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

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1 1 Prof. Dr. rer. nat. habil. S. Enders Faculty III for Process Science Institute of Chemical Engineering Department of hermodynamics Lecture Polymer hermodynamics 0331 L 337

2.1. First Law of hermodynamics Case a) Change of state for copper cube Copper cup has no possibility to exchange heat with the surrounding. he temperature may be 298K and kept constant.

2 2.1. First Law of hermodynamics Case a) Change of state for copper cube Copper cup has no possibility to exchange heat with the surrounding. he temperature may be 298K and kept constant. he system copper is in state 1 characterized with temperature 1 and pressure P 0 (and with mass m=8.96g). 2 Case b) Using Bunsen burner we heat-up the copper for a short time. he metal takes the heat via his surface. he heat will be distributed on the whole phase very rapidly. he surrounding (in this case the Bunsen burner) gives a certain amount heat, Q, to the system (in this case the copper). he related heat can be calculated using Q=C Cu m. Case c) he copper cube is again thermic isolated and his temperature is now 310 K. he system is now in the state 2 characterized with temperature 2, where temperature 2 is higher than temperature 1. he state 2 has more energy, because heat is added to the system.

3 2.1. First Law of hermodynamics 3 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. he internal energy can not be measured directly, only its change can be measured via accurate measurements of the exchanged energy. he change of internal energy is equal to the difference in energy U (internal energy of Copper in state 2 internal energy of Copper in state 1) 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.

4 2.1. First Law of hermodynamics 4 he 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 system P 1,V 1, 1,n ± heat J ± work W system P 2,V 2, 2,n change of state state 1 state 2

5 2.1. First Law of hermodynamics 5 du = dw + dq he 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 1, P 1, 2, P 2, V 1, U 1 V 2, U 2 route 2

6 2.1. First Law of hermodynamics 6 U f V n1 n2 n i du = dw + dq = (,,,.. ) caloric equation of state exact differential U U U du = d + dv + dn V n V, n, n 1 V,, n j j j 1 U U + dn + + n 2 2 n Vn,, i V,, n j 2 U U U du = d + dv + dn V n j i dn V, n, n i i V,, n j j j i i 1 can be estimated only via the second law of thermodynamic i

7 2.1. First Law of hermodynamics 7 C V U =, j Vn C V J J cv K molk he heat capacity, c V, is the heat which is necessary to warming one mol of the substance by 1K under isochore conditions. substance water (liquid) water (vapor) acetone c V [J/(mol K)] =298.15K P=0.101 MPa

8 2.1. First Law of hermodynamics 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). 8 introduction of a new thermodynamic function H = U + PV H = enthalpy = function of state dh = du + PdV + VdP dh = PdV + dq + PdV + VdP dh = dq + VdP for isobaric processes: dh dq = caloric equation of state

9 2.1. First Law of hermodynamics 9 (,,,.. ) H f P n1 n2 n i = caloric equation of state exact differential H H H dh = d + dp + dn P n Pn, n, 1, P, n j j j 1 H H + dn + + n 2 2 n, P, n i, P, n j 2 j i dn i 1 can be estimated only with the second law of thermodynamic H H H dh = d + dp + dn P n Pn, n, i i, P, n j j j i i

10 2.1. First Law of hermodynamics 10 C P H = P n, j C P J J cp K molk he heat capacity, c P, is the heat which is necessary to warm one mol of the substance by 1K under isobaric conditions. substance c P [J/(mol K)] ammonia bromine lead sulfate =298.15K P=0.101 MPa

11 11 he first law of thermodynamics does not say anything in which direction the process takes please. he 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. he work goes back to the ocean via friction. W. Ostwald: Perpetual motion of second kind Such a ship does not exist. Why???

12 12 N 2 O 2 Air wo different gases do not demixing voluntary, although it will be possible for energetic reasons. Why??? N 2 O 2

13 13 1 =298 K 2 =318 K dq 1? dq 2? 3?

14 14 detonating gas reaction At 600 C the following reaction will be run spontaneously: 2 H 2 +O 2 2 H 2 O Accompaniment: noisy bang, risk for explosion Nobody has observed that water vapor decomposes in his elements (oxygen and hydrogen) at 600 C.

15 15 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. here is no perpetual motion of the second kind. his 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).

16 16 All real processes in nature are irreversible processes. Example: Evaporation of superheated water (i.e. at 110 C) = irreversible Process thermometer cooling water notional fragmentation in different reversible processes column 1. Cooling of water to 100 C 2. Evaporation in equilibrium 3. Heating of water vapor to 110 C distillation flask distillate

17 Carnot Cycle (1824) 17 pressure pressure pressure pressure heat at 1 volume isothermal expansion isolator volume adiabatic expansion volume volume isolator adiabatic compression heat at 2 isothermal compression

pressure Polymer hermodynamics volume isotherm isotherm Carnot Cycle with ideal gas A B Isothermal expansion 18 V 2 V 2 WAB = nr1ln QAB = nr1ln V1 V1 B C adiabatic

18 pressure Polymer hermodynamics volume isotherm isotherm Carnot Cycle with ideal gas A B Isothermal expansion 18 V 2 V 2 WAB = nr1ln QAB = nr1ln V1 V1 B C adiabatic expansion ( ) W C Q BC = V 2 1 BC = 0 C D isothermal compression V 4 V 4 WCD = nr2ln QCD = nr2ln V3 V3 D A adiabatic compression ( ) W = C Q = S.N.L Carnot ( ) DA v 2 1 DA 0

19 Carnot Cycle with ideal gas 19 Law of Poisson pressure isotherm PV κ = constant volume isotherm V ( κ ) V V = constant = V3 V W = W + W + W + W AB BC CD DA V 2 V 4 V 2 W = nr 1 ln + 2 ln W = nr( 1 2 ) ln V1 V3 V1

20 20 Carnot Cycle with ideal gas pressure isotherm isotherm volume For the whole cycle is valid: du = 0 W = Q used heat = Q supplied heat = Q AB W Q Q + Q Q η = = = = 1+ Q Q Q Q benefit efficiency = effort work done = = η used heat AB CD CD AB AB AB AB CD

21 pressure Polymer hermodynamics volume isotherm isotherm Carnot Cycle with ideal gas W nr V 2 ( ) ln = 1 2 V1 η = efficiency supplied head = Q W QCD η = = 1+ Q Q AB AB AB 21 V nr ( ) η = = = 1 V 2 nr1 ln V ln V1 ( 1 2) he efficiency depends only on temperature difference. Q Q CD 2 CD AB AB Q Q = = Q 1 2 CD Q AB + =

Carnot Cycle with ideal gas 22 pressure isotherm isotherm Q CD Q AB + = 2 1 Clausius (1854) 0 ds = dq rev volume S = entropy S J J s K molk Rudolf Julius Emanuel Clausius (1822 1888), was a German

22 Carnot Cycle with ideal gas 22 pressure isotherm isotherm Q CD Q AB + = 2 1 Clausius (1854) 0 ds = dq rev volume S = entropy S J J s K molk Rudolf Julius Emanuel Clausius ( ), was a German physicist and mathematician and is considered one of the central founders of the science of thermodynamics. His most important paper, on the mechanical theory of heat, published in 1850, first stated the basic ideas of the second law of thermodynamics. In 1854 he introduced the concept of entropy. he entropy, denoted by the symbol S, is a state function.

23 Carnot Cycle with ideal gas 23 1) isothermal expansion: ds=dq AB / 1 2) adiabatic expansion: ds=0 S = constant 3) isotherme compression: ds=dq CD / 2 4) adiabatic compression: ds=0 S = constant

24 24 Statistical Interpretation of Entropy Number of possible configuration for 3 distinguishable particles in 3 rooms ε 3 ε 2 ε 1 probability: Number of possible configurations corresponds to the permutations of the number of particles: P=N! = 3*2*1 = 6 W N! = N! N! N! N! W is called thermodynamic probability and it is the number of possible configuration for a given macroscopic state. i

25 25 Example: distribution of 4 particles with identical energy in 4 volumes. A W A = 4!/(4!0!0!0!)= 4!/4! =1 B W B = 4!/(3!0!1!0!)= 4!/3! =4 C D W C = 4!/(2!1!1!0!)= 4!/2! =12 W D = 4!/(1!1!1!1!)= 4!/1! =24 he macroscopic state D can be realized by the most configurations. he macroscopic state D has the highest thermodynamic probability.

Statistical interpretation S = k lnw k Boltzmann-constant k=r/n AV =1.

26 Statistical interpretation S = k lnw k Boltzmann-constant k=r/n AV = J/K absolute values of entropy are known S S S kln W 2 Δ = 2 1 = W1 In statistical thermodynamics the entropy is defined as a number of microscopic configurations that result in the observed macroscopic description of the thermodynamic system. 26 Ice melting classic example of entropy increasing described in 1862 by R. Clausius as an increase in the disgregation of the molecules of the body of ice.

27 27 he entropy of an isolated system can never decrease, but always increases for irreversible processes or stays constant for reversible processes. isolated systems ds 0 irreversible reversible open and closed systems: ΔS total 0 irreversible reversible Δ Stotal =Δ Ssystem +ΔSsurrounding A. Einstein: he second law of thermodynamics is the premier law of all sciences. 0

28 28 Dickerson/Geis Chemie eine lebendige und anschauliche Einführung

29 29 N 2 O 2 ds 0 irreversible reversible air for isolated systems Reason: he spontaneous demixing of air leads to a smaller number of microscopic configurations and hence to a decrease of entropy. N 2 O 2 disagreement to the second law of thermodynamics

30 30 1 =298 K 2 =318 K n 1 =n 2,,, c p1 =c p2 dq 1 =-dq 2 =dq (i.e. water) First law of thermodynamics gives information of the resulting temperature 3. dq 1? 3? dq 2? Second law of thermodynamics gives information in which direction this process takes place. ds = ds + ds ds total total 1 2 dq1 dq2 1 1 = + = dq 298K 318K 298K 318K 318K 298K dstotal = dq 0 dq K * 318K

31 31 1 =298 K 2 =318 K dq 1? 3? dq 2? If two systems having different temperatures are in thermic contact results in the heat flow from the system with higher temperature to the system with lower temperature, never the other way around.

32 32 he hot coffee in the cup will be cooled down in a cool room. he opposite procedure that means the cafe will be heated and the room will be cooled down will never happen. Reason Second Law of hermodynamics

33 33 ds 0 irreversible reversible ds = dq rev We would produce only cold water. alternative formulation of second law of thermodynamics here is no perpetual motion of the second kind. his device uses the heat of a reservoir and transfer it into mechanical energy.

34 Example 34 A piece of steel with a mass of 40 kg and a heat capacity of c P,steel =0.5 kj/(kg*k) should be cooled down applying an oil bath having a heat capacity of c P,oil =2.5 kj/(kg*k). he steel has a temperature of 450 C. he amount of oil at 25 C to be available is 150kg. Is this procedure possible? dq steel = dq steel oil oil ( ) ( ) m c d = m c d m c = m c steel steel oil oil steel steel steel oil oil oil 40 kg *0.5 kj kg K = K ( ) K = 150 kg *2.5 kj kg K ( K)

35 Example 35 A piece of steel with a mass of 40 kg and a heat capacity of c P,steel =0.5 kj/(kg*k) should be cooled down applying an oil bath having a heat capacity of c P,oil =2.5 kj/(kg*k). he steel has a temperature of 450 C. he amount of oil at 25 C to be available is 150kg. Is this procedure possible? 2 dq mcpd mcpd 2 = = Δ = = P ln 1 ds S mc Δ S = steel 40 kg *0.5 kj kg K K ln = K Δ Soil = 150 kg *2.5 kj K ln = kg K K kj kj kj Δ Stotal = = 9.8 > 0 K K K kj K kj K

36 S = f(, P, n, n n ) 1 2 i state function - entropy S S S ds = d + dp + dn P n P, n, n.. n, n, n.. n 1 P,, n.. n 1 2 i 1 2 i 2 j 1 S S + dn + + n 2 2 n P,, n.. n i P,, n, n.. n 1 i i i dn i 1 36 assumption: pure substance S S ds = d + dp P P

37 state function - entropy 37 S = f(, P) S S ds = d + dp P P S dq =? reversible change of state ds = P isobaric change of state of ideal gases dq = C d Cd P P = ds S = P C P

38 state function - entropy S = f(, P) S S ds = d + P P dp S dq =? reversible change of state ds = P isothermal change of state ideal gas dq = PdV nr = PV dq PdV ds = = V nr nr P nr = dv = dp ds = 2 2 P P P P 2 S nr V = = P P dp nr = P dp 38

39 state function - entropy 39 S = f(, V ) S S ds = d + dv V V S dq =? reversible change of state ds = V isochoric change of state of ideal gases dq = Cd ds Cd S = V V V C = V

40 state function - entropy 40 S = f(, V) S S ds = d + dv V V S dq =? reversible change of state ds = V isothermal change of state of ideal gases dq = PdV dq PdV ds = = ds = dv P

41 state function - entropy 41 C V ds = P d dp C P ds = V d + dv he entropy of an ideal gas increases with increasing temperature and/or volume. he entropy of an ideal gas decreases with increasing pressure.

42 F U S Polymer hermodynamics df = du ds Sd du = dq + dw df rev rev state function with dq = ds dw = PdV du = ds PdV = ds PdV ds df = PdV Sd for dv = 0 df = Sd free energy = Helmholtz energy F [ J] Sd = PdV Sd 42

43 df = PdV Sd F = f( V, ) state function F = U S F F df = dv + d V V 43 F F = P = S V V

44 G H S rev rev state function dg = dh ds Sd H = U + PV dh = du + PdV + VdP du = dq + dw dg dg du = ds PdV dq = ds dw = PdV dh = ds PdV + PdV + VdP = ds + VdP = ds + VdP ds = VdP Sd G = free enthalpy = Gibbs-energy Sd G [ J] 44

45 state function 45 P G H S dg = VdP Sd G = f(, P) G G G G dg = d + dp = S = V P P P for isobaric process (dp=0) dg = Sd

46 d Polymer hermodynamics equilibrium conditions state function = dv = 0 d = dp = 0 46 df < 0 voluntary = 0 equilibrium > 0 constraint dg < 0 voluntary = 0 equilibrium > 0 constraint at equilibrium df=0 at equilibrium dg=0 minimum of free energy minimum of free enthalpy general equilibrium conditions

47 equilibrium conditions state function EPot = mgh 47 stable E Pot global minimum stable against all perturbations instable E Pot global maximum instable against all perturbations metastable E Pot local minimum stable against small perturbations

48 equilibrium conditions state function 48 Entropy S ds=0 equilibrium state variable

49 equilibrium conditions state function 49 free enthalpy G equilibrium dg=0 state variable

50 equilibrium conditions state function 50 free energy F equilibrium df=0 state variable

Summarizing, Key Point: An irreversible process is either spontaneous (ΔS universe > 0) or does not occur (ΔS universe < 0)

Summarizing, Key Point: An irreversible process is either spontaneous (ΔS universe > 0) or does not occur (ΔS universe < 0) Key Point: ΔS universe allows us to distinguish between reversible and irreversible