8.principle of entropy

8/13/2019 8.Principle of Entropy

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School of Physics and Astronomy

Junior Honours Thermodynamics GJA 2013-2014

Lecture TOPIC 8 1

(Finn: 5.1, 5.2, 5.3, 5.5)Synopsis: The Clausius inequality. Entropy, and principle of increase in entropy.

INTRODUCTION TO THE THERMODYNAMIC FUNCTION ENTROPY

The Clausius inequality

This considers the heat transfers to a substance as it is taken round an arbitrary cyclic processexchanging heat with any number of surrounding bodies. It can be derived by breaking downthe process into equivalent interactions with a large number of Carnot engines and refrigerators.

Heat exchange with 3 bodies

We look at this procedure for a system exchang-ing heat with 3 different bodies (diagram) andthen generalise the result. Consider two Carnotrefrigerators A and B and an Engine. The refrig-erators are adjusted to exactly the same amountof heat to their hot reservoirs as is taken fromthose reservoirs by the Engine: (Q1 = Q1A;Q2= Q2A) No such restriction can be put on theheat transfers to and from the cold reservoir andtherefore heat flowing from it is Q0A+Q0BQ0.

(Well see later that this is negative)

T1

reservoirhot

WA

T2

reservoirhot

cold reservoir T0

Q0A

WA

Q1A

A

Q0B

WB

Q2B

B

Q0

Engine

W

Q1 Q2

Now consider the complete ensemble as a composite system with the Engine driving therefrigerators. The net work done bythe composite system is W WA+ WB.

No net heat flows into/out of the hot reservoirs. The balance of heat flow and mechanicalwork for the composite system is (Q0A+ Q0B) Q0 = W (WA+ WB) (1st Law). Since theheat flow is from a single reservoir, the Kelvin-Planck statement of the Second Law would becontradicted ifW > WA+WB. Therefore we must haveW WA+WB andQ0A+Q0B Q0 0

From the Carnot refrigerator efficiencies we have, Q1AQ0A =

T1T0 from whichQ0A= Q1

T0T1 , and

similarlyQ0B =Q2 T0T2

. Thus in terms of the heat entering the cold reservoir,

Q0A+ Q0B Q0=

Q1T1

+Q2T2

T0 Q0 0

Q1T1

+Q2T2

Q0T0

0

So far we considered heat flows Qiaround a system of engines and refrigerators. Now considerheat inputs qi to the working substance of the engine. The engine absorbs heat q1 = Q1from the hot reservoirs, and returns heat to the cold reservoir, so there is a change of sign:q0 = Q0.

q0T0

+q1T1

+ q2T2

0

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The above argument can be generalised and refined as follows.

1. If the number of engines and refrigerators is arbitrary,i

qiTi

0

Note that the Ti are the temperatures of the reservoirs with which heat is exchanged

rather, not the temperature of the system.2. In the limit where the amounts of heat entering the working substance are very small at

each step, but the number of steps is very large, the summation may be replaced by anintegral:

d q

T 0, Clausius inequality

Note that T is still the temperature of the reservoirs, which may or may not be equal tothat of the system.

3. Reversible cycles. For the special case where the device operates entirely reversibly,Tsystem= Treservoirs during each heat exchange. Since all processes can also be reversed,

an equivalent conclusion is d qT

0

For both inequalities to be valid simultaneously, they must be restricted to the = case,so for reversible cycles,

d qR

T = 0, reversible cycle only

Note in this last expression T is also the temperature of the system.

Entropy - a new state variable

Let a system (think of it as a sample ofgas) perform a reversible cycle from ini-tial statei to an intermediate statef thenback to i, as in the sketched indicator di-agram. Since the cycle is reversible theequality sign in the Clausius inequality ap-plies giving,

d qR

T =

fi

d qRT

+

if

d qRT

= 0

from which

fi|path 1

d qRT

=

fi|path 2

d qRT

.

Because the value of the integral is path-independent, d qRT

is an exact differential of some statefunction, we call it call entropy S:

fi

d qRT

=

fi

dS= Sf Si = S

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The Principle of Increasing Entropy

Now reconsider the existence proof for entropy if one part (i to f) is irreversible.

The Clausius inequality leads to f

i

d q

T +

i

f

d qRT

0

from which

fi

d qT

f

id qR

T = Sf SiThis means that in an irreversible process, the change in entropy,hereSf Si, is greater than the integration of infinitesimal contri-butions (heat supplied from the surroundings) / (tempera-ture of the contributing part of the surroundings).

Example calculation of an entropy change

A sample of water at 20C is placed in thermal contact with a heat reservoir at 100C. Thesample, of heat capacityCP, is thereby heated irreversiblyat constant pressure from its initialequilibrium state at 20C to a final equilibrium state at 100C.

To calculate the entropy change of the water (thesystem) take it from its initial to its final equi-librium state by an alternative reversible pro-cess, using a different set of reservoirs as the sur-roundings : a series of a large number of heatreservoirs each at only a slightly higher temper-ature than the one before it, starting at Ti andending at Tf. An intermediate stage in the al-ternative process is the reversible transfer of anamount of heat CPdT from the reservoir effec-

tively at T (which will also be the temperatureof the water for that stage). The change in en-tropy of the water is CPdT/T, and the changein entropy of the reservoir is CPdT/T. For thecomplete process, the change in entropy of thewater is SSY S =CPln

3 73293

=CP 0.24141.

By this strategy, the change in entropy for a sys-tem undergoing an irreversible process betweenequilibrium states can be calculated. It is alsopossible, for this particular example, to calculate

the change in entropy SSURR

of the suround-ings which consist of a reservoir at 100C. It de-livers an amount of heat CP(Tf Ti) =CP(373 293) =CP 80 to the water. The surroundingscan be thought of as delivering that amount ofheat reversibly, by an alternative process, per-haps to an other reservoir at a temperature onlyslightly less than 373 K. The change in entropy ofthe surroundings is thus CP(373 293)/373 =CP 0.21448.

For this example (irreversible heating of water) note that STOT

= SSY S

+ SSURR

=0.02693CP >0

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Changes of entropy of thermally isolated systems in adiabatic processes

When a system is thermally isolated no heat is exchanged with the surroundings. Then for anirreversible process undergone by a thermally isolated system

dS >0 ( Sf Si = S >0 for a finite process)

It also follows that for a finite reversible adiabatic process S= 0.

Conclusion, so far:

The entropy of a thermally isolated system increases in any irreversible process

and is unaltered in a reversible process. This is the principle of increasing

entropy.

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Summary so far: 0th law (P, V) = states in m1st law : dU=d q+d w Same hotness Same T

energy is conserved if heat is included

2nd law(i) KP (ii) C (KP = C)&(C = KP) = C KP

Kelvin Planck & Clausius statements

CARNOTS THEOREM: any cycle< C & C independent of working substance

Arbitrary Engine between 2 Ts Two Carnot engines in series

q1T1

+ q2T2

0 thermodynamic temperature, : C= 1 21

Arbitrary closed cycle Ideal gas Carnot cycle

Clausius Inequalityi

qiTsub i

d q

Tsub 0 TIG

sub subsidiary body; often a reservoirReversible closed cycle

d qRTsys res

= 0

State functionS, S=

d qR

T & S

d q

T for general changes

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8.principle of entropy

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