Dr. Fritz Wilhelm page 1 of 24 C:\physics\230 lecture\ch22 Heat engines entropy 2nd law.docx; 2/18/2010

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1 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 Homework: See website.. Heat Engines and the Second Law o hermodynamics,. A Heat Pump or Rerigerator. Reversible and Irreversible Processes. Probability and Entropy, 4.4 he Carnot engine, 5.4a Problem: A Carnot engine between 500 and 50 C, 7.4b In a gasoline engine (Otto engine), 8.5 Entropy, 9.5b (.8) Macrostates and Microstates, 0.6 Macroscopic Entropy Deinition,.7 Entropy Change or a Quasi Static Reversible Process o an Ideal Gas,.7a Entropy Changes During Phase Changes, 4.7b he entropy change during a thermal conduction, 5.7c Entropy Change in Calorimetric Processes,5 Optional:.8 hermodynamic Equilibrium (Enthalpie etc.),8.9 Calculating the critical temperature o a real gas 9.

2 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00. Heat Engines and the Second Law o hermodynamics. Heat is the energy o a disordered movement o molecules. (.) In a heat engine we attempt to turn the disordered movement into a ordered movements which can be used to do work on a continous, cyclical basis. he ocean is a huge heat reservoir, in some areas with water temperatures o 5 degree Celsius. Wouldn t it be nice to build a ship with an engine that takes hot water out o the ocean, extracts work rom it, during which process the water cools down. It seems reasonable to think that the energy dierence Q = mc could be turned into work. Such a ship could run on hot water so to speak and leave a trail o ice behind it. Even though this sounds plausible, it cannot be done. he crucial point in this example is that we would have a machine which runs on the extraction o energy rom a single heat reservoir, the ocean. Its cycle would start and end in the same heat reservoir. (his is called a perpetual machine o the second kind. BW, a perpetual machine o the irst kind would run, do work, without extracting energy rom anything in the world.) he statement that we cannot build a perpetual motion machine o the second kind is the essence o the second law o thermodynamics. here are more sophisticated ways to say the same thing, some o these ways we are going to examine in more detail. A heat engine carries some working substance (gasoline vapor-air mixture, steam) through a cyclic process during which () the working substance absorbs heat Q h rom a high-temperature energy reservoir (exploded gasoline-vapor-air mixture, hot steam), () work is done by the engine (exploding gas pushes piston), and () energy is expelled by heat Q c to a dierent reservoir at a lower temperature. his cycle is continously being repeated. Hot reservoir at h Heat Q h enters system Engine- G Work W Heat Q C leaves done by system system (engine) Cold reservoir at c he system to which we apply our thermodynamic variables is a gas, which undergoes a cyclical process, orms a closed loop in the P-space. he change in internal energy ΔU is thereore 0. he heat exchange consists o stage () and (). In stage () heat Q h enters the system at the higher temperature h, and heat Q c leaves the system at c. Work is done by the system (engine), thereore W eng is negative. We indicate that by using W eng. (.) U = 0 = Qh Qc Weng (.) Weng = Qh Qc he thermal eiciency e is the ratio o the work to the heat input, it is the ratio o the work you get to the energy you pay or:

3 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 W Q Q Q (.4) e = = = Q Q Q eng h c c h h h his leads to a second way o stating the second law o thermodynamics: (.5) he thermal eiciency o a heat engine is always smaller than. Example: he human body can be modeled as a heat engine to some degree, burning carbohydrates and at and producing heat, energy, water and carbon dioxide in complex chemical reactions. hese reactions require catalysts or enzymes to occur at low temperatures, around 40 C. A typical sugar has an energy content o about 5000 cal/g. Let us assume that the human body has an eiciency o 0.. How much work can the body do with gram o sugar? How much unusable energy is produced? Weng Qh Qc (.6) e = = = 0. Qh Qh he body can do 500 cal or 009 J o work. It can lit kg through a vertical distance o 00 m cal cannot be used or doing work. Heat Q h ejected into the environment. h Cold reservoir at c rerigerator or heat pump. Heat Q c taken rom the cool reservoir to cool it more. Hot reservoir at h Work W delivered into the system to run the pump. A Heat Pump or Rerigerator, serves to cool down a heat reservoir. Heat (hot air) Q c is taken out o the cooler reservoir and ejected into the warmer reservoir as Q h. o perorm this cycle the necessary work-energy is delivered to the system, or example through the electricity running a rerigerator. his means that W is positive. We see that such a device is essentially the reverse process o a heat engine, in which work is being generated. (he book by Serway labels the heat amounts dierently, be careul.) U = 0 = Qc Qh + W (.7) W = Q Q he coeicient o perormance COP (in cooling mode) is deined as the ratio o heat taken out o the cold reservoir (the ridge) and the work necessary to do this (what you pay or.) h c (.8) COP(in cooling mode) = Q c W

4 Dr. Fritz Wilhelm page 4 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 hus, we see that it is impossible to construct a cyclical machine whose sole eect is to transer energy continously by heat rom a cooler object to another warmer object without the input o energy (or example electricity) or work. Note: An important aspect o all heat transer is that huge numbers o molecules are involved. In order to be able to talk about temperature these molecules must be in thermal equilibrium. Remembering Maxwell s distribution curves, we see that all o these processes are o a statistical nature. Problem: A certain rerigerator runs at 0.50 kw and has a COP o liters o water at 0 degree Celsius are placed into the reezing department. How long does it take until the water has obtained a temperature o 0 degree Celsius below reezing? he amount o heat extracted rom the water is the heat to lower the water temperature to reezing, then the latent heat o melting, and then the energy to lower the temperature o ice rom 0 to -0 degree Celsius. he heat capacity o water is.0 cal/gc. he latent heat o melting is 80 cal. he heat capacity o ice is 0.50 cal/gc. hereore the total heat that must be extracted is: cal cal cal 6 Qc = 000 g (.00 0C C ) = 5kCal = kJ =. 0 J gc g gc he rerigerator runs at 500 Watts. energy Q Power = c time = t Qc hereore: t = =640 seconds=44 minutes. What is wrong with this? P he power rating reers to the energy input into the ridge not to the heat extracted. We must use COP=Q c /W, or W=Q c /5. he time required is thereore 44/5 minutes =8.8 minutes.. Reversible and Irreversible Processes. Probability and Entropy. I you would look at a ilm showing two containers connected by a membrane, in which the gas would low in such a way that it collects in one container and leaves the other empty, you would immediately know that the ilm is running backwards. Processes which, i let to themselves can only unold in one direction, are GAS acuum called irreversible. I there are only two gas molecules we know that it is possible that both molecules are in one and the same container. We just know that this is less likely, less probable. he more molecules we have, the less probable it is that all molecules assemble in one container. I we call any situation in which we have certain amounts o molecules in both containers a state, we can say: A state in which we have an approximately equal number o molecules in each container is more likely than a state in which there are many more molecules in one container than in the other.

5 Dr. Fritz Wilhelm page 5 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 Or, put dierently: It is more probable that a state with an uneven distribution o molecules unolds in a direction in which we have an even distribution o molecules, than the other way around. Or, the natural ree movement o states is towards states o higher probability. I we replace the word probability in this sentence by entropy (S), we have a working deinition o this important term. the natural ree movement o states is towards states o higher entropy S. (his is yet another orm o the second law o thermodynamics. ) All macroscopic processes are irreversible, they occur in the direction o higher probability (entropy) o states. S > 0 Reversible processes are characterised by no change in their entropy. S = 0 he entropy increases in all irreversible processes ( S > 0 )..4 he Carnot engine (see igure) is a (theoretical) cyclical series o reversible processes, which allow us to calculate the absolute highest theoretical eiciency o a heat engine. heoretical here means an idealized situation which can never be achieved in reality. Another statement or the second law would be that a Carnot engine eiciency is impossible to achieve with a real engine. Carnot chose this particular cycle because we can calculate the work and heat exchange during each portion o the cycle. Each step o this cycle can be rendered ininitesimal, quasistatic.we know the laws or heat exchange in adiabatic and in isothermal processes. γ γ adiabatic: P = constant or = constant (.9) isothermal: W= nr ln i For the eiciency we have rom (.4) Qc (.0) e = Qh For the whole cycle we have ΔU=0 and Carnot Engine.: according the second law: Weng = Qh Qc For the adiabatic branches we have P isothermal expansion A rom A to B. Q U = nr where we remember that the h enters the D to A system. adiabatic internal energy change is completely compression. calculated through the temperature change. D B he work done on the system is equal to B to C negative the work done by the engine, which is the dierence between the absolute values o C adiabatic Q h and Q c. isothermal compression rom C to D. Q c leaves the system. Note that the adiabatic lines (P γ =constant) in a P diagram always intersect the isothermal lines (P=constant). Adiabatic lines are steeper than isothermal lines. For the adiabatic expansion rom B to C, the work is equal to the change in internal energy, which is proportional to Δ.

6 Dr. Fritz Wilhelm page 6 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 U BC = nr( C h ) = QBC + WBC < 0 (.) U DA = nr( h c ) = QDA + W DA > 0 he equal and opposite amount o work is done on the branch with adiabatic contraction rom D to A, between the same initial and inal temperatures. he total work on the adiabatic lines is thereore 0. We need only to consider the work done along the isothermal branches. his is consistent with our deinition (.) or the work in a cyclical process in which the gas does work. U = 0 = Q Q W ; h c eng A to B: Isothermal expansion, ΔU=0 because Δ=0; heat Q h enters the system (Q h >0).he gas expands, raises the piston, and thus, does work. We can calculate Q h because it is equal to the work done during that process W. We need to make sure that both Q h and Q c are positive quantities. U = 0 = Q W AB h AB B B (.) nr B Qh = WAB ( ) = Pd = d = nrln > 0 A A A C to D: At this lower isothermal branch heat Q c leaves the system (Q c <0)at the lower temperature. he gas is being compressed, work is done on it. U = 0 = Q W CD c CD D nr D (.) Qc = W ( ) = Pd = = nrln < 0 Q c C C = nr ln > 0 nr ln Qc (.4) e = = Q h nr ln D D C C D B A We need to get rid o the ln expressions. We have ormulas to relate the volume endpoints to the higher and lower temperatures because o the adiabatic ormulas: For the adiabatic expansion between B to C, rom temperatures to ( h to c ) we get (.5) = γ γ B C For the adiabatic expansion between A and D, rom temperatures to we get: γ γ (.6) A = D Dividing equation (.5) by (.6) we get: C

7 Dr. Fritz Wilhelm page 7 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 (.7) γ γ B C = γ γ A D B C (.8) = A D his means or the eiciency: C nr ln Qc D (.9) e = = = Q h B nr ln A his is a quantitative statement o the second law o thermodynamics. It is the highest eiciency o a heat engine because it is calculated or a reversible (quasi static) process. (.0) he highest possible eiciency in a heat engine is the Carnot eiciency e =.4a Problem: A Carnot engine runs between a high temperature o 500 C and a low temperature o 50 C. he isothermal expansion (upper branch) occurs between a volume o liters and.00 liters. he isothermal contraction (lower branch) occurs between.00 l and.00 l. here are mols o a gas involved. Calculate the work done by the engine, the heat taken in Q h, and the heat ejected Q c. Calculate the eiciency o the cyle. Draw a P diagram and show the area corresponding to the total work being done. Show the beginning and end points o the isothermal and adiabatic curves. Show during which branch o the cycle heat enters and leaves the engine. (Note that we deined work W as the work done on the system. In your calculations here you will ind negative work or the path rom A to B to C to D. he negative o this value is the work done by the engine.) W = 5.9 kj, Qh = 8.9 kj; Qc =.7 kj; e = 0.58 U = 0 Q= Qh Qc = W Qh = nr ln = = 8.9kJ W = Q e = 5.9kJ h

8 Dr. Fritz Wilhelm page 8 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00.4b In a gasoline engine (Otto engine) the compressed air-gasoline mixture is ignited through spark plugs. In a Diesel engine the mixture gets compressed to such a pressure that the gasolineair mixture ignites by itsel. In both cases the pistons cycle between the same compressed and decompressed volumes o air. In comparison to the Carnot cycle that means that the volumes B and C are the same, ; so are the volumes A and D,. Also, he adiabatic processes occur between A and B, as well as C and D. (here are 4 dierent temperatures at the intersecting points. he isotherms are located below the adiabatic curves.) A Adiabatic expansion rom to. Power stroke, volume expansion, temperature drops rom A to B. Work: the gas pushes the piston downward Q h enters the engine. Spark+combustion. No stroke. D Adiabatic compression rom to B Q c leaves the engine as exhaust valve is opened. Pressure drops or short period o time. C Q (.) c e= ; W = Q Q Q h eng h c We can calculate Q by noting that on the respective branches vertical W=0 and thereore (.) U = Q = nc ( ) > 0 And BC C B C (.) U = Q = nc ( ) > 0 DA h A D Note that we have our dierent temperatures at the intersection points. Furthermore we know that on the adiabatic branches we have: γ γ (.4) A = B And γ γ (.5) = γ (.6) ( ) ( ) D C ( B C) ( ) γ γ A D = B C = γ = A D γ

9 Dr. Fritz Wilhelm page 9 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 Qc B C (.7) e = = = = γ Qh A D hereore we get the eiciency o an Otto engine in terms o the compression ratio, the ratio between the higher volume and the lower volume: >. Note that the smaller volume is and the larger volume is. We express the eiciency o such an engine directly by the compression ratio: γ (.8) γ e = = = = B C γ A D his compression ratio is much higher in a Diesel engine than in an Otto engine. Example: he pressure in a Diesel engine varies rom to 0 atmospheres during the adiabatic compression. Find the eiciency o the engine. o get the compression ratio we use the adiabatic equations between the vertical branches: (.9) P γ.4 γ PA P A 0 = P = = 8.50 γ = = PB PB γ γ A B he compression ratio is 8.5, thereore we get the eiciency rom (.8) (.0) e = = = 57.5% γ Entropy. As we mentioned beore, systems let to themselves (isolated systems) tend to evolve in a direction o greater disorder. Entropy S, is a measure o this disorder. All macroscopic processes occur in such ways that the total net entropy o all systems interacting with each other increases. his process goes hand in hand with the notion that natural processes evolve in the direction o lowest possible energy. he increase in disorder o one system always outweighs the increase in order o another system. For example, the organisation o molecules into one particular pattern o a plant, comes at the cost o order in its environment. Entropy in the macroscopic universe is simple minded, it only increases, somehow like time. All o these macroscopic processes are irreversible. he disorganisation can be described in terms o the change in entropy rom one state to the other: S = k B ln(number o all possible microstates) he entropy can be looked at as an energy that cannot be retrieved without creating larger energy losses somewhere else. hus, the impossibility to do work at a 00% eiciency is related to the unavoidable increase in entropy.

10 Dr. Fritz Wilhelm page 0 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00.5b Macrostates and Microstates. Consider again the ree expansion o a gas rom a compartment o volume A to the compartment o volume B, both o which are equal in size. I we have our molecules to start with in one container, ater a while we will most likely have molecules in each compartment. In the example o a ree adiabatic expansion we considered just two o ive possible macro states. We have an initial and a inal state with 0 particles being in one compartment and our in the other. hese are the states with lowest probability. he probability o inding two molecules in each container is the highest probability o any o these so-called macro-states. It does not matter which o the our molecules is in either compartment. All that matters or the deinition o the macro-state is the number o molecules. We can see that there are the ollowing 5 macrostates possible. Here are the macro-states: 0 in A and 4 in B in A and in B in A and in B in A and in B 4 in A and 0 in B In macro-states the overall act o a certain number o molecules in a particular state counts, not the composition in terms o individual particles. Such overall situations correspond to the macroscopic quantities like pressure, temperature, and volume, and total number o particles. (All that matters is how many particles are in a state, not which particular ones o the indistinguishable particles.) he most likely situation (macro-state) corresponds to that with an equal number o molecules in each compartment, in A and in B. I we have our molecules to begin with, we igure out in how many dierent ways (micro-states) this macro-state can be achieved. Each such path, possibility, where we now label each molecule, counts as a micro state. I we distinguish the individual molecules by labeling them each with a number through 4 there are 4 =6 dierent conigurations (micro-states) available. here are 6 dierent ways in which the macrostate molecules in each compartment can be achieved. Each one o the particles can be in one o dierent containers (A or B) at a time. We can see that the most probable macrostate is the one with equal numbers o molecules in each volume. We can see rom the next igure that the largest number o micro-states corresponds to this macro-state. here are many more possible paths (microstates) available to achieve this most probable macro-state. Entropy change between two states is deined as: (.) S = k B ln(number o micro-states) We deine the concept o microstates as the number o paths in which a particular coniguration o a state can be obtained. In our ollowing example the two dierent initial and inal states are a) the state in which our particles are contained in one side o the container and

11 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 b) the state in which we have an equal number o particles in both containers. he law o entropy states that any natural process evolves towards the state o highest probability, which is the state in which we have an equal number o particles in each state. In our case this is the macrostate C. We simply count in how many dierent ways state b) (any macrostate C) can be reached rom state a) (Macrostate A or E), given that all particles are identical. his is a purely statistical count. Macrostate C Macrostate B Macrostate D Macrostate A Macrostate E ( n ) microstates orm 5 macrostates We are distributing N particles into two possible chambers. here are N ways (microstates, micro-paths) to do that. It is thereore N times more probable to have the 4 molecules distributed equally ( in each) in both containers (macro-state) than having all molecules in one container. he entropy o this state is higher by N (.) S = kbln = kbn ln than the state in which all molecules are in one single box. In general, this becomes: (.) S = k B ln(number o micro-states)

12 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 o distribute n moles o a gas in two chambers we have nn A microstates. he entropy change is thereore: (.4) S = k ln( nn A B ) = nn AkBln = nr ln his is the entropy change or a ree, adiabatic expansion o n moles o an ideal gas into equally sized compartments. his law can also be expressed in terms o the probability to encounter a particular macrostate. In our example we have 6 equivalent microstates which correspond to the 5 macro-states o having particles in each chamber. o encounter any particular one o these 6 microstates in one single obervation has a probability o -4 =/6=/P. I we have 0 particles to distribute we have 0 =04 0 microstates. For a hundred particles we have =04 0 he probability to ind a single particular state occupied in the case o N particles is equal to -N. he entropy is deined in terms o the probability to encounter a microstate /P through: S = kbln P= kbln P = kbln(probability to ind a particular macrostate occupied) (.5) J [ S ] = K P is the probability to orm a macrostate through the many microstates. he inverse: /P is the probability to ind such a state occupied. All we need to remember in this context is that the act that events evolve in the direction o greater probability is the same as saying that the entropy increases..6 Macroscopic Entropy Deinition. For our purposes another deinition o entropy is more relevant: he entropy change in a reversible process is given by: (.6) ds = dq reversible As natural processes are irreversible, the actual entropy change is larger than the one given by equation(.6). We can calculate a lower limit or the entropy change by substituting irreversible processes in nature by reversible processes or which we can deine the thermodynamic state variables at every point o the path through quasistatic thought-processes. dqreversible (.7) ds ; S AB B A dq reversible

13 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 he entropy S(P,,) is, like the internal energy U, only dependent on the initial and inal state o a thermodynamic process, which again means that the entropy change or a complete cycle is equal to 0. Furthermore we can calculate the entropy change or an irreversible process by calculating the change o entropy or a reversible process as long as the initial and inal states o the irreversible and reversible processes are the same. Let us calculate the change o entropy or a gas expanding reely, and adiabatically into a vacuum, doubling its original volume. As we have seen earlier, this is an irreversible process with: (.8) Q= 0; = 0; W = 0; U = 0 We need to ind a series o reversible (quasi-static) processes with the same initial and inal states as in the ree adiabatic expansion. he volume doubles, the temperature remains the same. We can achieve the same expansion o a gas by allowing it to do work against a piston (during which work is done by the system, or negative work is done on the system) in an isothermal quasistatic, reversible process. As entropy change is path-independent, only the initial and inal values matter, not the path. dqr ( P,, ) (.9) ds = For an isothermal process the change in internal energy is 0 and we have nr nr (.40) dq = dw = Pd = d rom P = r dqr nr d (.4) S = S Si = = nr ln nr ln = = i i he actual entropy change would be greater than this minimal value or a reversible process. Note that this result is the same as the one we got through our microscopic approach in (.4)..7 Entropy Change or a Quasi Static Reversible Process o an Ideal Gas. Recall that quasi-static means that the system is in thermodynamic equilibrium at all times during the process. Any state change o an ideal gas involves the three state variables P,,. Let us here calculate the entropy change o a gas that expands rom an initial volume i to a inal volume while undergoing a temperature change rom i to. he three state variables are always connected by the ideal gas law. P=nR. For the entropy change ΔS we need to know the heat change Q which is implicitely deined by the irst law o thermodynamics ΔU=Q+W. For a reversible process we can write this in terms o ininitesimal changes: (.4) δq = du δw = du + Pd his leads to i

14 Dr. Fritz Wilhelm page 4 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 (.4) Q U W S = = he internal energy U o an ideal gas depends only on the initial and inal temperature and we have shown earlier that (.44) du = ncd. he change in work is always given by (No path speciied): d (.45) W = Pd dw = Pd = nr his is not necessarily an isothermal process, can change. Now we can perorm the integration, noting that in order to obtain the entropy change we must divide the change in heat by the temperature, δqr S = ; δqr = du δw = ncd + Pd i nr δqr d nr (.46) P = ds = = nc + he ininitesimal change o the entropy is a total derivative, which is why we use ds and not δ S d (.47) δq d d = = + r ds nc nr I we integrate this rom (i) to () we see that we get our result without speciying a path: (In our calculation o the entropy change we do not need to speciy any path, just the initial and inal state variables.) dqr d d (.48) S = = nc nr nc ln nr ln + = + i i i We conirm that the entropy unction depends only on initial and inal states, and not on the path taken. As all macroscopic processes o heat exchange are irreversible we need o ind equivalent processes which are quasi-static and/or reversible. his oten boils down to replacing macroscopic temperature, pressure, or volume changes by microscopic ones, i.e. we use integration over ininitesimal (reversible) steps..7a Entropy Changes During Phase Changes: he entropy change o the melting substance in a melting process is clearly equal to the latent heat ml divided by the temperature at which the melting occurs: ml (.49) Smelting = melting i i

15 Dr. Fritz Wilhelm page 5 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 he heat necessary comes rom the environment which is at a higher temperature, the net entropy change is positive..7b he entropy change during a thermal conduction is equal to the heat loss at h minus the heat-gain at the lower temperature c. As the amount o heat Q lost at the higher temperature h is equal to the amount o heat gained at the lower temperature, and as the temperature appears in the denominator, the entropy change is always positive: Qh Q c (.50) S = + = Q > 0 h c c h Problem: Calculate the entropy change ocurring during the heat low through a steel bar o length 0cm and cross section.0cm, one end o which is at a temperature o the steam point o water, the other at the reezing point o water: k steel =79.5W/mK. 4 ka (.5) Q = = 00 W = 0.6 W L 0. he entropy transer per second is thereore: 4 W (.5) 0.6W = K 7K K.7c Entropy Change in Calorimetric Processes: When two dierent substances at dierent temperatures are in close contact with each other, heat will low rom the hotter substance to the cooler substance until a inal equilibrium temperature is reached. his is clearly an irreversible process. However, we can ind an equivalent process in which or example the cooler substance is put into contact with a heat reservoir at an ininitesimally higher temperature. A small amount o heat will low rom the heat-reservoir to the cooler substance in an approximately reversible process. his will be repeated until the cooler substance has reached the inal temperature. he same process is used to cool down the hotter substance by putting it in contact with successive heat-reservoirs o ininitesimally decreasing temperatures. First substance at lower temperature: m; c; (.5) Second substance at higher temperature: m; c; Final, equilibrium temperature: mc ( ) = mc ( ) heat gained+q> 0 heat lost Q< 0 (.54) mc mc = mc mc mc + m c = m c + mc (.55) m c + mc = mc + mc

16 Dr. Fritz Wilhelm page 6 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 Now we use the integral orm o the entropy ormula to calculate the entropy change or both substances: he irst substance gains heat at the lower temperature, thereore Q >0 is positive. he second substance gives up heat at the higher temperature, thereore Q <0 is negative. dq dq mc mc mc mc (.56) S = + d d d d = + = (.57) S= mc ln mc ln he entropy is obviously always positive in such heat exchanges. Example: 50 g o aluminum (c=0.5cal/gcº) at 00ºC are immersed into 00g o water at 0ºCelsius. Find the entropy change or the combination o aluminum plus water. We need to equate the heat loss o aluminum with the heat gain o water: (.58) malcal ( 00 ) = mc w w ( 0) =7.8º (.59) dq = mcd he entropy change or aluminum and requires an integration: 00.8 dqr malcal 00.8 cal cal (.60) S Al = = d malcal ln = = = 7 C C i 7 he same or water: 00.8 d 00.8 cal (.6) SW= mc w w = 00 ln =.6 9 C 9 he total entropy change is 0. cal/k. Problem: 00g o ice at 0 C are immersed into 00g o water at 60 C. Calculate the entropy change o the mixture ater thermodynamic equilibrium has been established. Qice = ml = 00 80cal = 8000cal cal mw cw = 00g = 8000cal = 40C his heat is extracted rom the hot water: gc = 0 C he temperature o the water drops rom 60 to 40 C due to the melting o the ice. Now we need to calculate the inal temperature o the water mixture: 00 0 = 00 0 ( ) ( ) =. C We could also have calculated this right rom the beginning: w,

17 Dr. Fritz Wilhelm page 7 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 (.6) ( 0) ( 60 ) m L + m c = mc ice icewater w w cal cal cal 00 80cal+ 00g = 00g 60C 00g gc gc gc =. C For the entropy change we get: Qice d d S= + mc mc + = (.6) Optional: cal.00cal cal 86. = + 00g ln + 00g ln = 7K gk 7 gk cal cal = ( ) =.84 K K.8 hermodynamic Equilibrium: A system o physical process is said to be in equilibrium i it does not change in time. he equilibrium is stable i a change rom its equilibrium position causes the system to revert to its previous equilibrium point. It is unstable i any such change causes the system to move urther away rom its equilibrium point. It is indierent i a change does not inluence the equilibrium in either way. In mechanics, the equilibrium points are given by the extreme points o the potential energy. A minimum o the potential energy curve corresponds to a stable equilibrium point. he unction in thermodynamics which corresponds to the potential energy in mechanics is the entropy S, provided that we deal with an isolated system, which does not exchange energy or mass with the environment. In this situation, a system is in equilibrium i it is at the point o maximum entropy. Any change away rom this point will entail processes which tend to increase the entropy to the maximum. In a system that can exchange energy with the environment, but not mass, we need to also consider the entropy change o the environment. We have done this to some extent in a ew o the previous problems. In the last example (.6) we could consider ice as the system and the water as the environment. Such processes evolve by themselves only i (.64) S + S 0 system We saw that we needed to calculate the sum o entropy changes o the environment and the system. We are at the equilibrium point i (.65) Ssystem + Senv = 0 his point was reached in the previous example when the water is at a temperature o. C, where we treated ice and water as one system. As long as the temperature is above or below that temperature the system+environment keeps changing until it has reached the thermodynamic equilibrium. I, in addition to heat exchange ΔQ, work ΔW is involved in the process, we can write or the system plus environment: env

18 Dr. Fritz Wilhelm page 8 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 Q U W (.66) S = = or U W S = 0 I work can only be done by pressure on a piston, we have ΔW=-PΔ, and we get: (.67) U + P S = 0 I we keep the volume constant, like in a chemical process inside a closed lask, we get: (.68) U S = 0 I, in addition, we maintain a constant temperature,equation (.67) becomes: (.69) ( U S ) = 0; and are constant Free energy F or Helmholtz potential Free Energy F (Helmholtz potential): (.70) Under isothermal and iso volumetric conditions the thermodynamic equilibrium is given by the minimum Free Energy F : F = U S; F = 0 with and constant. Note that the term ree is used when the product S is subtracted. For constant pressure and constant temperature, equation (.67) is equivalent to (.7) ( U + P S ) = 0 temperature and pressure are constant Free Enthalpie or Gibbs potential Free Enthalpie G (Gibbs potential): (.7) Under isothermal and isobaric conditions (, P constant) we obtain thermodynamic equilibrium i the ree Enthalpie G is at a minimum, G = 0: G=U+P-S his means or example, that a chemical reaction, with and P, constant will spantaneously occur i its ree enthalpie will decrease in the process. Processes like these occur in organisms all the time. In some processes heat exchange is not possible, but work is. his means that ΔS=ΔQ=0 in (.7). We get the unction called Enthalpie H: Δ(U+P)=0.

19 Dr. Fritz Wilhelm page 9 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 (.7) Under adiabatic-isobaric conditions ( S = 0, P constant) we obtain thermodynamic equilibrium i the Enthalpie H is at a minimum, H = 0: H=U+P I volume is maintained constant, we see that enthalpie and internal energy are the same. We achieve equilibrium i U = 0, the internal energy is at a minimum. In chemical reactions in which the number o moles o the products are variables, we need to include those in our ormulas. his is let to chemistry. Optional:.9 Calculating the critical temperature o a real gas: An interesting application o the enthalpie unction occurs in the understanding o gas liquiication. An ideal gas would never liquiy! A real gas however can be liquiied. In a ree expansion o a real gas the temperature does change. Its equation o state is given by the van der Waals equation mentioned in chapter 9.: a na P + ( mol b) = R, P + ( nb) = nr mol a is an internal pressure deriving rom the attractive orce o mol the molecules to each-other: F=PA (pressure times area). he potential energy Uattr involved in this attraction can be guessed at by: (.74) -a U attr = -Fx = -PAx = -P = b is the volume occupied by at single inite mole o particles. he van der Waals equation can also be written as: (.75) P R a = b Let us study this in an experiment illustrating the Joule-homson eect: he gas in the let chamber is under the constant pressure P and gets slowly pushed into the volume to the right which is under the constant pressure P. he porous wall prevents any turbulence. he whole volume o gas to the let gets pressed into the right chamber. P P We assume that the pressure in the chamber to the right is smaller than the pressure in the chamber to the let. he net work done on the gas is porous wall thereore: P P which is an increase in the internal energy o the gas. In real experiments we observe an increase in the

20 Dr. Fritz Wilhelm page 0 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 temperature or CO and air, and a decrease or H. In order to conveniently describe the process we use the enthalpie unction which contains the additional P term, H = U + P. he enthalpie remains constant because P P = U U (.76) P + U = P + U H H Let us calculate the enthalpie or a van der Waals gas. he internal energy or one mole o such a gas depends on the number o degrees o reedom : (.77) U = R he internal pressure results in a negative work term (work is positive i it is exerted on the system!) a a = which must be added to the internal energy. he additional P term in the a enthalpie also contains the. Again, we must use the negative term because it corresponds to inside pressure. hus we get the total enthalpie unction a R a R a H = R + ( ) ; with P = b b P (.78) modiied internal energy U a R a H = R + b he enthalpie unction is a total dierential, which is now just a unction o and. We simpliy the unction actoring out R: a (.79) H = R + b he total change in this unction must be 0, i H is constant. (.80) H H dh = d + d = 0 a H = R + b H (.8) = R + ; b H a b a Rb a = R + = R + = + b ( b) ( b) ( b) From (.80) we solve or d: H Rb a b a + ( ) ( ) b b R (.8) d = d = d = d H R + + b b

21 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 We approximate -b by to get: b a ( br a) (.8) R R br a d d = = d + + R + During an expansion o the gas, d is positive; the denominator is always positive. he numerator can be positive or negative according to the value or. he equation (.84) br a = 0 deines the inversion temperature i : a (.85) i = Rb he critical temperature c o a real gas is the temperature above which the gas cannot be liquiied, no matter how large the pressure. hat critical temperature is: CO : 04.K O :54.4K (.86) N :6.K H :.K He : 5.K he critical temperature or a real gas is given by: 8a (.87) c = 7Rb It corresponds to the inlection point in the isotherm o a P diagram or a real gas. Below are three isotherms with increasing temperatures or the isotherms rom let to

22 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 right. P (.88) P P = 0 = =constant =constant (.89) R a P = b P R a = + = 0 ( ) ( b) P R 6a = = 0 4 ( ) ( b) We obtain the critical temperature by dividing the last two equations into each other: R a α) = b (.90) β ) ( ) R 6a β = 4 α ( b) = b= = b b Substituting c =b into α gives us the desired relationship:

23 Dr. Fritz Wilhelm page o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 (.9) = b = critical volume R a R a a 4b 8a = = c = = 7b 4b 7b 7b R 7Rb ( b b) he inversion temperature and the critical temperature are thereore related: (.9) = 6.75 i Furthermore, inserting critical volume b and critical temperature into P R a = b we get the critical pressure below which no phase-change is possible: 8a R (.9) 7Rb a 8a a 4a a a Pc = = = = b b 9b 7b b 9b 7b 7b Example: For CO we measure a critical temperature o 04.K and a critical pressure o 7. bar or 7.E5 Pascals. 8a 7Rb (.94) c = a = c = b 04. = b 7Rb a a b m Pc = b = = b= 5 7b 7Pc mol (.95) 6 Pa m a = 0.7 ; mol c

24 Dr. Fritz Wilhelm page 4 o 4 C:\physics\0 lecture\ch Heat engines entropy nd law.docx; /8/00 See the ile in Excel or this calculation Isothermal lines or mole o carbon-dioxide. he dashed black line corresponds to the isothermal o C, above which carbon dioxide cannot be liquiied, no matter how high the pressure. he an der Waals P graph has an inlection point at C. Below the solid line passing through A, C, and E, gas and liquid coexist. When we move rom right to let on the a=0c line we reach the point o liquiication at A. he pressure increases with decreasing volume until the point B. hen the pressure decreases to the point D.It increases again beyond this point. R 8.4 a 0.64 b 4.7E-05 cm^ del.00e-06 0C a 7 0Celsius b 9 C c 04 40C d