1 / 80

Chapter 3 The Second and Third Laws of Thermodynamics

Chapter 3 The Second and Third Laws of Thermodynamics. 3.1 The direction of spontaneous change 3.2 The Carnot Cycle and Carnot’s Theorem 3.3 The Concept of Entropy 3.4 The Calculation of Entropy Changes 3.5 Physical significance of entropy and The third Law of thermodynamics

walter-gallegos
Télécharger la présentation

Chapter 3 The Second and Third Laws of Thermodynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3 The Second and Third Laws of Thermodynamics • 3.1 The direction of spontaneous change • 3.2 The Carnot Cycle and Carnot’s Theorem • 3.3 The Concept of Entropy • 3.4 The Calculation of Entropy Changes • 3.5 Physical significance of entropy and The third Law of thermodynamics • 3.6 the Helmholtz and Gibbs energies • 3.7 Criteria for spontaneity • 3.8 the fundamental equation of thermodynamics • 3.9 Calculation of G and A

  2. New Words and Expressions • Spontaneous 自发的 • Efficiency of heat engine 热机效率 • Carnot theorem 卡诺循环 • Entropy 熵 • Inequality of Clausius 克劳修斯不等式 • The principle of the increase of entropy 熵增原理 • Physical significance of entropy 熵的物理意义

  3. New Words and Expressions • Reversible phase transformation process 可逆相变过程 • Irreversible phase transformation process 不可逆相变过程 • Condensed phase 凝聚相 • Perfect crystalline pure materials 完美晶体 • Conventional molar entropy 规定摩尔熵 • Standard molar entropy change of reaction标准摩尔反应熵 • Helmholtz function • Gibbs function

  4. 3.1 The direction of spontaneous change The first law of thermodynamics is concerned with the conservation of energy and with the interrelationship of work and heat. A second important problem with which thermodynamics deals is whether a chemical or physical change can take place spontaneously. This particular aspect is the concern of the second law of thermodynamics.

  5. Natural(Spontaneous) Processes Some things happen naturally, some things don’t. A gas expands to fill the available volume, a hot body cools to the temperature of its surroundings, and a chemical reaction runs in one direction rather than another. Some aspect of the world determines the spontaneous direction of change, the direction of change that does not require work to be done to bring the change about.

  6. We can confine a gas to a smaller volume, we can cool an object with a refrigerator, and we can force some reactions to go in reverse(as in the electrolysis of water). However, none of these processes happens spontaneously; each one must be brought about by doing work.

  7. Classics Statements of the Second Law of thermodynamics • No process is possible in which the sole result is the transference of heat from high-temperature heat source to low-temperature. (Clausius) No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. (Kelvin)

  8. 不可能把热从低温物体传到高温物体，而不引起其它变化不可能把热从低温物体传到高温物体，而不引起其它变化 Classics Statements of the Second Law of thermodynamics

  9. 3.2 The Carnot Cycle and Carnot’s Theorem • Carnot’s great theoretical contribution was to realize that the production of work by an engine of this type depends on a flow of heat from a higher temperature Th (h standing for hotter) to a lower temperature Tc (c standing for colder). He considered an ideal type of engine, involving an ideal gas,

  10. in which all processes occurred reversibly, and showed that such an engine would have the maximum possible efficiency for any engine working between the same two temperatures.

  11. The Carnot Cycle

  12. The Carnot Cycle Let us consider these four steps in further detail. 1. Step AB is the reversible isothermal expansion at Th. UAB = 0, -qAB = WAB = R ThlnV1/V2 (for 1 mol) 2. Step BC is the reversible adiabatic expansion qBC = 0, UBC = WBC =CV(Tc - Th)

  13. The Carnot Cycle 3. Step CB is the reversible isothermal compression at Tc UCD= 0, -qCD= WCD = R TclnV3/V4 4. Step DA is the reversible adiabatic compression qDA = 0, UDA = WDA =CV(Th – Tc)

  14. The Carnot Cycle The net values: Unet = 0 qrev= R ThlnV2/V1+ R TclnV4/V3 =R(Th -Tc)ln V2/V1 (since V4/V3= V1/V2) wrev = R(Th -Tc)ln V1/V2 Thus efficiency, , of a heat engine:  = work performed/heat absorbed =w/qh = (qh +qc)/qh =1+qc/qh (3.1)

  15. Efficiency of heat engine • The efficiency of a reversible Carnot engine can be defined as the work done by the system during the cycle, divided by the work that would have been done if all the heat absorbed at the higher temperature had been converted into work. for a Carnot engine rev = w/qh = (qh +qc)/qh = R(Th -Tc)ln V1/V2/ RTh ln V1/V2 =(Th - Tc )/Th. (3.2)

  16. 则 Carnot’s Theorem • Carnot’s Theorem • The efficiency of all reversible cycles operating between the temperatures Th and Tc is the same, namely (Th - Tc )/Th. • Form Eqs. 3.1 and 3.2 it follows that, for the reversible engine, (Th - Tc )/Th=(qh +qc)/qh • or qh/ Th + qc/ Tc=0 (3.3) 根据卡诺定理：

  17. Carnot’s Theorem equation 3.3 applies to any reversible cycle that has isothermal and adiabatic parts, and it can be put into a more general form to apply to any reversible cycle. This result leads to important quantitative formulations of the second law of thermodynamics.

  18. 3.3 The Concept of Entropy

  19. 或 对微小变化 • it is convenient to write dqrev/T as dS, • dS=dqrev/T or S= dqrev/T (3.4) • SI unit: J K-1

  20. The property S is known as the entropy of the system, and it is a state function. The change of entropy is the same whatever path is followed.

  21. For an irreversible cycle, • qirrh/ Th + qirrc/ Tc <0 (3.5) • Consider the entire cycle ABA is in part reversible • (process AB irreversible, process BA reversible),

  22. 则有 Irreversible Processes

  23. 对于微小变化： (3.6) Inequality of Clausius This is known as the inequality of Clausius, after Rudolf Clausius, who suggested this relationship in 1854.

  24. Suppose the system is isolated from its surroundings, then dq=0, and the Clausius inequality implies that • Sr 0 (3.7) • This tells us that in an isolated system the entropy cannot decrease when a spontaneous change takes place. • Stotr 0 , or dS + dSsurr  0 (3.8) • dSsurr=dqsurr/Tsurr=dqsys/ Tsurr, entropy change of surroundings

  25. 3.4 The Calculation of Entropy Changes • We have seen that entropy is a state function, which means that an entropy change SAB when a system changes from state A to state B is independent of the path. This entropy change is given by • SAB= dqrev/T •  dqirr /T< SAB •  dq/T is equal to the entropy change if the process is reversible and less than the entropy change if the process is irreversible.

  26. (a) PVT change process • 恒压过程（isobaric process） S=CPdT/T= nCP,mlnT2/T1 • 恒容过程（isochoric process） S=CVdT/T= nCV,mlnT2/T1 • 恒温变化过程（isothermal process） S=  dqrev/T = qrev/T but, for solid or liquid, 一定温度下，而P，V变化不大时，液、固体的熵变很小， S0。

  27. Ideal Gases • S at Constant Temperature S=RV/dV= nR lnV2/V1 (3.9) • S at Constant pressure S=CPdT/T= nCP,mlnT2/T1 (3.10) • S at Constant volume S=CVdT/T= nCV,mlnT2/T1 (3.11)

  28. Ideal Gases • S for changing T,P and V S=CVdT/T+ RV/dV =nCV,mlnT2/T1+nR lnV2/V1 = nCp,mlnT2/T1-nR lnP2/P1 =nCp,m lnV2/V1+nCV,m lnP2/P1 (3.12)

  29. Ideal Gases • Example 3.1 Calculate the entropy change when argon at 250C and 1.00atm in a container of volume 500cm3 is allowed to 1000cm3 and is simultaneously heated to 1000C. • Answer The amount of Ar present • n=PV/RT=0.0204mol • The entropy change in the first step expansion from 500cm3 to 1000cm3 at

  30. Ideal Gases 298K, then in the second step, from 298K to 373K at constant volume, so S=S1+S2 =nRln2.00+nCV,m ln373K/298K =0.02048.314ln2.00+ 0.0204(20.70-8.314) ln373/298 =0.175(JK-1)

  31. Mixing process of ideal gas • Mixing process of ideal gas （at constant T and P) S=n1Rln(V1+V2)/V1+n2R ln (V1+V2)/V2 or S=-(n1Rlny1+n2R ln y2 ) (3.13) y1= V1/ V1+V2 , y2= V2/ V1+V2 isothermal mixing process of different gas. But for isothermal mixing process of same gas, S=0.

  32. When liquids are mixed, the entropy change is sometimes given by Eq.3. 13, which we derived for ideal gases. The condition for this equation to apply to liquids is that the intermolecular forces between the different components must all be equal.

  33. Example 3.2 Exactly one liter of a 0.100M solution of a substance A is added to 3.00 liters of a 0.050M solution of a substance B. Assume ideal behavior and calculate the entropy of mixing.

  34. Solution 0.100M solution of a substance A is present and the volume increases by a factor of 4; S(A) =0.100*8.3145*ln4.00 =1.153JK-1 0.150 mol of B is present and the volume increases by a factor of 4/3;

  35. S(B) =0.150*8.3145*ln4/3 =0.359 JK-1 The net S is therefore 1.153+0.359=1.512 JK-1

  36. (b)entropy Changes in phase transformation process • reversible phase transition • At the normal transition point (at constant temperature, and pressure)any transfer of heat between the system and its surroundings is reversible because the two phases in the system are in equilibrium. Because at constant pressure, q= H(相变焓）, the change in molar entropy of the system is • trsS= trsH/Ttrs (3.14)

  37. Entropy of Fusion fusS= fusH/Tfus (3.15) • Entropy of Vaporization vapS= vapH/Tvap (3.16) • In irreversible phase transformation process, must design reversible paths to calculate S. • 寻求可逆途径的依据： （1）途径中的每一步必须可逆。 （2）途径中每步S的计算有依据

  38. 3.5 Physical significance of entropy and The third Law of thermodynamics • Physical significance of entropy • Entropy is a measure of disorder; an crease in entropy means an increase in disorder. • Entropy of fusion increase on melting, and the conversion of a solid into a gas is also accompanied by an entropy increase, for the same reason.

  39. Entropy Changes in Chemical Reactions: in general, for a gaseous chemical reaction there is an increase of entropy in the system if there is an increase in the number of molecules. • Ionization processes always involve an entropy decrease, because the bound water molecules have restricted freedom of motion.

  40. The Third Law of Thermodynamics • At T=0, all energy of thermal motion has been quenched, and in a perfect crystal all the atoms or ions are in a regular, uniform array. • The Third Law of Thermodynamics: The entropy of all perfect crystalline substances is zero at T=0. • S0, as T0 provided all the substances involved are perfectly ordered.

  41. Absolute Entropies on the basis that S(0)=0 , at constant pressure: • S (T2)= S(T1)+CP/TdT (3.17) • The standard reaction entropy, rS , is denoted, like the standard reaction enthalpy. rS =( Sm) products ( Sm)reactants (3.18)

  42. Example 3.3 To calculate the standard reaction entropy of H2 (g) +1/2O2(g) 2H2O(l), at 298K Answer rS = Sm(H2O,l){ Sm (H2,g)+1/2 Sm (O2,g)} =69.9-{130.7+1/2(205.1)} =-163.4JK-1mol-1

  43. 3.6 the Helmholtz and Gibbs energies • dU=q + W = q - PexdV+ W’ • dS  q /T • dS  (dU + PexdV- W’ )/T • if constant T and V, PexdV=0, • d(U-TS)T,V  W’ • A  U-TS, which is the Helmholtz energy. • (d A) T,V  W’ • if W’ =0, no add work, • (d A) T,V 0 (< spontaneous, = equilibrium)

  44. if constant T and P, • dS  (dU + PdV- W’ )/T • d(H-TS)T,P  W’ • G  H-TS, which is the Gibbs energy. • (d G) T,P  W’ • if W’ =0, no add work, • (d G) T,P 0 (< spontaneous, = equilibrium)

  45. Some remarks on the A and G • the Helmholtz energy----Maximum work • (dA)T  W (- PexdV+ W’ ) • so, (dA)T = Wmax • A is sometimes called the maximum work function or the work function. • G, the Gibbs energy---- Maximum additional work (d G) T,P  W’ (d G) T,P = W’max

  46. 3.7 Criteria for spontaneity 1. dS- q /T  0 or (dS)U,V  0 (3.19) dS>q /T (不可逆） dS=q /T （平衡） dS〈q /T （不可能） 2. (d A) T,V, W’ =0 0（〈自发， = 平衡）(3.20) (d A) T,V 〉0 （不自发） (3.21) 3. (d G) T,P, W’ =0, ,0 （〈自发， = 平衡）(3.22) (d G) T,P, 〉0 （不自发） (3.23)

  47. 3.8 the fundamental equation of thermodynamics • For an infinitesimal process in involving only PV work , we can combine the first and second laws , and to derive the fundamental equation. • dU=TdSPdV • dH= TdS+VdP • dA=SdTPdV • dG=SdT+VdP (3.24)

  48. eqn 3.24 applies to any change-reversible or irreversible- of a closed system of constant composition that does no additional(non-PV)work.

More Related