GROUP (4)
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3.6 Entropy and Reversible Heat
 Consider the evaporation process of water vapor in a
cylinder-piston system which is in contact with heat reservoir
at temperature T;if the process is carried out in a reversible
manner, then,the maximum work done by the system is given
by:Wrev = – PH2O (T) ΔV. If the process is carried out in an
irreversible manner; then:
Wirr = – (PH2O(T) – ΔP)ΔV
 Thus,the absolute value of the maximum work done by the
system will be given by |Wrev|,i.e.
(- Wmax) = (-Wrev) > (-Wirr)
since the sign of the work done by the system is negative.
 Now applying the First Law of Thermodynamics to the
reversible and irreversible processes we have:
ΔUrev = qrev+ Wrev = qrev – (-Wrev)
and:
ΔUirr = qirr + Wirr =qirr – (-Wirr)
 Since the initials states and the final states of the reversible
and irreversible processes are the same,then:
ΔUrev = ΔUirr
thus:
qrev - qirr = (-Wrev) – (-Wirr) > 0
therefore: qmax - qirr = ( -Wmax) – (-Wirr) > 0
 Thus, if the process is carried out reversibly:
ΔStotal = ΔSsys + ΔSsur
= qrev / T + (-qrev / T) = zero
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 For an irreversible process:
ΔS sys = qrev / T
or: ΔSsys =(qrev + qirr - qir) / T = qirr / T+(qrev-qirr) /T
or:
ΔSsys = qirr / T+ΔSirr
ΔSsur = – qirr / T
 Where qirr is the amount of heat entering the system leaving
the heat reservoir ( qrev - qirr) represents to amount of work
degraded in the system and enters back to the heat reservoir
Therefore:
ΔStotal = ΔSsys + ΔSsur
Thus:
ΔStotal = qirr /T+ ΔSirr – qirr /T
or:
ΔStotal = ΔSirr
ΔSirr is the entropy created as a result of the irreversibility;
and:
ΔSsystem =qirr / T + ΔSirr
 Since |qrev| > |qirr|, the value of ΔSirr will be a positive
 Consideration of the condensation process shows that the
work done on the system has a minimum value when the
condensation is conducted reversibly , and correspondingly:
wrev < wirr
and since heat is transferred out of the system to the heat
reservoir, the sign of q is negative,and accordingly:
(- qrev ) < (- qirr )
 Thus for reversible processes:
ΔSsys = qrev / T
ΔS sur = - qrev / T
therefore:
ΔStotal = ΔSsys+ΔSsur
or:
ΔStotal = qrev / T- qrev / T = zero
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For irreversible processes:
ΔS sys = qirr / T
or:
ΔSsys = qrev / T + qirr /T – qirr /T
= qirr /T + (qrev – qirr) / T
= qirr / T+ΔS irr
ΔS sur = - qirr /T
thus:
ΔSsys = ΔSsys + ΔS sur = ΔS irr
also(- qrev) < (- qirr )and the signs of q,s are negative
Therefor, ΔSirr will be appositive value
 The last equation indicates that the entropy change of a
system can be measured, orcalculated through a reversible
path for which the heat flow is qrev and ∆Sirr = 0.
 The important feature to be noted from the previous
equationsis that, in going from an initial to a final state, either
by evaporation or condensation, the entropy
difference
between the final state and the initial state,(Sf – Si) is
independent of whether theprocess is conducted reversibly or
irreversibly, and thus independent ofthe path of the process;
therefore the entropy of the system is a state property.
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3.7 The Reversible Isothermal Compression
of an Ideal Gas
 Application of The First Law of Thermodynamics to
mentioned isothermal system gives:
∆U = 0
thus:
q+w=0
or:
q=-w
 But for the mentioned reversible process:
∂W = - Pext. dν
∂W= - (P +∂P) dν= - P dν
 However, for ideal gases:
P = nRT / ν,
thus:
∂W = (- nRT / ν)∂ν
By integration, the work done on the system will be
W = -nRTln (ν2 / ν1)
Since ν2< ν1, the work done on the system will be positive.
Applying the First Law of Thermodynamics yields:
q = nRTln (ν2 / ν1)
where q will be negative value;also we have:
∆Ssys = q / T = nR ln (ν2 / ν1)
thus :
∆Ssur = - q / T= -nR ln (ν2 / ν1)
since ν2 < ν1, the result of the reversible compression of an
ideal gas is decreasing the entropy of the system and
increasing the entropy of the environment by an equal amount
thus:
∆Sirr = 0
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3.8 The Reversible and Irreversible
Adiabatic Expansion of an Ideal Gas
 For reversible adiabatic process:
∆S sys = ∆Ssur= 0
also:
∆Sirr = 0
For irreversible adiabatic process:
∆Ssur = 0
∆S sys = ∆Sirr > 0
This is because work degradation occurs and transfer to heat
inside the system causing higher final temperature than the
final temperature obtained in the case of reversible process.
 For irreversible adiabatic processes, the greater the heat
produced in the gas (system) due to degradation, the greater
the degree of irreversibility, the higher the final temperature
and the internal energy, and the greater the entropy
increases.
 Thus, during an irreversible expansion, the work done by the
gas still equals the decrease in the internal energy of the
system (gas); however, the decrease in u is less than ∆u of
the reversible expansion due to the heat appearing in the gas
as the result of degradation.

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3.9 Summary Statements
 From the discussion as far, three points have emerged;those are:
(1) when a system undergoes a spontaneous (irreversible) process,
the entropy of the system increases,
(2) when a system undergoes a reversible process, no entropy is
created, i.e. ∆Sirr= 0, entropy is simply transferred from
sss
one part of the system to another part, and
(3) entropy is a state function
3.10 The Properties of Heat Engine
 Heat engine is a device which converts heat into work it is
interesting to note that the first steam engine ( which is a
device that transfers heat by using steam as medium was
built in 1769 and was operational for considerable number of
yearsbefore introducing the principles of reversible and
irreversible processes).
 Based on trials, the principle of impossibility of creating virtual
motion machine of the second type,i.e.Theheatenginewhich is
capable of absorbing an amount of heat from a heat reservoir
and transferring it fully into useful work has been emerged.
 Thus, a heat engine can only be created by working a
medium (say ideal gas) through a cyclic process; the simplest
heat engine as shown in figure (3.4). In this device operation,
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the heat engine withdraw an amount of heat, q2, from a heat
reservoir at high temperature t2, and some of this heat is
converted to work, w, and the remainder of heat, q1, is
transferred to a low- temperature heat reservoir at t1, thus:
│q2│>│q1│
and:
t2 > t1
 A typical example of heat engine is the familiar simple steam
engine where superheated steam is passed from a boiler (the
high-temperature heat reservoir) into a cylinder where
itperformswork by expanding against a piston (the engine); as
a result, the steam temperature decreases and at the
completion of the piston stroke, the spent steam is exhausted
to theatmosphere(the low- temperature heat reservoir).
 A flywheel then returns the piston to its original position; thus
completing the cycle and preparing for the next working
stroke.The efficiency of an engine is defined as :
𝐰𝐨𝐫𝐤 𝐨𝐛𝐭𝐚𝐢𝐧𝐞𝐝
−𝐰
𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲 =
=
𝑯𝒆𝒂𝒕 𝒊𝒏𝒑𝒖𝒕
𝒒𝟐
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Figure(3.4): schematic representation of of a heat engine
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 Carnot in 1824 explained the factors governing the efficiency
of an engine by considering acycle that consists of 4reversible steps shown in figure (3.5) .This cycle consist of
two expansion steps : an isothermal step (A → B) and an
adiabatic step (B→C)and two compression steps: an
isothermal one (C→ D) and an adiabatic step (D → A).
For this cycle, which is known as Carnot cycle work done by
the system is given by :w = - (q2+q1), where q2 is the heat
absorbed by the engine from the high- temperature heat
reservoir at t2 during the reversible expansion step (A → B)
and q1 is the heat transferred from the working substance ,
heat engine , to the low- temperature heat reservoir at t1
during the reversiblecompression step (C→D).Thus, for
Carnot cycle:
efficiency= (q2 + q1) /q= 1 + (q1 / q2)
 Let us now question the following: is it possible to have
another
more
efficiency
heat engine
working
between t2 and t1 following Carnot cycle?
 To answer this question by yes or no, let us propose that this
is possible, thus assume that we have two devices: the first
has the parameters q1, q2 and w, and the second has the
parameters q′1, q′2, and w′such that q2 = q′2 and w >w′, so
q1> q′1; then the first device is more efficient than the
second. Let the first engine to run in the forward direction of
Carnot cycle, thus it performs as heat engine;then,the work
done by the system is given by : w = - (q2 + q1).
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Figure(3.5): the Carnot cycle
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Let the second engine to run in the reverse direction of
Carnotcycle, thus it function as heat pump, then :
w′+ q′1= - q′2
or:
w′ = - q′2 - q′1= - [q′2 + q′1]
where w′ is the work done on the system.
Since q2 is(+ve) and q′2 is (-ve) , and w > w′, then:
w + w′ =-(q′1+q1 )
 This means that an amount of work (w + w′) has been
obtained from a quantity of heat (q′ 1+q1) by absorbing it from
the low- temperature heat reservoir without leaving a change
in any other thermodynamic system. Though this is in
agreementwith the First Law of Thermodynamics, it conflict
with human experience fact of impossibility of creation of a
perpetual motion machine of the second type. Thus, the
twomachines must be of the same efficiency.
 The above discussion gives rise to a preliminary formulation
of the Second Law of Thermodynamicswhich is known as the
principle of Thomson: this principle states that “It is
impossible, by means of cyclic process, to take heat from
a reservoir and convert it into work without, in the same
operation, transferring part of that heat to a cold
reservoir” .
 Now consider that we have two other engines; the operating
parameters of the first one q2,q1,w and the operating
parameters of the second one q′2, q′1, w′ such that :
w = w′
q2 > q′2
Since:
q1 / q2 = q′1 / q′2
thus:
q1 > q′1
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 This means that the second engine is more efficient than the
first. Let the second engine to run in the forward direction of
Carnot cycle, thus it performs as heat engine, then:
w′ = - ( q′2 + q′1)
 Let the second device to sun in the reverse direction of
Carnotcycle;thus it performs as heat pump;
thus:
W = - (q2 + q1)
therefore:
q2 - q′2 = q1-q′1= q.
which means that an amount of heat q is pumped from the
low- temperature heat reservoir to the high- temperature heat
reservoir
 This corresponds to the spontaneous heat flow from cold
bodies to hot bodies which opposes the human experience.
Thus the two devices must have the same efficiency.
 The above discussion have laid to the formulation of
theprinciple of Clauses which is one of the statements of
the second law of thermodynamics; it states that:“It is
impossible to transfer heat from a cold reservoir to hot
reservoir without, in the same process, converting a
certain amount of work into heat”.
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