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Topic 2 The first law of thermodynamics 2012. 02.15 2.1 Introduction of thermodynamics Therme = Heat Dynamic = Power The study of energy and its transformation Four laws: The zeroth law---temperature The first law---internal energy enthalpy The second---entropy S The third law---S=0 Energy transformations in our life History and Progress Stage I:end of 17th –mid 19th Century Character of “heat” Steam engine Heat engine theory Large amounts of phenomenon Stage II: mid19th-1870s The first and second laws of thermodynamics Stage III:1870s-beginning of 20th century The third law of thermodynamic Statistic thermodynamic Boltzmann Stage IV: 1930sQuantum statistical mechanics Non-equilibrium thermodynamics Basic concepts System and the surroundings States and properties of systems: thermodynamic parameters Process and paths Thermodynamic equilibrium Systems and surroundings An open system can exchange matter and energy with its surroundings A closed system can exchange energy with its surroundings, but it cannot exchange matter boundary An isolated system can exchange neither energy nor matter with its surroundings Process and path P1, V1, T1 Initial state(T1,P1,V1) state (T2,P2,T2) Initial state final Path 1: T1,P1,V1→T1,P2,Va (Constant-temperature process) →T2,P2,V2 (Constant-pressure process) Path2: T1,P1,V1→Tb,P2,V1 (Constant-volume process) →T2,P2,V2 (Constant-pressure process) Path3:T1,P1,V1→T2,P1,Vc (Constant-presure process) →T2,P2,V2 (Constanttemperature process) 1 2 ( )T P2, Va, T1 3 ( )V P2, V1, Tb ( )P ( )P P2, V2, T2 ( )P P1, Vc, T2 ( )T Final state State and path functions State properties: temperature, pressure, volume density, viscosity Extensive properties: dependent on amount Intensive properties: independent on amount State function is a property that is independent of how a sample is prepared, completely differential ,single valued Properties that relate to the preparation of the state are called path functions Question: T, P, V, ρ, Vm…… W, Q Extensive or intensive? State function or path function? Thermodynamic equilibrium No pure inter- or Intra- energy transformation Thermal equilibrium Temperature is equal everywhere Mechanical equilibrium No work Phase equilibrium No phase transition Chemical equilibrium No component change 2.2 The zeroth law of thermodynamics--temperature If two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. If A and C are each in thermal equilibrium with B, A is also in equilibrium with C. Temperature as a quality of heat, by Galileo and Newton The temperatures are equal for all systems in thermal equilibrium. Temperature scale Thermometers 2.3 The first law of thermodynamics The derivation: a long zigzag story…… A design of 1st generation perpetual motion machine J.R. Mayer James Prescott Joule Lord Kelvin States in words: The total energy of system plus surroundings remains constant. The total energy of an isolated system is conserved. (The first generation) Perpetual motion machines do not exist Mathematical Expression: E(total energy)= U(internal energy) + T (kinetic energy/energy of motion) + system Q/W V (potential energy) surrounding In equalibrium state: T=0, V=0, E=U U = Q + W closed system dU = δQ + δW closed system U--- Internal energy of the system Q--- Heat W--- Work by the system, -; to the system, + About U: Internal energy Unit :J/kJ U UT U e UV U R U N U ES U U sys U final U initial •State function U f (T , V ) U U dU ( )V dT ( )T dV T V About Heat The energy of a system changes as a result of a temperature difference Heat flow from the hotter body to the colder one Unit: J/kJ Q>0 endothermic Q<0 exothermic cooler heater Insulating shield Q=0 Thermal shield Heat capacity U U dU ( )V dT ( )T dV T V U The heat capacity at CV ( )V T constant volume dU=dQ=Qv, at constant volume, no additional work QV=CvdT Calorimetry Qv=Cv.△T About Work: Work is done when an object is moved against an opposing force Giving out work: W<0; Given work: W>0 Type of work dW comments units Expansion -psurrdV Psurr is the external pressure dV is the change in volume Pa m3 Surface expansion γdσ extension Electrical γ is the surface tension dσ is the change in area N.m-1 m2 fdl f is the tension dl is the change of length N m Фdq Ф is the electric potential dq is the change in charge V C 2.4 Expansion work/p-V work The work arising from a change in volume W= -Psurr.ΔV (?) W Wi PSurrdV PSurrdV V2 V1 Think about: If Q=0, ΔU=-pdV ? When would it be right? Example: Calculation of expansion work δWsurr,1 psurr dV 0 Free expansion Psurr=0 等外压膨胀(psurr=常数)Expansion against constant pressure Wsurr,2 psurr (V2 V1 ) 多次等外压膨胀 Wsurr,3 p '(V ' V1 ) p "(V " V ') p2 (V2 V ") 准静态膨胀 reversible expansion W Wi PSurrdV PSurrdV V2 V1 Reversible expansion The system is always infinitesimally close to equilibrium, and an infinitesimal change in conditions can reverse the process to restore both system and surrounding to their initial state. A reversible process is obviously an idealization Reversible isothermal process in a perfect gas 2 V1 P2 W pdV nRT ln nRT ln V2 P1 1 Features of reversible expansion quasi-static (quasi-equilibrium) Carried out infinitesimally slowly so that the driving force is only infinitesimally greater than the opposing force Return to the original system state without having had a change, either in the system or the surroundings The work done by a system during a reversible process is the maximum work we can get. The work done on a system in a reversible process is the minimum work we need to do to achieve that state change. 判断下列哪个过程是可逆过程? (1)摩擦生热; (2)室温、标准压力下一杯水蒸发为同温同压的水; (3) 373K,P0下,一杯水蒸发为气体; (4)手电筒中干电池放电使灯泡发亮; (5)N2和O2混合; (6)恒温下将1mol水倾入一溶有不挥发溶质的大量溶液中 2.5 Enthalpy 焓 △U= Qv (Closed system, only expansion work exchanging, constant volume) Qv , heat transactions at constant volume Qp, heat transactions at constant pressure ΔU= Q+W=Qp- PsurrΔV Psurr= P1= P2 ∴ΔU= Qp – P2 V2 + P1V1= U2-U1 ∴QP=(U2+ P2 V2)-(U1+ P1V1) Enthalpy H≡U+PV,∴Qp=H2-H1=ΔH •State function, extensive Comparing H and U H=U+PV State function Same units Heat capacity Heat capacity at constant pressure Qp H Cp ( )p T dT Heat capacity at constant volume QV U CV ( )V T dT H Qp CpdT U QV CV dT Closed system, in equilibrium, PV work only Relating ΔH and ΔU W’=0 , ΔH=Qp and ΔU=Qv For process only involve solids or liquids Δ(PV) is negligible ΔH ≈ ΔU For gases: Δ(PV)=nRΔT Qp QV nRT For example: H2O(l,0℃)→H2O(l,50℃) System: Water; Water+ electric line W, Q, △U ? U = Q + W Home work: A: P66 2.7(a,b) Y:P12 1, 5, 9 Preparation for next class; Y: 1.6-1.9 A: 2.6, 3