1. Chapter 2 T h e Laws of Thermodynamics

2. 1 The first law In physics,the principle of conservation of energy is of fundamental importance. Therefore,besides the work which is performed by or on a system one has also to consider the heat exchanged with the surroundings.Thus we can assign an internal energy U to each macroscopic system.

3. For an isolated system which does not exchange work or heat with its surroundings,the internal energy U is identical to the total energy E of the system known from mechanics or electrodynamics. The change of the internal energy for an arbitrary (reversible or irreversible)change of state is given by the sum of the work and heat exchanged with the surroundings.We write First law: (2.1) （吸收热量）

4. Once again we explicitly remark that,e.g., the work has the form only for reversible processes;for irreversible processes it may be that.The same holds for the exchanged heat: =Cv dT is only valid for reversible processes,while Equation (2.1)is always true. There exist many formulations for the first law of thermodynamics,which all have the same meaning, for example,there is no perpetuum mobile of the first kind.

5. Example2.1: Internal energy and total differential As an example we calculate the internal energy of an ideal gas.In the section "kinetic theory of the ideal gas"we have already derived the following equation: where was the mean kinetic energy per particle.The total mean energy of the system,I.e.,with (2.2)

6. Let us consider a container with an ideal gas at constant volume.If the temperature is changed by dT we have Hence it holds that (2.3) For dilute gases the specific heat is constant,so that we are able to integrate Equation(2. 3),

7. If we additionally consider that the total heat capacity is proportional to the particle number, Cv=Nc v we find where c v is the constant specific heat per particle of the ideal gas.By comparison with Equation(2.2)one obtains (2.4)

8. Example2.2: Adiabatic equations for the ideal gas A process in which there is no heat exchange is called an adiabatic process.According to the first law, with and it holds that for a reversible adiabatic process. From equation 2.2 we know that for an ideal gas

9. Therefore we obtain a relationship between dT and dV for adiabatic changes of the volume of an ideal gas: If we insert the ideal gas law for p(V,T)we obtain (2.6)

10. (2.7)

11. Equations(2.6)and (2.7)are the adiabatic equations of the ideal gas.Note that they differ logically from the ideal gas law,since here we have considered a specific process (an adiabatic process ):exactly as for processes with constant temperature (isotherms), constant pressure (isobars),or constant volume (isochores)we can eliminate a variable of the ideal gas equation. As we will see,for adiabatic,reversible processes the total entropy of the system is constant (they are isoentropes).Because of. the adiabates (isoentropes) in a pV diagram are steeper than the isotherms. (2.7)

13. The first law holds independently of whether a change of state is reversible or irreversible: From the example of the isothermal expansion of an ideal gas ， Because: Against force-rev No against -irr

14. 2 Carnot's process and entropy This cycle,with an ideal gas as the working material,was presented by Carnot in 1824.The Carnot process is performed in four successive reversible steps,which we will illustrate in a pV diagram (Figure 2.1):

15. Figure 2.1 Carnot process in the PV diagram

16. Step I: Isothermal expansion from volume V1 to volume V2 at constant temperature T h For the isotherm it holds that From Example 2.1we know that the energy of an ideal gas,which is the working material in our case,cannot be changed at constant temperature.Consequently it holds that (2.13) (2.12 ）

17. From this we can calculate: This is the amount of heat exchanged with the heat bath in the first step. Since V2>V1, >0;i.e.,the amount of heat is added to the gas at the expense of the heat bath. (2.14)

18. Step 2.Adiabatic expansion of the isolated working material from V2 to V3 · Here the temperature changes from Th to Tc.The indices h and C denote hot and cold,i.e.,Th 〉 Tc : Since =0(for adiabatic processes)the work performed in the expansion is taken from the internal energy, For an ideal gas,Cv=3Nk/2. (2.16) (2.15)

19. Step3.We now compress the system isothermally from V3 to V4 at the (constant) smaller temperature Tc.Analogously to Step 1we have The work performed during the compression is,because =0 at T=const.,submitted to the heat bath in form of heat: (2.17) (2.18) (2.19)

20. This is the amount of heat absorbed by the heat bath in this step.Since V4 〈 V3,it follows that 〈 0;i.e.,the gas loses this amount of heat. Step4.Finally we restore the system to the initial state via an adiabatic compressionfrom V4 to V1 · The temperature increases again from Tc to Th: （ 2.20 ）

21. Since =O it follows Let us first check the total energy balance of the process.We have (2.22) (2.21 ）

22. In addition,we have the following equations for the amount of heat exchanged with the heat bath: On the other hand we have,according to or that (2.24) (2.23)

23. Then,however, that This equation is of great importance,for it is valid not only for our special Carnot process, but according to all experience,for any reversible cyclic process.The quantity is also known as reduced heat.If we decompose the Carnot process into infinitesimal parts, we may obviously write instead of Equation above (2.26) (2.25)

24. In other words:1/T is the integrating factor of the nonexact differential. To show that Equation above is contour-independent for an arbitrary cycle (and not only for Carnot ‘ s) 。 we divide the arbitrary cycle (Figure 2.2)into a sequence of infinitesimal Carnot-like parts(N → ∞ ),as illustrated in Figure 2.3. Figure 2.3 Decomposition of an arbitrary cycle into many small carnot processes.

25. the entropy S,which is defined via Of course,Equation(2.28)can also serve as a measuring instruction.To this end,one has to measure the amount of heat reversibly exchanged by the system at a given temperature T.However,by this method only entropy differences are determined,not absolute values of entropy. (2.28)

26. Obviously,a Carnot engine is an engine which transforms heat into work. We now want to calculate the efficiency of this engine.As efficiency we define the ratio between the heat transformed into work and the total heat absorbed, (2.30)

27. If we insert We have (2.32) (2.31)

28. 3 Entropy and the second law The state quantity entropy was introduced by R.Clausius in.Since (2.33) Especially for isolated systems,we have =0.Therefore,in an isolated system the entropy is constant dS=0.Every experience confirms that this extremum is a maximum. R.Clausius

29. Second law--For isolated systems in equilibrium it holds that dS=0 S=S max and for irreversible processes it holds that dS 〉 O (2.35) S- 杂乱程度

30. The entropy is obviously an extensive quantity,since the internal energy as well as the amount of heat are extensive quantities. Therefore,when heat is exchanged at temperature T,the entropy is a quantity analogous to the volume,when compression work is performed against a pressure P. Since

31. So, we once again denote the first law for reversible changes of state in explicit form, we observe that the entropy just fits in the set of extensive state quantities(S,V,N,q,...)which describe the change of the internal energy under the influence of intensive,locally definable field quantities(T,p, μ,φ,...).

32. If the function U(S,V,N,q,...) is given,we can determine Thus,U= U(S,V,N,q,...) is also called the fundamental relation.

33. Example 2.3 Entropy of the ideal gas For a reversible change of state the first law reads dU=T dS-p dV (2.38) for dN=0.With the equations of state for an ideal gas we can solve Equation(2.38)for dS.

34. Starting from a state T 0,V 0 with entropy So,we integrate this equation, and,if we substitute V ∝ T/p, （ 2.40)

35. 4 lnsertion:Microscopic interpretation of entropy and of the second law In equilibrium the entropy thus assumes its maximum value and does not change anymore.Ludwig Boltzmann was the first to show,in his famous H- theorem (pronounced Eta-Theorem)in1872,that this tendency can also be founded on a statistical description and classical mechanics. Boltzmann1844 一 1906

36. We are now able to go one step further and try to establish a connection between the entropy and the number of microstates compatible with a macrostate. the number of microstates Ω compatible with a given macrostate is a quantity very similar to the entropy of this macrostate. The lager Ω,the more probable is the corresponding macrostate, and the macrostate with the largest number Ω max of possible microscopic realizations corresponds the thermodynamic equilibrium.

37. For two statistically independent systems the total number of microstates is obviously the product of the number for the individual systems, The larger Ω,the more probable is the corresponding macrostate,and the macrostate with the largest number Ω max of possible microscopic realizations corresponds to thermodynamic equilibrium.

38. If we now assume that there is a one-to-one correspondence between entropy and Ω,for instance,there is only one mathematical function which simultaneously fulfills and Therefore it must hold that,and this is in fact the fundamental relationship between thermodynamics and statistical mechanics. Furthermore,according to our conclusions,the entropy has to assume a maximum for the equilibrium state,just as the number of compatible microstates Ωassumes a maximum in equilibrium.

39. We therefore find,by counting all possible values (Ω),the macrostate with Ω max as the equilibrium state.Exactly at this point,however,we assumed that all microstates ( q v,p v )compatible with our macrostate have "equal rights”; i.e. the equal probability of all microstates. Anyhow,we see that the equal probability of all compatible microstates is an assumption which can only be justified by an experimental examination of the consequences.Up to now, no one has invented an experiment which disproves this assumption.

40. With respect to the energy law we may for example say: a)There is no perpetuum mobile of the second kind.A perpetuum mobile of the second kind is an engine which does nothing but perform work while cooling a heat bath.Thus it is an engine which transforms heat into work with100% efficiency. From the microscopic point of view the following formulation is especially instructive: b)Each isolated macroscopic system wants to assume the most probable state,i.e.,the state which is characterized by the largest number of microscopic realization possibilities.

41. Now we want to use both laws to derive some consequences for the state variables T,p, μ, φ,...in an isolated system in equilibrium.. Therefore the state variables U i, S i,V i and N i with i =1,2 do not have constant values;however,it must be the case that U 1 +U 2 =U =const. S 1 +S 2 =S =const. V 1 +V 2 =V=const. N 1 +N 2 =N=const.

42. Let us now remember the first law for a reversible change of state for both partial systems: Here T i,p i,and μ i are the temperatures,pressures and chemical potentials in the two partial systems. We now have dU I +dU 2 =0. it follows,with dS l =-dS 2, dV 1 =- dV 2...,that

43. Equation above is true only if it separately holds that T 1 = T 2, P 1 =P 2, μ 1 = μ 2, we can conclude that if an isolated system is in equilibrium,it has everywhere the same constant temperature,the same pressure and the same chemical potential etc. In most cases,pressure balance occurs most quickly,i.e.,mechanical equilibrium is followed by thermal equilibrium. Establishing chemical equilibrium may take very long in some cases.

44. If a system is in thermodynamic equilibrium,i.e.,if it has everywhere the same temperature, the same pressure and the same chemical potential,one speaks of global equilibrium. 5 Global and local equilibrium If one can divide the whole system into small partial systems which still contain very many particles and which are individually approximately in thermodynamic equilibrium. If the total system can be divided into such parts one speaks of local equilibrium.

45. 6 Thermodynamic engines Cyclic heat engines play an extraordinarily large role in technics. The experience summarized in the second law asserts that the work performed irreversible processes is smallest and the heat largest,

46. If we now have a cyclic engine which leads the working material back to the initial state after one cycle,it holds according to the first law that And thus Hence of all possible processes,reversible processes produce the largest amount of utilizable work and requires the smallest amount of work. The best efficiency of transforming heat into work is reached by an engine working in a reversible way.

47. Now we want to calculate the efficiency of a general reversible cyclic process.To this end we schematize the substantial parts of a heat engine in Figure 2.8. Each engine needs a heat reservoir(T=T h ) from which to extract heat energy and a second reservoir(T= Tc ) to absorb the waste heat of the process,i.e.,to cool the engine. Figure 2.8 Heat engine

48. According to the first law it holds that We have already defined the efficiency ηas the fraction (see section about Carnot's process),which tells us how much heat energy is transformed into work ( ): Since the engine shall work reversible,it holds that

49. The signs in Equation above correspond to the directions given in the figure.Since (performed work)we have For the efficiency it always holds that (2.56)

50. If the engine works in the reverse direction in Figure 2.x.. Figure 2.x Cold engine

51. An important point of our consideration is that Equation (2.56)always holds,independent of the working material and the technical realization of the engine, for if there were two reversible cycles with different efficiencies,one could construct a perpetuum mobile of the second kind. One would then be able to connect the two processes as shown in Figure 2.9.Engine A works here in the reverse direction,i.e., as a heat pump expelling the energy W A and the heat Q CA from the cold reservoir as beat Q hA into the hot reservoir.

52. The energy W A is here generated by the process B,which we shall suppose works with higher efficiency,so that there remains an amount of work WB-WA. Figure 2.9 Perpetuum mobile of the second kind

53. If η A and η B denote the efficiencies of the engines(with η B 〉 η A ), then if we only consider absolute values and choose the signs according to the figure, If we now adjust the engine in such a way that Q hA =Q hB =Q h then there will be no change in the hot reservoir on a long time scale, since the same amount of heat is taken off as pumped back. Then

54. > since η B 〉 η A.Thus the heat is effectively drawn off from the cold reservoir.Hence,the engine performs the work

55. This is exactly a perpetuum mobile of the second kind, which permanently performs work and merely cools a heat reservoir. The vain efforts lasting for centuries to construct such an engine,which does not contradict the energy law but rather the entropy law,resulted in the recognition that Δ Q c =W B -W A =0,or for all reversible processes at given T h and T c.

56. Exercise 2.6 Mixture temperatures Calculate the range of possible final temperature T f in equilibrium for a system consisting of two partial systems A and B,if A and B have initial temperatures T A,T B and heat capacities, which are independent of temperature.

57. Solution First case:Totally irreversible process δW=0, A change in the temperature of the partial systems is connected with an exchanged mount of heat via Thus,for the final temperature T f follows that

58. Since C A and C B do not depend on temperature or Here T f is the "mixture temperature" for an irreversible process,e.g.,when fluids of different temperatures are poured together.The entropy Changes are

59. If for instance, T A 〉 T B, the T f 〈 T A and ΔS A 〈 0,and T f 〉 T B and ΔS B 〉 0,respectively.However, Second case:Reversible process with a heat engine between A and B: Now it holds that

60. from which it immediately follows after integration that or It always holds that

61. 7 Euler's equation and the Gibbs-Duhem relation the extensive internal energy U has to be interpreted as a function of the extensive state variables S,V,N 1,...,N K U( αS,αV, α N 1,..., α N K )= α U(S,V,N 1,...,N K ) T(αS, αV, α N 1,..., α N K )= T(S,V,N 1,...,N K )

62. If we insert this into Equation (2.67)and consider that according to Equation If we consider an infinitesimal increase of the system( α =1+ ε withε 《 l),we can expand the left hand side in a Taylor series. According to

63. So,it follows that i.e.,Euler’s equation is valid If we calculate the total differential of Euler's equation, it holds that

64. If we compare this with Equation above,obviously the condition must always be fulfilled(plus additional terms,if other state variables are necessary).One calls Equation the Gibbs-Duhem relation. It means that the intensive variables T,p, μ 1,..., μ k which are conjugate to the extensive variables S,V,N 1,...,N K are not all independent of each other.

65. Exercise : There is a large vessel with water of the temperature 0 0 C,which is attached to a heat bath and which is in thermal equilibrium with the heat bath. The heat bath’s temperature is 100 0 C. Now, please determine the entropy difference of the water and of the heat bath and of the whole system,respectively.