Chapter XVI The Second Law of Thermodynamics

Chia sẻ: Thanh An | Ngày: | Loại File: PDF | Số trang:31

lượt xem

Chapter XVI The Second Law of Thermodynamics

Mô tả tài liệu
  Download Vui lòng tải xuống để xem tài liệu đầy đủ

The first law of thermodynamics gives the quantiative relations between the internal energy of a system and the quantities of heat and work that the system exchange with surroundings. It express the conservation of energy. The first law of thermodynamics is true, but not enough ! Why ? Many thermodynamic processes which don’t violate the 1st law, but don’t happen in nature !

Chủ đề:

Nội dung Text: Chapter XVI The Second Law of Thermodynamics

  1. GENERAL PHYSICS II Electromagnetism & Thermal Physics 5/6/2008 1
  2. Chapter XVI The Second Law of Thermodynamics §1. Reversible Carnot cycles §2. The second law of thermodynamics §3. Entropy and quantitative formulation of the second law §4. Heat engines and refrigerators 5/6/2008 2
  3. The first law of thermodynamics gives the quantiative relations between the internal energy of a system and the quantities of heat and work that the system exchange with surroundings. It express the conservation of energy. The first law of thermodynamics is true, but not enough ! Why ? Many thermodynamic processes which don’t violate the 1st law, but don’t happen in nature ! • Examples: In an isolating system hot heat flow cold a heat transfer from the cold part to the hot part doesn’t happen ! No one has succeded in building a machine that converts heat completely into mechanical energy It is necessary to have a supplemental law which says about the directions of thermodynamic processes → the 2nd law 5/6/2008 3
  4. §1. Reversible Carnot cycle: We are interested in the directions of thermodynamic processes. First we introduce the concept of reversible and irreversible processes. 1.1 Revisible and irrevisible processes:  Definition: A reversible thermodynamic process is a process which can takes place in the direct as well as in the reverse way. In the reverse process the system passes through all the immediate states that it undergoes in the direct process. Otherwise the process is called irreversible.  It is observed by experiment that in nature thermodynamic process are all irreversible process. But for a theoretical consideration it is convenient to deal with reversible processes, as idealized models. 5/6/2008 4
  5. 1.2 Reversible Carnot cycle: The reversible Carnot cycle is an idealized model for processe in heat engines  Consider a cyclic process which is represented by the closed path P → Q → P. The mechanical work done in the cycle is In the P-V diagram, this work is pesented by the area within the closed path. 5/6/2008 5
  6.  The Carnot cycle consists of 4 steps: 1) A → B: isothermal expansion to volume VB at temperature T2 2) B → C: adiabatic expansion to volume VC , tempeature drops to T1 < T2 3) C → D: isothermal compression to volume VD at constant temperature T 1 4) D → A: adiabatic compression to the original state of the system  Exchanges of heat and work in each step: P-V diagram of a Carnot cycle • During s-1: the system absorbs heat Q 2 (Q2 > 0) and does work WAB (WAB > 0) An ideal gas is used for VB working substance Q2  AB W  2 ln nRT VA 5/6/2008 6
  7. • During s-3: the system rejects heat Q 1 , Q1 < 0 and the work is done on the system, WCD < 0: VC Q1  CD W  nRT1 ln  VD • For two adiabatic steps s-2 & s-4: 1  T  T 1 T2VB 1  1VC 1 & T2VA   1VD  VB 1 VC 1   VB VC 1   1  VA VD  VA VD P-V diagram of a Carnot cycle Q1 1  VC / VD ) T ln( T1 | Q2 | | Q1 |      ln(V / V )   Q2 2  B A T T2 T2 T1 The ratio of the heat rejected and the heat absorbed is just equal to the ratio of the temperatures. 5/6/2008 7
  8. What is the results of a cycle ABCDA ? - The system absorbs heat Q 2 from the surroundings at temperature T2 (hot reservoir) , deposits heat Q1 in the surroundings at temperature T1 (cold reservoir). - The internal energy of the system remains unchanged. - The net mechanical work done by the system in one cycle is represented by the P-V diagram of area enclosed and is positive. From the 1st law: a Carnot cycle Wnet = |Q2| - |Q1| > 0 (since T2 > T1 ) By continously repeating the cycle it is possible to do mechanical work by continously taking heat energy from a hot reservoir and depositing a smaller amount of heat energy in a cold reservoir. Such a device is called a heat engine. 5/6/2008 8
  9. Consider the cycle in the reverse direction A→D→C→B→A: The result is as follows: Mechanical work is put into the system in order to take heat energy from the cold reservoir and deposit heat in a higher-temperature reservoir. In this case the device is a refrigerator. We see that heat energy can’t flow from the cold reservoir to the hot reservoir without action of an external mechanical work. 1.3 Efficiency of a Carnot engine: For a heat engine the efficiency represents the fraction of heat that is converted to work Mechanical work done by engine Efficiency = Heat energy supplied from the hot reservoir 5/6/2008 9
  10. From the general definition we can calculate the efficiency of a Carnot engine: W | Q2 |  Q1 | | | Q1 | T1 eCarnot    1  1 Q2 | Q2 | | Q2 | T2 Remark: The efficiency of Carnot engine depends only on the temperatures of two heat reservoirs The efficiency is large when the temperature difference is large, and the efficiency is very small when the temperatures are nearly equal The efficiency can never be exactly unity unless T1 = 0. This means heat can never be converted completely into mechanical work. Although the Carnot engine is an idealized model, but the conclusions derived from it give an good orientation for investigation of other heat engines that we usually encounter. 5/6/2008 10
  11. §2. The second law of thermodynamics: Experiment evidence suggests that it is impossible to build a heat engine that converts heat completely to work, that is, an engine with 100% thermal efficiency. This impossibility is the basis of the following statement of The second law of thermodynamics. 2.1 “Engine” statement of the second law: “It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending in the same state in which it began”. This statement is also called the Kelvin-Planck statement of the second law. 5/6/2008 11
  12.  Some interpretation: The difference between mechnical energy and heat energy is that they relate to two different kinds of motion: organized and random motions, respectively. One can convert the organized motions of molecules to random motions, but it is difficult to control the random motions of molecules, thus one can’t convert completely random motions to organized motions, one can convert a part of it.  Can we build an automobile or power plan by cooling the surrounding air? It is impossible in practice. The reason is as follows: the engine receives heat from one body (hot reservoir), but at the same time needs another body (cold reservoir) in which it deposits the remaining heat (heat can’t be converted 100%). 5/6/2008 12
  13. 2.2 “Refrigerator” statement of the second law: The second law has also an alternative statement: “It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body”. This is called the “refrigerator” statement, or also the Clausius statement of the second law of thermodynamics. This statement means that in order to transfer heat from a cooler to a hotter body it ís absolutely neccesary to spend a work which does on the system. 5/6/2008 13
  14. 2.3 The equivalency of two statements of the second law: Two statements my seem to have no relation each to other. But, in fact, they are completely equivalent.  Suppose that we have a refrigerator which can transfer heat from a cooler to a hotter body without any external work (the “refrigerator” statement is violated). Then we can use this refrigerator in conjuntion with a heat engine, pumping the heat rejected by the engine back to the hot reservoir → violate the “engine” statement.  Suppose that we have a heat engine which can run by receive heat from the hot reservoir (the “engine” statement is violated). Then we can use this heat engine in conjunction with a refrigerator. The output work from the engine can be used for driving the refrigerator that pumps heat from the cool reservoir and deliver to the hot reservoir without requiring any input of work → violate the “refrigerator” statement. Violation of Violation of the “refrigerator” statement the “engine” statement 5/6/2008 14
  15. §3. Entropy and quantitative formulation of the second law: We know two equivalent formulation of the second law. But we haven’t had any equation to express the law. With the help of the concept of entropy we can write an equation for the second law 3.1 Definition: Consider an infinitesimal reversible isothermal expansion of an ideal gas. For this process: dU  , 0 dQ  pdV  dW nRT dV dQ nR dV V T V We denote by S the new quantity which has in the considered process the following change dQ dS  (for an infinitesimal T reversible process) 5/6/2008 15
  16. The entropy change in a reversible process from the intial to final state: 2 2 dQ S    S2  1  dS  S 1 1 T  the isothermal reversible expansion from the volume V1 to V2 : For 2 1 Q   2 1    S S S dQ T 1 T For this process we have also: dQ 2 V2 dV 2  V S S S     ln    2 1  nR nR   1 T V1 V 1  V Here gas expands, V2 > V1 → Δ > 0 , the entropy increases. When S the volume increases, the free space for molecular motions is larger → the “randomness” is also larger. Thus we can use entropy as a quantittive measure of disorder.  Units of entropy in SI : J / 0K 5/6/2008 16
  17. 3.2 Entropy changes in a reversible process:  First consider a Carnot cycle. It is a idealized reversible process. Recall for a reversible Carnot cycle we have the following equation: |Q 2 | |Q1|  T2 T1 Q1 Q 2 Since Q1 0 Q1  | Q1 |    0 T1 T2 In two adibatic steps there is no heat exchange → the net change in entropy in one cycle is   S1  S2  S   0 There is no net entropy change in one reversible Carnot cycle ! 5/6/2008 17
  18.  Now consider an arbitrary reversible cyclic process for an ideal gas: It can beconsidered to be equivalent to a large number of “thin” Carnot cycles. For any of “thin” Carnot cycle we have Q1 Q 2   0 T1 T2 The total entropy change is Qi T 0 i i The approximation becomes better as if the number of subcycles is larger, and every subcycle is thiner, so we obtain for an arbitrary reversible cyclic process: dQ  0 T 5/6/2008 18
  19. P  This result leads also to the consequence: The entropy change of a system in a reversible B process from A to B is independent of the path. 2 dQ S   doesn’t depend on the path ! A 1 T V 3.3 A quantitative statement of the second law:  practise, reversible processes are only idealized situations. In  the real world, friction can never be completely eliminated, and in the In reverse direction of a process, it is difficult for the system to collect back completely the energy which was lost by friction in the direct process → real thermodynamic processes are irrevisible. 5/6/2008 19
  20. m m m m Consider an example of irreversible process: The system consists of two identical metal blocks. The whole system is thermally isolated from the surroundings. The initial state a): the temperatures of two blocks are different T1 & T2 The final state b): the two blocks in contact → T = (½)(T1 + T2 ) 1  dQ T T dQ T dT T dT   T   T  (T1  2 )  T    S   mc mc  ln  mc ln  2mc ln  mc       2  T1 T T2 T T1 T T2 T 1  T 2  T  T1T2      5/6/2008 20


Đồng bộ tài khoản