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MME 6701: Lecture 02
The Laws of
Thermodynamics
A. K. M. B. Rashid
Professor, Department of MME
BUET, Dhaka
Today’s Topics
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Introduction
The concept of energy and the First Law
The Second Law and the entropy function
Combined statement of the First and Second laws
Concept of temperature and the Zeroth Law
The Third Law and the absolute entropy
Problem solving
1
Introduction

Quantitative description of thermodynamics
 Qualitative observation of the natural process to develop three
balanced equations (the laws)
 These equations, together with experimental data and
information about the process will be used to relate the change
in system properties to a change in its thermodynamic state

Thermodynamics mainly deals with energy and its
transformation to various forms
 Based on the laws of experience governing the gross
behaviour of matter in the system

The laws of thermodynamics are highly condensed
expressions of a broad body of experimental evidence

These laws deal two fundamental principles:
1. conservation of energy
2. degradation of energy

The laws are empirical.
No claim is made that they can be deduced from any fundamental
philosophical principles.
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Concept of Energy

No such thing as absolute energy, only the relative energy;
the difference in energy can only be measured.

Energy is a state function

All energies of a system are extensive properties

Principal energies to consider:
1. Kinetic energy, EK = ½ mv2
2. Potential energy, EP = mgh
3. Internal energy, U
Energy and its conservation
Law of conservation of energy
“The total energy of an isolated system, which is the sum
of all external and internal energy, is constant, though it
may change from one from to another”
 During process, energy can only be changed by
 transforming from one form to another
 transferring across the boundary
3
The first law
“There exists a property of the universe, called its energy,
which cannot change no matter what processes occur”
 The total energy of the universe cannot change for any process.
 During a process, energy of system changes and this change is
accounted for in terms of internal energy.
 The change in internal energy of a system can only be done by
transferring energy across its boundary.
 The change in internal energy of a system for a process must be
equal to the sum of all energy transfers across the boundary of
the system during the process.
 Three kinds of energy transfer during a process:
Q
1. heat flows into the system, Q
2. mechanical work done on the
system, W
3. non-mechanical work done
on the system, W
W
U
W
Sign Convention
 If Q is positive, heat (and thus energy) flows into the system
 If W (or W) is positive, the surroundings does work on the system
 For a finite process:
DU = Q + W + W
4
 If changes in external energies were to be considered:
DEK + DEP + DU = Q + W + W
where DEK and DEP are external kinetic and potential energy
of the system.
 For an infinitesimal change in the condition of the system
dU = dQ + dW + dW
Example 3.1
A closed system initially at rest on the surface of the earth
undergoes a process for which there is a net energy transfer to
the system by work of magnitude 200 Btu. During the process
there is a net heat transfer of energy from the system to its
surroundings of 30 Btu.
At the end of the process, the system has a velocity of 200 ft/s
at an elevation of 200 ft. The mass of the system is 50 lb, and
the local acceleration of gravity g = 32.0 ft/s2.
Determine the change of internal energy of the system for the
process in Btu.
5
Answer:
DKE = ½ mV2 = ½ (50 lb) (200 ft/s)2 = 1000000 lb (ft/s)2


1 lbf
  1Btu 
 39.9 Btu
2   778 ft - lbf 

 32.2 ft - lb/s  
 1000000 lb - ft 2 /s2 
DPE = mgh = (50 lb) (32 ft/s)2 (200 ft) = 322000 lb (ft/s)2
1 lbf
  1Btu 
 322000 lb - ft 2 /s 2  
 12.8 Btu
2   778 ft - lbf 

 32.2 ft - lb/s  
Q = - 30 Btu
W = + 200 Btu
W’ = 0
DKE + DPE + DU = Q + W + W
DU = -30 + 200 – 39.9 – 12.8
= +117.3 Btu
 The First Law of Thermodynamics:
DU = Q + W + W’
 How the values of Q, W and W’ be evaluated?
 What magnitudes may the heat and work effects have,
and what criteria govern these magnitudes?
 The answers to these questions require an examination
of the nature of processes
which leads to the development of the 2nd law
 The 1st and the 2nd laws lay the foundation of
describing the thermodynamic behaviour of matter.
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Spontaneous processes
 A process that involves the spontaneous movement
of a system from an unstable state to a stable state
is called a spontaneous process.
Example:
1. The mixing of two gases
2. The equalization of temperature
 All natural processes are spontaneous process.
 This type of process is said to be irreversible, causing
a permanent change in the system.
Reversible processes
 The rate of change and the degree of irreversibility
of a spontaneous process depends on how far the
system was from its stable state.
 the more distant the system is from its equilibrium,
the higher the spontaneity and quicker the rate of
change towards equilibrium and the more
permanent/irreversible the change will be.
7
 We can imagine a process for which this degree of
irreversibility is minimum.
 the ultimate lowest limit of this minimisation is a
process for which the degree of irreversibility is zero.
 For this zero irreversible process, spontaneity is no
longer applicable.
 the process passes through a continuum of equilibrium
states and the system is never away from equilibrium.
 this type of process is called the reversible process.
Consequence:
 No permanent change in the universe.
Implication:
 Although such a path is imaginary, it is possible to conduct
an actual process in such a manner that it is virtually
reversible.
 Since changes in state functions are independent of the
process connecting two states, one can always choose the
simplest (i.e., a reversible) process, even though the real
process is an irreversible one.
8
Quantification of irreversibility
W
h
1. The heat reservoir in the weight-heat reservoir
system is at the T2. The weight is allowed to fall,
performing work, and the heat produced enters the
reservoir to raise its temperature to T3.
2. The weight-heat reservoir, at temperature T2, is in
contact with another heat reservoir at T1 (<T2) and the
heat is allowed to flow out to lower the weight-heat
reservoir temperature to T1.
3. The heat reservoir in the weight-heat reservoir
system is at temperature T1. The weight is allowed to
fall, performing work, and the heat produced enters
the heat reservoir to raise its temperature to T3.
Process 3, summation of process 1 and 2, is more
irreversible than either process 1 or process 2.
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 Examination of these processes suggests that the
amount of heat produced Q, and the temperature T at
which the heat is produced are important to measure
the degree of irreversibility of the process.
 To compare between process 1 and process 3, the
quantity
Q1/T2 < Q3/T1
which agrees with the conclusion that process 1 is less
irreversible than process 3.
The quantity Q/T is, thus, can be taken as a
measure of the degree of irreversibility of the process.
The second law
 Under a given condition, a system can undergo
many processes in which the energy can be conserved
(according to the First Law)
 But the processes that occur naturally
have a natural direction of change.
All of these natural processes occur due to the same cause
“some kind of energy in them spreads out.”
 A state function, called entropy, is defined to measure
the spontaneous dispersion of energy
how much energy is spread out in a process at a specific
temperature.
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“There exists a property of the universe, called its
entropy, which always tend to change spontaneously
in the same direction no matter what process occurs”
 By convention, entropy is defined so that, when summed
for the system and its surroundings, it always increases.
 Entropy increase predicts what physical and chemical
events will happen spontaneously.
 That's why an entropy increase (or equivalently, the
second law) is often called "time's arrow".
The entropy function
 Entropy is a state function and an extensive variable.
 Entropy can be transferred across the boundary of a
system as well as can be produced inside the system.
 Total change in entropy for the system
DSsys = DSt + DSp
(3.3)
 By second law, for all systems during any process
DSp  0
(3.4)
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 The overall change in entropy for the universe,
that is, the system plus surroundings:
DSun = DSsys + DSsur
= [ DSt + DSp ] + [ DS’t + DS’p ]
DSun = DSp + DS’p  0
(3.6)
 Thus, for any kind of process,
the entropy of the universe can only increase.
Self-Assessment Question 3.4
Establish the criterion of spontaneity of a system based on
entropy.
DSun  0

The process occurs spontaneously
DSun  0

The process will not occur
DSun = 0

The system is in equilibrium
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Intuitive meaning of
entropy production
 Changes in real world are always accompanied by friction, and
some energy is dissipated.
 Entropy production is quantitative measure of this dissipation.
 The further a system is from its equilibrium state, the faster it
tends to change, the greater the frictional or dissipative effects,
and the larger its rate of entropy production.
 The first law deals with the conservation of energy, but the
second deals with the degradation of energy.
 Reversible processes have zero entropy production.
 Irreversible processes produce some entropy and cause some
permanent change in the universe.
Self-Assessment Question #3.5
A system is changed from state A to state B using a reversible
adiabatic process. What would be the total entropy change for
the system?
Problem 3.9
Indicate the relative magnitudes of the entropy transfer versus
entropy production in the following processes:
(a) A thermally insulated container has two compartments of
equal size. Initially one side is filled with a gas and the other
is evacuated. A valve is opened and the gas expands to fill
both compartments.
(b) A gas contained in a steel cylinder is slowly expanded to
twice its volume.
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Relation between entropy and heat
 Sadi Carnot (1824) – inventor of heat engines.
Heat reservoir at
high temperature, T1
Q1
Heat
Engine
Q2
Heat reservoir at
low temperature, T2
W
Schematic representation of the working principle of a heat engine
Process I: The system changes
isothermally from (TI, P0) to (T1, P1) as
the pressure of the system is reduced
allowing the volume of the system to
increase from V0 to V1. A quantity of
heat QI flows from the high heat
reservoir into the system.
Process II: The pressure of the system
is further reduced to P2 using adiabatic
process. The temperature is reduced to
TIII and the volume is increased to V2.
Process III: A quantity of heat QIII is
allowed to exchange from the system
into the low heat reservoir and the
system is isothermally compressed to
(P3, V3).
States and Processes
of a Carnot cycle
Process IV: The volume of the system
is further compressed adiabatically
back to the starting state (P0, V0). The
temperature is reduced to T0.
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The Carnot cycle
P
A
T
Q1
TI
IV
B
A
TI
IV
I
D III
TIII
II
B
I
TIII
Q3 C
II
III
C
D
V
(a)
V
(b)
The Carnot cycle in (a) P-V coordinate, and in (b) T-V coordinate.
The reversible paths used to complete the cycle from state A to state B are:
I – isothermal expansion,
II – adiabatic expansion,
III – isothermal compression
IV – adiabatic compression.
 The overall efficiency of heat engine
 
Work obtained
-W


Heat input
QI
TI - TIII
TI
(3.8)
W = total work done on the system (heat engine)
QI = heat absorbed by the system during stage I
TI = temperature of hot reservoir
TIII = temperature of cold reservoir
 Using the first law, for the overall cyclic process:
DU = 0 = W + QI + QIII (using W’=0)
W = - (QI + QIII)
(3.9)
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 Substituting Eq.(3.9) into
Eq.(3.8) will result
Pressure
QI  QIII

QI
QI
QIII

TI
TIII
B
TI - TIII
TI
A
 0
Volume
A cyclic process ABA broken into
a large number of Carnot cycles
 For the overall cyclic path ABA,
o dQrev / T = 0

 Q/T = 0
 Thus, Q/T has the property of a state function.
Clausius called this thermodynamic function as the entropy of the system, S.
 Thus, the relation between heat absorbed and
entropy transfer into a system becomes
Qrev
DS =
T
 For infinitesimal change of process
dQrev = TdS ;
DS =

dS =

Qrev =
 TdS
dQrev
T
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Entropy change for
different processes
 Since entropy is a state function, for the process AB,
DSrev [AB] = DSirr [AB]
DSrev,t + DSrev,p = DSirr,t + DSirr,p
DSirr,t = DSrev,t - DSirr,p
 dQirr/T <  dQrev/T
 For an isothermal process, T is constant, and
[ Qrev ]T > [ Qirr ]T
(3.14)
 For all possible isothermal processes:
Heat absorbed by a reversible process is the maximum.
The work function
 For a finite process, the work done is defined as:
W 

dW 
F(x) . dx

F = force
dx = displacement
dx
Pext
Area, A
Volume, V
P
Movement of piston
dW = (Pext . A) . (-dx)
dW = Pext (-dV)
17
 For reversible expansion of the gas, the value of dP
would be so small that for all practical purposes,
Pext = P
 Thus, the work done on the system
dWrev = - P dV
dWirr = - Pext dV
dWrev = -PdV
Wrev = -  PdV
 Since for irreversible processes, Pext < P,
work done by a reversible process is always the maximum.
Some common non-mechanical work
1. Gravitational work, mf h
where mf is the force acting on the system and h the height to
which the system is lifted against gravity.
2. Electrical work, EdQ
where E is the potential difference and dQ the charge.
3. Surface work, dA
where  is the surface tension and A the area.
4. Centrifugal work, m2r
where m is the mass of the system,  the angular velocity, and
r the distance from axis of rotation.
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Combined statement of
the first and second law
 The first law:
 From the second law:
 From the work function:
dU = dQ + dW + dW’
dQrev = T dS
dWrev = - P dV
 The combined statement for any reversible process:
dU = TdS – PdV + dW’
 the basis of derivations of the conditions for equilibrium in
thermodynamic systems.
 a key relationship in solving practical thermodynamic problems.
Example 3.2
One mole of an ideal gas is kept at 10 atm pressure and 100
K temperature. The PVT relationship for an ideal gas can be
expressed by the relationship PV=nRT, where the value of R,
the ideal gas constant, can be taken as 0.082 litre-atm/mol-K.
Calculate the amount of mechanical work done on the system
if it undergoes a reversible
(i)
isothermal expansion to 1 atm,
(ii) isobaric expansion to 10 litres, and
(iii) isochoric process.
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Isothermal expansion to 1 atm
The Ideal Gas Law:
PV = n RT

WT = - (nRT/V) dV
W = - P dV

= - nRT (dV/V)
(for isothermal process)
W = - nRT ln (V2/V1) = nRT ln (V1/V2)
W = - (P1V1) ln (V2/V1) = (P1V1) ln (V1/V2)
W = - nRT ln (P1/P2) = nRT ln (P2/P1)
Given Data:
n = 1 mole, P1 = 10 atm, P2 = 1 atm,
T1 = T2 = 100 K,
R = 0.082 litre-atm/mol-K
V1 = nRT1/P1 = (1 mol)(0.082 litre-atm/mol-K)(100 K)/(10 atm) = 0.82 litres
V2 = nRT2/P2 = (1 mol)(0.082 litre-atm/mol-K) (100 K) / (1 atm) = 8.20 litres
W = - nRT ln (V2/V1)
= -(1 mol) (0.082 litre-atm/mol-K) (100 K) ln (8.2 litre/0.82 litre)
= -18.88 litre-atm
Isobaric expansion to 10 litres


W = - P dV = - P dV
(for adiabatic process)
W = - P (V2 - V1) = n R (T1 – T2)
Given Data:
n = 1 mole,
P1 = P2 = 10 atm,
T1 = 100 K,
V2 = 10 litres
V1 = nRT1/P1 = (1 mol)(0.082 litre-atm/mol-K)(100 K)/(10 atm) = 0.82 litres
W = – P (V2 - V1) = – (10 atm) (10 - 0.82 litre) = - 91.8 litre-atm
Isochoric process
For isochoric process, dV = 0

W = - P dV = 0
20
Example 3.3
An ideal gas is held in a piston-cylinder assembly undergoes
a reversible adiabatic expansion for which the relationship
between pressure and volume is given by PV = constant.
The initial pressure is 3 bar and the initial and final volumes
are 0.1 m3 and 0.2 m3 respectively.
Determine :
(1) the amount of mechanical work done to the system, and
(2) the change in internal energy of the system for the
process if (a)  = 1.5, (b)  = 1.0, and (c)  = 0.
Work done for adiabatic process
PV = Constant, k
P = k / V
W = - P dV = - (k/V) dV = - k V - dV

W = -
W =


k V21-  – k V11- 
1-
=
k V11-  – k V21- 
( P1V1 ) V11-  – ( P2V2 ) V21- 
1-
1-
=
P1V1 – P2V2
1-
21
 = 1.5
Given Data:
P1 = 3.0 bar,
V1 = 0.1 m3,
V2 = 0.2 m3 ,
 = 1.5
PV = Constant, k
P2 = P1 ( V1/ V2 )  = (3.0 bar) (0.1 m3 / 0.2 m3) 1.5 = 1.06 bar
W =
(3.0 bar) (0.1 m3) – (1.06 bar) (0.2 m3)
105 N/m2
1 kJ
1.5 – 1
1 bar
103 N.m
W = - 17.6 kJ
The process is adiabatic, thus Q = 0
DU = Q + W = W = - 17.6 kJ
 = 1.0
PV = constant,
P = constant / V

DU = W = - P dV = (P1V1) ln (V1/V2) = - 20.79 kJ
 =0
P = Constant
(Simple isobaric process)
DU = W = – P (V2 – V1) = - 30.0 kJ
22
Table 3.1
Reversible mechanical work done on the system
containing n mole of an ideal gas during various kinds
of processes
Process
Work Done On System
Isothermal
W = - n RT ln (V2/V1) = (P1V1) ln (V1/V2)
Adiabatic
W = (P1V1 – P2V2) / (1 – )
Isobaric
W = - P (V2 – V1) = n R (T1 – T2)
Isochoric
W =0
The third law
 The first and the second laws deal with only the changes
in energy and entropy, respectively.
 The zero levels for both energy and entropy are arbitrary.
 Does there exist any particular choice for the zero
level of energy or entropy?
23
The concept of temperature
 Temperature is a property of matter that provides a
universal measure of the tendency of system to
exchange heat.
 Attempts to quantify this aspect of behaviour of matter
lead to the development of the Zeroth Law of
Thermodynamics:
“If two systems are separately in thermal equilibrium
with a third, then they must also be in thermal
equilibrium with each other”
 Experiments studying cryogenic behaviour of matter
established that
 there is a lower limit to temperature, known as the
absolute zero
 the energy and entropy of a body are decreased with
falling temperature
 If the energy of each element at absolute zero is assigned
a zero value, the energy of a compound at absolute zero
is, however, found to be not zero.
 the energy change accompanying chemical reactions is
not zero at absolute zero
 The choice of the absolute zero for the zero entropy is
fruitful and leads to the development of the third law of
thermodynamics.
24
Work of T.W. Richards (1902):
“For many reactions, the changes in entropy and heat
capacity approach zero at low temperature”
Work of Nernst (1906): Nernst heat theorem
 Generalisation of Richard’s finding:
“For all reactions involving substances in the condensed
state, the change in entropy is zero at the absolute zero”
 For reaction A + B = AB, the change in entropy at 0 K,
DS = SAB – (SA + SB) = 0
 if the entropies of elements SA and SB are assigned zero values
at the absolute zero, then the entropy of the compound SAB is
also zero
 constitutes one statement of the third law of thermodynamics
Work of Planck:
“Entropy of any homogeneous substance, which is in
complete internal equilibrium, may be taken as zero at 0 K”
 Entropy of substance having non-equilibrium structure will
not be zero at 0 K.
The Third Law:
“There exists a lower limit to the temperature that can be
attained by matter, called the absolute zero of temperature,
and the entropy of all homogeneous substances which are
in complete internal equilibrium is the same at that
temperature and may be taken as zero”
25
Validation of the third law
A direct validation of the Third Law is difficult.
1. The temperature 0 K could not be attained.
even if attained by extrapolation, the kinetics of
reaction at these low temperatures is very sluggish.
2. A direct measure of entropy change, DS = Qrev/T,
is possible only for reversible reactions.
reactions at 0 K is hardly reversible.
Indirect Validation of the Third Law:
 Consider the cyclic process:
Sn (grey)  Sn (white)
292 K
Sn (grey)
0K
III
IV
Sn (grey)  Sn (white)
Heat one mole of grey tin
from 0 K to 292 K
II. Let the grey tin transforms
into white tin at 292 K:
II
I
I.
Sn (white)
III. Cool the white tin from
292 K to 0 K
IV. The white tin transforms
into grey tin at 0 K (by an
imaginary process)
26
 The net entropy change in this cyclic process
DScyc = DSI + DSII + DSIII + DSIV = 0
T
DS =

0
CP
dT
T
Debye’s formula :
For the atomic heat capacity of a crystalline substance at
temperatures approaching 0 K
4
CV = 12p R
5
T
q
3
= 464.6 T
q
3
cal/deg.mol
[ 0 < T < q/10]
q = Debye characteristic temperature of the substance
Experimental data:
DSI = 9.11 cal/mol-K
DSII = 1.85
DSIII = -11.04
DSIV = - 0.08
DSI + DSII + DSIII  0
DSII = - (DSI + DSIII)
DSIV = 0
 This is a pervasive observation!!
and, therefore, confirms the third law.
 The entropy of all substances is the same at 0 K.
by convention, this is considered to be as ZERO !
27
3.17
5 kg of steam contained within a piston-cylinder assembly undergoes an
expansion from state 1, where the specific internal energy (the internal energy
per unit mass) is u1 = 2709.9 kJ/kg, to state 2, where u2 = 2659.6 kJ/kg. During the
process, there is a heat transfer of energy to the steam with a magnitude of 80 kJ.
Also, a paddle wheel transfers energy to the steam by work in the amount of 18.5
kJ. There is no significant change in the kinetic or potential energy of the steam.
Determine the amount of energy transfer by work in kJ from steam to the piston
during the process.
3.25
20 litres of hydrogen gas at 27 C and 50 atm expands reversibly and
isothermally to 200 litres. In doing so, it absorbs 400 kJ of thermal energy from
its surroundings. Determine the change in internal energy of hydrogen gas for
the process.
3.27
One mole of an ideal gas is heated at a constant pressure of 1 atm from
0 to 100 C. (a) Calculate in calories the work involved. (b) If the gas were
expanded isothermally at 0 C from 1 atm to some other pressure, what must be
the final pressure if the system performs work which is equal to the work of
part a? Use R = 1.987 cal/mol-K.
3.30
Calculate the work performed by the system for the reversible adiabatic
expansion of 2 moles of an ideal gas at 300 K and 20 atm to a final pressure of 2 atm.
Ignoring non-mechanical work, what would be the change in internal energy for this
process? Use the relationship PV = constant, where  = 5/3.
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