamental property relation connecting these properties for
closed PVT systems.
 All other equations relating properties of such systems derive from it.
the first law for a closed system of n moles of a substance, if written for the
special case of a reversible process:
(6.1)
7
fundamental property relation
Additional thermodynamic
properties

8. Fundamental Property Relations
The above equations are equivalent fundamental property relations. They are
derived for a reversible process. However, they contain only properties of the
system, which depend only on the state of the system, and not the path by
which it reached that state. These equations are therefore not restricted in
application to reversible processes.
Application is to any process in a closed PVT system resulting in a
differential change from one equilibrium state to another.
The system may consist of a single phase (a homogeneous system), or it may
comprise several phases (a heterogeneous system); it may be chemically inert,
or it may undergo chemical reaction
Internal Energy: (6.1)
8

9. All other thermodynamic properties may be evaluated from these by
simple mathematical operations.
Fundamental Property Relations
Maxwell’s equation
For one mole (or to a unit mass) of a homogeneous fluid of constant
composition (i.e. n = 1) they simplify to
U = U(S, V) H = H(S, P) A = A(T, V) G =
G(T, P)
9

10. Enthalpy as a Function of T and P
The most useful property relations for the enthalpy and entropy of homogeneous
phase result when these properties are express as function of P and T
𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃
𝑑𝐻
𝑑𝑇 𝑃
= 𝑇
𝑑𝑆
𝑑𝑇 𝑃
𝑑𝐻
𝑑𝑃 𝑇
= 𝑇
𝑑𝑆
𝑑𝑃 𝑇
+ 𝑉
at const. P
at const. T
The pressure derivative
of entropy results
directly from Eq. (6.17):
The corresponding derivative for enthalpy changes with respect to P at constant T:
The definition of heat capacity
at constant pressure is:
10

11. Entropy as a Function of T and P
With expressions for the four partial
derivatives given,
we can write the required functional
relations as:
These are general equations relating enthalpy and entropy to temperature
and pressure for homogeneous fluids of constant composition.
11
The pressure derivative of entropy results directly from Eq. (6.17):
See slide no
10

12. Ideal Gas State
𝑑𝐻 = 𝐶𝑃𝑑𝑇 + 1 − 𝛽𝑇 𝑉𝑑𝑃
𝑑𝑆 = 𝐶𝑃
𝑑𝑇
𝑇
− 𝛽𝑉𝑑𝑃
Alternative Forms for Liquids
12
Since

13. Internal Energy (U) as a Function of P
Alternative Forms for Liquids
13
See slide no
10
(3.3
)
(3.4
)

14. Example 6.1: Determine the enthalpy and entropy changes of liquid
water for a change of state from 1 bar and 25oC to 1000 bar and 50oC.
The following data for water are available:
𝑑𝐻 = 𝐶𝑃𝑑𝑇 + 1 − 𝛽𝑇 𝑉𝑑𝑃
𝑑𝑆 = 𝐶𝑃
𝑑𝑇
𝑇
− 𝛽𝑉𝑑𝑃
Solution:
14
T
P
Do yourself
Determine the enthalpy
and entropy changes for
the same changes of
states in the other way
e.g., first keep T fixed
and change P and then
change T keeping P
const.
Is there any difference?
Why or why not?

15. Internal Energy and Entropy as Functions of T and V
The most useful property relations for the internal energy and entropy of
homogeneous phase result when these properties are express as function of
V and T
𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉
and,
𝑑𝑆
𝑑𝑇 𝑣
=
𝐶𝑣
𝑇
15
(6.8
)

16. The fundamental property relation for G=G(P,T),
Gibbs Energy as a Generating Function
𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇 (6.11) 𝐺 = 𝐻 − 𝑇𝑆 (6.4)
After
substitution,
The advantage of this equation is that all term are
dimensionless. It has enthalpy term in the right side instead
of entropy.
The Gibbs energy, G or G/RT, when given
as a function of its canonical variables T and
P, serves as a generating function for the
other thermodynamic properties through
simple mathematics, and implicitly
represents complete property information.
The Gibbs free energy (or Gibbs energy) is a thermodynamic potential that can be used
to calculate the maximum reversible work that may be performed by a thermodynamic
system at a constant temperature and pressure.
It follows from the mathematical identity
16

17. The great practical value of the ideal-gas state is now evident. It provides the base
for calculation of real-gas properties.
The generic residual property is defined by:
Residual Properties
The residual volume, for example, is:
From a practical perspective this equation divides property calculations into two
parts:
• first, simple calculations for properties in the ideal-gas state;
• second, calculations for the residual properties, which have the nature of
corrections to the ideal-gas-state values.
17

18. The residual Gibbs
energy
Residual Properties
18
(6.37
)
from

19. 19
(const T )

20. 𝑍 − 1 =
𝐵𝑃
𝑅𝑇
RESIDUAL PROPERTIES FROM THE VIRIAL EQUATIONS OF STATE
The two-term virial equation of state (Volume explicit)
𝐵 𝑎𝑛𝑑
𝑑𝐵
𝑑𝑇
all the residual properties can be obtained
20
(6.49
)

21. RESIDUAL PROPERTIES FROM THE VIRIAL EQUATIONS OF STATE
Equations (6.46), (6.48), and (6.49) from previous slides, are incompatible with
pressure-explicit equations of state and must be transformed such that P is
no longer the variable of integration. In carrying out this transformation, the
molar density ρ is a more convenient variable of integration than V, because ρ
goes to zero, rather than to infinity, as P goes to zero.
Thus, the equation PV = ZRT is written in alternative form as
𝑃 = 𝑍𝜌𝑅𝑇 𝑑𝑃 = 𝑅𝑇(𝑍𝑑𝜌 + 𝜌𝑑𝑍) at const T
𝑑𝑃
𝑃
=
𝑑𝜌
𝜌
+
𝑑𝑍
𝑍
𝐺𝑅
𝑅𝑇
=
0
𝜌
(𝑍 − 1)
𝑑𝜌
𝜌
+
1
𝑧
(𝑍 − 1)
𝑑𝑍
𝑍
21

22. Pressure-explicit three-term virial equation
RESIDUAL PROPERTIES FROM THE VIRIAL EQUATIONS OF STATE
Equations (6.46), (6.48), and (6.49) from previous slides, are incompatible with
pressure-explicit equations of state and must be transformed such that P is no
longer the variable of integration. In carrying out this transformation, the molar
density ρ is a more convenient variable of integration than V, because ρ goes to
zero, rather than to infinity, as P goes to zero.
Thus, the equation PV = ZRT is written in alternative form as
Pressure-explicit two-term virial equation
22

23. Vander Waals CUBIC EQUATIONS OF STATE
“a” term is to account for interactions
among the molecules and the b term
accounts for the finite size of the
molecules.
23

24. Vander Waals CUBIC EQUATIONS OF STATE
24

25. GENERIC CUBIC EQUATIONS OF STATE
25

26. GENERIC CUBIC EQUATIONS OF STATE
26

27. Vander Waals CUBIC EQUATIONS OF STATE
𝑍 − 1 =
𝑏𝜌
1 − 𝑏𝜌
−
𝑎𝜌
𝑅𝑇
𝑍 =
1
1 − 𝑏𝜌
−
𝑎𝜌
𝑅𝑇
0
𝜌
(
𝑏
1 − 𝑏𝜌
−
𝑎
𝑅𝑇
) 𝑑𝜌 =
𝑏𝑙𝑛(1 − 𝑏𝜌)
−𝑏
−
𝑎𝜌
𝑅𝑇
𝐺𝑅
𝑅𝑇
= 𝑍 − 1 − lnZ − ln 1 − bρ −
𝑎𝜌
𝑅𝑇
𝐺𝑅
𝑅𝑇
= 𝑍 − 1 − lnZ − ln 1 −
𝑏𝑃
𝑍𝑅𝑇
−
𝑎𝑃
𝑍𝑅2𝑇2
𝐺𝑅
𝑅𝑇
= 𝑍 − 1 − lnZ − ln 1 −
𝛽
𝑍
− qI
q =
𝑎
𝑏𝑅𝑇
I →
β =
𝑏𝑃
𝑅𝑇
∈= 𝜎, I =
𝛽
𝑍 +∈ 𝛽
∈≠ 𝜎, I =
1
𝜎 −∈
𝑙𝑛
𝑍 + 𝜎𝛽
𝑍 +∈ 𝛽
27

28. Vander Waals CUBIC EQUATIONS OF STATE
𝑆𝑅
𝑅
= ln Z − β
𝑍 =
1
1 − 𝑏𝜌
−
𝑎𝜌
𝑅𝑇
𝜕𝑍
𝜕𝑇 𝜌
=
𝑎𝜌
𝑅𝑇2
𝐻𝑅
𝑅𝑇
= −
𝑎𝜌
𝑅𝑇
+ 𝑍 − 1
𝐻𝑅
𝑅𝑇
= −
𝑞𝛽
𝑍
+ 𝑍 − 1
28

29. Solution:
29

30. For two phases α and β of a pure species coexisting at equilibrium,
where Gα and Gβ are the molar or specific Gibbs energies of the individual phases.
Two phase systems
At equilibrium pressure and temperature
30
Integration of Eq. (6.9) for this change yields the latent
heat of phase transition:
Clapeyron equation

31. Clapeyron equation
∆𝐻𝑙𝑣 = 𝑅∆𝑍𝑙𝑣
𝑑𝑃𝑠𝑎𝑡
/𝑃𝑠𝑎𝑡
𝑑𝑇/(
1
𝑇2)
= −𝑅∆𝑍𝑙𝑣
𝑑𝑙𝑛𝑃𝑠𝑎𝑡
𝑑(1/𝑇)
Equations (6.86) through (6.88) are equivalent, exact forms of the Clapeyron
equation for pure-species vaporization. 31

32. 32

33. Temperature Dependence of the Vapor Pressure of Liquids
33

34. 6.6 THERMODYNAMIC DIAGRAMS
34

35. 35
P-H
Diagram
for
R-134
a

36. 36

37. 37
6.7 TABLES OF THERMODYNAMIC
PROPERTIES

38. 38

39. Example 6.9
Superheated steam originally at P1 and T1 expands through a
nozzle to an exhaust pressure P2. Assuming the process is
reversible and adiabatic, determine the downstream state of the
steam and ΔH for P1 = 1000 kPa, t1 = 250°C, and P2 = 200 kPa.
Solution:
The process is both reversible and adiabatic, so, there is no change in the entropy
of the steam.
For the initial temperature of 250°C at 1000 kPa, no entries appear in the tables for
superheated steam.
Interpolation between values for 240°C and 260°C yields, at 1000 kPa,
39

40. Solution:
Because the entropy of saturated vapor at 200 kPa is greater than S2, the final
state must lie in the two-phase liquid/vapor region. Thus t2 is the saturation
temperature at 200 kPa, given in the superheat tables as t2 = 120.23°C.
Entropy equations becomes,
Enthalpy:
40