# In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0

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1 Chapter 4 Entropy and second law of thermodynamics 4.1 Carnot cycle In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0 since the internal energy U is a state function. A reversible cyclic process can be represented by a closed loop in the P diagram (Fig. 4.1). Figure 4.1: Reversible cyclic process. he shaded area of the loop is the total work done by the system in one cycle. It follows from the first law of thermodynamics that, since U = 0, the work done by the system during a cycle is equal to the heat absorbed, i. e., W = Q = Pd = Area enclosed. In a cyclic process, work is converted into heat and in the end of each cycle the system returns to its original state. We will introduce the heat engine which is based on the properties of a cyclic process. 23

2 24 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS Let us consider the operation of a heat engine in more detail (Fig. 4.2). Part of the heat that is transferred to the system from a heat bath with temperature 1, Q 1, is converted into work, W, and the rest, Q 2, is delivered to a second bath with 2 < 1 (condenser). Following the first law of thermodynamics, Q 1 Q 2 = W. Carnot process: his a reversible cycle process bounded by two isotherms and two adiabatic lines. he Carnot process can be realized with an arbitrary working substance, but we shall consider here an ideal gas (Fig. 4.3). Figure 4.2: Heat engine. Figure 4.3: Ideal gas as a working substance for the Carnot process. In the P diagram the Carnot process can be represented as shown in Fig Figure 4.4: he Carnot process in the P diagram. One Carnot cycle consists of four consecutive thermodynamic processes, which are 1. A B: isothermal expansion at = 1 ; the volume changes as A B, while heat Q 1 is absorbed from a bath and the system performs work.

3 4.2. SECOND LAW OF HERMODYNAMICS B C: adiabatic expansion, during which and while = B C, 3. C D: isothermal compression at = 2 ; the system ejects heat Q 2 to the bath (Q 2 < 0 is a convention). 4. D A: adiabatic compression, during which 2 1, D A, = 0. In one cycle operation, the system receives an amount of heat Q 1 from a hot reservoir, performs work, and rejects waste heat Q 2 to a cold reservoir. From the first law of thermodynamics we have: 0 = du = (+δw) = Q+W = Q 1 +Q 2 +W, where W is the work performed by the system, equal to the area enclosed in the loop (shaded area in Fig. 4.4). he efficiency of the Carnot engine is defined as η performed work absorbed heat = W Q 1 = Q 1 +Q 2 Q 1 = Q 1 Q 2 Q 1. η is 100% if there is no waste heat (Q 2 = 0). However, we will see that this is impossible due to the second law of thermodynamics. 4.2 Second law of thermodynamics Definition by Clausius: here is no thermodynamic transformation whose sole effect is to deliver heat from a reservoir of lower temperature to a reservoir of higher temperature. Summary: heat does not flow upwards. Definition by Kelvin: here is no thermodynamic transformation whose sole effect is to extract heat from a reservoir and convert it entirely to work. Summary: a perpetuum mobile of second type does not exist.

4 26 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS In order to prove that both definition are equivalent, we will show that the falsehood of one implies the falsehood of the other. For that purpose, we consider two heat reservoirs with temperatures 1 and 2 with 1 > 2. (1) If Kelvin s statement were false, we could extract heat from 2 and convert it entirely to work. We could then convert the work back to heat entirely and deliver it to 1 (there is no law against this) (Fig. 4.5). hus, Clausius statement would be negated. Figure 4.5: Q 2 = W = Q 1 process. (2) If Clausius statement were false, we could let an amount of heat Q 1 flow from 2 to 1 ( 2 < 1 ). hen, we could connect a Carnot engine between 1 and 2 such as to extract Q 1 from 1 and return an amount Q 2 < Q 1 back to 2. he net work output of such an engine would be Q 1 Q 2 > 0, which would mean that an amount of heat Q 1 Q 2 is converted into work, without any other effect. his would contradict Kelvin s statement. From the microscopic point of view - heat transfer is an exchange of energy due to the random motion of atoms; - work s performance requires an organized action of atoms. In these terms, heat being converted entirely into work means chaos changing spontaneously to order, which is a very improbable process. Usually, - one configuration corresponds to order; - many configurations correspond to chaos.

5 4.3. ABSOLUE EMPERAURE Absolute temperature In order to introduce the concept of absolute temperature, let us discuss the consequences of the second law of thermodynamics. i) he second law of thermodynamics implies that a Carnot engine cannot be 100% efficient since otherwise all heat absorbed from a warm reservoir would be converted into work in one cycle. ii) No engine working between two given temperatures can be more efficient than a Carnot engine. A Carnot engine is a reversible engine between two baths. he previous statement means as well that an irreversible engine cannot be more efficient than a reversible one. In order to show this, we consider two engines C and X (with X not necessarily a Carnot one) working between baths at 1 (warm) and 2 (cold). We run the Carnot engine C in reverse, as a refrigerator C, and feed the work output of X to C (Fig. 4.6). Figure 4.6: Linked engine X and Carnot refrigerator C. he total work output of such a system is W tot = ( Q 1 Q 2 ) ( Q 1 Q 2 ). If we adjust the two engines so that Q 1 = Q 1, no net heat is extracted from the heat bath at 1. In this case, an amount of heat Q 2 Q 2 is extracted from the heat bath at 2 and converted entirely to work, with no other effect. his would violate the second law of thermodynamics unless Q 2 Q 2. We divide this inequality by Q 1 and, using the fact that Q 1 = Q 1, get Q 2 Q 1 Q 2 Q 1 and 1 Q 2 Q 1 1 Q 2 Q 1.

6 28 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS From this follows that the efficiencies of the engines compare as η C η X. Consequence: all Carnot engines have the same efficiency since X can be, as a special case, a Carnot engine. he Carnot engine is universal, i.e., it depends only on the temperatures involved and not on the working substance. Definition: absolute temperature θ of a heat reservoir (bath) is defined such that the ratio of the absolute temperatures of two reservoirs is given by θ 2 θ 1 Q 2 Q 1 = 1 η = Q 2 Q 1, where η is the efficiency of a Carnot engine operating between the two reservoirs. Since Q 2 isstrictly greater thanzero, Q 2 > 0, accordingtothesecondlawofthermodynamics, the absolute temperature is bounded from below: θ > 0. his means that the absolute zero θ = 0 is a limiting value that can never be reached since this would violate the second law of thermodynamics. If an ideal gas is considered as a working substance in a Carnot engine, the absolute temperature coincides with the temperature of the ideal gas defined as = P Nk B = θ. 4.4 emperature as integrating factor he absolute temperature can be considered as an integrating factor that converts the inexact differential into an exact differential /. Let us consider the following theorem stated by Clausius: In an arbitrary cyclic process P, the following inequality holds: P 0; where the equality holds for P reversible.

7 4.4. EMPERAURE AS INEGRAING FACOR 29 Proof. Let us divide the cycle P into n segments so that on each segment itstemperature i (i = 1,...,n)isconstant. Weconsidernowareservoir at temperature 0 > i ( i) (Fig. 4.7) and introduce Carnot engines between the reservoir at 0 and i. Figure 4.7: Cycle P connected to the reservoir at 0 via Carnot engines. For each Carnot engine, W i +Q (0) i +Q C i = 0 Q (0) i 0 + QC i i = 0 (first law of thermodynamics) and (definition of absolute temperature). For cycle P, with W + n Q i = 0 (first law of thermodynamics), Q i = Q C i. hen, the total heat absorbed from the reservoir at 0 is n n Q (0) = Q (0) Q C n i i = 0 = 0 i and the work performed by the system is ( ) n W = W + W i = n Q i + n ( Q (0) i Q i ) = n Q i i Q (0) i = Q (0).

8 30 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS If Q (0) > 0, the combined machine would convert completely the heat from reservoir at 0 into mechanical work. his would violate Kelvin s principle and therefore Q (0) 0, i.e., For n, he theorem of Clausius is proved. 0 n P Q i i For a reversible process, we can revert the energy, which gives us 0. his implies that for a reversible process. P P = 0 Corollary: for a reversible open path P, the integral dq depends only on the end points and not on the particular path. 4.5 Entropy P Since dq = 0 for a reversible process, is an exact differential: Eq. (4.1) implies that there exists a state function S. ds =. (4.1) Definition: he state function S (potential), whose differential is given as is called entropy. ds =

9 4.5. ENROPY 31 he entropy is defined up to an additive constant. he difference between entropies of any two states A and B is B S(B) S(A), where the integral extends along any reversible path connecting A and B, and the result of the integration is independent of the path. What happens when the integration is along an irreversible path? Since I R is a cycle (see Fig. 4.8), it follows from Clausius theorem that 0 I herefore, in general I R R = S(B) S(A). A B A S(B) S(A), and the equality holds for a reversible process. Figure 4.8: I R (irreversiblereversible) cycle. In particular, for an isolated system, which does not exchange heat with a reservoir, = 0 and therefore S 0. his means that the entropy of an isolated system never decreases and remains constant during a reversible transformation. Note: i) hejointsystemofasystem anditsenvironment iscalled universe. Definedinthis way, the universe is an isolated system and, therefore, its entropy never decreases. However, the entropy of a non-isolated system may decrease at the expense of the system s environment. ii) Since the entropy is a state function, S(B) S(A) is independent of the path, regardless whether it is reversible or irreversible. For an irreversible path, the entropy of the environment changes, whereas for a reversible one it does not. iii) Remember that the entropy difference S(B) S(A) = only when the path is reversible; otherwise the difference is larger that the integral. B A

10 32 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS 4.6 he energy equation We note that for a reversible process in a closed system the first law of thermodynamics can be written as ds = du δw = du +Pd. Let us consider and as independent variables (f.i., in a gas): S = S(,), U = U(,) caloric equation of state, P = P(,) thermic equation of state. Since usually the internal energy U is not directly measurable, it is more practical to consider ds = /. We will make use of one of the heat equations derived in section 3.5: ( ) [( ) U U = d + [( ) U = C d + +P Dividing both sides of eq. (4.2) by we get On the other hand, = ds = C d + 1 ds = ] +P d ] d (4.2) [( ) ] U +P d. (4.3) ( ) ( ) S S d + d. (4.4) Comparing eq. (4.3) and (4.4), we obtain the following expressions: ( ) S ( ) S = C, = 1 [( ) ] U +P. We use the commutativity of differentiation operations, to write 1 [ ( ) ] U ( C = 1 2 ) = [( ) U S = [ 1 S, (( ) U ] +P + 1 [[ )] +P, ( ) ] U + ( ) ] P ;

11 4.7. USEFUL COEFFICIENS 33 we cancel identical terms: and get ( ) U = ( ) P ( ) U = U P energy equation. (4.5) In the energy equation, the derivative of the internal energy is written in terms of measurable quantities. Examples: for an ideal gas ( ) U = nr P = 0, which means that the internal energy of the ideal gas is not dependent on its volume: U = U() = C +const. Gay-Lussac experiment: U = 3 2 Nk B gas of monoatomic molecules, U = 5 2 Nk B gas of diatomic molecules. 4.7 Useful coefficients As ( we saw in the last section, the energy equation relates the experimentally inaccessible U ) to the experimentally accessible ( ) P. We will show now that ( ) P can be related to other thermodynamic coefficients as well. For that purpose, we will make use of the chain rule ( ) ( ) ( ) x y z = 1, (4.6) y z z x x y which can be derived by considering the differential of function f(x,y,z): df = f f f dx+ dy + x y z dz, ( ) x y ( ) y z z x = f/ y f/ x, = f/ z f/ y,

12 34 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS Using (4.6), we get ( ) P where ( ) z = f/ x x y f/ z. 1 = = (/ ) P = α, (4.7) ( /) P (/ P) (/ P) k α = 1 ( ) P k = 1 ( ) P k S = 1 ( ) P S (coefficient of thermal expansion), (isothermal compressibility), (adiabatic compressibility). We substitute in [( ) ] U = ds = C d + +P d expressions (4.7) and (4.5) and get the following useful relations. 1) he absorbed heat is expressed in terms of directly measurable coefficients as where and are independent variables. 2) If and P are used as independent variables, then = ds = C d + α k d, (4.8) ds = C P d αdp (4.9) 3) If and P are used as independent variables, d can be rewritten in terms of d and dp as ( ) ( ) d = d + dp = 1 P P α + k α dp. he corresponding ds equation is then ds = C P α d + ( CP k α α ) dp. (4.10) here is also an important connection between C P and C (which follows from eq. (4.8) and (4.9)): [( ) ] 2 C P C = α2 P = ( k ) > 0. (4.11) P

13 4.8. ENROPY AND DISORDER 35 By introducing the isochore pressure coefficient it is possible to rewrite C P C as β = 1 P ( ) P β 2 k P 2., Indeed, β = 1 P α k α = βpk C P C = α2 k = β2 P 2 k 2 k = β 2 k P Entropy and disorder We want now to establish a connection between thermodynamics and statistical mechanics, which we will be learning in the second half of the course. 1. We give a definition of the multiplicity of a macrostate as ( ) number of microstates multiplicity of = that correspond to the a macrostate macrostate. 2. he entropy can be defined in terms of the multiplicity as S k B ln(multiplicity). We shall see in statistical mechanics that this definition is equivalent to the phenomenological concept we have learnt previously in this section. 3. With the above definition of entropy at hand, we can formulate the second law of thermodynamics in this way: If an isolated system of many particles is allowed to change, then, with large probability, it will evolve to the macrostate of largest entropy and will remain in that macrostate. 4. (energy input by heating) S any system, where is the temperature of the reservoir. 5. If two macroscopic systems are in thermal equilibrium and in thermal contact, the entropy of the composite system equals the sum of the two individual entropies.

14 36 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS 6. Since entropy is proportional to the multiplicity of a macrostate, it can be viewed as a measure of disorder : Order Disorder strong correlation absence of correlation small multiplicity large multiplicity As an illustration, let us consider a class in the school, where all kids sit always in the same place. his situation can be defined as ordered since there is a strong correlation between the kids positions. For example, if Peter sits in the first row, you know straight away who is sitting behind and besides him: there is only one possibility (multiplicity is very low). Looking now at a class in the kindergarten, we will find that the children there are all mixed, sitting and moving all the time. Now, if Peter sits in the first row, you have no idea where the other kids are sitting. here are no correlations between the kids positions and there are many options to have the children in the class. Such a system of kids is in a disordered state, with very high multiplicity. he entropy of the kindergarten class is high! 4.9 hird law of thermodynamics (Nernst law) While with the second law of thermodynamics we can only define the entropy (up to a constant), the third law of thermodynamics tells us how S() behaves at 0. Namely, it states that the entropy of a thermodynamic system at = 0 is a universal constant, that we take to be 0: lim S() = 0. 0 S( 0) is independent from the values that the rest of variables take. Consequences: 1) he heat capacities of all substances disappear at = 0: ( ) S lim C = lim = 0, 0 0 ( ) S lim C P = lim = P Note that an ideal gas provides a contradiction to the previous statement since in that case C = const and C P = const: C = 3 2 Nk B, C P = C +Nk B = 5 2 Nk B, C P C = Nk B 0. (4.12)

15 4.9. HIRD LAW OF HERMODYNAMICS (NERNS LAW) 37 But, for 0 the ideal gas is not anymore a realistic system because a real gas undergoes at low temperatures a phase transition - condensation. 2) lim β = lim ( ) = 0 3) he absolute = 0 (zero point) is unattainable. In order to show the validity of the last statement, let us analyze what happens when we aretryingtoreachlowtemperaturesbysubsequentlyperformingadiabaticandisothermal transformations. If a gas is used as a working substance (Linde method), such a sequence of transformations looks in the P diagram as shown in Fig Figure 4.9: Linde method: sequence of adiabatic and isothermal transformations in a gas. Let us consider now the processes involved. A B: isothermic compression. Work is performed on the gas and an amount of heat Q 1 < 0 is given to the reservoir in a reversible process. As a result, S = Q 1 diminishes. B C: adiabatic expansion. he gas performs work. Since = 0, the entropy remains constant and the temperature diminishes.

16 38 CHAPER 4. ENROPY AND SECOND LAW OF HERMODYNAMICS In principle, by repeating this process many times, one could reach the = 0 point. But, according to the third law of thermodynamics, all entropy curves must end at 0 for 0, which means that only through an infinite number of steps is it possible to reach = 0.

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