Materials
Thermodynamics
INTRODUCTION AND
BASIC CONCEPT
Why study Thermodynamics?
Pervasive
Applies to every volume element in every
instant in time
Established
J. Willard Gibbs (1883) Equilibrium of
Heterogeneous Substances – completed the
apparatus of phenomenological thermodynamics
Comprehensive
Basic info how
matter behaves
Enables maps
generation of
equilibrium states
Apparatus capable of handling
most complex kinds of: system,
applications & influences
Identifies properties of systems important in
wide range applications. Provide relationships
between these database properties, predict
matter behaviour
Widely used in science & industry to answer
real-world questions about the behaviour of
matter
DEFINITION OF THERMODYNAMICS
 Thermodynamics: The science of energy.
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 Energy: The ability to cause changes.
 The name thermodynamics stems from the Greek words therme (heat) and
dynamis (force/power).
 Thermodynamics: The physical sciences that deal with
energy and the transfer, transport, and conversion of
energy.
 Thermodynamics application:




Microscopic organisms
Household appliances
Transport vehicles
Power generation
Laws of Thermodynamics
The ZEROTH LAW
definition of temperature.
of thermodynamics, which underlies the
The FIRST LAW
of thermodynamics, which mandates
conservation of energy, and states in particular that heat is a form of
energy.
The SECOND
LAW of thermodynamics, is about the
quality of energy. It states that as energy is transferred/transformed,
more and more of it is wasted. There is a natural tendency of any
isolated system to degenerate into a more disordered state (entropy).
The THIRD
LAW of thermodynamics, which concerns on
entropy of an object at absolute zero temperature, and implies that it is
impossible to cool a system all the way to exactly absolute zero.
 The first law of thermodynamics: An expression of the conservation of
energy principle.
 The first law asserts that energy is a thermodynamic property.
 Conservation of energy principle:
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During an interaction, energy can change from one form to another but the
total amount of energy remains constant.
 Energy cannot be created or destroyed.

Conservation of energy principle for
the human body. (diet industry)
Energy cannot be created or
destroyed; it can only change forms
(the first law).
The second law of thermodynamics:
It asserts that energy has quality as well as quantity, and actual processes
occur in the direction of decreasing quality of energy.
A cup of hot coffee left on a table eventually cools,
But a cup of cool coffee in the same room never gets hot by itself.
Heat flows in the direction of
decreasing temperature.
A boiling pan of water is unlikely to
ever become a block of ice.
A smashed plate could never
reassemble itself, as this would reduce
the entropy of the system in defiance of
the second law of thermodynamics.
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Closed
System
Mass Flow
dm
dt
 0
Open
System
Mass Flow
System and Surroundings
• System – quantity of matter or
region in space chosen for study
Surroundings
Closed System
• Surroundings – mass or region
outside the system
System Boundary
Surroundings
Control Surface
Open System
Mass flow
( Fixed space or volume)
Open systems have mass flow across their boundaries.
Surroundings
Heat
System
• Closed system (control mass) – no
mass can enter or leave a system
• Also in special case system can be
considered as isolated system
Work
Surroundings
Mass Flow
• Open system (control volume) – usually
encloses a device that involves
mass flow such as compressor,
turbine or nozzle.
Heat
System
Power
Mass
Flow
• Closed system (Piston and Cylinder)
m
Piston
• Open system (Gas Turbine Engine)
Fuel Flow In
Combustor
Gas at pressure, p
Shaft Work Output
Air Flow In
Compressor Work Out
Exhaust Gases Out
• Any characteristic of a system called property.
• Some familiar properties are pressure, P, temperature, T,
volume, V, and mass, m.
• Properties are considered to be either intensive or extensive.
• Intensive properties – those that are independent of the
size of the system, such as temperature, pressure and density.
• Extensive properties – those whose values depend on the size
or extent of the system such as mass, volume and total energy.
• Specific properties is the extensive properties per unit mass.
m
½m
½m
V
½V
½V
T
T
T
P
P
P



Extensive
properties
Intensive
properties
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
6.2
Exothermic
Endothermic
6.2
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
Energy, Pressure, Volume, Temperature
E = Efinal - Einitial
P = Pfinal - Pinitial
V = Vfinal - Vinitial
T = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2 is
the same even though they took different
paths.
6.3
Non State functions are properties that are determined by how
that condition was achieved.
Work, Heat
W ≠ Wfinal - Winitial
q ≠ qfinal - qinitial
Work done by hiker 1 and hiker 2 is
different because they took two
different paths.
6.3
First Law of thermodynamics – energy
can be converted from one form to another,
but cannot be created or destroyed.
Esystem + Esurroundings = 0
or
Esystem = -Esurroundings
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
[system]
[surroundings]
6.3
E = q + w
E  change in internal energy of a system
q  heat exchange between the system and the surroundings
w  work done on (or by) the system
w = -PV when a gas expands against a constant external pressure
Sign Conventions for Work and Heat
Process
Sign
Work done by the system on the surroundings
-
Work done on the system by the surroundings
+
Heat absorbed by the system from the surroundings (endothermic)
+
Heat absorbed by the surroundings from the system (exothermic)
-
6.3
Closed system does not allow mass transferred but, allow
heat transferred between the system and surrounding.
Because of that, all energy exchange between a closed
system and its surrounding appears as heat and work.
Therefore the second term of Equation can be stated as:
Work
E surrounding   Q  W
Q and W always refer to the system, and the choice of
the signs depends on the way energy or work
transferred.
If Q and W transfer into the system from the
surroundings, then, the sign is +ve
w = Fd
w = -PV
V > 0
-PV < 0
F
P ꞏ V = 2 ꞏ d3 = Fd = w
d
wsys < 0
Work is
not a
state
function!
w = wfinal - winitial
initial
final
6.3
A sample of Nitrogen gas expands in volume from 1.6 L to
5.4 L at constant temperature. What is the work done in
Joules if the gas expands (a) against a vacuum and (b)
against a constant pressure of 3.7 atm?
w = -PV
(a)
V = 5.4 L – 1.6 L = 3.8 L
P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 Joules
(b)
V = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = -14.1 L•atm
w = -14.1 L•atm x
101.3 J
= -1430 J
1L•atm
6.3
Thermodynamics - General Terms
Algebra relation between various state variables to
form a thermodynamics system.
1 (at constant n and T)
Boyle’s law: V 
P
Charles’ law: V  T
(at constant n and P)
Ideal Gas Equation:
PV = nRT
2.4 Ideal Gas Model
The relationship among the state 23
variables, temperature, pressure,
and specific volume is called the equation of state.
We now consider the equation of state for the vapour or gaseous phase
of simple compressible substances.
The Ideal-Gas Equation of State
Recall that the combination of Boyle’s and Charles’ laws
T 
P  R 
v
where R is the constant of proportionality and is
called the gas constant and takes on a different
value for each gas.
The ideal gas equation of state may be written several ways.
V  mv
mR  ( MN ) R  NR
V
v
N
Pv  RT
V
P  RT
m
PV  mRT
PV  NR T
V
P  RT
N
P = absolute pressure (MPa or kPa)
molar specific volume in m /kmol
Pv  R T vT == absolute
temperature (in K)
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R=
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8.314 kJ/kmol K
Some values of the Universal Gas Constant, Ru
8.314 kJ/(kmol K)
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8.314 kPa m3/(kmol K)
1.986 Btu/(lbmol R)
1545 ft lbf/(lbmol R)
10.73 psia ft3/(lbmol R)
The ideal gas equation of state is used when;
1) The pressure is small compared to the critical pressure
(P<Pcr) or
2) When the temperature is twice the critical temperature and
the pressure is less than ten times the critical pressure (T=
2Tcr, P<10Pcr)
The critical point is that state where there is an
instantaneous change from the liquid phase to the
vapor phase for a substance. Critical point data is
given in Table A-1.
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E = q + w
At constant pressure, P:
q = H and w = -PV
E = H - PV
H = E + PV
6.4
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0
Hproducts > Hreactants
H > 0
6.4
Is H negative or positive?
System absorbs heat
H > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 0°C and 1 atm.
H2O (s)
H2O (l)
H = 6.01 kJ
6.4
Is H negative or positive?
System gives off heat
H < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 25°C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) H = -890.4 kJ
6.4
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
H = 6.01 kJ
If you reverse a reaction, the sign of H changes
H2O (l)
•
H2O (l)
H2O (s)
H = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then H must change by the same factor n.
2H2O (s)
2H2O (l)
H = 2 x 6.01 = 12.0 kJ
6.4
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
H = 6.01 kJ
H2O (l)
H2O (g)
H = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burns in air?
P4 (s) + 5O2 (g)
266 g P4
x
P4O10 (s)
1 mol P4
123.9 g P4
x
H = -3013 kJ
-3013 kJ
= -6470 kJ
1 mol P4
6.4
The specific heat (s) of a substance is the amount of heat (q) required to
raise the temperature of 1 gram of the substance by 1 degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to
raise the temperature of a given quantity (m) of the substance by 1
degree Celsius.
The Specific Heats of Some Common
Substances
Substances
Specific Heat (J/gꞏ°C)
Al
0.900
Au
0.129
C (graphite)
0.720
C (diamond)
0.502
Cu
0.385
Fe
0.444
Hg
0.139
H2O
4.184
C2H5OH (ethanol)
2.46
C = ms
Heat (q) absorbed or released:
q = ms
q = C
 = Tfinal - Tinitial
6.5
Specific heats
Definition: The energy required to raise the temperature of a unit mass of a
substance by 1 degree.
Specific heat constant volume, Cv
the energy required to raise the
unit mass of a substance by 1 degree
as the volume is maintained constant.
Specific heat of constant pressure, Cp
pressure is maintained constant.
Cp  Cv
 express the specific heats in terms of thermodynamic properties.
 Conservation of energy principle ein - eout = esystem
C
C
v
p
  u
 
  T
  h
 
  T






change in u with T at constant V
v
change in h with T at constant P
p
 common units: kJ/kgC or kJ/kgK
 molar basis, Cv & Cp: kJ/kmol C or kJ/kmolK
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Internal energy, enthalpy & specific heats of ideal gas
 The differential changes in the internal energy & enthalpy of an ideal gas
can be expressed as
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du  C v ( T ) dT
dh  C p ( T ) dT
 The integration in these equation can be performed, yielding
u 2  u1  Cv ,av (T2  T1 )
(kJ/kg)
h2  h1  C p ,av (T2  T1 )
u  Cv ,av .T
h  C p ,av .T
U  mCv ,av .T
H  mC p ,av .T
(kJ/kg)
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There are three ways to determine u and h for ideal
gases:
1. Using tabulated u and h data – when it is readily
available
2. Using Cv and Cp as a function of temperature – perform
integration –computerized
3. Using average Cv and Cp – convenient when property
tables not available. Accurate if T is small
Refer Table A-2
Specific heat relations of ideal gases
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 Derive relationship between specific heats.
We know that
h  u  pv
h  u  RT
Differentiate w.r.t T
Which means that
dh du

R
dT dT
C p (T )  Cv (T )  R
If CP and Cv remain constant with temperature;
C p  Cv  R
Specific heat ratio, k
(kJ/kgK)
k
Cp
Cv
Internal energy, enthalpy & specific heats of solids & liquids
Incompressible substance
a substance whose specific
volume(density) is constant
 Assumption: constant-volume should be taken to imply that the energy
associated with the volume change is negligible compared with other
forms of energy.
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It can be mathematically shown that:
C p  Cv  C
Specific heat values for common solids & liquids are given in Table A-3
u  Cav (T2  T1 )
 h   u  vdP  C av  T  v  P
How much heat is given off when an 869 g iron bar cools
from 940C to 50C?
Given: Specific heat (s) of Fe = 0.444 J/g • 0C
t = tfinal – tinitial = 50C – 940C = -890C
q = mst
= 869 g x 0.444 J/g • 0C x –890C = -34,000 J
6.5
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