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Polymer Thermodynamics Lecture 0331 L 337 2. First and Second Law of Thermodynamics

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Polymer Thermodynamics Lecture 0331 L 337 2. First and Second Law of Thermodynamics 1 Prof. Dr. rer. nat. habil. S. Enders Faculty III for Process Science Institute of Chemical Engineering Department of Thermodynamics Lecture Polymer Thermodynamics 0331 L 337 2. First and Second Law of Thermodynamics Polymer Thermodynamics 2 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics Case a) Change of state for copper cube Copper cup has no possibility to exchange heat with the surrounding. The temperature may be 298K and kept constant. The system copper is in state 1 characterized with temperature T1 and pressure P0 (and with mass m=8.96g). Case b) Using Bunsen burner we heat-up the copper for a short time. The metal takes the heat via his surface. The heat will be distributed on the whole phase very rapidly. The surrounding (in this case the Bunsen burner) gives a certain amount heat, Q, to the system (in this case the copper). The related heat can be calculated using Q=CCu⋅m⋅∆T. Case c) The copper cube is again thermic isolated and his temperature is now 310 K. The system is now in the state 2 characterized with temperature T2, where temperature T2 is higher than temperature T1. The state 2 has more energy, because heat is added to the system. Polymer Thermodynamics 3 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics In order to distinguish clearly between the total energy (internal, potential and kinetic) of a system and energy, which can be exchanged with the surrounding, we introduce the term internal energy U. The internal energy can not be measured directly, only its change can be measured via accurate measurements of the exchanged energy. The change of internal energy is equal to 9 the difference in energy ∆U (internal energy of Copper in state 2 – internal energy of Copper in state 1) 9 the sum of energies exchanged with the surrounding during the change of state In our example we can measure the internal energy measuring the heat Q during the change of state. Polymer Thermodynamics 4 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics The equivalence of change in internal energy on one side and the change in heat and work on the other side has to be reality, because the law of energy conservation. before afterward ± heat J system system P2,V2,T2,n P1,V1,T1,n ± work W change of state state 1 state 2 Polymer Thermodynamics 5 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics dU= dW+ dQ The exchanged amount of work and heat of a system is identical to the change of internal energy of the system. or: A Perpetual motion of the first kind do not exist.. route 1 T2, P2, T1, P1, V1, U1 V2, U2 route 2 Polymer Thermodynamics 6 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics dU = dW+ dQ UfTVnnn= (, ,12 , ..)i caloric equation of state exact differential ⎛⎞∂∂∂UUU ⎛⎞ ⎛⎞ dU=++⎜⎟ dT ⎜⎟ dV⎜⎟ dn1 ⎝⎠∂∂∂TVnVn,, ⎝⎠ Tn 1 jj⎝⎠TVn,,j≠1 ⎛⎞∂∂UU⎛⎞ can be estimated +++⎜⎟ dn2 … ⎜⎟ dni only via the second ∂∂nn2 i law of thermodynamic ⎝⎠TVn,,j≠2 ⎝⎠TVn,,ji≠ ⎛⎞∂∂UU ⎛⎞ ⎛⎞ ∂ U dU=++⎜⎟ dT ⎜⎟ dV∑ ⎜⎟ dni ⎝⎠∂∂TVVn,, ⎝⎠ Tn i ∂ ni jj⎝⎠TVn,,ji≠ Polymer Thermodynamics 7 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics ⎛⎞∂U ⎡⎤JJ ⎡ ⎤ CV = ⎜⎟ Cc ∂T VV⎢⎥ ⎢ ⎥ ⎝⎠Vn, j ⎣⎦K ⎣molK ⎦ The heat capacity, cV, is the heat which is necessary to warming one mol of the substance by 1K under isochore conditions. T=298.15K substance c [J/(mol K)] V P=0.101 MPa water (liquid) 75.15 water (vapor) 33.56 acetone 125 Polymer Thermodynamics 8 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics dU= dW+=−+ dQ PdV dQ for isochoric processes: dU= dQ caloric equation of state Lots of processes are going on at constant pressure (i.e. atmospheric pressure). introduction of a new thermodynamic function HUPV= + H = enthalpy = function of state dH= dU++ PdV VdP dH=− PdV ++dQ PdV +VdP dH=+ dQ VdP caloric equation of state for isobaric processes: dH= dQ Polymer Thermodynamics 9 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics HfTPnnn= (, ,12 , ..)i caloric equation of state exact differential ⎛⎞∂∂∂HHH ⎛⎞ ⎛⎞ dH=++⎜⎟ dT ⎜⎟ dP⎜⎟ dn1 ⎝⎠∂∂∂TPnPn,, ⎝⎠ Tn 1 jj⎝⎠TPn,,j≠1 ⎛⎞∂∂HH⎛⎞ can be estimated +++⎜⎟ dn2 … ⎜⎟ dni only with the second ∂∂nn2 i law of thermodynamic ⎝⎠TPn,,j≠2 ⎝⎠TPn,,ji≠ ⎛⎞∂∂HH ⎛⎞ ⎛⎞ ∂ H dH=++⎜⎟ dT ⎜⎟ dP∑⎜⎟ dni ⎝⎠∂∂TPPn,, ⎝⎠ Tn i ∂ ni jj⎝⎠TPn,,ji≠ Polymer Thermodynamics 10 2. First and Second Law of Thermodynamics 2.1. First Law of Thermodynamics ⎡ JJ⎤⎡ ⎤ ⎛⎞∂H Cc CP = ⎜⎟ PP⎢ ⎥⎢ ⎥ ∂T ⎣ K ⎦⎣molK ⎦ ⎝⎠P,n j The heat capacity, cP, is the heat which is necessary to warm one mol of the substance by 1K under isobaric conditions. substance cP [J/(mol K)] ammonia 35.52 T=298.15K P=0.101 MPa bromine 75.71 lead sulfate 104.3 Polymer Thermodynamics 11 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics The first law of thermodynamics does not say anything in which direction the process takes please. The first law of thermodynamics leads only to dU= dW+ dQ Marine propulsion can be established according the first law of thermodynamics by constructing an engine that use heat of the ocean and transformed this heat completely into work. The work goes back to the ocean via friction. W. Ostwald: Perpetual motion of second kind Such a ship does not exist. Why ??? Polymer Thermodynamics 12 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics N2 O2 Two different gases do not demixing voluntary, although it will be possible for energetic reasons. Air Why ??? N2 O2 Polymer Thermodynamics 13 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics T =318 K T1=298 K 2 dQ1 ?dQ2 ? T3 ? Polymer Thermodynamics 14 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics detonating gas reaction At 600°C the following reaction will be run spontaneously: 2 H2+O2 → 2 H2O Accompaniment: noisy bang, risk for explosion Nobody has observed that water vapor decomposes in his elements (oxygen and hydrogen) at 600°C. Polymer Thermodynamics 15 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics Max Planck (1897) It is impossible for every device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. There is no perpetual motion of the second kind. This device uses the heat of a reservoir and transfer it into mechanical energy. All real processes in nature are irreversible processes. Reversible process is a process or operation of a system or a devise that a net reverse in operation will accomplished the converse of the original functions. For a reversible process the net change at each stage of the system and the surrounding is zero and it can be reversed at each stage. Reversible processes are idealized limiting cases of real processes and they are infinity slowly, because all changes of state are differential small. All processes going on with finite speed are irreversible (all real processes). Polymer Thermodynamics 16 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics All real processes in nature are irreversible processes. Example: Evaporation of superheated water (i.e. at 110°C) = irreversible Process thermometer notional fragmentation in different reversible processes cooling water 1. Cooling of water to 100°C 2. Evaporation in equilibrium column 3. Heating of water vapor to 110°C distillate distillation flask Polymer Thermodynamics 17 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics Carnot Cycle (1824) pressure pressure volume volume isothermal adiabatic expansion expansion ↑heat at T1 isolator pressure pressure volume volume adiabatic isothermal compression compression isolator ↓ heat at T2 Polymer Thermodynamics 18 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics Carnot Cycle with ideal gas A → B Isothermal expansion ⎛⎞VV22 ⎛⎞ isotherm WAB=− nRT11ln⎜⎟ Q AB = nRT ln ⎜⎟ ⎝⎠VV11 ⎝⎠ pressure isotherm B → C adiabatic expansion WCTTQBC=− V( 21) BC =0 volume C → D isothermal compression ⎛⎞VV44 ⎛⎞ WnRTCD=−22ln⎜⎟ QnRT CD = ln ⎜⎟ ⎝⎠VV33 ⎝⎠ D → A adiabatic compression WCTTQDA= v( 21−=) DA 0 S.N.L Carnot (1796-1832) Polymer Thermodynamics 19 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics Carnot Cycle with ideal gas Law of Poisson isotherm PV κ = constant pressure isotherm ()κ −1 ⎛⎞VV⎛⎞ TV =→=constant ⎜⎟41⎜⎟ ⎝⎠VV32⎝⎠ volume WW=+++AB W BC W CD W DA ⎡⎤ ⎛⎞VV24⎛⎞ ⎛ V2 ⎞ Wn=+→R ⎢⎥TT1 ln⎜⎟2 ln ⎜⎟ WnRT=−()1 T2 ln ⎜⎟ ⎣ ⎝⎠V1 ⎝⎠V3 ⎦ ⎝ V1 ⎠ Polymer Thermodynamics 20 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics Carnot Cycle with ideal gas isotherm benefit pressure efficiency = isotherm effort work done volume ==η used heat For the whole cycle is valid: dU= 0 →− W = Q used heat== QABCD supplied heat Q W Q QQ+ Q η ===ABCDCD =+1 QQABABAB Q Q AB Polymer Thermodynamics 21 2. First and Second Law of Thermodynamics 2.1. Second Law of Thermodynamics Carnot Cycle with ideal gas ⎛⎞V2 WnRTT=−()12ln ⎜⎟ ⎝⎠V1 isotherm η = efficiency pressure isotherm supplied head = QAB W Q η ==+1 CD volume QQAB AB ⎛⎞V nR() T− T ln 2 QQTQ 12 ⎜⎟ CD=−2 CD =− AB ⎝⎠V1 ()TT12− T2 η ===−1 QTAB 12TT1 ⎛⎞V TT nRT ln 2 11 Q Q 1 ⎜⎟ CD +=AB 0 ⎝⎠V1 T21T The efficiency depends only on temperature difference. Polymer Thermodynamics 22 2. First and Second Law of Thermodynamics 2.1.
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