DOCSLIB.ORG

Workshop Chem 201

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Workshop Chem 201 Workshop Chem 201 Thermodynamics: general concepts of energy, entropy, enthalpy and spontaneity of reactions Peer mentor: Gustavo Rodrigues de Moraes Sessions: Monday & Friday 2 – 4 PM [email protected] Discord: https://discord.com/invite/7C7Aew Zoom room: https://cccconfer.zoom.us/j/98202589993 STEM Center: tinyurl.com/Mesa-STEM-Center Heat – Chapter 6 • First Law of Thermodynamics: The energy of the universe is constant. ∆E = q + w where, ∆E = change in the system’s internal energy, Measuring heat! q = heat, w = work Enthalpy • What is it? Enthalpy, the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. Enthalpy is an energy-like property or state function. Enthalpy is defined by the letter H We will basically study reactions under constant pressure and constant volume H = E + PV E = internal energy of the system P = pressure of the system V = volume of the system Enthalpy • Thus, ∆H of the system is equal to the energy flow as heat… ∆H = Hproducts - Hreactants • When Hproducts > Hreactants, ΔH is positive: heat is absorbed by the system, and the reaction is endothermic. • When Hproducts < Hreactants, ΔH is negative: overall decrease in enthalpy is achieved by the generation of heat, and the reaction is exothermic ∆E = (−) Exothermic reaction ∆E = (+) Endothermic reaction Heat is increasing in the surroundings Heat is increasing in the system Increasing number of microstates Decreasing number of microstates Hess’s Law • Hess’s Law stated that enthalpy is the same whether the reaction takes place in one step or in a series of steps. ∆H1 = ∆H2 + ∆H3 Practice time 1. Calculate the enthalpy of the following chemical reaction: CS2(ℓ) + 3O2(g) ---> CO2(g) + 2SO2(g) GiVen: C(s) + O2(g) ---> CO2(g) ΔH = −393.5 kJ/mol S(s) + O2(g) ---> SO2(g) ΔH = −296.8 kJ/mol C(s) + 2S(s) ---> CS2(ℓ) ΔH = +87.9 kJ/mol 2. Calculate the enthalpy for this reaction giVen the following thermochemical equations: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ 5 C2H2(g) + ⁄2O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = −1299.5 kJ C(s) + O2(g) ---> CO2(g) ΔH° = −393.5 kJ 1 H2(g) + ⁄2O2(g) ---> H2O(ℓ) ΔH° = −285.8 kJ 3. If C2H5OH + 3 O2(g) à 2 CO2(g) + 3 H2O(l) ∆H = -1418 kJ/mol How much heat is released when 10.00g of ethanol is burned? Entropy (∆S) • What is it? Entropy is the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work (Britannica) . Entropy can be viewed as a measure of molecular randomness or disorder. • The Second Law of Thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe. Entropy ∆Suniv = ∆Ssys + ∆Ssurr ΔSuniv is positive: • Entropy of the universe increases • Process is spontaneous in the direction written ΔSuniv is negative: • Process is spontaneous in the opposite direction ΔSuniv is zero: • Process has no tendency to occur • System is at equilibrium Spontaneity • Processes are spontaneous when they result in an increase in disorder. • Entropy changes in the surroundings are determined primarily by the flow of energy into or out of the system as heat. • ∆Ssurr = (+) energy is being added up to the surroundings. Reaction is exothermic/ exergonic. Surroundings: greater disorder. • ∆Ssurr = (−) energy is being absorbed from the surroundings. Reaction is endothermic/ endergonic. Surroundings: more organized. H2O1(l) à H2O(g) 18 mL of liquid water becoming to 31 L of gaseous water • A system tends to undergo changes that lower its energy because when a system at constant temperature moves to a lower energy, the energy it gives up is transferred to the surroundings, leading to an increase in entropy there. • Temperature has a huge impact on spontaneity: • Low temperatures: energy transfer plays an important role • High temperature: energy transfer is not so important ∆� ∆� = − ���� � 4. Arrange the compounds in order of increasing entropy (∆S) and justify your order. a) CH3OH(s), b) CH3OH(l), c) CH3OH(g), d) CH3CH2OH(g) 5. Predict the sign of ∆S∘ for each of the following reactions. a. The thermal decomposition of solid calcium carbonate: CaCO3(s) à CaO(s) + CO2(g) b. The oxidation of SO2 in air: 2SO2(g) + O2(g) à 2SO3(g) Free Energy (∆G) ∆G = ∆H − (T × ∆S) ∆G = variation in free energy ∆H = variation in enthalpy T = temperature in Kelvins ∆S = variation in entropy ∆! ∆S =− univ " Reactions are spontaneous when ∆G is negative. Since energy is being given off. Gibbs equation ∆" #( ) • ��� = � #$ Keq > 1 Keq ≈1 Keq < 1 Product favored Equilibrium Reactant favored ∆G∘ = (−) ∆G∘ = 0 ∆G∘ = (+) Free energy is released Reversible Free energy is required Exergonic Endergonic Ex: batteries, combustion Ex: extremely weak acids, extremely insoluble salts • ∆� = � × � × ��� Thermo scenarios Practice time o 6. Using the thermodynamic tables, calculate ΔH rxn for each chemical reaction. a. 2Mg(s) + O2(g) → 2MgO(s) b. CaCO3(s) → CaO(s) + CO2(g) c. AgNO3(s) + NaCl(s) → AgCl(s) + NaNO3(s) 7. At what temperature would the following reaction occur spontaneously? 2Cr2O3→ 4Cr + 3O2 (∆H°= 1256.4 kJ; ∆S°= 587 J/K) 8. Given the dissolving of CaCl2(s) 2+ − CaCl2(s)→Ca (aq)+2Cl (aq) ΔG° = CaCl2(s) = -748.1 kJ/mol, ΔH = CaCl2= -795.8 kJ ∆� − ∆�° o � = Δs = � • Ca2+= -53.1 J/Kmol, • CaCl2= 104.6 J/Kmol, • Cl- = 56.5 J/Kmol Calculate the temperature of this reaction. 9. Calculate ΔGo using only the following ΔG values for the reaction: CS2(l)+2O2(g)→CO2(g)+2SO2(g) o C(s)+O2(g)→CO2(g) with ∆G =−394.39kJ/mol o S(s)+O2(g)→SO2(g) with ∆G = −300.13kJ/mol o C(s)+2S(s)→CS2(l) with ∆G = 67.1kJ/mol o o 10. Assess the feasibility of the reaction below by evaluating (∆S rxn, ∆H rxn o and ∆G rxn) for this reaction at 25°C. If the reaction is spontaneous, is it enthalpically favored, entropically favored or both? (NH)4Cl(s)→NH3(g)+HCl(g) 11. Determine the Kp at 298.15 K for the reaction: 2CO(g) + O2(g) ⇌ 2CO2(g) 12. ∆G° for the reaction below is 31.4 kJ/mol at 298.15 K. Use the thermodynamics quantity to decide in which direction the reaction is spontaneous when the concentrations of − + H2SO4(aq), HSO 4(aq), and H3O (aq) are 0.2 M, 0.1 M, and 0.1 M respectively. + − H2SO4(aq) + H2O(l) ⇌ H3O (aq) + HSO 4(aq) Useful resources: • Thermo videos: https://ch301.cm.utexas.edu/thermo/index.php Exercises: https://www2.southeastern.edu/Academics/Faculty/wparkinson/help/ enthalpy/test.html Copyright • https://www.britannica.com/science/enthalpy • https://www.britannica.com/science/entropy-physics • https://www.chemteam.info/Thermochem/HessLawIntro1a.html • https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Ch emistry_- _The_Central_Science_(Brown_et_al.)/05._Thermochemistry/5.E%3A_The rmochemistry_(Exercises) • https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1202 /Unit_7%3A_Principles_of_Thermodynamics/7.E%3A_Exercises_on_Entrop y_and_Gibbs_Energy • Zumdahl, Steven S.. Chemistry (p. 225). Cengage Learning. Kindle Edition. .
Load more

Recommended publications
  • Equilibrium Thermodynamics
    Equilibrium Thermodynamics Instructor: - Clare Yu (e-mail [email protected], phone: 824-6216) - office hours: Wed from 2:30 pm to 3:30 pm in Rowland Hall 210E Textbooks: - primary: Herbert Callen “Thermodynamics and an Introduction to Thermostatistics” - secondary: Frederick Reif “Statistical and Thermal Physics” - secondary: Kittel and Kroemer “Thermal Physics” - secondary: Enrico Fermi “Thermodynamics” Grading: - weekly homework: 25% - discussion problems: 10% - midterm exam: 30% - final exam: 35% Equilibrium Thermodynamics Material Covered: Equilibrium thermodynamics, phase transitions, critical phenomena (~ 10 first chapters of Callen’s textbook) Homework: - Homework assignments posted on course website Exams: - One midterm, 80 minutes, Tuesday, May 8 - Final, 2 hours, Tuesday, June 12, 10:30 am - 12:20 pm - All exams are in this room 210M RH Course website is at http://eiffel.ps.uci.edu/cyu/p115B/class.html The Subject of Thermodynamics Thermodynamics describes average properties of macroscopic matter in equilibrium. - Macroscopic matter: large objects that consist of many atoms and molecules. - Average properties: properties (such as volume, pressure, temperature etc) that do not depend on the detailed positions and velocities of atoms and molecules of macroscopic matter. Such quantities are called thermodynamic coordinates, variables or parameters. - Equilibrium: state of a macroscopic system in which all average properties do not change with time. (System is not driven by external driving force.) Why Study Thermodynamics ? - Thermodynamics predicts that the average macroscopic properties of a system in equilibrium are not independent from each other. Therefore, if we measure a subset of these properties, we can calculate the rest of them using thermodynamic relations. - Thermodynamics not only gives the exact description of the state of equilibrium but also provides an approximate description (to a very high degree of precision!) of relatively slow processes.
    [Show full text]
  • Statistical Thermodynamics - Fall 2009
    1 Statistical Thermodynamics - Fall 2009 Professor Dmitry Garanin Thermodynamics September 9, 2012 I. PREFACE The course of Statistical Thermodynamics consist of two parts: Thermodynamics and Statistical Physics. These both branches of physics deal with systems of a large number of particles (atoms, molecules, etc.) at equilibrium. 3 19 One cm of an ideal gas under normal conditions contains NL =2.69 10 atoms, the so-called Loschmidt number. Although one may describe the motion of the atoms with the help of× Newton’s equations, direct solution of such a large number of differential equations is impossible. On the other hand, one does not need the too detailed information about the motion of the individual particles, the microscopic behavior of the system. One is rather interested in the macroscopic quantities, such as the pressure P . Pressure in gases is due to the bombardment of the walls of the container by the flying atoms of the contained gas. It does not exist if there are only a few gas molecules. Macroscopic quantities such as pressure arise only in systems of a large number of particles. Both thermodynamics and statistical physics study macroscopic quantities and relations between them. Some macroscopics quantities, such as temperature and entropy, are non-mechanical. Equilibruim, or thermodynamic equilibrium, is the state of the system that is achieved after some time after time-dependent forces acting on the system have been switched off. One can say that the system approaches the equilibrium, if undisturbed. Again, thermodynamic equilibrium arises solely in macroscopic systems. There is no thermodynamic equilibrium in a system of a few particles that are moving according to the Newton’s law.
    [Show full text]
  • Szilard-1929.Pdf
    In memory of Leo Szilard, who passed away on May 30,1964, we present an English translation of his classical paper ober die Enfropieuerminderung in einem thermodynamischen System bei Eingrifen intelligenter Wesen, which appeared in the Zeitschrift fur Physik, 1929,53,840-856. The publica- tion in this journal of this translation was approved by Dr. Szilard before he died, but he never saw the copy. At Mrs. Szilard’s request, Dr. Carl Eckart revised the translation. This is one of the earliest, if not the earliest paper, in which the relations of physical entropy to information (in the sense of modem mathematical theory of communication) were rigorously demonstrated and in which Max- well’s famous demon was successfully exorcised: a milestone in the integra- tion of physical and cognitive concepts. ON THE DECREASE OF ENTROPY IN A THERMODYNAMIC SYSTEM BY THE INTERVENTION OF INTELLIGENT BEINGS by Leo Szilard Translated by Anatol Rapoport and Mechthilde Knoller from the original article “Uber die Entropiever- minderung in einem thermodynamischen System bei Eingriffen intelligenter Wesen.” Zeitschrift fur Physik, 1929, 65, 840-866. 0*3 resulting quantity of entropy. We find that The objective of the investigation is to it is exactly as great as is necessary for full find the conditions which apparently allow compensation. The actual production of the construction of a perpetual-motion ma- entropy in connection with the measure- chine of the second kind, if one permits an ment, therefore, need not be greater than intelligent being to intervene in a thermo- Equation (1) requires. dynamic system. When such beings make F(J measurements, they make the system behave in a manner distinctly different from the way HERE is an objection, already historical, a mechanical system behaves when left to Tagainst the universal validity of the itself.
    [Show full text]
  • Thermodynamic Processes: the Limits of Possible
    Thermodynamic Processes: The Limits of Possible Thermodynamics put severe restrictions on what processes involving a change of the thermodynamic state of a system (U,V,N,…) are possible. In a quasi-static process system moves from one equilibrium state to another via a series of other equilibrium states . All quasi-static processes fall into two main classes: reversible and irreversible processes . Processes with decreasing total entropy of a thermodynamic system and its environment are prohibited by Postulate II Notes Graphic representation of a single thermodynamic system Phase space of extensive coordinates The fundamental relation S(1) =S(U (1) , X (1) ) of a thermodynamic system defines a hypersurface in the coordinate space S(1) S(1) U(1) U(1) X(1) X(1) S(1) – entropy of system 1 (1) (1) (1) (1) (1) U – energy of system 1 X = V , N 1 , …N m – coordinates of system 1 Notes Graphic representation of a composite thermodynamic system Phase space of extensive coordinates The fundamental relation of a composite thermodynamic system S = S (1) (U (1 ), X (1) ) + S (2) (U-U(1) ,X (2) ) (system 1 and system 2). defines a hyper-surface in the coordinate space of the composite system S(1+2) S(1+2) U (1,2) X = V, N 1, …N m – coordinates U of subsystems (1 and 2) X(1,2) (1,2) S – entropy of a composite system X U – energy of a composite system Notes Irreversible and reversible processes If we change constraints on some of the composite system coordinates, (e.g.
    [Show full text]
  • INTRODUCTION to Thermodynamic Properties Introduction to Thermodynamic Properties
    INTRODUCTION TO Thermodynamic Properties Introduction to Thermodynamic Properties © 1998–2018 COMSOL Protected by patents listed on www.comsol.com/patents, and U.S. Patents 7,519,518; 7,596,474; 7,623,991; 8,219,373; 8,457,932; 8,954,302; 9,098,106; 9,146,652; 9,323,503; 9,372,673; and 9,454,625. Patents pending. This Documentation and the Programs described herein are furnished under the COMSOL Software License Agreement (www.comsol.com/comsol-license-agreement) and may be used or copied only under the terms of the license agreement. COMSOL, the COMSOL logo, COMSOL Multiphysics, COMSOL Desktop, COMSOL Server, and LiveLink are either registered trademarks or trademarks of COMSOL AB. All other trademarks are the property of their respective owners, and COMSOL AB and its subsidiaries and products are not affiliated with, endorsed by, sponsored by, or supported by those trademark owners. For a list of such trademark owners, see www.comsol.com/trademarks. Version: COMSOL 5.4 Contact Information Visit the Contact COMSOL page at www.comsol.com/contact to submit general inquiries, contact Technical Support, or search for an address and phone number. You can also visit the Worldwide Sales Offices page at www.comsol.com/contact/offices for address and contact information. If you need to contact Support, an online request form is located at the COMSOL Access page at www.comsol.com/support/case. Other useful links include: • Support Center: www.comsol.com/support • Product Download: www.comsol.com/product-download • Product Updates: www.comsol.com/support/updates •COMSOL Blog: www.comsol.com/blogs • Discussion Forum: www.comsol.com/community •Events: www.comsol.com/events • COMSOL Video Gallery: www.comsol.com/video • Support Knowledge Base: www.comsol.com/support/knowledgebase Part number: CM021603 Contents The Thermodynamic Properties Data Base .
    [Show full text]
  • Thermodynamics Cycle Analysis and Numerical Modeling of Thermoelastic Cooling Systems
    international journal of refrigeration 56 (2015) 65e80 Available online at www.sciencedirect.com ScienceDirect www.iifiir.org journal homepage: www.elsevier.com/locate/ijrefrig Thermodynamics cycle analysis and numerical modeling of thermoelastic cooling systems Suxin Qian, Jiazhen Ling, Yunho Hwang*, Reinhard Radermacher, Ichiro Takeuchi Center for Environmental Energy Engineering, Department of Mechanical Engineering, University of Maryland, 4164 Glenn L. Martin Hall Bldg., College Park, MD 20742, USA article info abstract Article history: To avoid global warming potential gases emission from vapor compression air- Received 3 October 2014 conditioners and water chillers, alternative cooling technologies have recently garnered Received in revised form more and more attentions. Thermoelastic cooling is among one of the alternative candi- 3 March 2015 dates, and have demonstrated promising performance improvement potential on the Accepted 2 April 2015 material level. However, a thermoelastic cooling system integrated with heat transfer fluid Available online 14 April 2015 loops have not been studied yet. This paper intends to bridge such a gap by introducing the single-stage cycle design options at the beginning. An analytical coefficient of performance Keywords: (COP) equation was then derived for one of the options using reverse Brayton cycle design. Shape memory alloy The equation provides physical insights on how the system performance behaves under Elastocaloric different conditions. The performance of the same thermoelastic cooling cycle using NiTi Efficiency alloy was then evaluated based on a dynamic model developed in this study. It was found Nitinol that the system COP was 1.7 for a baseline case considering both driving motor and Solid-state cooling parasitic pump power consumptions, while COP ranged from 5.2 to 7.7 when estimated with future improvements.
    [Show full text]
  • 12/8 and 12/10/2010
    PY105 C1 1. Help for Final exam has been posted on WebAssign. 2. The Final exam will be on Wednesday December 15th from 6-8 pm. First Law of Thermodynamics 3. You will take the exam in multiple rooms, divided as follows: SCI 107: Abbasi to Fasullo, as well as Khajah PHO 203: Flynn to Okuda, except for Khajah SCI B58: Ordonez to Zhang 1 2 Heat and Work done by a Gas Thermodynamics Initial: Consider a cylinder of ideal Thermodynamics is the study of systems involving gas at room temperature. Suppose the piston on top of energy in the form of heat and work. the cylinder is free to move vertically without friction. When the cylinder is placed in a container of hot water, heat Equilibrium: is transferred into the cylinder. Where does the heat energy go? Why does the volume increase? 3 4 The First Law of Thermodynamics The First Law of Thermodynamics The First Law is often written as: Some of the heat energy goes into raising the temperature of the gas (which is equivalent to raising the internal energy of the gas). The rest of it does work by raising the piston. ΔEQWint =− Conservation of energy leads to: QEW=Δ + int (the first law of thermodynamics) This form of the First Law says that the change in internal energy of a system is Q is the heat added to a system (or removed if it is negative) equal to the heat supplied to the system Eint is the internal energy of the system (the energy minus the work done by the system (usually associated with the motion of the atoms and/or molecules).
    [Show full text]
  • 16. the First Law of Thermodynamics
    16. The First Law of Thermodynamics Introduction and Summary The First Law of Thermodynamics is basically a statement of Conservation of Energy. The total energy of a thermodynamic system is called the Internal Energy U. A system can have its Internal Energy changed (DU) in two major ways: (1) Heat Q can flow into the system from the surroundings and (2) the system can do work W on the surroundings. The concept of Heat Energy Q has been discussed previously in connection with Specific Heat c and the Latent Heats of fusion and vaporization at phase transitions. Here, we will focus on the Heat Q absorbed or given off by an ideal gas. Also, the Work W done by an ideal gas will be calculated. Both heat and work are energy, and it is important to understand the difference between the heat Q and the work W. Thermodynamic Work W Previously, we understood the concept of Mechanical Work in connection with Newton's Laws of Motion. There the focus was work done ON THE SYSTEM. For example, the work done on a mass M in raising it up a height h against gravity is W=Mgh. Thermodynamic Work is similar to Mechanical Work BUT in Thermodynamic Work the focus is on the work done BY THE SYSTEM. Historically, Thermodynamics was developed in order to understand the work done BY heat engines like the steam engine. So when a system (like a heat engine) does work ON its surroundings, then the Thermodynamic Work is POSITIVE. (This is the opposite for Mechanical Work.) If work is done on the heat engine, the Thermodynamic Work is NEGATIVE.
    [Show full text]
  • Physics 3700 Carnot Cycle. Heat Engines. Refrigerators. Relevant Sections in Text
    Carnot Cycle. Heat Engines. Refrigerators. Physics 3700 Carnot Cycle. Heat Engines. Refrigerators. Relevant sections in text: x4.1, 4.2 Heat Engines A heat engine is any device which, through a cyclic process, absorbs energy via heat from some external source and converts (some of) this energy to work. Many real world \engines" (e.g., automobile engines, steam engines) can be modeled as heat engines. There are some fundamental thermodynamic features and limitations of heat engines which arise irrespective of the details of how the engines work. Here I will briefly examine some of these since they are useful, instructive, and even interesting. The principal piece of information which thermodynamics gives us about this simple | but rather general | model of an engine is a limitation on the possible efficiency of the engine. The efficiency of a heat engine is defined as the ratio of work performed W to the energy absorbed as heat Qh: W e = : Qh Let me comment on notation and conventions. First of all, note that I use a slightly different symbol for work here. Following standard conventions (including the text's) for discussions of engines (and refrigerators), we deal with the work done by the engine on its environment. So, in the context of heat engines if W > 0 then energy is leaving the thermodynamic system in the form of work. Secondly, the subscript \h" on Qh does not signify \heat", but \hot". The idea here is that in order for energy to flow from the environment into the system the source of that energy must be at a higher temperature than the engine.
    [Show full text]
  • Beta Type Stirling Engine. Schmidt and Finite Physical Dimensions Thermodynamics Methods Faced to Experiments
    entropy Article Beta Type Stirling Engine. Schmidt and Finite Physical Dimensions Thermodynamics Methods Faced to Experiments Cătălina Dobre 1 , Lavinia Grosu 2, Monica Costea 1 and Mihaela Constantin 1,* 1 Department of Engineering Thermodynamics, Engines, Thermal and Refrigeration Equipments, University Politehnica of Bucharest, Splaiul Independent, ei 313, 060042 Bucharest, Romania; [email protected] (C.D.); [email protected] (M.C.) 2 Laboratory of Energy, Mechanics and Electromagnetic, Paris West Nanterre La Défense University, 50, Rue de Sèvres, 92410 Ville d’Avray, France; [email protected] * Correspondence: [email protected] Received: 10 September 2020; Accepted: 9 November 2020; Published: 11 November 2020 Abstract: The paper presents experimental tests and theoretical studies of a Stirling engine cycle applied to a β-type machine. The finite physical dimension thermodynamics (FPDT) method and 0D modeling by the imperfectly regenerated Schmidt model are used to develop analytical models for the Stirling engine cycle. The purpose of this study is to show that two simple models that take into account only the irreversibility due to temperature difference in the heat exchangers and imperfect regeneration are able to indicate engine behavior. The share of energy loss for each is determined using these two models as well as the experimental results of a particular engine. The energies exchanged by the working gas are expressed according to the practical parameters, which are necessary for the engineer during the entire project, namely the maximum pressure, the maximum volume, the compression ratio, the temperature of the heat sources, etc. The numerical model allows for evaluation of the energy processes according to the angle of the crankshaft (kinematic–thermodynamic coupling).
    [Show full text]
  • Carnot Cycle
    Summary of heat engines so far - A thermodynamic system in a process Reversible Heat Source connecting state A to state B ∆Q Th RHS T-dynamic - In this process, the system can do work and system emit/absorb heat ∆WRWS - What processes maximize work done by the system? Rev Work Source - We have proven that reversible processes (processes that do not change the total entropy of the system and its environment) maximize Process A →B work done by the system in the process A→B of the system Notes Cyclic Engine -To continuously generate work , a heat engine should go through a Heat source Heat sink cyclic process . ∆Q ∆Q Th h c Tc - Such an engine needs to have a heat source and a heat sink at different temperatures. ∆WRWS Working body - The net result of each cycle is to e. g. rubber band extract heat from the heat source and distribute it between the heat Rev Work Source transferred to the sink and useful work. Notes Carnot cycle As an example, we will consider a particular cycle consisting of two isotherms and two adiabats called the Carnot cycle applied to a rubber band engine. As we will see, Carnot cycle is reversible Here T and S are temperature and entropy of the working body Heat source Heat sink T AB ∆ ∆ T Th Qh Qc Tc h ∆WRWS Working body Tc e. g. rubber band D C Rev Work Source S SA SB Notes 1. Carnot Cycle: Isothermal Contraction (A) (B) T ∆ = 0 h U Th ∆Qsource LB LA Heat source Heat source T h Th T A B Force= Γ Th 2 2 D C cBT (L − L0 ) (L − L0 ) A→B h A B Tc ∆WRWS = ∆Qsource = Th∆S = − S 2 S S L0 L0 A B Notes 2.
    [Show full text]
  • ESCI 341 – Atmospheric Thermodynamics Lesson 11 – the Second Law of Thermodynamics
    ESCI 341 – Atmospheric Thermodynamics Lesson 11 – The Second Law of Thermodynamics References: Physical Chemistry (4th edition), Levine Thermodynamics and an Introduction to Thermostatistics, Callen THE SECOND LAW OF THERMODYNAMICS There are two common ways of stating the Second Law of Thermodynamics. Statement #1: “In an isolated system, entropy never decreases ( Sisol 0 ).” Statement #2: “It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.” Statement #2 is called the Kelvin-Planck statement, and asserts that you cannot convert a given amount of heat into an equal amount of work (there is no such thing as a 100% efficient engine.) Even though the two statement look very different, they are actually equivalent! PROOF THAT STATEMENT #1 AND STATEMENT #2 ARE EQUIVALENT Imagine the cyclic process shown on the thermodynamic diagram below, where 휷 and η are arbitrary thermodynamic variables. Leg 1: A reversible, isothermal process from State 1 to State 2. The change in entropy of this process is SQT1 rev (1) Leg 2: A reversible, adiabatic process to State 3 (there is no change in entropy during this process). Leg 3: An irreversible, adiabatic process back to State 1. The entropy change during this leg we will call S3. Since entropy is a state variable, and since we end up at the same point as we started, the total entropy change of the cycle must be zero. Therefore, we have SSQT31 rev .
    [Show full text]