3.6Entropy and Reversible Heat

GROUP (4)

3.6Entropy 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

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 qirris 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:

wrevwirr

andsince 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

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 )andthesigns ofq,s are negative

Therefor,ΔSirrwill 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 isqrevand∆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.

3.7 The Reversible IsothermalCompression

ofan 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 increasingthe entropy of the environment by an equal amount

thus: ∆Sirr= 0

3.8The Reversible and Irreversible AdiabaticExpansion 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∆uof the reversible expansion due to the heat appearing in the gas as the result of degradation.

3.9 Summary Statements

From the discussion as far, three points have emerged;those are: (1) when a system undergoes a spontaneous (irreversible) process, sss 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 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,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 :

Carnot in 1824 explained the factors governing the efficiency of an engine by considering acycle that consists of 4- reversible 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), whereq2 is the heat absorbed by the engine from the high- temperature heat reservoir at t2 during the reversibleexpansion step (A → B) and q1 is the heat transferred fromthe working substance , heat engine , to the low- temperatureheat reservoir at t1during the reversiblecompression step (C→D).Thus, forCarnot 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 thatq2 = q′2 and w >w′, soq1> 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).

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]

wherew′ is the work done on the system.

Since q2is(+ve) and q′2 is (-ve) , and ww′, 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 sameefficiency.
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 oneq2,q1,wand the operating parameters of the second oneq′2, q′1, w′ such that :

w = w′

q2> q′2

Since: q1/q2 = q′1/q′2

thus: q1> q′1

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”.