I have not proofread these notes; so please watch out for typos, anything misleading or just plain wrong.

Transcription

1 hermodynamics I have not roofread these notes; so lease watch out for tyos, anything misleading or just lain wrong. Please read ages in Chater 8 of Kittel and Kroemer and ay attention to the first 6 oints of the summary on age 257. See also ages and one of the Maxwell relations on age 71. hermodynamics deals with equilibrium states and transformations between them. he first law of thermodynamics 1 is a generalization of the mechanical (and electromagnetic) conservation of energy to include heat. Energy is conserved if we include heat and it can be exressed mathematically as du dq + dw q + w. (1) q is the heat transferred to the system in the infinitesimal rocess and w is the work done on the system by an external agent. One is not allowed to write them as differentials of functions of state variables, dq and dw, since it makes no sense to say that a body has a certain amount of work or heat in contrast to the internal energy U. Fermi states Q can be interreted hysically as the amount of energy that is received by the system in forms other than work.. What is imlied is that work is identified using knowledge of mechanics, fluid dynamics, electromagnetism, etc. All tyes of work are freely convertible into each other, because the entroy transfer is zero. See K&K. When can one write dw d? he rocess must be quasi-static (arbitrarily close to equilibrium states so that ressure can be defined; too raid an exansion and inhomogeneities will cause a uniform ressure to be undefined.) and reversible: a reversible rocess is defined as one which may be exactly reversed by an infinitesimal change in the external conditions. For examle if there is friction it will not be reversible. What is the roblem? Even if we return to to the original state aarently along the same ath and the area enclosed in the 1 Zeroth law of thermodynamics If of three bodies A, B, and C, A and B are searately in equilibrium with C then, A and B are in equilibrium with each other. 1

2 lane vanishes the work done is ositive. (Monatomic) Ideal Gas Review: From ( ) du d + dq the first Law and the equation of state N k B. We recall 2 dq d C ( ) du d. Define ν N N A to be the number of moles where N A is the Avogadro number. Recall that R N A k B is the gas J constant and has the value K mole. It is roughly 2 calories er mole er Kelvin. We have C d + d dq and d + dp ν R d. (2) We obtain for one mole of an ideal gas We have used (C + R) d d dq C C + R. (3) C ( ) dq d ( ) U + ( ). (4) Use the fact that U only deends on the temerature to identify the first term with C and use the equation of state to obtain R/ for the second term. Exlain simly why C > C for an ideal gas. Observe that the boxed equation is true for molar heat caacities of an ideal gas; re-write it for N atoms. A transformation of a thermodynamical system is said to be adiabatic if it is reversible and if the system is thermally insulated so that no heat can be exchanged between it and its environment during the transformation. he change in entroy of the system is zero. Adiabatic Equation of State: For one mole of a monatomic ideal gas we have du C d, and d R d. Since dq 0 we obtain C d + R since no heat is exchanged. Integrating we have d 0 (5) R/C K γ K (6) where γ C C v. Obtain this result directly from the statistical mechanical exression for the entroy. Work can be comletely converted into heat, but the inverse is not true: It is very easy to do mechanical work w > 0 uon a device M, and then to extract from it an 2 In roblems ay attention to whether the heat caacity is that of 1 mole or a given number of articles or a given mass of the substance, etc. 2

3 equivalent amount of heat q > 0, which goes to increase the internal energy of a system B. Heat cannot be comletely converted into work. A device that generates work from heat must necessarily remove the entroy from the heat and this must be rejected as waste heat. his is the content of the second law. For the inverse rocess the work should not be rovided at the exense of the heat engine itself; otherwise one could not continue the rocess of heat-to-work conversion indefinitely. hus one wants the engine to be in the same macrostate at the end of the rocess as it was at the beginning (i.e., to have gone through a cycle) so that it is ready to start again converting more heat into work in the next cycle. Kelvin statement: A transformation whose only final result is to transform into work heat extracted from a source which is at the same temerature throughout is imossible. Clausius statement: A transformation whose only final result is to transfer heat from a body at a given temerature to a body at a higher temerature is imossible. It is not difficult to show that these are equivalent. hermodynamically, one defines entroy as dq/ by first defining a Carnot cycle and then considering an arbitrary cyclic rocess as Clausius did. Reservoir A large body at a fixed temerature that can exchange heat without doing any work is called a source of heat; this could be a large body of water that is incomressible. If a heat reservoir at temerature absorbs heat Q, its resulting entroy change S is given by S Q. 3 his is an imortant statement. Heat engines o what extent can one extract internal energy from a heat reservoir in the form of heat and convert it into macroscoic work (with the engine returning to the same macrostate at 3 A statistical mechanical justification: Assume that a reservoir absorbs energy in the form of heat and its (internal) energy changes U Q. he change in the number of states available to the system is obtained by (one can use k B ln[g(u,, N)] to be closer in sirit to statistical mechanics) ( ) ( ) S 2 S S S(U + Q., N) S(U,, N) Q + U U 2 U,,N U,,N ( Q) 2 2! + (7) he first term yields Q ; by the definition of the heat reservoir the changes in energy Q are small comared to the energy of the reservoir and its temerature does not change. he second term is (1/ ) and this is U negligible. 3

4 the end of the rocess)? Read K&K section on Carnot cycles. Follow K&K and define all quantities of heat and work as ositive and aend signs as aroriate. Consider two isotherms at temeratures θ h and θ l and two adiabatic lines that intersect them. 4 Let the system absorb an amount of heat Q h and reject an amount of heat Q l at the two reservoirs. Let W be the total work done and this can be comuted for an ideal gas. We have W Q h Q l. he efficiency is defined to be η W Q h 1 Q l Q h. Concetual aside: Carnot s work allowed one to conclude that q f(θ 1,θ 2 ) vanishes for a (reversible) cycle oerating between temeratures θ 1 and θ 2 measured on any scale. One can argue that this allows one to deduce the existence of an absolute temerature! herefore, q vanishes for a reversible cycle. From this Clausius showed that it is true for an arbitrary cycle thus establishing the existence of a state function that he christened entroy. Elementary argument he following is done to cature the flavor of the clever arguments made by nineteenth century hysicists and is not for exams. We show that all reversible Carnot cycles have the same efficiency. Consider a (reversible) Carnot engine H that extracts an amount of heat Q 1 from reservoir R 1 and rejects Q 2 into reservoir R 2 erforming an amount of work W Q 1 Q 2 and another (not necessarily reversible) engine H that extracts an amount of heat Q 1 from reservoir R 1 and rejects Q 2 into reservoir R 2 erforming an amount of work W Q 1 Q 2. Can η > η? Run H as an engine and run the Carnot cycle in reverse using the work roduced by the engine H. Now adjust the stroke length of H so that it delivers an amount of work exactly equal to W. (his is not obvious is ossible and Fermi s book avoids this with a more comlicated argument. ) hus we have Suose η > η. hen we have W Q 1 Q 2 Q 1 Q 2. (8) 1 Q 2 Q 1 > 1 Q 2 Q 1 Q 1 Q 2 Q 1 < Q 1 Q 2 Q 1 Q 1 > Q 1. (9) Since Q 2 Q 2 + [Q 1 Q 1 ] we also have Q 2 > Q 2. hus we have extracted a ositive amount of heat Q 2 Q 2 from the colder reservoir R 2 and delivered an equal amount Q 1 Q 1 to the hotter reservoir R 1 contradicting the Clausius statement. hus we must have η η. 4 Note that we are assuming the existence of these for an arbitrary fluid and this is a ostulate in the Carathéodory formulation. 4

5 Note that H was a Carnot cycle and reversible. If H were also reversible we can similarly deduce that η η and hence η η. All reversible Carnot cycles have the same efficiency. If H is irreversible η η. Draw a schematic heat engine(m) extracting heat q from a reservoir at temerature and converting it into work w > 0. without affecting the entroy of the environment. thermodynamics. he change in entroy is S S R + S M the second law and therefore, w 0. he work changes a macroscoic degree of freedom We have q w from the first law of ( q) + 0 < 0. iolates Roughly seaking energy that is distributed (randomly) among many degrees of freedom cannot be transformed into energy associated with work done by a single degree of freedom (dislacement of a iston for examle) without decreasing the entroy. he engine requires an auxiliary system the simlest being a second heat reservoir at some temerature l lower than h. One then obtains a realizable, heat engine (that is not ideal) which not only absorbs heat Q h from the high-temerature reservoir at temerature h, but also rejects heat Q l to some second reservoir at some lower temerature l and doing work W. he first law imoses the condition Q h Q l + W. he change in entroy S Q h h + Q l l 0. (10) Elementary substitution yields Q h h + Q h W 0 W ( 1 Q h l l l 1 h ) W Q h 1 l h. (11) his gives a bound on the efficiency η W Q h given in Equation 8.11, η η C with η C 1 l h. he maximum efficiency is obtained in a reversible engine for which the change in entroy vanishes and is known as the Carnot efficiency of an engine. All reversible engines acting between the same two temeratures have the same efficiency and no engine oerating between those two temeratures can have greater efficiency. Amazing. Draw the Carnot cylce in the S lane and deduce the efficiency of a Carnot cycle. Refrigerators: A refrigerator is a device which, oerating in a cycle, removes heat Q l from a reservoir at lower absolute temerature l and rejects Q h to a reservoir at higher 5

6 absolute temerature h. It can be reresented by a diagram. Exercise: Show that it is imossible to construct a erfect refrigerator that removes heat Q from a reservoir at l and rejects all of it into a reservoir at a higher temerature. he first law, alied to the refrigerator reads W + Q l Q h. he change in entroy is S Q l l + Q h h 0 Q l Q h l h. (12) he efficiency is defined to be γ R Q c W Note that this may be greater than unity! Q l Q h Q l l h l ; this is Equation he Helmholtz free energy is defined by the Legendre transformation F U S, to obtain F (, ) from U(S, ) where and S are conjugate. Similarly we define the Gibbs free energy by G U S + and the enthaly by H U +. We have the following useful differential relations: du ds d + µ dn (13) df S d d + µ dn (14) dg S d + d + µ dn (15) dh ds + d + µ dn (16) hat F (, ) the Helmholtz free energy is a minimum at constant and is a useful and imortant idea. It is the equivalent for the canonical ensemble of the statement that entroy tends to a maximum for the microcanonical ensemble ( closed system.) Consider a system in equilibrium with a heat reservoir R at a common temerature and allow only energy exchange. hen ds total ds R + ds (17) where S is the entroy of the system. For the reservoir we have ds R du R / and therefore ds total ds + du R ds du 1 (du ds) 1 df (18) and hence it follows that the increase in entroy for the total isolated system corresonds to the decrease in Helmholtz free energy of the system under consideration. What is free about the free energies? he Helmholtz free energy is the maximum work that can be extracted from a system, crudely seaking. his is the energy that is 6

7 free to be extracted to do useful work. When a system is in equilibrium at temerature the work that can be extracted is clearly not U as one might think from energy conservation. his is done carefully in Landau and Lifshitz 20. Given below is a less general and simlified version. Let a body be in equilibrium in equilibrium with a reservoir whose temerature. Only energy exchange is allowed. he body can do work on some object, assumed thermally isolated both from the reservoir and from the body. he reservoir in thermal equilibrium with the body and the object on which work is done, form a closed system. he temerature of the reservoir, is by definition, constant in the rocesses considered. If the reservoir were absent, the work done by the body on the thermally isolated object, for a given change in state of the body would be comletely defined, and equal to the change in the energy of the body. Since the body exchanges energy with the reservoir one can ask for the maximum work which the body can do for a given change in its state. hus the total change U in the energy of the body in some (not necessarily small) change in its state consists of two arts from the first law of thermodynamics: (a) the work W > 0 done by the body on the external source and (b) the heat gained from the reservoir. he suffix R indicates quantities ertaining to the reservoir, while those for the body have no suffix. 0. he entroy of the reservoir changes by S R < 0 since it gives u heat. he heat absorbed by the body is S R > 0. hus (the body does ositive work W on the external source and so its internal energy decreases and it receives ositive heat from the reservoir) U W S R. he law of increase of entroy states that S + S R 0; the entroy of the thermally isolated source of work does not vary. We therefore find S R We write this as W F W U W U + S. (19) since F U S. he maximum work that can be done by a body in contact with a heat reservoir is the decrease in the Helmholtz free energy of the system. If there is no contact with the external source so the body undergoes only sontaneous fluctuations so that W 0 we have shown that the Helmholtz free energy is a minimum. Maxwell s relations: Start from du ds d 5 and use the relationshi for equality of mixed artial derivatives. One knows that if the second artial derivatives of 5 You should remember this! 7

8 f({x i }) exist and are continuous the Hessian matrix ( ) U S. his is one of four Maxwell s rela- ( ) and U S U S we obtain tions. S and ( ) S 2 f x i x j is symmetric. Since we have ( ) U S (20) Similarly from the enthaly H U + and the Gibbs free energy, G U S + the Helmholtz free energy F U S three more relations can be derived. Here is one more examle for good measure: df S d d and therefore, ( ). (21) Make sure you can derive the other two. In the good old days you were given mnemonics to recall these four relations (only for exams). What is the oint? From (S, ) and (, ) there are 24 first artial derivatives. One can show that there are three indeendent first artial derivatives. hey can be chosen to be related to three measurable quantities: α 1 ( ) the coefficient of thermal exansion, κ 1 ( ) the isothermal comressibility, and C the heat caacity at constant ressure. Problems involving artial derivatives are eased by the use of the following artial derivative identities, stated without mathematical rigor. Suose z is a function of x and y then ( ) z 1 ). (22) x ( ) x y z y ( ) y z x Note the sign and try to rove this relation. ( x z y ( ) z x y 1. (23) 8

9 A standard derivation of the difference in heat caacities C C for any system. We recall some useful exressions. C C ( ) H ( ) U Please do not think that C is ( U/ ). Exlain why this is wrong without equations. (24) (25) Choose and to be the basic variables and consider the entroy S S(, ). Be clear that in the canonical ensemble all thermodynamic variables can be exressed as a function of and. We use S since dq ds and the heat caacities are related to dq. We start from ds Find ( S/ ) from above: d + + d. (26) ( ) ( ) S. (27) Multilying by, we obtain C C ( ) ( ) S. (28) Now we use Maxwell s relations and artial derivative identities to write the right-hand side in terms of measurable quantities. ( ) is related to α the coefficient of thermal exansion. Once you see a artial derivative of S look for Maxwell s hel. From Equation(21) we have ( S/ ) ( / ). We can leave it at this oint saying that this derivative can be obtained from the equation of state, a relation between,, and, f(,, ) 0. However, this is not the conventional form. One uses the artial derivative identity ( ) ( ) ( ) 1. (29) Now the second term is again related to the coefficient of thermal exansion and the third term to the isothermal comressibility κ defined earlier. Combining these together we obtain the standard form C C What is the right-hand side for a monatomic ideal gas? α2 κ. (30) One can obviously seek other such relations that may rove useful in different contexts and for variables other than and in roblems in magnetism, suerconductivity, etc. You 9

10 will learn these when you need them. Problem: Exress ( ) S in terms of measurable quantities. Remarks on Enthaly H U + thalein means ut heat into, in Greek. It is an extensive variable and is a useful measure of the energy in the system that includes the internal energy and the work done. It is extremely imortant in thermochemistry. Since dh ds + d if one heats the system at constant ressure the heat dq increases the enthaly of the system. Since dh du + d at constant ressure the change in enthaly includes the change in internal energy and the work done for the volume to change by d. Differential scanning calorimetry is an old technique and it is done isobarically. Freshman chemistry roblem: Somewhat tricky for hysicists! he latent heat of vaorization of 1 mole of water ( g) is aroximately 40.7 kj/mole. What is the change in internal energy? Answer: 37.6 kj/mole. Hint: he heat sulied (latent heat) goes into increasing the internal energy and in doing work. What is the change in entroy in the above roblem? Enthaly is also useful in food chemistry and energy as for examle in the concet of enthaly of combustion er gram of material. For examle, when one octane (c 8 H 18 burns the change in enthaly is 5471 kj/mol. If a man requires 12 MJ er day (about 2870 Calories) and given that glucose has a secific enthaly of 16 kj/g he would need 750 g of glucose a day! hird Law: (Nernst) he entroy change associated with any isothermal, reversible rocess aroaches zero as 0. (Fowler and Guggenheim) It is imossible by any rocedure to reduce any system to absolute zero in a finite number of oerations. We will discuss this law if time ermits. 10