Simple thermodynamic systems

Transcription

1 Simle thermodynamic systems Asaf Pe er 1 October 30, Different statements of the second law We have seen that Clausius defined the second law as the entroy of an isolated system cannot decrease during a natural rocess. From this one can immediately see that Clausius s original form, namely that heat cannot by itself ass from a colder to a hotter body immediately follows. For if there was a machine that transforms an amount dq of heat from system A (with temerature 1 to system B (with 2 > 1, then the entroy change would be ( 1 S = dq 1 < 0, (1 2 1 which violates the rincile of entroy increase. Lord Kelvin had another definition of the second law: a rocess whose only effect is the comlete conversion of heat into work, cannot occur. Let us show that Kelvin s and Clausius s definitions are in fact equivalent. Suose we have some machine M that extracts heat Q from a heat bath, and converts itcomletelyintowork, W = Q. heworkisdoneonabody, B (e.g., byliftingit. Sincethe only net effect of the rocess is conversion of heat into work, the machine, M must oerate in such a way so that at the end of the rocess it returns to its original state (namely, its oeration must be cyclic. Since it returns to its original state, its entroy doesn t change in the rocess. Similarly, the machine is doing a work on the body B thereby changing some of its external arameters (e.g., lifting it u in the air, but does not change its internal roerties such as its entroy. hus, the total change in entroy is just the change in entroy of the heat bath, S HB = Q < 0 (2 which violates Clausius s definition of the second law. hus, Clausius s and Kelvin s definitions are, in fact, equivalent. In reality, only a fraction of the heat extracted from a heat bath can be converted into work. A machine (or device that converts heat into work is called heat engine. 1 Physics De., University College Cork

2 2 2. Heat engines We have seen that there is no such thing as a erfect heat engine, which converts 100% of the heat taken from a heat bath into work: such a machine violates the second law of thermodynamics. However, we can think of (and build machines which convert art of the heat taken from a heat bath into work: these could work rovided the total entroy in the rocess increases. hus, such a machine (demonstrated in Figure 1 would in general oerate in the following stes: 1. Extract some heat Q 1 from a heat bath at temerature Use this heat to do some work, W. 3. Dum the remaining heat Q 2 into another heat bath at lower temerature, 2 (< 1. hus, Q 2 = Q 1 W. he net entroy change during the cycle is which must of course be S 0, or S = Q Q 2 2, (3 Q 2 Q (4 Since W = Q 1 Q 2, then the larger Q 2 is, the less work W can be done by the cycle. hus, generally, we would like to lower Q 2 as much as ossible. We define the efficiency of a heat engine by η W Q 1, (5 which is the ratio of the work done by the system to the amount of heat taken from the heat bath. Of course, ideally we would want η = 1 - all heat taken is converted to work. As we discussed above, the second law tells us that this is imossible. Using equation 4 we find: Q 2 Q Q 1 W Q η 2 1 η = (6 Obviously, the maximum theoretical efficiency occurs when the rocess is reversible, and the inequality becomes equality. An ideal rocess with maximum efficiency is known as Carnot cycle. hus, a Carnot cycle has 4 different stages:

3 3 Fig. 1. A sketch of a heat engine. Heat Q 1 is taken from heat bath at temerature 1, and used to do work W on a body. he remaining heat, Q 2 = Q 1 W is deosited in a second heat bath at temerature 2 < Extract heat Q 1 from the hotter heat bath at temerature 1 : this is done at constant temerature, and thus this rocess is isothermal. 2. he working substance of the device is cooled reversibly and adiabatically from 1 to 2 : this rocess is done at constant entroy, and is thus an isentroic exansion. 3. Dum an amount of heat Q 2 into the cooler heat bath at 2 (again, by an isothermal rocess. 4. he working substance is heated adiabatically and reversibly from 1 to 2 by comression; thus, it ends u at the same state as it was at the beginning of the cycle. Carnot cycle can be carried using various working substances, such as gases, liquids or steam. It is demonstrated in Figure 2. It is ossible to oerate a heat engine backward, namely to use external work to extract heat from a cooler heat bath and dam it into a hotter heat bath. In this case, the heat engine is called refrigerator. In everyday used refrigerators, the work is electrical work. he efficiency of the rocess can be found in a similar way to that of heat engine, and is η = heat extracted Work required = Q 2 W (7 (Q 2 is the heat extracted from the cooler heat bath. Clearly, the efficiency is maximal when the temerature difference between heat baths is small.

4 S 1 C B P [arb. scale] A = 1 B Entroy, S [arb. scale] S 2 D A 2 D = 2 C [arb. scale] [arb. scale] Fig. 2. A Carnot cycle. Left- in a diagram. Right - in temerature - entroy sace. he shaded region in equal to the work done Otto cycle Another imortant examle of heat engine is the Otto cycle, which is in everyday use in etrol engines. his cycle contain the following stages: 1. A mixture of air and etrol is comressed adiabatically and reversibly from volume 2 to volume he comression leads to an exlosion of the mixture: the exlosion sharly increases the ressure and the temerature, while the volume ( 1 is held fixed (amount of heat Q 1 is added. 3. he hot gas ushes the iston back, doing a mechanical work: this exansion from 1 back to 2 can be considered reversible and adiabatic. 4. he outlet valve is oened, and the hot exhaust gases rush into the atmoshere; this can be considered as cooling at constant volume 2 (extracting heat Q 2. his cycle is seen in Figure 3. Let us calculate the maximum efficiency of the cycle. he amount of heat absorbed by the gas in ste (2: Q 1 = C ( C B (8

5 P [arb. scale] C 2 B D A [arb. scale] Fig. 3. Otto cycle in a diagram. he shaded region is equal to the work done. he amount of heat given from the gas to the environment in ste (4: he work done: and the efficiency: Q 2 = C ( D A (9 W = Q 1 Q 2 = C ( C B D + A (10 η = W Q 1 = C B D + A C B = 1 D A C B. (11 If the gas is considered as ideal, this can be simlified by noting that during the adiabatic comression and exansion (stes (1 and (3, one have c c γ = D γ D ; Cc γ 1 = D γ 1 D ; B γ B = A γ A ; B γ 1 B = A γ 1 A ; (12 We also know that B = C = 1, and A = D = 2, and thus which lead to C γ 1 1 = D γ 1 2 B γ 1 1 = A γ 1 C = D. (14 B A Using the result of Equation 14, we can write the efficiency in Equation 11 as η = 1 D D (1 D ( γ 1 A A = 1 D 1 = 1 = 1. (15 C B C (1 B C 2 C 2, (13

6 6 3. More on heat caacities Using the second law we can obtain the thermodynamic relation between the heat caacities, exressed in terms of measurable quantities. In real life, C is easy to measure, but hard to calculate; while C is harder to measure, but easier to calculate. If we take and to be indeendent variables, we can write the first law of thermodynamics in the form dq = de +d = ( E d +( E = C d + [ + ( E d +d (16 ] d where C = ( Q/ = ( E/ is the heat caacity er unit volume; d is the work done against external force, and the last term, ( E/ d is the change in internal energy caused by changing the mean sacing between the molecules. For ideal gas, the internal energy deends only on the temerature, and so the last term vanishes; but life is not always ideal. Similarly, C = ( [ Q = C + + ( ]( E. (17 In order to roceed, let us have a closer look into the square brackets. Starting from the fundamental thermodynamic relation, de = ds d ds = 1 de + d, (18 and recalling that and are indeendent variables, one finds ( ( S S ds = d + d = 1 [( ( E E d + Equate coefficients of d and d, one finds: ( S ( S ( = 1 E [( ; = 1 E +]. ] d + d (19 (20 Using again the trick of cross-differentiation, 2 S/( = 2 S/(, one finds [ ( ] 1 E = { [( ]} 1 E + (21

7 7 Since the mixed second order derivatives of E are the same, we get ( ( E = + Use this result in Equation 17, one finds [ ( ]( E C = C + + P = C + ( ( (22 (23 Equation 23 allows us to write the difference C C in terms of quantities which are relatively easily measured (unlike terms which contain derivatives of the internal energy. If we aly this equation to a erfect gas ( = nr then we find C C = nr, (24 which is of course in agreement with the thermodynamic relation derived earlier this course. In the general case, we can roceed by noting that at constant ressure we can use the relation d = ( / d +( / d = 0, from which ( ( ( =. (25 Introducing the exansion coefficient, β 0 1 ( (26 and the isothermal comressibility, K 1 ( which are easily measured quantities, we get from Equation 23 (27 C C = β2 0 K. (28 All quantities on the right hand side of Equation 28 are easily measured, which enable a quick relation of C and C. Since K is always ositive, C C. Both heat caacities are equal if β 0 = 0, as is, e.g., in water at 4 Celsius.

8 8 4. he Joule effect (or Joule exansion We have considered several times the free (adiabatic exansion of gas into a vacuum, that occurs, e.g., when the membrane that searates the gas from the vacuum breaks; the entire (gas + vacuum system is assumed isolated from the rest of the world. he first to study this rocess in details was James Joule, and the rocess is known after him, Joule exansion (or Joule effect. Here, we would want to generalize the treatment to real (nonideal gases. While for ideal gases the internal energy is a function of temerature only, E = E(, for real gases E = E(,. When the gas exands, its energy is conserved; since its volume changes, we therefore exect to see a (small change in its temerature. Let us quantify this. Start with de = 0, to write ( E d + ( E d = 0, (29 which imlies ( ( E E d = d (30 hefirstthingtonoteisthatboth( E/ and( E/ areositive; thefirstisobvious, while the second originates from an increase in the otential energy of articles with increase in their average searation. Since in an exansion d > 0, Equation 30 imlies that d < 0, namely the temerature decreases in free exansion. A quantitative measure of this exansion is given by the Joule coefficient, ( ( E α J = ( E. (31 E FromthediscussionfollowingEquation16,weknowthatthedenominatoris( E/ = C. For the numerator, we reeat the arguments and calculations as in Equations to write de = ds d ds = 1 de + d, (32 and recalling that and are indeendent variables, ( ( S S ds = d + d = 1 [( E Equate coefficients of d and d, one finds: ( S ( S ( = 1 E [( ; = 1 E d + +]. ( ] E d + d (33 (34

9 9 Using again the trick of cross-differentiation, 2 S/( = 2 S/(, [ ( ] 1 E = { [( ]} 1 E + Since the mixed second order derivatives of E are the same, we get ( ( E = + Overall, we can write the Joule coefficient in Equation 31 as α J = 1 [ ( ] P P C Examle. he equation of state of real gases can be written as ( R 1+ B (35 (36 (37 (38 (er mole. hus, α J = 1 C R 2 2 db d. (39 For Argon gas, at S..P. (standard temerature and ressure; = 273 K, = 1 Atm, C = (3/2R and db/d = 0.25cm 3 K 1 mol 1. hus, α J Kcm 2 mole. When the gas exands, its temerature changes. he temerature change is found using Equation 31: 2 ( = 2 = 1 = d. (40 1 E For constant α J, doubling the volume ( 2 1 = cm 3 mole 1, = 0.6 deg. 5. he Joule-homson effect By the Joule-homson effect we refer to a change in the temerature of a gas which occurs during adiabatic exansion of gas through a throttle from high ressure 1 to lower ressure 2. We will now calculate this change in temerature. Similar to the Joule effect, this is an irreversible rocess. Assume that before the exansion, the gas occuies volume 1, after the exansion - 2. We can think of the system as having two istons, one at each side of the throttle, that kees the ressures 1 and 2 constants (see Figure 4.

10 10 Fig. 4. he Joule-homson exansion through a throttle. he two istons ensure constant ressures on the two sides of the throttle. he work done on the gas by the istons is W = d d = 1 (0 1 2 ( 2 0 = (41 he device is thermally isolated. hus, from the first law E = Q+W = W, or E 2 E 1 = E = E Recall the definition of the Enthaly, H E +; Equation 42 is simly H 1 = H 2. hus, the Joule-homson rocess conserves the enthaly - it is an isenthalic rocess. We can now reeat similar calculations to those used in calculating the temerature in the Joule effect (Equations 30-36, but with H relacing E and relacing. We use dh(, = 0; he temerature and the ressure change during the rocess, and so ( ( H H d + d = 0. (43 In analogue to Equation 31 we can define the Joule-homson coefficient, ( ( H α J = H ( H (42. (44 Similar to the discussion on the Joule effect, our goal now is to exress α J using measurable quantities. he denominator in Equation 44 is ( ( ( ( H E Q = + = = C, (45

11 11 from which α J = 1 [( E + C ( ( ]. (46 We can write the numerator using the fundamental relation: Let us substitute in: ds = dh = de = ds d ds = 1 de + d = 1 dh d. (47 ( ( S H d+ ( S d d+ ( H Equating coefficients of d, d, one gets (analogue to Equation 34, ( S ( S ( = 1 H [( ; = 1 H d. (48 cross-differentiate ( 2 S/( = 2 S/( and simlify, one gets ( ( H =, (50 ]. (49 from which we obtain α J = 1 C [ ( ]. (51 For an ideal gas, α J = 0; thus, the temerature of the gas doesn t change. However, even for non-ideal gas, α J = 0 rovided that ( = (52 Equation 52 defines a curve in the lane (see Figure 5: this curve is known as inversion curve. Also lotted in the figure are isenthals, curves of constant enthaly. Inside the inversion curve, increases with along isenthal, and thus α J = ( / H > 0: the gas cools in exansion. Outside the inversion curve, α J < 0, and so the gas warms u in exansion. Above a certain temerature, i, known as the inversion temerature, the gas is always outside the inversion curve, and thus it is always warmed in Joule-homson exansion. For most gases, i is above room temerature.

12 12 Fig. 5. Inversion curve in the - lane is shown in bold. he solid lines are isenthals (=curves of constant enthaly, H(, = Const. Inside the inversion curve, increases with along isenthal, while outside it decreases. Calculation of temerature change during Joule-homson exansion is done using 2 ( = 2 1 = d (53 his is difficult to evaluate analytically, since non-ideal gases often do not have a simle relation between and. But this equation can easily be solved numerically. 1 H 6. Adiabatic cooling We know from the third law, that when the temerature of a system aroaches zero, so does the entroy. hus, a cooling rocess is a rocess in which the entroy decreases. In a reversible rocess, the entroy doesn t change, S = 0. hus, adiabatic cooling

13 13 rocess deends on the the fact that the entroy of a system deends not only on the temerature, but on other arameters as well (e.g., the volume, magnetic field, etc.. We shell label these arameters by α. During a reversible adiabatic rocess (in a thermally isolated system, S = S(,α = Const. (54 hus, the rocess must change the arameter α to comensate for the decrease in temerature, so that over all the entroy will remain constant. Examle - 1. Adiabatic exansion of the gas is a good examle of adiabatic cooling. For α we take the ressure. his is a 2-ste rocess: 1. Isothermal comression of gas from 1 to 2 : during this rocess, the gas is in contact with a heat bath of temerature 1 (line A B in Figure Heat bath is removed, and the gas is now thermally isolated. It is now allows to exand reversibly and adiabatically back to the lower ressure 1 (line B C in Figure 6. At the end of the rocess, the gas have the same ressure as initially, but have a lower temerature. his is the basis of Clausius rocess for liquefying gases. (see Figure 6. Fig. 6. Entroy-temerature diagram for cooling of a gas by adiabatic exansion

14 14 Examle - 2. Adiabatic demagnetization is another owerful method of achieving low temeratures. It is based on the examles discussed earlier about a aramagnetic solid in a magnetic field. he entroy (=amount of disorder deends on the temerature and on the magnetic field. hus, if we reversibly decrease the magnetic field, the magnetic moment becomes more disordered - hence the temerature must decrease if S is to be constant. We can estimate the magnitude of this effect on a aramagnetic solid of N sin 1/2 molecules (we already discussed this system in class: S = 1 (E F = N (E 1 F 1 = N ( lnz 1 +k B lnz 1 (55 β Recall that the artition function for a single article was found to be Z 1 = e µb/k B +e µb/k B = 2cosh µb k B, (56 from which we find ( S = Nk B [ln 2cosh µb µb k B k B ] µb tanh, (57 k B which is a function of B/; hence during ste 2 (adiabatic reversible de-magnetization, this ratio remains constant, and the final temerature is B f f = B i i f = B f B i i. (58 For i 1 K, a change in the magnetic field by a factor of 100, can cool aramagnets down to 10 mk. his is demonstrated in Figure 7. And final question: Can this rocess give us absolute zero?

15 15 Fig. 7. Entroy-temerature diagram for a aramagnetic salt, for different alied magnetic fields