See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/305160401 THERMODYNAMICS Book · July 2015 CITATIONS READS 0 29,040 1 author: Francis Wanjala Nyongesa University of Nairobi 20 PUBLICATIONS 97 CITATIONS SEE PROFILE Some of the authors of this publication are also working on these related projects: Ceramic water filters View project Electrophoretic Deposition (EPD) of Materials and its applications in Dye Sensitized Solar Cells and in Photocatalytic purification of water View project All content following this page was uploaded by Francis Wanjala Nyongesa on 11 July 2016. The user has requested enhancement of the downloaded file. UNIVERSITY OF NAIROBI DISTANCE LEARNNG STUDY MODULE Nyongesa, F. W., Ph.D University of Nairobi Reviewer Awour, J. B., Ph.D University of Nairobi CODL, University of Nairobi SPH 302: Thermodynamics DISTANCE LEARNNG STUDY MODULE Dr. Nyongesa F. Wanjala, Ph.D., MSc., UoN. Department of Physics University of Nairobi Reviewer Dr. Awuor John Buers, Ph.D., MSc., UoN. Department of Physics University of Nairobi CODL, University of Nairobi, 2015 1 SPH 302: Thermodynamics Forward The study of thermodynamics was inaugurated by 19 -century engineers, who wanted to know the ultimate limitations the laws of physics impose on the operation of steam engines and other devices that convert heat to mechanical energy and vice versa. Today the scope of thermodynamics has immensely increased with innumerable applications in chemistry and engineering. th Thermodynamics is the study of relationships involving heat, mechanical work and other aspects of energy transfer that takes place in devices such as refrigerators, heat pumps, internal combustion engines etc. These relationships are governed by the four laws of thermodynamics which are now some of the most important fundamental laws in nature. This module attempts to provide a clear and modern view of the essential Principles of Thermodynamics and its applications, relevant to Science and Engineering. The module is organized in 6 lectures. Lecture 1 begins by defining basic thermodynamic concepts and the Zeroth law which introduces the concept of thermal equilibrium and temperature. Lecture 2 deals with the 1ST Law of thermodynamics which introduces the concept of internal energy. Lecture 3 deals with the 2ND Law of thermodynamics which gives the direction of natural thermodynamic processes and defines the thermal efficiency of devices that convert heat into work and vice versa. Lecture 4 deals with Entropy and the concept of reversibility. Lecture 5 introduces thermodynamic potentials and Maxwell/s equations. Lastly, Lecture 6 deals with phase changes and phase equilibrium in thermodynamic systems. Each lecture begins with clear statement of study objectives, pertinent definitions and simple qualitative explanations of principles together with illustrations. This is followed by worked examples drawn from everyday life to help clarify the principles and indicate possible applications. To develop your creativity and understanding of the ideas, we have included a number of problems at the end of each lecture. The questions are both Qualitative and Quantitative (Problem sets) with the former intended to develop creative thinking and to provoke discussions. Step-by-step solutions to selected problems are given in the Appendix. F. W. Nyongesa 2 SPH 302: Thermodynamics COURSE UNIT OUTLINE Introduction ……………………..…………........ Page 4 Lecture 1: Thermodynamic Concepts and the Zeroth Law 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Introduction ………………………………………………………… Lecture Objectives ………………………………………………………… Introduction to Thermodynamics ……………………………………. Thermodynamic Concepts ……………………………………. Behaviour of Gases ……………………………………. Zeroth Law of Thermodynamics ……………………………………. Temperature and Temperature scales ……………………………………. Problem set 1 ……………………………………………………….. 6 6 6 7 13 17 18 22 Lecture 2: First Law of Thermodynamics 2.1 2.2 2.3 2.4 2.5 2.6 Introduction ………………………………………………………. Lecture Objectives ………………………………………………………. First Law of Thermodynamics …………………………………………..… Application of 1ST law of Thermodynamic …………………………………… Heat capacities of Ideal gas …………………………………… The Adiabatic Process of ideal gas …………………………………… Problem set 2 ……………………….……………………………… 24 24 24 27 33 36 45 Lecture 3: Heat Engines and Second Law of Thermodynamics 3.1 3.2 3.3 3.4 3.5 3.5 Introduction ……………………………………………………… Lecture Objectives ……………………………………………………… Heat Engines and second law of thermodynamic ..…………….. Heat Pumps ……………………………………………………… The Carnot Engine ……………………………………………………… Internal Combustion engine ……………………………………………. Problem set 3 ……………………….…………………………….. ..........52 ………52 ………53 ………53 ………60 ………66 ………74 Lecture 4: Entropy and Second Law of Thermodynamics 4.1 4.2 4.3 4.4 4.4 4.5 Introduction …………………………………………………….. Lecture Objectives …………………………………………………….. Entropy: One way process ………………………………… Entropy and Second Law of Thermodynamics ……………………….. Entropy and Performance of Heat Engines ……………………….. Third Law of Thermodynamics …………………………………. Problem set 4 ……………………….…………………………….. ………79 ………79 ………79 ………82 ………85 ………88 ………92 Lecture 5: Thermodynamic potentials and Maxwell’s Equations 5.1 5.2 5.3 5.4 5.5 Introduction …………………………………………………….. Lecture Objectives …………………………………………………….. Maxwell’s Relations …………………………………………………….. Maxwell’s Relations using Cyclic rule ………………………………… Applications of Maxwell’s Relations ………………………………… Problem set 5 ……………………….………………………….... ………89 ………90 ………95 …….100 …….102 …….104 3 SPH 302: Thermodynamics Lecture 6: Phase Changes and phase Equilibrium 6.1 6.2 6.3 6.4 6.5 6.6 Introduction ……………………………………………………. Lecture Objectives ……………………………………………………. Systems, Phases and Components ………………………………… Phase Transitions …………………………………………... First order phase changes …………………………………………... Second order phase changes …………………………………………... Problem set 6 ………………………………………………………………. Appendix A: Solutions to Problem sets Appendix C: Useful Constants ……..105 ……..106 ……..120 ……..124 ……..126 ……..130 ……..133 …………………………….…….. …...134 …………………………………… ……138 4 SPH 302: Thermodynamics INTRODUCTION TO THE COURSE This course unit introduces you to Thermodynamics which is a basic science that underlines the principles of all energy conversion devices and in particular, the transformation of Thermal Energy into Mechanical Work and vice versa in materials and machines. Thermodynamics has innumerable applications not only in physics but also in chemistry, biology and engineering and as such, it is an important course in all fields of science. Aims of the course At the end of this course unit, you should be able to explain and apply the laws of thermodynamics to solve problems related to energy conversion processes in mechanical systems. Prerequisites To follow this course unit, you will require knowledge of partial differential equations and Structure and Properties of Matter. Mode of delivery and Study Skills required This course is conducted as a mix of several teaching methods; Distance and e-learning with limited face-to-face. This study module is also published on the web. You are advised to set aside at least 5 hours for each lecture. At the end of every the lecture, you should attempt all the problem sets. You will be called upon to submit solutions to selected questions from the problem sets for assessment. Mode of Assessment Test and Assignments - 30% Written Examinations - 70% Reference materials 1. 2. 3. 4. Finn C. B. J., Thermal Physics, 1986. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics., S. Chand & Company Ltd, New Delhi. F.O. Akuffo, A. Brew-Hammond, F. Makau Luti and J.G.M. Massaquoi. ANST, UNESCO (1997), Nairobi, Kenya. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons, Inc., NY., Pg 479489, 517-561. 5 SPH 302: Thermodynamics Lecture 1 THERMODYNAMIC CONCEPTS AND ZEROTH LAW OF THERMODYNAMICS Lecture Outline 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 Introduction Lecture Objectives Introduction to Thermodynamic Thermodynamic Concepts Behaviour of Gases The Zeroth Law of Thermodynamics Temperature and Temperature Scales Summary A cup of hot coffee on a table at room temperature gets colder with time as it losses heat to the environment (room). Eventually, it attains thermal equilibrium with its environment i.e., when its temperature does not change over time. This is the concept of the Zeroth Law of Thermodynamics. 6 SPH 302: Thermodynamics 1.1 Introduction This lecture introduces you to the study of thermodynamics and what it entails. We shall start by familiarizing ourselves with terminologies that are often used when dealing with the subject of thermodynamics and it is therefore very important that these terminologies become part and parcel of your vocabulary. We shall also introduce the ideal gas laws and the equation of state. Thereafter, the Zeroth law of Thermodynamics is discussed which introduces the concept of temperature and thermometric properties. 1.2 Objectives At the end of this lecture, you should be able to: 1. Explain what thermodynamics entails 2. Define the basic concepts relating to a thermodynamic system. 3. Define thermodynamic equilibrium and distinguish between the various types of equilibrium 4. Explain an Ideal gas and the ideal gas Laws 5. Explain the Zeroth Law of Thermodynamics and its importance in temperature measurements 6. Explain the concept of temperature and the various temperature scales 1.3 Introduction to Thermodynamics Every time you drive a car, turn on an air conditioner, use a refrigerator or any electric appliance, you reap the practical benefits of thermodynamics. The term Thermodynamics is a combination of two words: Thermo associated with thermal/heat energy Dynamics motion or mechanical work Literally, Thermodynamics is a science which is based upon the general laws of nature which govern the conversion of heat into mechanical work and vice-versa. In terms of classification, Thermodynamics is classified as part of classical mechanics, which is a branch of physics where we study the behaviour of a system by considering only the large scale response (macroscopic properties) or the bulk properties (such as density, volume, pressure etc.) of the system that we can observe and measure in experiments. This is unlike modern physics which attempts to explain the behaviour of matter from a microscopic or atomic viewpoint (using quantum and statistical mechanics) by taking into account the atomic constitution of matter i.e., structure etc. The study of thermodynamics was inaugurated by 19 th -century engineers, who wanted to know the ultimate limitations the laws of physics impose on the operation of steam engines and other machines that generate mechanical energy 7 SPH 302: Thermodynamics or heat energy. Today the scope of thermodynamics has immensely increased with innumerable applications in chemistry and engineering. Definition Thermodynamics is the study of relationships involving heat, mechanical work and other aspects of energy transfer that takes place in thermodynamic systems such as refrigerators, heat pumps, internal combustion engines etc. These relationships are governed by the laws of thermodynamics which have become some of the most important fundamental laws not only in physics and other sciences but also nature in general. The four principles (referred to as "Laws of thermodynamics") are: Zeroth Law of Thermodynamics, which introduces the concept of temperature and thermodynamic equilibrium where two objects attain the same temperature when brought in thermal contact. First Law of Thermodynamics, which mandates conservation of energy, and represents the relationship between heat and mechanical work. Second Law of Thermodynamics, which depicts the manner in which energy changes take place by stating natural processes take place in a direction such as to increase the disorder (Entropy) of the system OR equivalently, the Entropy of an isolated macroscopic system never decreases. Third Law of Thermodynamics, which explains the nature of bodies in the neighbourhood of absolute zero temperature. The entropy of a perfect crystal tends to zero at absolute zero temperature, implying that it is impossible to cool a system all the way to absolute zero. 1.4 Thermodynamic Concepts To define terminologies related to thermodynamics, we shall consider a thermodynamic device such as a refrigerator or a gasoline engine where energy transfer takes place in the form of heat or mechanical work into or out of the device. In a refrigerator for example, we do work (to the refrigerator) by extracting heat from the refrigerator and expelling it to the outside (the room). The refrigerator in this case is our working system and the room is the surrounding. The fridge will only operate efficiently if the door is closed i.e., the system has to be isolated from the surrounding so that we can account for any form of energy input into or out of the system. By considering our refrigerator as a thermodynamic system let us now define basic thermodynamic terminologies/concepts. Thermodynamic System:– Any portion of the material universe which can be isolated completely and arbitrarily from the rest for consideration of the changes which may occur within it under varying conditions. Often a system 8 SPH 302: Thermodynamics may be considered as composed of smaller systems, which together make up the larger system. Take Note A system is like a proverbial rat. To nail it, you must bring it to the open i.e., isolate it from the its surrounding (hiding places). An example of a system is biological organism (Fig. 1-1) or a mechanical device (Fig. 1-2), or a specialized material e.g., steam expanding in a turbine. A thermodynamic system can interact (and exchange heat or matter) with its surroundings through the boundary wall. Heat and work are the two means of transferring energy into or out of a system. There are three classes of systems namely closed, open and isolated systems. SURROUNDING Heat Fig. 1-1 Human digestive system is an open system which exchanges food (matter) with the surrounding (body) Boundary Gas system Fig. 1-2. A gas in cylinder (closed) system can only exchange heat with the surrounding Open systems:- A system where the exchange of both heat and matter can occur through the boundary. Example is a human digestive system. Closed systems:- It is a system where only the exchange of heat between it and the surrounding is possible through the boundary. There is no matter exchange. Example is a refrigerator or a gas enclosed in a cylinder (Fig. 1.2). Throughout the course, we will be interested with the exchange of heat energy only, restricting ourselves to closed systems. 9 SPH 302: Thermodynamics Isolated System:- This is a system which is thermally insulated and has no communication of heat with the surrounding though work may be done on it. Such changes are called adiabatic changes. Example is a thermos flask. Boundary:– This is a “wall” that separates the system from the surrounding. The boundary may allow for exchange of heat and matter between the system and the surrounding depending on its nature. If the boundary inhibits the system from changing its volume or shape so that no mechanical work is performed on it, the wall is said to be rigid. The surroundings:- Refers to the rest of the universe outside the system. Diathermal wall:- A wall that allows for exchange of heat between the system and the surrounding. Good examples are metallic walls. Two systems separated by a diathermal wall are said to be in contact. In other words, two systems are in thermal contact if heating one of them results in macroscopic changes in the other. Adiabatic or Adiathermal wall:- A wall that does not allow for exchange of heat between the system and the surrounding. A good example is the walls of a vacuum flask. A system enclosed by an adiabatic wall is called an isolated system. Thermal equilibrium:- This implies the system is in a steady state condition i.e. the temperature is uniform throughout the system and remains constant in time such that there is no flow of heat through the system. Mechanical equilibrium:- This means that there are no unbalanced forces acting within the system. Chemical equilibrium:- This means that there are no chemical reactions occurring within the system. Thermodynamic (complete) equilibrium:- This represents a condition in which the system experiences thermal, mechanical and chemical equilibrium. In this case, the bulk physical properties or thermodynamic variables (Pressure, Volume and Temperature) of the system are uniform and do not change with time. Questions Think of and name examples of systems hat fall in the categories of equilibrium defined above. State variables:- These are directly measurable variables which are sufficient to describe the bulk behaviour of the system. In the case of a gas system, these include pressure (P), temperature (T), volume (V), internal energy (U), Entropy 10 SPH 302: Thermodynamics (S) and composition (). In a magnetic solid, these are the magnetic field (H), the magnetization (M) and temperature (T). For a homogeneous system, consisting of a single substance, the composition () is fixed. Usually, we require at least two state variables to specify the state of a system. For example, of the 3 measurable variables P, V and T, in a gas system, only two are independent and any one may be expressed in terms of the other two e.g., P is a function of V and T or P = P(V, T) V is a function of P and T or V = V(P, T) Hence, the state of a gas is equally well specified by quoting (P,V); (P,T) or (V,T). State variables are also known as thermodynamic variables or coordinates. Intensive & Extensive Variable State variables can be extensive or intensive. Extensive variables are macroscopic parameters that are proportional to the mass or size of the substance present in the system and as such they correspond to some measure of the system as a whole. Examples include internal energy (U), entropy (S), mass (m), volume (V), length, heat capacity (C) etc. as shown in Table 1 below. Table 1.1 Intensive and extensive variable in some systems System Gas or Fluid Film Cell Intensive variable Pressure, P Surface Tension, () E.m.f, () Extensive variable Volume, V Area , A Charge, Z An intensive variable is discrete (local) in nature i.e., it is independent of mass or size of the system but it is characteristic of the substance present in the system. Examples of such variables are pressure, temperature, viscosity, density, magnetic induction etc. State functions:- These are functions that are dependent only on the value of state variables P, V, and T at a given equilibrium state and not on the process taking place in the system. Examples include the internal energy (U), the enthalpy (H), and the entropy (S). On the other hand, work and heat are dependent on nature of the process (path) and not on the final value of the state variable and are thus not state functions. Heat & Work:Heat is the transfer of thermal energy between a system and its environment due to the temperature difference. Work is the transfer of mechanical energy Heat is equivalent to work in that both represent ways of transferring energy. Neither heat nor work is an intrinsic property of a system: that is, we cannot say that a system contains a certain amount of heat or work unlike properties 11 SPH 302: Thermodynamics such as pressure, temperature, and the internal energy. Thus, Heat and work are not properties of the state of the system; they are not state functions. Instead we say that a certain amount of energy can be transferred, either into or out of the system, as heat or as work. Both heat and work are thus associated with the thermodynamic process. Activity 1.1 1. Explain the difference between heat and thermal energy 2. Compare and contrast between heat and work 3. Explain why heat and work are not state functions Process:- It is a mechanism of bringing about a change from one equilibrium state of a system to another. A good example is when pushing a piston and compressing a gas in a cylinder from one equilibrium state (P1,V1) to a new equilibrium state (P2,V2). A process may be reversible or irreversible. Reversible process:- Reversible implies that in any such change, the system must be capable of being returned to its original state with the surrounding unchanged. This requires two conditions 1. 2. The processes must be quasi-static i.e., it should take place very slowly such that it is always in a succession of equilibrium states at any given time There must be no hysteresis i.e., no dissipative forces such as friction are present to dissipate energy. P (Mpa) a P2 , V2 Reversible Irreversible b P1 , V1 V m-3 Figure 1-3. Reversible and irreversible processes. Irreversible process cannot be plotted on an indicator diagram. Working substance:- This is a gas or fluid enclosed in the system that either receives energy transfer in the form of heat or work from the surrounding or transfers energy in the same form to the surrounding. 12 SPH 302: Thermodynamics Usually, the working substance should remain unchanged at the end of the process i.e., it should not introduce new parameters during the process. Because of this, we usually consider a working substance that is ideal or otherwise called an ideal gas and not a real gas. In the next section we look at the behaviour of ideal gases. Activity 1.2 1. Explain the difference between the following thermodynamic terms: (i) Open and closed system (ii) Diathermal and adiathermal wall (iii) State variable and state function (iv) Reversible and irreversible processes (v) Intensive and extensive variables 2. Explain what is thermodynamic equilibrium 1.5 Behaviour of Gases In a gas, the molecules are well separated and they fly about randomly in the container. We have no way of ascertaining the initial positions and velocities of each individual molecule. In lack of a microscopic description involving the individual positions and velocities of the molecules of the gas, we must be satisfied with a macroscopic description involving just a few measurable variables (such as mass of the gas, number of moles, volume, density, pressure and temperature of the gas) that characterize the average conditions in the volume of a gas. We will deal with the study of macroscopic properties of gases and the relationship between these properties with the average measurable variables of the gas system. In order not to introduce other macroscopic variables in our gas system that may arise due to interactions between the gas molecules, we shall consider an “ideal” situation or otherwise called an ideal gas. 1.5.1 What is an Ideal gas? An ideal gas is an abstraction and it is a gas whose properties represent the limiting behaviour of real gases at sufficiently low densities. In an ideal gas, it is assumed that (i) (ii) There are no intermolecular attractions. This implies that the internal energy of the gas is entirely kinetic and would be dependent on temperature The molecules themselves occupy negligible volume. 13 SPH 302: Thermodynamics The behaviour of an ideal gas is dependent on three factors namely its temperature, pressure and volume. These parameters obey some simple ideal gas laws namely: Boyle’s law At constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure i.e. P 1/V. This simply means applying pressure compresses a gas as shown in Fig 1-2 below. Thus: PV = constant ………….……………..(1.1) Boyle’s Law Or P1V1 = P2V2 [For any two states] P 0 V m 3C Figure 1-5. Boyle’s law on a P-V diagram Take Note Boyle’s law shows that the PV curve is a hyperbola. The locus of points ( P1V1 ), ( P2V2 ), ( P3V3 ),….., ( PnVn ) in thermal equilibrium is called an isotherm. Pressure law At constant volume, the pressure (P) of a fixed mass of a gas is absolute temperature (T) as shown in Fig. 1-6. Thus: P cons tan t T ………………………….………..(1.2) Pressure Law 14 SPH 302: Thermodynamics P oC K -273oC 0K Figure 1-6. Pressure law on a P-V diagram Take Note Under ideal situations, the pressure goes to zero at absolute zero of temperature (0 K or –273oC) Charles’ Law At constant pressure, the volume (V) of a fixed mass of a gas (i.e. fixed number of moles, n) is proportional to its absolute temperature (T) i.e. VT (Fig. 1-7). V cons tan t T …………………………….…….…(1.3) Charles Law V -273oC 0K oC K Figure 1-6. Charles Law on a P-V diagram Take Note Charles law implies that an ideal gas can be compressed to zero volume at absolute zero temperature (0 K or –273oC). This cannot happen with real gases 15 SPH 302: Thermodynamics Additionally, at low pressures, V Number of moles (n) of the gas i.e. V n ……………………………………………....(1.4) Combining equations (1.1) to (1.4) gives PV = nRT [Equation of state] …….………....(1.5) Equation of State Where R is the universal gas constant = 8.314 J mol-1K-1. A gas that obeys equation (1.15) at all temperatures and pressure is called an ideal gas. Eqn. (1.15) is usually referred to as the equation of state of an ideal gas. Activity 1.3 1. List the properties of an ideal gas 2. State the three gas laws 3. Starting from first principles, derive the equation of state Worked Examples 1. What is the pressure of 7 kg of nitrogen gas confined to a volume of 0.4 m3 at 20C ? Solution nRT 7 10 3 g M = 250 moles From PV = nRT P = where n = = 28 g/mol V m P = 15.2 105 Nm-2. 2. Early in the morning, the tires of an automobile are cold (280K) and their air is at a pressure of 3.0 atm. Later in the day, after a long trip, the tires are hot (330K). What is the pressure? Assume volume of the tires remain constant. Solution P P From 1 2 T1 T2 T P2 P1 2 3.5 atm. T1 Take Note Pressure gauges for automobile tires are usually calibrated to read overpressure i.e., excess above atmospheric pressure. Thus the gauge would read 2.0 atm in the morning and 2.5 atms later in the day. 16 SPH 302: Thermodynamics 1.5.2 Non Ideal gas A real gas (e.g. oxygen) only approximate the ideal gas behaviour or equation (1.4) at low enough densities (i.e., at high temperatures and low pressures) but deviates from the law at low temperatures and high pressures. This is because a real gas is characterized by: (i) (ii) there are intermolecular attractions The gas molecules themselves occupy a finite volume and as such, the gas cannot be compressed to zero volume as stipulated by Charles law. Real gases liquefy at low pressures. The Equation of state of a real gas that takes into account the above factors is called Van der Waals gas equation and is given by a P 2 V nb nRT ……………………..…(1.6) Where a & b are constants with a being dependent on molecular interactions, = V/n is the molecular volume Activity 1.4 1. Explain why a real gas deviates from the ideal gas behaviour 17 SPH 302: Thermodynamics 1.6 The Zeroth law of thermodynamics It is a common observation that if you place a cup of hot coffee or glass of ice water on a table at room temperature, the coffee will get colder and the ice water will get warmer, the temperature of each approaching that of the room. In each case, the object will tend towards thermal equilibrium with its environment i.e., when its temperature does not change over time. Fig 1-8. A steaming cup of coffee loosing heat to the surrounding as it approaches thermal equilibrium Such approach to thermal equilibrium must involve energy exchange (in this case heat) between the system and the environment. This is the concept of the Zeroth law of thermodynamics which is stated as follows: Zeroth Law: "If body A and body C are each in thermal equilibrium with a third body B, then, A is also in thermal equilibrium with C." such arrangement is shown in Fig. 1-9. A B C Figure 1-9. Bodies in thermal equilibrium It follows from the Zeroth law, a whole series of systems could be found that would be in thermal equilibrium with each other, say, A, B, C, D, etc. All these systems possess a common property called temperature, which determines the direction of heat flow. Take Note: Significance of Zeroth Law The Zeroth Law introduces the concept of temperature and provides a means of determining temperature –Through thermal equilibrium. 18 SPH 302: Thermodynamics 1.7 Temperature and Temperature scales The temperature of a system is a property that determines whether or not, that system is in thermal equilibrium with other systems. We can also define Temperature as a measure of the degree of hotness. The temperature of a substance is determined by measuring the change in the thermometric property of a substance that varies linearly with change in temperature. For example, a substance like mercury expands linearly in proportion to the increase in temperature of the surrounding or substance whose temperature is to be measured. Most thermometers are based on this principle. Common types of thermometers include: (i) Liquid thermometers:- Based on principle of change in volume of a liquid with change in temperature e.g., mercury thermometer. (ii) Resistance thermometer:- Based on the principle of the change in the resistance of a conductor is linearly dependent on temperature e.g., platinum resistance thermometers (iii) Thermoelectric thermometers or thermocouples:(iv) Bimetallic thermometers:1.7.1 Scales of Temperature If X is the thermometric property that varies linearly with change in temperature then, the temperature, TX of a substance on an X scale can be given by a linear relationship of the form TX aX Where a is a constant whose value is fixed as the reference point. The customarily chosen reference point is the temperature at which ice, water and vapour coexist in equilibrium, known as the triple point, TP of water and is assigned a value of 273.16. We may thus write X TX 273.16 X TP ………………..…….……. (1.7) Which implies a zero of temperature i.e., T x= 0 at X = 0. We can now apply Eqn (1.7) to several thermometers depending on the thermometric property chosen. For a liquid-in-glass thermometer, X is the length (L) of the liquid column, so Eqn (1.7) gives L TX 273.16 L TP ………………..…….…...…. (1.8a) 19 SPH 302: Thermodynamics For a gas at constant pressure, X is the volume (V) of the gas, and Eqn (1.7) gives V TV 273.16 V TP ………………..…….…..…. (1.8b) Likewise, for a platinum resistance thermometer, X is the electrical resistance (R) of the platinum wire such that Eqn (1.7) gives R TR 273.16 R TP ………………..……..….…. (1.8c) The perfect gas scale The perfect gas scale does not depend on the particular properties of a particular gas but uses pressure and volume of a gas to indicate temperature. In this case, quite a wide range of temperature can be covered. In this scale P K Tgas 273.16 P TP ………………..…….….…. (1.8d) Where K, the Kelvin, is the unit of temperature on the ideal gas scale. 1.7.2 Commonly used Temperature Scales There are three commonly used temperature scales namely: (i) The Kelvin scale The Kelvin, (K) is the unit of temperature on the ideal gas scale. On this scale, the temperature between ice point and steam point is 100K (ii) The Celsius Scale This is a convenient scale whose range is within the range of commonly encountered temperatures. On this scale, the temperature between ice point and steam point is 100 oC. 20 SPH 302: Thermodynamics (ii) Fahrenheit scale This scale has a range of 180 equal parts with the ice point at 32 oF and Steam point is 212 oF. Relationship between temperature scales A simple Relationship between the three temperature scales i.e., Celsius, Fahrenheit and Kelvin is given by C F 32 K 273 100 180 100 Worked Example 1.1 A platinum resistance thermometer has a resistance R = 90.35 when its bulb is placed in a triple point cell. Determine its temperature when the bulb is placed in an environment where its resistance is 96.28 . Solution From the equation, TR R 273.16 R TP 96.28 TR 273.16 K 280.6 K 90.35 Worked Example 1.2 The temperature of the surface of the sun is about 6500 oC. What is this temperature on the Kelvin scale? Solution From the equation, C F 32 100 180 K 6,500 273 6,773 K 21 SPH 302: Thermodynamics 1.8 Lecture Summary In this lecture, we learned the following 1. Thermodynamics is the study of relationships involving heat, mechanical work and other aspects of energy transfer that takes place in thermodynamic systems such as refrigerators, heat pumps, internal combustion engines etc. 2. The ideal gas obeys the gas law PV = nRT exactly. No such gas exists. 3. The internal energy of an ideal gas is entirely kinetic and depends only on its temperature 4. The Equation of state of a real gas is given by a P 2 V nb nRT 4. Statement of the Zeroth Law : "If body A and body C are each in thermal equilibrium with a third body B, then, A is also in thermal equilibrium with C." 5. The temperature of a system is a property that determines whether or not, that system is in thermal equilibrium with other systems. The Zeroth law tells us that if we introduce heat into a system, this heat has to flow to the surrounding in order to maintain thermal equilibrium. A closed system (such as gas enclosed in a cylinder) on the other hand can only transfer this heat as work to the surrounding i.e., by expanding and doing work to the environment in order to attain thermal equilibrium. The 1st Law of Thermodynamics explores this possibility in which heat input is converted to mechanical work. In the next lecture, we shall discuss the relationship involving heat, mechanical work and internal energy, given by the 1ST law of thermodynamics. 22 SPH 302: Thermodynamics Questions for Discussion 1. Explain why a real gas behaves like an ideal gas at low densities but not at high densities 2. At the airport of La Paz, Bolivia, one of the highest in the world, pilots find it preferable to take off early in the morning or late at night, when the air is very cold. Explain Why? Activity PROBLEM SET 1 The normal boiling point of liquid oxygen is -183oC. What is this temperature on the Kelvin scale? 1.9 Further Reading 1. Finn C. B. J., Thermal Physics, 1986. 2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics., S. Chand & Company Ltd, New Delhi. 3. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons, Inc., NY., Pg 479-489. 4. Aduda B. O., (2004). Structure and Properties of Matter, Distance Learning study module, University of Nairobi, 2004. Memorable quotes Imagination is more important than knowledge. Knowledge is limited, imagination encircles the whole world. –Albert Einstein. 23 SPH 302: Thermodynamics Lecture 2 FIRST LAW OF THERMODYNAMICS Outline 2.1 2.2 2.3 2.4 2.5 2.6 2.7 Introduction Objectives 1ST Law of Thermodynamics Applications of 1ST Law of Thermodynamics Heat Capacities of an Ideal Gas The Adiabatic Process Summary Fig 2.1 (a) This Donkey is an engine: It requires fuel (food) which it ‘burns’ in order to work. This is the concept of 1ST law of thermodynamics. In this picture, the work required of the donkey is more than it can provide. 24 SPH 302: Thermodynamics 2.1 Introduction In this lecture we shall study the first law of thermodynamics and explore possible thermodynamic processes through which heat can be converted into work. We shall also discuss the adiabatic processes, and derive the equation of state for the adiabatic process. 2.2 Objectives At the end of this lecture, you should be able to: 1. State and explain first law of thermodynamics and recognize perpetual motion machines of the first kind 2. State the significance and limitations of the 1ST law of thermodynamics 3. Apply the 1ST law to determine work done in isothermal, isobaric, isochoric and adiabatic processes. 4. Derive the equation of state for the adiabatic process 2.3 First Law of Thermodynamics The 1st law of thermodynamics is an expression of the principle of conservation of energy which states that: “Energy can be transformed (changed from one form to another), but cannot be created or destroyed”. The 1ST law thus denies the possibility of creating or destroying energy and broadens this principle to include energy exchange by both heat transfer and mechanical work between the system and the surrounding and introduces the concept of the internal energy of a system. Fig. 2.2. This warden at KWSTI escaping from the wrath of the zebra is a model of 1ST law of thermodynamics. His stored energy is converted into mechanical work (running). 25 SPH 302: Thermodynamics To state energy relationships (i.e., heat, work and internal energy) or the 1ST law precisely, we shall consider a closed system consisting of a mechanical device with a fixed mass of a compressed (ideal) gas enclosed in a cylinder with movable piston (Fig. 2-1). Now, If heat dQ is supplied to the system, this heat goes to increase the internal energy (dU) of the gas as well as enable the gas do some mechanical work (dW) on the surrounding i.e. push on the piston. If the process is reversible then: Boundary SURROUNDING dQ in 1ST Law …………………..……….. (2.1) dQ = dU + dW Movable piston Gas system Area A dx Fig. 2-3. A gas in cylinder system But, the Work done by the gas can also be given by dW = Force distance = PA dx Where P = Pressure the gas applies to the piston; A = surface area of the piston and F = PA dW = PA dx = PdV since Adx = volume (dV) Hence, 1ST law (Eqn. 2.1) can also be written as dQ = dU + PdV 1ST Law …….…..…….…….…. (2.2) Eqn. (2.1) is the 1ST law of thermodynamics. It states that:“Whenever we employ some process involving heat and work to change the state of a system, the change in the internal energy of the system is given by dU = dQ - dW and it does not depend on the path followed by the process”. 26 SPH 302: Thermodynamics Take Note Sign convention of dQ and dW If the surrounding delivers heat to the system, then dQ is +ve. If the system delivers heat to its surroundings then dQ is -ve. Likewise, if the surrounding performs work on the system, then dW is -ve, and +ve otherwise. 2.3.1 Internal Energy (U) The internal energy of the system (assume an ideal gas) is the sum of the following forms of energy a) Kinetic energy due to translational, rotational and vibrational motion of the molecules, all of which depend only on temperature. b) potential energy due to intermolecular forces, which depends on the separation between the molecules, and c) The energy of electrons and nuclei. Take Note: Significance of First Law The first law provides a method for determining the internal energy (dU) of a system. Take Note: Limitations of The First Law The first law does not say how much of the heat energy can be converted into work. These limitations are addressed in the 2 ND law of thermodynamics. 2.3.2 Work as a path dependent function The internal energy (dU) depends only of temperature (T) of the system and it is independent on the path taken and is termed a state function. On the other hand, the heat (dQ) supplied to the system as well as the work (dW) done by the gas system, both, not only depend on the initial and final states of the system but also on the path taken. As such, dQ and dW are not state functions. 27 SPH 302: Thermodynamics Proving that dW is path dependent To prove that work done is dependent on the path taken in the process, let us consider an ideal gas being taken from state A to D as shown in figure 2.2. We see that there are infinite number of ways of reaching point D i.e., either through path ABD or path ACD or through the curved path A D. Along path ABD, the work done is dWABD = WAB + WBD = 0 + Pf (Vf – Vi) = Pf (Vf – Vi) ……………………..…….(a) P Pf Pi B D A C Vi Vf V Fig.. 2-2. Work as a path dependent function Along path ACD, the work done is dWACD = WAC + WCD = Pi (Vf – Vi) = Pi (Vf – Vi) + 0 ……….…… ……..…….(b) We see that dWABD dWACD implying that (dW) is a path dependent function. Thus Work done depends on the actual path taken, that is, the way P varies with V. Take Note: A state function (such as internal energy, U) is a quantity which in equilibrium depends only on the thermodynamic variables rather than on the history of the system 28 SPH 302: Thermodynamics 2.3.4 Thermodynamic Processes There are FOUR main thermodynamic processes namely: Isothermal, Isobaric, Isochoric and Adiabatic. (i) Isothermal process:- Processes that occur at constant temperature e.g., gas expanding in a cylinder from state P1V1 to state P2V2 (along path a b) as shown in Fig. 1-4 below. P (Mpa) a P1 Isochoric (dV = 0) P2 Adiabatic (dQ=0) b d Isothermal (dT=0) Th c P3 TC Isobaric (dP = 0) V1 V2 V m-3 Figure 1-4. Isothermal, Isobaric, isochoric and adiabatic process on a P-V diagram (ii) Isobaric process:- Processes that occur at constant pressure e.g., gas expanding in a cylinder from state P2V1 to state P2V2 (along path d b) as shown in Fig. 1-4. (iii) Isochoric process:- Processes that occur at constant volume (dV = 0) e.g., from state P1V1 to state P2V1 (along path a d) as shown in Fig. 1-4. (iv) Adiabatic process:- Processes where no heat enters or leaves the system (dQ = 0) such as along path a c. Such processes take place rapidly and include rapid expansion or rapid cooling. 29 SPH 302: Thermodynamics 2.4 Applications of First Law of Thermodynamics Since work done is path dependent, it would be of interest to explore possible thermodynamic processes through which we can convert heat energy into mechanical work by applying the 1ST Law of thermodynamics and find out which process gives maximum work output. This is important in a thermodynamic system as we shall see later in lecture 3. (a) Work done in isothermal process (dT = 0) If an ideal gas is taken through the isothermal process (dT = 0), then, from the ideal gas law (PV = nRT); we have that PV = constant PV diagram is hyperbola as shown in Fig 2.3. The curved path 1 2 is the isothermal process P PV = c isothermal 1 2 W V1 V2 V Figure 2-3. Work done in isothermal process The work done in this process is the shaded area and is given by V2 W PdV V1 But P nRT . V Substituting for P we have V2 dV V V V W nRT ln 2 V1 W nRT 1 ……………………......… (2.3) W in isothermal process 30 SPH 302: Thermodynamics Also from Boyle’s law, we have that P1V1 P2V2 W nRT ln P1 P2 V2 P1 V1 P2 ……………………..……(2.4) W in isothermal process Take Note: Significance of Isothermal process No Since dU depends on temperature T, then in isothermal process, dU = 0. The form of 1ST law becomes dQ PdV . In isothermal process, the heat supplied = work output by the ideal gas. This represents maximum work output from the system (b) Work done in isochoric process (dV = 0) Along the isochoric process (path 1 2 in Fig. 2-4), the volume remains constant (dV = 0) and no work is done in the process. Thus, the heat supplied to the gas only raises its internal energy and accordingly, 1st law becomes ………………………..……..(2.5) dQ = dU P 2 PV T2 1 PV T1 V1,2 V Figure 2-4. Work done in isochoric process (c) Work done in isobaric process (dP = 0) In isobaric process, the pressure remains constant (dP = 0) and the work done along path 1 2 in Fig 2.5 is given by dW = P(V2 – V1) ……………………….(2.6) 31 SPH 302: Thermodynamics P 2 PV T2 1 PV T1 V1 V2 V Figure 2-5. Work done in isobaric process Take Note: Significance of Isobaric process For the isobaric process, the form of 1st Law can be given by dQ = dU + PdV = dH No where H is called the Enthalpy and is the heat evolved in a chemical reaction (that occur at constant pressure). Enthalpy is a state function. (d) Work done in Adiabatic process (dQ = 0) In this case, no heat leaves or enters the system such that the form of first law becomes dU = - dW ……………………..………….(2.7) We note however that, pressure and temperature change and we need to evaluate the work done in terms of changes in (P) and (T). We shall discuss more of the adiabatic process in section 2.6. (e) Other kinds of work (i) Extension of a wire: For a wire extended by a torsional force F through a distance dx, the work done is dW = - Fdx …………………………………..…..(2.8) (ii) Electrolytic cell Work done by external circuit in charging a cell (increasing its charge by dZ) is given by dW = - dz ……………………………..…… (2.9) where = emf of the cell 32 SPH 302: Thermodynamics Worked Example 2.1 10 moles of an ideal gas are compressed isothermally and reversibly from a pressure of 1 atm. to 10 atm. at 300K. How much work is done? Solution V2 W= PdV nRT V1 V2 V dV nRT ln 2 . V1 V V1 P W nRT ln 1 P2 But P1V1 = P2V2 = 57.4 KJ Worked Example 2.2 An ideal gas is taken through the cyclic process ABCA as shown in Fig 2-6 below. Determine (i) The net heat transferred to the system during one complete cycle (ii) The net heat input for the reversed cycle ACBA. P (kPa) B 8 6 4 2 A 6 C 8 10 V m3 Solution (i) From first law, we have that dQ = dU + dW But dW = PdV = area under the P-V curve and dU = 0 in a cyclic process. dQ = PdV = 12 8 22 12 KJ (ii) dQ is the same as in (i) Worked Example 2.3 In Figure 2-7 below, an Ideal gas system can be taken from state a to c either through abc or adc. In processes ab and bc, 600 J and 200J are given to the system respectively. Calculate (i) dU along ab 33 SPH 302: Thermodynamics (ii) (iii) dU along abc Total heat added in adc P104Pa 8 b c 200J 600J 3 a d 2 5 V10-3 m3 Fig.2-7. P-V diagram of ideal gas Solution From 1ST law (dU = dQ – dW) we have (i) Along ab, dW = 0: dU = dQ = 600 J (ii) Total dQ along a b c = (600 + 200) J = 800 J dW along ab = 0 since dV = 0 dW along bc = PdV = P(5 - 2) 10-3m3 = (8 104Pa)(3 10-3m3) = 240 J. Total dW along a b c = 240 J dU along a b c = dQ - dW = (800 - 240) J = 560 J (iii) Along a d c, dU = 560 J (NB. Since dU is a state function, it is independent of the process such that the total dU along a b c is the same as dU along a d c. dW along a d c = P1(V2 - V1 ) = 90 J since along d c, dW = 0 dQ = dU + dW = (560 + 90)J = 650 J. Activity 2.1 Calculate the work done by a Van der Waals gas whose equation of state is given by P a V nb nRT in expanding from a V2 volume V1 to a volume V2. 34 SPH 302: Thermodynamics Activity 2.2 A sample of an ideal gas is in a vertical cylinder fitted with a piston. 5.79 kJ of heat energy is transferred to the gas so that the state of the gas changes from point A to point B along the semicircle shown in Figure 2-8 below. Determine the change in internal energy of the gas. P (KPa) 500 400 300 A B 200 f 100 0 4 1.2 3.6 6 V m-3 Fig 2-8 35 SPH 302: Thermodynamics 2.5 Heat Capacities of An Ideal Gas We can change the state of a system or body by transferring energy to or from it in the form of heat or by doing work on the system. In either case, the temperature of the system changes. The amount of heat transferred to the system is measured by the heat capacity. The heat capacity (symbol C) is the amount of heat required to cause a unit rise in the temperature of a substance. It is given by C dQ dT ……………………..……... (2.10) On the other hand, the specific heat capacity (c) is the heat capacity per unit mass or the amount of heat per unit mass required to cause a unit rise in the temperature of a substance. It is given by c C dQ dQ m mdT nMdT …………… (2.11) Where m = mass of the substance = nM with M being the molecular mass; n = number of moles. The change in temperature (dT) corresponding to the transfer of a particular quantity of heat energy dQ will depend on the circumstances under which the heat was transferred i.e., either at constant volume or at constant pressure etc. Hence, we speak of Molar heat capacity (Heat capacity per unit mole) at const Volume (CV) or Molar heat capacity at const pressure (CP). The Molar heat capacity at constant Volume (CV) is define by dQ CV ndT V …………………………….(2.12) Where n = number of moles in the gas Likewise, the Molar heat capacity at constant pressure (CP) is given by dQ CP ndT P ……………….……… …(2.13) Take Note: Usually the heat capacity increases with temperature and is given by the expression: constants CP a bT cT 2 dT 3 where a, b and c are No 36 SPH 302: Thermodynamics 2.5.1 Relationship between Heat Capacities Cp and CV We have seen from above that if the heat is introduced into the system at dQ S.T dQ = nCPdT. In this case therefore, ndT P constant pressure then, C P the form of 1ST law of thermodynamics can be written as …………….. …..…(2.14) nCpdT = dU + nRdT. While for a constant volume (isochoric) process (dV = 0), the form of 1 ST law becomes dQ = nCvdT = dU ……………..……….... …..(2.15) Substituting for dU in Eqn (2.14) we have nCpdT = nCvdT + nRdT Cp - Cv = R …………….…….…(2.16) Mayer’s relation This relation is known as Mayer’s relation Take Note dU is a state function dependent on T always. As such, total internal energy gained either when heat is supplied at constant volume or at constant pressure is the same. Thus, for simplicity, we can always write dU as dU = nCvdT This expression holds for any ideal gas and for every kind of process. Why Cp is greater than Cv Notably, heat input at constant pressure CP is greater than heat input at constant volume CV. This is because, when heat is supplied at constant pressure, this heat goes in doing two things a) It raises the temperature of the gas (i.e., increase in internal energy) and b) It does work in expanding the gas against the external pressure i.e., nCPdT = dU + dW On the other hand, when the gas is heated at constant volume, no work is done (dW = 0) and all the heat supplied only raises its internal energy. Hence CP > CV. 37 SPH 302: Thermodynamics o An exception where CP < CV is water. When water is heated between 0 C and 4 o C, the volume decreases such that water has its greatest density (least volume) at 4oC as shown in Fig 2.9. The reverse occurs on cooling below 4oC which is why ice floats on water. This anomalous behaviour of water has an important effect on animal and plant life in lakes. A lake cools from the surface down and at 4 oC, the cooled water at the surface flows to the bottom (due to its greater density). As the surface cools below 4 oC (freezes), the surface water or ice (now less dense) floats on the warmer water below. The water below thus remains at 4 oC thereby sustaining plant and animal life below the surface during the cold/winter seasons. V (cm3) ice 0 water 2 4oC To C Fig. 2.9. The anomalous expansion of water Questions Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume for an ideal gas 2.5.2 The Gamma () Relation of Heat Capacities The Gamma relation is the ratio of heat capacities, Cp to Cv. This relationship can be obtained from the kinetic theory of gases as follows: From the kinetic theory of gases, the internal energy (U) of a monatomic gas (n =1) operating at constant volume (dV = 0) is given by (refer to SPH 203 for complete derivation), U 32 RT or dU 32 RdT But we have seen that for isochoric process, dU CV dT always 38 SPH 302: Thermodynamics RdT CV dT 3 2 CV 32 R ……………………………. (2.17) Also, from Equation (2.16), we have that CP - Cv = R. Combining (2.16) and (2.17) we have CP CV R 32 R R CP 52 R ……………………………………..(2.18) Further, taking the rations of CP to CV we have CP 1.67 CV ……………………..…(2.19) The Gamma Relation This is the gamma relation of heat capacities Take Note Molecules which contain more than 1 atom posses rotational K.E in addition to translational K.E. When this is taken into account, it can be shown that: For a diatomic gas (n = 2), CP 7 1.4 CV 5 For a polyatomic gas, CP 4 1.33 CV 3 Activity 2.3 Derive the Mayer’s Relation and the Gamma relation of heat capacities from first principles 2.6 The Adiabatic Process In an adiabatic process there is no heat input into the system (dQ = 0) but instead, a gas does work (-dW) to the environment by expanding such that the 39 SPH 302: Thermodynamics internal energy (dU) decreases as the temperature drops as shown in the Figure 2.10 below. Processes that take place suddenly or quickly are adiabatic. P (Mpa) a P1 Adiabatic, PV = C Isothermal, PV = C Th b TC P2 V1 V2 V m-3 Fig. 2-10. The adiabatic ab process of ideal gas Further, from fig 2.10, we see that in adiabatic process, all the 3 state variables P, V and T change during the process unlike in other processes like isothermal, isobaric or isochoric where one parameter remains constant. Just like in isothermal process of ideal gas where we derived the relationships between state variables P and V i.e. PV = const, it would be interesting to see how the state variables are related in the adiabatic process. We shall look at the following relationships: T-V, P-V and T-P for adiabatic process as well as the work done in this process. (a) The T-V relation The form of 1ST law for adiabatic process can be given by (NB dU = nCVdT for any process of an ideal gas) ………………..……..(2.20) nCVdT = - PdV we note that change in temperature (dT) is accompanied by change in volume (dV). Thus, we can obtain the T-V relation by eliminating P (using P = nRT/V) from the above equation (2.20). Doing so we obtain nCV dT nRT dV V 40 SPH 302: Thermodynamics dT R dV 0 T CV V But C CV C R P P 1 1 CV CV CV dT dV 1 0 T V for n = 1 Integrating we have ln T 1ln V C ln T ln V 1 C ln TV 1 C TV 1 C 1 T1V1 OR …….………(2.21a) T2V2 T-V Relation 1 (b) The P-V relation The above equation (2.21) can be converted into a relation between P and V by eliminating T (using T = PV/nR) i.e., PV 1 V c nR Or PV c ……………………..(2.22) Poisson’s Law P1V1 P1V1 where C = nR = constant Equation (2.22) is known as Poisson’s Law. It tells us that if we allow the gas to change its volume with no other constants, the path followed is adiabatic and can be represented on a PV diagram by a parabola-like curve given by PV const with slope = . 41 SPH 302: Thermodynamics Significance of PV c Since > 1, it follows that the adiabatic curve (PV = c) has a steeper slope ( times) than the isothermal curve (PV = c) at any point at which the two curves intersect on a P-V indicator diagram as shown in the figure 2-11 below. P (Mpa) Adiabatic, PV = c Pi Isothermal, PV = c Pf Vi V m-3 Vf Fig. 2-11. Isothermal and adiabatic curves on a P-V diagram. The adiabatic curve is times steeper than the isothermal curve. The slopes of the isothermal and adiabatic can be obtained by differentiating the Equations: PV = C and PV c respectively. pdV VdP 0 and or PV 1dV V dP 0 slope Doing so we obtain dP P dV V for isothermal dP P dV V for adiabatic dP of adiabatic is steeper than the isothermal. dV Activity 2.4 1 C , derive the P-V Starting from the T-V relation: TV relationship for the adiabatic process and show that the adiabatic curve is times steeper than the isothermal curve on a P-V diagram. 42 SPH 302: Thermodynamics (c) T-P relation We can obtain a T-P relation by eliminating V in Equation (2.22) using the ideal gas relation (PV = nRT). From Equation (2.22) we see that But V PV c nRT P c P T c P 1 nRT P or T P c P since nR = c P 1 c T OR TP OR 1 c ……………………..(2.23) T-P Relation (d) Work done in adiabatic Process The work done in adiabatic process when the gas expands from P1V1 to P2V2 as shown in Fig 2-12 can be evaluated using the equations of the adiabatic process, PV c as follows P (Mpa) A P1 Adiabatic isothermal T1 B P2 a V1 T2 b V2 V m-3 Fig 2-9. Work done in Adiabatic process 43 SPH 302: Thermodynamics V2 V2 V1 V1 W PdV C W C 1 dV V i.e., by replacing for P 1 1 V 1 V 1 2 1 Where c is a constant that needs to be replaced. But we note that P1V1 P2V2 c . Replacing for the constant C we get 1 C C 1 P2V2 P1V1 W 1 1 V2 V1 1 V2 1 V1 1 PV P2V2 W 1 1 1 ………………….….….[2.24] W in Adiabatic Process This work is the Area (ABba) under the adiabatic curve. Alternative Derivation The above equation (2.24) can also be obtained from the 1ST law of thermodynamics, noting than in this case, dQ = 0. Thus V2 W PdV dU nCV dT V1 W nCV T2 T1 CV T1 T2 i.e., TH > TC (dT is –ve) for n = 1 ………..….…………..(a) NB. In this case, the change in volume (dV) is manifested in change in temperature (dT) since temperature drops in the process, T 1 < T2 such that dT is –ve. Substituting for CV using the relationships of the specific heat (CP – CV = R and CP ) we get CV C R CV CV P R CV 1 ………………..……..(b) 44 SPH 302: Thermodynamics Replacing for CV in Equation (a) we get W R T T2 1 1 ………….…. (2.25) Work done in Adiabatic process Further, the above equation can also be expressed in terms of P and V by substituting for T using the ideal gas Equation (PV = nRT) Thus: from T PV R T2 we get P2V2 R PV P2V2 W 1 1 1 and T1 P1V1 R ……………...(2.24) Work done in Adiabatic process 2.6.1 Examples of adiabatic process (i) Air compressor when air is let out from an air compressor like those used in gasoline (petrol) stations or in paint-spraying equipments, it feels colder than the outside air. This is because the air expands rapidly (nearly adiabatic) leading to temperature drop. (ii) Opening a bottle of carbonated beverage Similar process occurs when you open a bottle of your favourite carbonated beverage such as coke cola. In this case, the gas enclosed in the beverage bottle expands very rapidly as it cools and in the process cooling the surrounding air such that the water vapour condenses forming a miniature cloud near the mouth of the bottle. Worked Example A quantity of nitrogen gas at an initial pressure P 1 = 2 10-5 Nm-2 and an initial volume V1 = 4.0 m3 is compressed rapidly to a volume V2 = 2.5 m3. (a) what is the final pressure ? (b) if the nitrogen is now cooled to the original temperature, find the final pressure ( for nitrogen = 1.4) Solution (a) A rapid compression is an adiabatic process. Thus from P1 V 1 = P 2 V2 V P2 = P1 1 V2 = 3.86 105 Nm-2. (b) If final temperature = initial temperature then dT = 0 (isothermal) Thus using P1V1 = P2V2 P2 = V1 P1 = 3.2 105 Nm-2. V2 45 SPH 302: Thermodynamics 2.7 Summary In this lecture, we have learned that 1. The formulation of 1St law of thermodynamics is given by dQ = dU + dW where dQ = Heat input, dU = Internal energy and dW = work done 2. The work done in various thermodynamic processes is given by V (i) Isothermal process (dT = 0): dW nRT ln 2 V1 (ii) Isochoric process (dV = 0): dW = 0 T (iIi) Isobaric process (dP = 0): dW nRT ln 1 T2 1 PV P V or 1 1 1 2 2 R T T W 1 1 2 (iv) Adiabatic process W 3. Internal Energy of an ideal gas is given by dU = nCvdT 4. The heat capacities at constant pressure and volume are given by dQ dQ and CP Cv dT P dT v 5. The relationship between heat capacities is given by [Mayer’s Relation] CP CV R CP 1.67 CV [The gamma Relation] 6. The relationships in the adiabatic process are given by T-V Relation: P-V Relation: T-P Relation: TV 1 C PV C TP 1 c In the next lecture, we shall study the second law of thermodynamics which provides specific guidelines for energy transfer and conversion processes by defining the thermal efficiency of processes that convert heat into work. 46 SPH 302: Thermodynamics Questions for Discussion 1. Suppose that U is the balance of a bank account, Q the money paid in as cash, and W the money drawn out by Cheque. (a) Write an equation which relates an initial balance UI, a final balance UF, Q and W. (b) If the bank manager could keep records of only one of these three variables, which would he choose, and why? (c) To what extend does the analogy between this situation and the 1st law of thermodynamics hold well? 2. When ice melts at 0o C, its volume decreases. Is the internal energy change greater than, less than, or equal to the heat added? How can you tell. 3. Household refrigerators always have tubing on the outside, usually at the back or bottom. When the refrigerator is running, the tubing becomes quite hot. Where does the heat come from? 4. In constant-volume process, dU = nCVdT. But in a constant-pressure process, it is not true that dU = nCPdT. Explain why 5. When you blow on the back of your hand with your mouth wide open, your breath feels warm. But if you partially close your mouth to form an “O” and then blow on your hand, your breath feels cool. Explain why. 6. Air escaping from an air hose at a gas station always feels cold. Explain 7. How can you best use a spoon to cool a cup of coffee? Stirring-which involves doing workwould seem to heat the coffee rather than cool it 8. How does a layer of snow protect plants during cold weather? During freezing spells, citrus growers in Nairobi often spray their fruit with water, hoping it will freeze. How does that help? 9. It is difficult to “boil” eggs in water at the top of a high mountain because water boils there at a relatively low temperature. What is a simple, practical way of overcoming this difficulty? 10. Will a 3-minute egg cook any faster if the water is boiling furiously than if it is simmering quietly? 11. Explain why the specific heat at constant pressure is greater than the specific heat at constant volume. Can CP ever be less than CV? If so, give an example. 12. Why is the difference between CP and CV often neglected for solids? 13. Real gases always cool when making a free expansion, whereas an ideal gas does not. Explain. 14. Discuss the similarities and especially the distinctions between heat, work, and internal energy. 47 SPH 302: Thermodynamics PROBLEM SET 2 1. Ice at 0o C and at a pressure of 1 atm has a density of 916.23 Kgm-3 while water under these conditions has a density of 999.84 Kgm-3. How much work is done against the atmosphere when 10Kg of ice melt into water. 2. Two moles of a monatomic ideal gas at a temperature of 300K expand reversibly and isothermally to twice the original volume. Determine (i) the work done by the gas, (ii) the change in the internal energy. (iii) the heat supplied and 3. Three moles of an ideal gas have an initial temperature of 127 o C. While the temperature is kept constant, the volume is increased until the pressure drops to 40% of its original value. (a) Draw a P-V diagram for this process (b) Calculate the work done by the gas 4. A gas is contained in a cylinder fitted with a frictionless piston and is taken from state a to state b along the path acb as shown in Fig. 1. 80J of heat flow into the system and the system does 30J of work. P P c b a d P1 P2 V Fig. 1 (i) (ii) (iii) a b d c V1 V2 V Fig. 2 How much heat flows into the system along the path adb if the work done by the gas system is 10J. When the system is returned from state b to a along the curved path, the work done on the system is 20J. Find the heat transfer. If Ua = 0 and Ud = 40J, find the heat absorbed in the process ad and db. 5. An ideal gas is taken through the cycle a b c d a as shown in Fig. 2. If P1 = 3 atm, P2 = 1 atm, V1 = 1 litre and V2 = 2 litres, find the work done in this cycle. 6. A system is taken from state a to b along the three paths shown in the figure 3 below. (a) Along which path is the work done by the system greatest? The least (b) If Ub > Ua, along which path is the value of the heat transfer Q the greatest? For which path is the heat absorbed or librated by the system? 48 SPH 302: Thermodynamics P b 1 2 a 3 0 V Fig.3 7. An ideal gas is taken through the cycle ABCDA as shown in Fig. 4 below. Determine (i) The work done on the gas per cycle (ii) The net heat energy added to the system per cycle (iv) The work done per cycle for 1.00 mol of the gas at 0 oC. P B 3Pi C A Pi D Vi V m-3 3Vi 8. An ideal gas expand from state i to f as shown in Fig 5 below (i) Determine the work done on the gas (ii) the gas is now compressed from state f back to state i along the same path. Determine the work done. P (Pa) i 6 106 4 106 f 2 106 0 1 2 3 4 V m-3 49 SPH 302: Thermodynamics 9. A sample of an ideal gas is taken through the process ABCDA as shown in Fig 6 below. Process AB is adiabatic while process BC is isobaric with 100 kJ of heat entering the system. From C to D, the process is isothermal while from D to A, the process is isobaric with 150 kJ of energy leaving the system. Determine the difference in the internal energy between points A and B i.e., UB – UA. P (atm) dQ = 100 KJ B 3.0 C A 1.0 D dQ = 150 KJ 0 0.09 0.2 0.4 1.2 V m3 10. An ideal monatomic gas ( = 5/3) expands reversibly from a state V1, P1 to a volume V2. Calculate the work done by the gas if the change takes place (i) Isothermally (ii) adiabatically (iii) such that PV is a constant. 11. For an ideal gas and starting from the same initial point, show the following processes both on a P-V and T-S diagrams (i) PV = constant (ii) PV = constant (iii) P = constant, and (ii) V = constant. 12. Show from 1ST principles that the following relationships hold for a reversible adiabatic expansion of an ideal gas (i) TV 1 const (ii) PV const (iii) T 1- 1 P cont 13. Find the change in the internal energy of one mole of a monatomic ideal gas in an isobaric expansion at 1 atm from a volume of 5 m3 to 10 m3 ( = 5/3). 14. Show that the adiabatic curve for an ideal gas is steeper by a factor of than the isothermal curve at any point at which the two curves intersect on the P-V indicator diagram. 15. During a reversible adiabatic expansion of an ideal gas, the pressure and volume at any moment are related by PV const where c and are constants. Show from first principles that the work done by the gas in expanding from state (P 1, V1) to a state (P2, V2) can be given P1V1 P2V2 1 by W 16. L Lo 2 The equation of state of a rubber band if given by F aT where Lo is the Lo L original length and a is constant = 1.310-2 NK-1. How much work is done when the band is 50 SPH 302: Thermodynamics stretched isothermally and reversibly from its original length of 10cm to 20cm, the temperature being 20o C? 17. Calculate the work done by a Van der Waals gas whose equation of state is given by a P 2 V nb nRT in expanding from a volume V1 to a volume V2. V (a) at constant pressure and (b) at constant temperature 18. Fig. 7 shows an energy cycle with three reversible processes to which 16g of oxygen gas (M r = 32) are subjected. Calculate (a) the heat taken from the gas during the isovolumetric cooling, (b) the internal energy gained by the gas during the isobaric heating (process (2)), (c) the work done by the gas during process (2), (d) the heat supplied during process (2) (two possible methods), and (e) the work done on the gas while it is compressed isothermally. (C v,m for oxygen is 21 mol-1 K-1 ). P (Mpa) isovolumetric (1) 0.10 (3) isothermal (2) isobaric Th = 500K TC = 300K V m3 Figure 7 2.8 References 1. Finn C. B. J., Thermal Physics, 1986. 2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics., S. Chand & Company Ltd, New Delhi. 3. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons, Inc., NY., Pg. 517- 536. Memorable quotes Without leaps of imagination or dreaming, we loose the excitement of possibilities. Dreaming, after all, is a form of planning. –Gloria Steinem. 51 SPH 302: Thermodynamics Lecture 3 HEAT ENGINES AND SECOND LAW OF THERMODYNAMICS Outline 3.1 3.2 3.3 3.4 3.5 3.6 3.7 Introduction Objectives Heat Engines and the Second law Heat Pumps The Carnot Engine Internal Combustion Engines Summary This motorcycle is a heat engine that takes in heat from the burning of petrol at a high temperature, does work and ejects the remainder of heat as waste to the atmosphere (low temperature reservoir) through the exhaust. 52 SPH 302: Thermodynamics 3.1 Introduction To operate a machanical device such as a vehicle, we need to convert heat into mechanical energy/work. Although it is easier to convert mechanical energy completely into heat (e.g., when you step on the brakes of a car), it is impossible to convert heat (internal energy) completely into mechanical energy. Heat engines (steam engines, automobile engines, jet engines etc.) are only partly successful at converting internal energy (heat) into mechanical energy (work). A closely related process occurs in the animal kingdom where food is ‘burned’ and partially converted into mechanical energy when muscles are at work. The rest of the unused food is excreted. On the other hand, refrigerators are partially successful in transporting heat from a cooler to hotter bodies (environment). The reasons to nd these questions lie in the directions of thermodynamic processes given by the 2 law of thermodynamics, The second law of thermodynamics deals with the direction in which natural processes occur. It is often said that the second law gives a preferred direction to the “arrow of time”, teaching us that systems naturally evolve with time in one direction but not in the other i.e., heat can only flow from high temperature reserviour and not vice versa. The second law of thermodynamics also provides specific guidelines for energy transfer and conversion processes by defining the thermal efficiency of heat engines thereby imposing limitations on the efficiency of processes that convert heat energy into mechanical energy and vice versa. In this lecture we shall discuss the principle of operation of heat engines and analyze their efficiency in accordance with second law. The Carnot engine is discussed as the most ideal heat engine whose efficiency is a criteria of perfection for all thermal power cycles. You will also be introduced to ideal cycles upon which internal combustion engines are based. Description of the ideal processes in refrigerators and air conditioners and their cycle efficiencies are also treated. Lastly, we shall look at why real engines differ from the ideal cycles. 3.2 Objectives At the end of this lecture, you should be able to: 1. Describe the working principle of a heat engine and recognize perpetual motion machines of the 2ND kind 2. Explain the 2nd Law of thermodynamics and state its significance. 3. Describe a Carnot engine, determine its thermal efficiency and appreciate the importance of the Carnot efficiency as a criteria of perfection for all thermal power cycles. 4. Explain the working principle of a refrigerator and internal combustion engines and determine the coefficient of performance. 53 SPH 302: Thermodynamics 3.3 Heat Engines and 2nd Law of Thermodynamics A heat engine is any device that converts heat into work by means of a cyclic process. Example of heat engines includes steam engines, automobile engines, jet engines etc. In any heat engine, the working substance undergoes a cyclic process i.e., it must able to be returned to its initial configuration at the end of each cycle, so that the process can be repeated all over again. A good example is the steam in a steam engine. During each cyclic process, a heat engine absorbs heat (Qh) from a high temperature heat reservoir (at Th oC) such as a boiler in the case of a steam engine or from the combustion of gasoline in the case of an automobile engine. The engine then converts this heat partially into work (W) and ejects the remainder as waste heat (Q2) into a low temperature reservoir (at T c oC) such as a condenser in the case of steam engine or the exhaust of the automobile engine as depicted in Fig. 3-1. High temperature reservoir at Th Qh E W Qc Low temperature reservoir at TC Fig. 3-1: Operation of a heat Engine Take Note A heat reservoir is simply a body that remains at constant temperature, even when heat is removed from or added to it. The second law asserts that a heat engine cannot convert all the heat it receives from a heat source into work. Some of the heat received must be rejected into a heat sink. This is the law of nature: “If you must eat, then you must shit”. This assertion is clearly stated in the “Kelvin-Plank’s statement” of the second law. The Kelvin-Plank or “Engine” Statement of 2ND Law This statement places emphasis on the operation of heat engines and states that 54 SPH 302: Thermodynamics “it is impossible to construct a device which operating in a cycle produces no other effect than extract heat from a high temperature reservoir and convert the heat completely into mechanical work”. According to this statement, it is impossible to construct a heat engine which produces work continuously by taking heat from a single reservoir and ejecting nothing into the low temperature sink as depicted in Fig. 3.2a. If the second law were not true, we could power an automobile by simply cooling the surrounding air. Therefore, a perpetual motion machine of the second kind does not exist. Example of a perpetual machine would be a car without an exhaust or a cooling system as shown in Fig 3-2b or an animal that does not shit. Such machine contradicts the Kelvin-Plank’s statement of the 2ND law. High temperature reservoir at Th Qh Engine E Fuel Tank W (b) A perpetual motion machine showing a car without exhaust or cooling system which is naturally unrealistic. Fig. 3-2: (a) A hypothetical heat Engine: NOT ALLOWED by the Kelvin-Plank statement The 2ND law also provides a basis for us to define a criterion of performance for all heat engines – Thermal efficiency. Efficiency of heat engines The “efficiency” or “effectiveness” of a heat engine is a measure of the extent to which a heat engine is able to convert the heat supplied to it into work. In other words, it is the ratio of the work output (i.e., what has been accomplished) to the heat input (i.e., what you pay for to get it done). i.e., W Qh ……………………………………….(3.1) Where W = work performed and Qh = heat obtained from the hot reservoir The work (W) done by a heat engine can be obtained from the 1 St law of thermodynamics (dW = dQ – dU). Since in such a cyclic process, there is no change in the internal energy of the system (i.e., dU = 0), it follows that dW = dQ 55 SPH 302: Thermodynamics i.e., the net heat input is converted to work output [Principle of heat engine]. Therefore ………………….….(3.2) W = Qh - Qc W Qh Qc Qh Qh 1 Qc Qh ………….……….….(3.3) If the working substance is an ideal gas, then it can also be shown that Qh QC Th TC 1 Th TC ……..……………..…….….(3.3) Experimental evidence suggests that even under ideal conditions, QC 0 (never) and as such, it is impossible to construct a heat engine with 100% thermal efficiency as stipulated in the 2ND law of thermodynamics. This is because a heat engine always ejects substantial amount of heat (QC) into the environment. Activity 3.1 Show from first principles that Qh QC Th TC 3.4 Heat Pumps A heat pump is a heat engine operated in reverse i.e., it is an engine which requires work input in order to transfer heat from a colder reservoir to a hot reservoir as shown in Fig 3.3. This is the principle involved in refrigerators, heat pumps and air conditioners. Heat pumps are widely used to heat buildings during the cold weather. They operate by transferring heat from the outside (which is at lower temperature QC) to the insider which is at a higher temperature Q h. In comparison, a heat pump is merely a refrigerator turned inside out so that its cold end is outdoors and its warm end indoors. 56 SPH 302: Thermodynamics HOT RESERVOIR (Outside) at Th HOT RESERVOIR (Outside) at Th Qh Qh R R W QC QC COLD RESEVIOR (Inside Refrigerator) at TC COLD RESEVIOR (Inside Refrigerator) at TC Fig. 3-3: (a) Operation of Refrigerator (b) Hypothetical Refrigerator: Clausius -NOT ALLOWED The second law also tells us that it is impossible to construct a heat pump which can transfer heat from a cold reservoir to a hotter region with no input of mechanical work”. This is stated in the “Refrigerator” or “Clausius Statement” of the second law. The Clausius Statement of 2ND Law “No cyclic process can transfer heat from a cold reservoir to a hotter region with no input of mechanical work”. This simply means: “For you to work, you must eat” Thus the second law denies the possibility of reversing the natural tendency for heat to flow from a hotter to a colder body without external interference (in the form of work). The second law also allows us to define a criterion of performance of heat pumps. This criterion or the “effectiveness” of a heat pump is called the HP coefficient of performance (COP ) and is defined by COP HP Heat transferre d to high temp reservoir Work input COP HP Qh Qh W Qh QC ………………(3.4) For heat pumps, COPHP >1 and it depends on the outside temperature. If the outside temperature is say 4oC or higher, COPHP 4 i.e., amount of energy transferred to the building is about 4 times greater than the work done by the motor in the heat pump. However, as the outside temperature decreases (TC 0), the heat delivered becomes equal to the work required and the COPHP can fall below unit. Under such circumstances, it becomes difficult for the heat pump to 57 SPH 302: Thermodynamics extract sufficient energy from the outside air and the heat pump has no economic advantage over electric heaters. Under such circumstances, it would be cheaper to burn the original fuel directly for the heat, rather than generate electricity to operate a heat pump. Thus, the use of heat pumps that extract energy from the air is only satisfactory in moderate climate but it is not appropriate in areas where cold (winter) temperatures are very low. An alternative is to bury the external coils deep in the ground so that the energy is extracted from the ground, which is warmer than the air above. 3.4.1 The Refrigerator A refrigerator is a heat engine which requires work input to transfer heat from a colder reservoir to a hot reservoir as shown in Fig 3.4. Practical refrigerators use gas and liquid Freon (dichlorodifluoromethane) as working fluid (refrigerants), and their cycle differs from the Carnot cycle. Refrigerants (Freon etc.) have a boiling point near room temperature when at high pressure, but a boiling point below at 0o C when at low pressure. In each cycle, Liquid Freon at low pressure and low temperature enters the cooling coils in the refrigerator box and absorbs heat while evaporating into Freon gas (see Fig 3.4a). This gas flows to the compressor (outside the refrigerator box), where its pressure, temperature and density are increased by the push of a piston. The high-pressure gas then circulates through the condenser coils, which are exposed to the atmospheric air as the Freon gas loses its heat and condenses into a liquid (see Fig 3.4b and c). This high-pressure liquid then passes through an expansion valve (a small orifice) where its pressure is reduced to match the low pressure in the cooling coils. This return of the fluid to the cooling coils completes the cycle. Figure 3-4(a) and (b) Principle of a mechanical refrigeration cycle 58 SPH 302: Thermodynamics Figure 3.4(c) The coils on the back of a refrigerator transfer energy by heat to the air outside (d) Inside a refrigerator The effectiveness of a refrigerator is sometimes called the coefficient of performance (COPR). For an ideal Carnot refrigerator, (COPR) is defined by Heat transferre d from cold reserviour Work input Q QC TC C ..................(3.5) W Qh QC Th TC COP R Carnot Refrigerator It can also be shown that COP HP 1 COP R ……………………(3.6) Activity 3.2 Show from first principles that COP HP 1 COP R 59 SPH 302: Thermodynamics From an economic point of view, the best refrigerator is one that removes the greatest amount of heat QC from the inside of refrigerator for the least amount of work (W). The value of COPR is 5. However, if (Th – TC) or Th (the room temperature) is very high, then COPR 0 and more work will be required to transfer a given quantity of heat from the refrigerator. That is why it becomes difficult to obtain cooling at very low temperatures in the refrigerator, since more work is needed. Note also that the refrigerator supplies heat to the heat pump at the back of the refrigerator. The function of such a heat pump is to deliver heat to some reservoir which is at a higher temperature than its surrounding. Take Note Work is always needed to operate a refrigerators. If no work were needed, COPR would be infinite and such device is a workless refrigerator; it is a mythical beast, like the unicorn and the free lunch. Worked Example 3.1 Suppose that QC = 200J, W = 100J. Determine the COP of the refrigerator Solution: In this case, Qh = 200 + 100 = 300J COP R QC 200 Qh QC 100 2 or COP R 200% Take Note In the case of a heat engine, efficiency cannot be more than 100% but in the case of a refrigerator, COPR can be higher than 100%. Worked example 3.2 In a refrigerator, the cooling chamber is maintained at 290K while the outside temperature is 305K. The motor (located outside) has compression cylinders operating at 320K and the expansion coils inside the chamber operating at 280K. If the motor operates reversibly (i) (ii) (iii) (iv) Give a schematic diagram illustrating the sequence of events Calculate the efficiency of the motor Calculate the COP of the refrigerator How much work must be done for each transfer of 5,000J of heat from the chamber? 60 SPH 302: Thermodynamics Solution In this case, the refrigerator is powered by the motor which supplies work to it as shown below. (i) Room at Th = 305K HOT RESERVOIR at ThE = 320K QhE Qh R W E QCE QC Inside Fridge at TC = 290K (ii) COLD RESEVIOR at TCE = 280K The refrigerator is powered by the motor (E). Hence the efficiency of the motor is (iii) COP R (iv) From Q QCE 320 280 W hE 12.5% QhE QhE 320 QC QC TC W Qh QC Th TC 19.3 QhE ThE T QhE hE 5000 5,258.6 J QCE TCE TCE W QhE QCE 258.6 J 61 SPH 302: Thermodynamics 3.4.2 The Air Conditioner The air conditioner employs a similar refrigeration cycle but in this case, the refrigerator box becomes a room or an entire building. The evaporator coils are inside, the condenser is outside, and fans circulate air through these as shown in the figure 3.6 below. Cold air to the room Hot air to outside Fig 3-5: An air conditioner working on the same principle as a refrigerator The parameters of interest are the rate of heat removal (heat current H from the room) and the power input P ( =W/t) to the compressor. H is the heat removed per sec = (QC/t). Hence coefficient of performance is COP AC QC Ht H TC W Pt P Th TC ……………..…(3.7) A typical value of COPAC is about 2.5 and is dependent on the inside and outside temperatures. 62 SPH 302: Thermodynamics 3.5 The Carnot Engine Since the conversion of work to heat is an irreversible process, a heat engine only partly reverses this process. For maximum efficiency, the thermodynamic process within the engine must be reversible i.e., the engine can, in principle be operated in reverse converting work into heat at the same rate as in the forward direction. Thus, when the engine takes heat from the hot reservoir at Th, The working substance should also be at Th i.e., no heat should be lost. Similarly, when the engine discards heat to the cold reservoir at TC, the engine itself must be at TC. Thus, every process involving heat transfer must be isothermal at either Th or TC. Conversely, for any process in which the working substance is intermediate between Th and TC, there must be no heat transfer (adiabatic process) between the engine and either reservoir otherwise, such a process could not be reversible. Additionally, thermal and mechanical equilibrium must be maintained at all times. The only engine with maximum possible efficiency is a hypothetical, idealized reversible heat engine called the Carnot Engine. A Carnot engine is a hypothetical engine which is assumed to be reversible with no dissipative effects such as turbulence and friction. The Carnot engine operates between two reservoirs with an ideal gas as the working fluid enclosed in a cylinder with a piston. Take Note It is important to emphasize that reversibility is a standard of perfection. In real life, no process is totally reversible; reversibility is approached but it is never achieved in practice. A reversible process is thus an abstraction because all natural processes result in energy loss in terms of turbulence, friction or other dissipative effects. A reversible engine therefore represents the perfect engine in the sense that it converts the highest fraction of heat into work. 3.5.1 The Carnot Cycle The Carnot cycle consists of a sequence of 4 steps (with two isothermal and two adiabatic processes) as depicted in the P-V diagram by cycle abcd in Fig. 3.6. These are namely: Path a b: Heat Qh enters the system and the gas expands isothermally (dT = 0) from volume V1 to V2 doing work on the piston Path b c: Gas expands adiabatically (dQ =0) as temperature decreases from Th to TC. 63 SPH 302: Thermodynamics SADI CARNOT, French Engineer (1796–1832) Carnot is considered to be the founder of the science of thermodynamics. Some of his notes found after his death indicate that he was the first to recognize the relationship between work and heat. Qh P W V1, P1 a Qh Isothermal expansion, Th V2, P2 b Adiabatic compression V4, P4 Adiabatic expansion d V3, P3 Isothermal compression Qc c W Temp cools with little work done Exhaust Qc Fig. 3-6. The Carnot Cycle on a P-V diagram V Path c d: Gas is compressed isothermally from volume V 3 to V1 thereby ejecting heat (QC) into the low temperature reservoir at TC. Path d a: Gas is compressed adiabatically from (V4, P4) to (P1,V1) as temperature rises from TC to Th. The total done in the cycle is the area abcd. 64 SPH 302: Thermodynamics The efficiency of a Carnot engine can be obtained using equation (3.1) i.e., Qh W Qh Qh QC ………..……………..(3.1) In this case we note that, for the isothermal expansion a b, dUab = 0 (since dT = 0). Hence we have that. V Qh dWab nRTh ln 2 V1 Similarly, V V QC dWcd nRTh ln 4 nRTC ln 3 V4 V3 V3 TC ln Q V4 1 C 1 Qh V T ln 2 h V1 ……………..………(3.8) But for the two adiabatic processes b c and d a, we have ThV2 1 TCV3 1 and ThV1 1 TCV4 1 From which V2 1 V3 1 V1 1 V4 1 Or V2 V3 V1 V4 Substituting these in equation (3.8) gives the efficiency as Q T 1 C 1 C Qh Th …….………..……(3.9) Carnot Efficiency This is the Carnot efficiency for a heat engine and it is the maximum possible efficiency attained by any engine operating between 2 reservoirs. We note that the efficiency () can only be = 1 (100%) iff TC = 0 K, which is not possible. No real engine can have a thermal efficiency greater than the Carnot efficiency. This assertion is stated in the Carnot’s theorem. 65 SPH 302: Thermodynamics Carnot's Theorem Carnot’s theorem is a result of the second law of thermodynamic. It is a principle that specifies limits on the maximum efficiency any heat engine can obtain, which solely depends on the difference between the hot and cold temperature reservoirs. Carnot's theorem states that: All real (irreversible) heat engines operating between any two heat reservoirs are less efficient than a Carnot’s engine operating between the same reservoirs. All reversible heat engines operating between the same two heat reservoirs have the same efficiency (as a Carnot’s engine operating between the same reservoirs). Activity 3.3 1. Show that the efficiency of a heat engine can be given as 1 QC T 1 C Qh Th 2. State the Carnot’s Theorem Take Note: Why the Carnot Efficiency is the best possible: It assumes that there is no temperature difference between the heat reservoir and the working fluid, so that the entropy gained by one exactly matches the entropy lost by the other, with no net change in entropy for the system as a whole. This condition is of course an ideal one, and cannot be met in practice by any real machine. Even theoretically; all real machines will be strictly worse than this. 66 SPH 302: Thermodynamics Worked Example 3.3 A man uses a Carnot engine operating in reverse as a heat pump to extract heat from the outside air (at -10o C) and inject it into his house (at 20o C). What is the amount of work that must be supplied to pump 4.2 KJ of heat from the outside to the inside? Solution Th = (20 + 273) K, TC = (-10 + 273) K QC TC 0.9 Qh Th Or Qh = (Qc /0.9) = 4.67 KJ Work supplied = Qh - QC = 0.47 KJ This is more economical heating method than the expenditure of 4.2 KJ of fuel in a conventional furnace or electric heater. 3.5.3 The steam engine The operation of a steam engine resembles the Carnot cycle but uses steam and water as working fluid instead of an ideal gas. It consists of a cylinder fitted with a piston that performs cyclic motions of expansion and compression (Fig. 3.7). condenser boiler cylinder water heat Fig. 3-7. A steam engine (schematic) Figure 3.7(b). This steam-driven locomotive that used to run from mombasa to kisumu obtains its energy by burning wood or coal. The generated energy vaporizes water into steam, which powers the locomotive. (This locomotive must take on water from tanks located along the route to replace steam lost through the funnel.) 67 SPH 302: Thermodynamics In each cycle of operation, the high-pressure steam enters the cylinder and pushes against the piston doing work. The low-pressure, spent steam is then exhausted from the cylinder and sent to condenser where an external coolant (air or flowing water) condenses the steam into liquid water which is pumped back to the boiler and the cycle is repeated. The Efficiency of a steam engine is defined by Steam 1 QC T 1 C Qh Th ………………(3.10) Such simple steam engine has efficiencies of about 5-18%. To maximize the efficiency, the designer must make the intake temperature Th as high as possible and TC as low as possible (= room temperature of water). Th is limited by the mechanical strength of the boiler (since pressure increases with temperature) and maximum value of Th possible is 500 oC (pressure = 235 atm in present day steam boilers). Most modern steam engines employ a turbine wheel instead of the cylinder and piston and can achieve efficiencies of up to 40%. Worked Example 3.3 A steam engine has a boiler that operates at 500 K. The cold reservoir’s temperature is that of the outside air, approximately 300K. Determine the maximum thermal efficiency of this steam engine. Solution Steam 1 1 QC T 1 C Qh Th 300 K 0.4 or 500 K 40% This is the max possible (theoretical efficiency). 68 SPH 302: Thermodynamics 3.6 Internal Combustion Engines Internal combustion engines are where the release of energy by combustion takes place within the confines of the engine. Examples of such engines include cars, trucks etc and they employ a fluid which remains gaseous throughout. These engines can broadly be divided into two categories (i) Spark ignition or compression ignition types with either 2 or 4 stroke (ii) Gas turbines We shall briefly describe the working principle of the spark ignition (petrol) engines and the compression ignition (diesel) engines. (a) The Gasoline Engine (Otto Cycle) In real engines, the cycles proceed rapidly with friction and turbulence with no quasistatic and the cycles are irreversible. To discuss a real engine cycle, we replace it by an idealized model (called the Otto cycle) in which the working substance is assumed to be an ideal gas and all processes are assumed to be reversible. Fig 3.8a below shows the idealized Otto cycle representing a gasoline engine operating in 4 steps (strokes) viz: Intake Stroke, e a: The intake valve opens and the mixture of air and gasoline vapour flows into the cylinder as the piston descends as shown in Fig 3-8a below. This is a quasistatic isobaric process (dP = 0) where the volume increases from V1 to V2. with V2 = rV1 where r is the compression ratio. Compression Stroke, a b: Intake valve closes and the mixture is compressed, approximately adiabatically from volume V2 to volume V1 as the temperature rises from Ta to Tb as shown in Fig 3-8b. This processes can be represented by the adiabatic expression; TaV2 1 TbV1 1 …………………….(a) Ignition and Power Stroke, b c d: Mixture is ignited by a spark plug (path b c) ejecting heat Qh into the system. This process occurs in a very short period of time and is not one of the strokes of the cycle. During this time, the temperature and pressure increases rapidly. However, the volume remains approximately constant because of the short time interval and as a result, approximately no work is done on the gas. The heat input along path b c therefore occurs at constant volume (isochoric) with Qh = CV(Tc – Tb) …………………….(b) During the power stroke (path c d), the mixture expands approximately adiabatically back to volume V2 pushing on the piston and doing some 69 SPH 302: Thermodynamics work. In this case, the temperature drops from T c to Td according to the adiabatic expression; TdV2 1 TcV2 1 ………………………(c) The work done in this process is the area under the curve cd. P P2 P c Qh b Qh P1 Adiabatics c Adiabatic W out W out b W in d QC d P0 QC e a V1 V2 e V Fig. 3-8(a). P-V diagram of Otto Cycle Representing the Gasoline Engine Cycle (a) Intake (b) Compression a V rV V Fig. 3-9(b). P-V diagram for a Diesel Cycle (c) Ignition & Power Stroke (d) Exhaust Figure 3-8b: The real cycles of a 4-stroke gasoline (petrol) engine 70 SPH 302: Thermodynamics Exhaust Stroke d a e: Exhaust valve opens allowing the gas to cool at nearly constant volume down to temperature T a as the pressure drops. During this interval, the volume remains approximately constant (isochoric) as heat Qc is ejected into the environment where Qc = CV(Td – Ta) ………………………(d) Along path a e, the combustion products are pushed out (isobarically) as the volume decreases from V2 to V1 leaving the cylinder for the next intake stroke. The thermal efficiency of this idealized engine is thus given by 1 QC C T Ta 1 V d Qh CV TC Tb But from equations (a) and (c) we get Td Ta V2 1 Tc Tb V1 1 Or V2 V1 The ratio 1 Tc Tb Td Ta V2 is called the compression ratio (rc) so V1 V 1 2 V1 1 1 1 rc 1 ………………………………………….(3.11) 71 SPH 302: Thermodynamics Take Note We can also express this efficiency in terms of temperature by noting from Equations (a) and (b) that. V2 V1 1 Ta Td Tb Tc Therefore, Equation (3.11) becomes 1 Ta T 1 d ……………..……(3.12) Tb Tc NB. During the Otto cycle, the lowest temperature is Ta and the highest is Tc such that the efficiency of the Carnot engine operating between these two temperatures will be given by 1 Ta , which is greater than the TC efficiency of the Otto cycle given by Equation (3.12) as expected.. Equation (3.11) shows that efficiency increases as the compression ratio (rC) increases. For typical compression ratios of 8 (for most engines), and with =1.4 (for air), we predict the theoretical efficiency of 1 1 56% for an engine 80.4 operating in the idealized Otto cycle. Although it is important to have a high compression as possible, it is noted that increasing the compression ratio also increases the temperature during the adiabatic compression of the air-fuel mixture causing the mixture to explode spontaneously during compression before ignition. This is pre-ignition or detonation and causes “knocking”, damaging the engine. Pre-ignition lower the value of rC to about 7 particularly for unleaded low octane petrol thereby lowering its efficiency. Activity 3.5 Explain why lead is usually added to petrol used in car engines 72 SPH 302: Thermodynamics Take Note In real engines, the cycles proceeds rapidly with friction, turbulence and no quasistatic (hence irreversible), loss of heat to cylinder walls and incomplete combustion of air-fuel mixtures leading to low efficiency values ( 20 30%). Incomplete combustion occurs if the mixture has less gasoline with the resulting products of CO and unburned hydrocarbons (air pollution) instead of CO2 and H2O. (b) The Diesel Engine It is similar to the petrol engine only that there is no fuel in the cylinder at the beginning of the compression stroke. Instead, air is adiabatically compressed (Fig 3-9) and just before the power stroke, the injectors eject fuel directly into the cylinder. Due to the high temperature, this fuel ignites spontaneously. No spark plugs are needed. Take Note Since no fuel is compressed, pre-ignition does not occur in diesel engines resulting in high compression ratios and high efficiency values (r 15 – 20, 65 – 70%). Diesel engines are often heavier and harder to start but they need no carburetor or ignition system. However, the fuel-injection system requires expensive high-precision machining. (c) Turbochargers and Intercoolers The power output of an engine is proportional to the mass of air forced into the engine’s cylinder. In turbocharged systems, air is compressed (nearly adiabatically with rise in temperature) before entering the engine. This gives a greater mass of air per unit volume. In intercoolers. The compressed air is cooled by passing through an internal cooling system where it loses heat to the surrounding (at constant pressure), and then compressed further. Due to high T and P values in these systems, turbochargers and intercoolers are more powerful than ordinary engines. 73 SPH 302: Thermodynamics Worked Example 3.4 The compression ratio of a diesel engine is 15 to 1 i.e., air is compressed to 1/15 of its initial volume. If the initial pressure and temperatures are 1.0 105 Pa and 27 oC (300K), Determine (a) The final pressure and temperature after compression (assume ideal gas conditions, = 1.4) (b) The work done by the gas if V1 = 1.0 litres (Cv = 20.8 J mol-1K-1). Solution V1 (a) Since 15 V2 And V P2 P1 1 V2 V T2 T1 1 V2 1 613o C 1 44.8 10 5 Pa 44 atm P1V1 0.0405 mol RT 1 1 PV P V 494 J W nCV T2 T1 1 1 1 2 2 (b) Using PV = nRT n NB. –ve sign since compression Take Note: Significance of the Second Law of Thermodynamics In this lecture, we have seen that the second law is important for the following reasons 1. The second law gives a preferred direction in which thermodynamic processes occur and It imposes limitations on the efficiency of processes that convert heat energy into mechanical energy and vice versa by defining the thermal efficiency of heat engines. In particular it stipulates that it is impossible to construct a heat engine with 100% thermal efficiency. 2. The Second law also establishes Entropy and absolute temperature as thermodynamic properties and provides us with objective criteria to determine reversibility 74 SPH 302: Thermodynamics 3.7 Summary So far, we have seen that where as the 1ST law denies the possibility of creating or destroying energy, the 2ND law limits the availability of energy and the way in which it can be used and converted i.e., it is the law of nature describing the direction of natural thermodynamic processes. A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work. The second law of thermodynamics can be stated in the following two ways: “it is impossible to construct a device which operating in a cycle produces no other effect than extract heat from a high temperature reservoir and convert the heat completely into mechanical work” [The Kelvin-Planck Statement]. “No cyclic process can transfer heat from a cold reservoir to a hotter region with no input of mechanical work” [The Clausius Statement]. Carnot’s theorem states that no real heat engine operating (irreversibly) between temperature TC and Th can be more efficient than a Carnot engine operating reversibly between the same two temperatures. The Thermal efficiency of a heat engine operating in the Carnot cycle is given by 1 The Coefficient of performance of refrigerator is given by COP R QC T 1 C Qh Th QC QC TC W Qh QC Th TC The efficiency of internal combustion engine is given by 1 1 rc 1 75 SPH 302: Thermodynamics Question for Discussion 1. The lady in the figure below is trying to cool herself by keeping the door of a refrigerator open. Will she achieve her objective? Explain. NB. The purpose of a refrigerator is to keep its contents cool. Beyond the attendant increase in your electricity bill, there is another good reason you should not try to cool the kitchen on a hot day by leaving the refrigerator door open. What might this reason be? 2. Does a refrigerator full of food consume more power if the room temperature is 28 oC than if it is 20 oC. Explain. 3. Why are heat pumps used for heating houses in mild climates but not in very cold climates? 4. Hot air rises. Why then does the temperature decrease when one climbs a mountain? 5. A room can be warmed by opening the door of an oven but it cannot be cooled by opening the door of a self-contained refrigerator. Discus and explain. 6. Briefly explain whether it is a violation of the 2ND law of thermodynamics to (a) Convert mechanical energy completely into heat (b) Convert heat completely into work 7. An electric motor has its shaft coupled to that of an electric generator. The motor drives the generator, and some current from the generator is used to run the motor. The excess current is used to light a home. What is wrong with this scheme. 8. Explain the difference in the operation of a gasoline engine and a diesel engine. 9. Why is the actual efficiency of a gasoline engine less than the theoretical efficiency? Explain. 76 SPH 302: Thermodynamics PROBLEM SET 3 1. A steam engine uses pressurized steam at 470 K, and exhausts it at 373K. What would be the efficiency of the machine if it were ideal? Explain why the actual efficiency will be lower than this. 2. A reversible Carnot cycle engine operates between temperatures of 1000K and 250K. If 1.5 kJ of heat are transferred to the engine at 1000K in one cycle, find (i) the efficiency of the engine (ii) the heat transferred from the engine at 250K 3. A Carnot engine takes 2000 J of heat from a reservoir at 500K, does some work, and discards some heat to a reservoir at 350K. Determine (i) The work done and the heat discarded by the engine (ii) The thermal efficiency 4. A gasoline engine takes in 16,100 J of heat and delivers 3,700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.6 104 Jg-1. Determine (i) The thermal efficiency (ii) The amount of heat discarded and the mass of fuel burned in each cycle (iii) The power of the engine (in HP) if the engine goes through 60.0 cycles per second. 5. A Carnot engine operates between a heat source at a high temperature Th, and a heat sink at a low temperature Tc. Which will have the greater effect on the efficiency, raising the value of Th by T, or lowering that of TC by the same value? 6. An inventor claims to have developed an engine which takes in 11 107 J at 400K, rejects 5 107 J at 200K and delivers 16.67 kW hours of work. Would you advice investing money in this project? Explain 7. An inventor claims to have developed a heat pump that draws heat from a lake at 3.0oC and delivers heat at a rate of 20 kW to a building at 35oC, while using only 1.9 kW of electrical power. How would you judge the claim?.[Resnick] 8. A Carnot engine working as a refrigerator between 260 K and 300 K receives 2100 J of heat from the reservoir at the lower temperature. Calculate (i) The amount of heat ejected to the high temperature reservoir (ii) The work done in each cycle to operate the refrigerator (iii) Its efficiency 9. 50 Kg of water at 0 oC has to be frozen into ice in a refrigerator. The room temperature is 20 oC. What is the minimum work input to the refrigerator to achieve this. [Latent heat of fusion of water is = 3.33 105 JKg-1]. 10. 0.02 moles of an ideal diatomic gas ( = 1.4) undergoes a Carnot cycle as shown in Figure 1 with temperatures of 227 oC and 27 oC. The initial pressure, Pa = 10.0 105 Pa, and during the isothermal expansion at the high temperature, the volume doubles. Determine (i) The pressure and volume at each of the points a, b, c and d 77 SPH 302: Thermodynamics (iii) (iv) Q, W and dU for each process in the cycle and for the entire cycle The efficiency of the cycle using the results of part (ii) above and compare it with the efficiency of the Carnot engine. P Q1 Isothermal at TC Adiabatic compression Adiabatic expansion Isothermal, TH Q2 Fig 1 V 11. A refrigerator does 153 J of work to transfer 568 J of heat from its cold compartment. (a) Calculate the COP of the refrigerator (b) How much heat is exhausted to the kitchen? [Resnick] 12. Two moles of a monatomic ideal are taken through the cycle shown in the Figure 2 below. Calculate. (i) The heat added to the gas (ii) The heat leaving the gas (iii) The net work done by the gas, and (iv) The efficiency of the cycle. [Ref: Resnick] P (atm) 10.4 B Adiabatic A 1.22 C 9.13 M3 78 SPH 302: Thermodynamics 13. A hypothetical engine, with an ideal gas as the working substance, operates in the cycle shown in the Fig. 3 below. Show that the efficiency of the engine can be given by 1 1 P P 3 1 1 1 V1 V3 Q1 P 1 P1 2 Q2 Adiabatic P3 3 V1 14. V3 V Figure 4 below shows an operation cycle for an idealized diesel engine where fuel is sprayed into the cylinder at pt B and the combustion occurs in the isobaric process B C. Show that the efficiency of the engine can be given by e 1 1 TD TA where symbols have their usual meaning TC TB P Q1 B C Adiabatics D Q2 E VB A VA Fig. 4. P-V diagram for an ideal diesel engine 79 SPH 302: Thermodynamics 15. The motor in a refrigerator has a power output of 210 W. The freezing compartment is at – 3.0oC and the outside air is at 26oC. Assuming that the efficiency is 85% of the ideal, calculate the amount of heat that can be extracted from the freezing compartment in 15 min. 16. In a refrigerator, the cooling chamber is maintained at 290K while the outside temperature is 305K. The motor (located outside) has compression cylinders operating at 320K and the expansion coils inside the chamber operating at 280K. If the motor operates reversibly (i) Give a schematic diagram illustrating the sequence of events (iii) Calculate the COP of the refrigerator (iv) How much work must be done for each transfer of 5000J of heat from the chamber? (v) Determine the entropy change inside and outside the chamber for this amount of refrigeration? 17. The T-S diagram of a reversible engine is shown in Figure 5 below. Calculate its efficiency. T (oC) B 400 C 300 A 500 1000 (J/K) References 4. Finn C. B. J., Thermal Physics, 1986. 5. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics., S. Chand & Company Ltd, New Delhi. 6. F.O. Akuffo, A. Brew-Hammond, F. Makau Luti and J.G.M. Massaquoi. ANST, UNESCO (1997), Nairobi, Kenya. 7. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons, Inc., NY., Pg. 545- 561. Memorable quotes “Wisdom is knowing when you can’t be wise” –Anonymous. 80 SPH 302: Thermodynamics Lecture 4 ENTROPY AND SECOND LAW OF THERMODYNAMICS Outline Introduction Objectives Entropy: One way process Entropy and 2ND Law of Thermodynamics 4.5 Entropy and Performance of Heat Engines: T - S diagrams 4.6 Third Law of Thermodynamics 4.7 Summary This professor’s room is untidy and disorderly. We say it has high entropy compared to his assistant’s room which is orderly representing low entropy. 81 SPH 302: Thermodynamics 4.1 Introduction We saw in lecture 1 that the Zeroth law of thermodynamics leads to the concept of temperature and in lecture 2 that the first law leads to the concept of internal energy. In lecture 3 we saw that the second law of thermodynamics determines the direction in which natural processes occur i.e., it gives a preferred direction to the “arrow of time”, teaching us that systems naturally evolve with time in one direction but not in the other. We also saw that the second law in conjunction with the fundamental thermodynamic relation places limits on a system's ability to do useful work. In this lecture, we shall see that the second law establishes still another concept; Entropy, a quantity in terms of which the second law of thermodynamics is expressed. Entropy is a thermodynamic property that can be used to determine the energy not available for useful work in a thermodynamic process, such as in energy conversion devices – engines and machines. Such devices have a theoretical maximum efficiency when converting energy to work. During this processes, entropy (or disorder) accumulates in the system, but has to be removed by dissipation in the form of waste heat. Entropy is a quantitative measure of disorder. The importance of Entropy is that it provides us with a criterion of reversibility and thereby enable us determine the direction of naturally occurring processes. We shall study Entropy and also discuss T-S diagrams and calculate efficiency of Carnot cycle from T-S diagrams. Lastly we shall discuss the Third law of thermodynamics and explain some of the consequences associated with the third law. 4.2 Objectives At the end of this lecture, you should be able to: 1. 2. 3. 4. Define entropy and state its significance State the 2ND law in terms of Entropy State the central Equation of Thermodynamics Explain T-S diagrams and obtain efficiency of Carnot engine from T-S diagrams 5. Explain the 3RD Law of thermodynamics and state some of the consequences of the 3Rd law 4.3 Entropy: One-Way Process There is a property of things that happen naturally in the world around us that is strange beyond belief. Yet we are so used to it that we hardly ever think about it. It is this: 82 SPH 302: Thermodynamics All naturally occurring processes proceed in one direction only: They never, of their own accord, proceed in the opposite direction. Consider the following example: Example 1: If you drop a stone, it falls to the ground. A stone resting on the ground never, of its own accord, leaps up into the air. Example 2: If you put a drop of ink in a glass of water, the ink molecules eventually spread out uniformly throughout the volume of water. They never, of their own accord, regroup into a drop-shaped clump. Such spontaneous one-way processes are irreversible. You cannot make them go backward by making any small change in their environment. Essentially, all naturally occurring processes are irreversible. Notably, it is not the energy that controls the direction of irreversible processes; it is a property called Entropy (S) of the system which is a measure of randomness or disorder of a system. One feature that distinguishes entropy from other such concepts as energy, momentum and angular momentum in that it does not obey a conservation law and there is no principle of conservation of entropy. In fact, since it is not a conserved quantity, entropy can be created at will. No matter what changes occur within a closed system, the energy of that system remains constant. Its entropy, however, always increases for irreversible processes. Thus, entropy has its central property called the Entropy principle or the principle of increasing Entropy which is stated as: “The Entropy of an irreversible processes always increases; It never decreases”. There are two equivalent ways to define the change in entropy of a system: (1) A macroscopic approach, involving heat transfer and the temperature at which the transfer occurs, and (2) a microscopic approach, involving counting the ways in which the atoms or molecules that make up the system can be arranged. We use the first approach in this lecture. Take Note 1. Entropy is a state function just internal energy (U), temperature (T), volume (V) and Pressure (P). 2. Entropy is an extensive property since it depends on the mass of the working substance. 3. It is difficult to form a physical concept of entropy since there is nothing physical to represent it. Moreover, it cannot be felt like temperature or pressure etc. 83 SPH 302: Thermodynamics 4.3.1 Defining Entropy change The concept of entropy is defined by the second law of thermodynamics, which states that the entropy of an isolated system always increases or remains constant. Thus, entropy is a measure of the tendency of a process, such as a chemical reaction, to be entropically favored, or to proceed in a particular direction. It determines that thermal energy always flows spontaneously from regions of higher temperature to regions of lower temperature, in the form of heat (see Fig. 4-1). These processes reduce the state of order of the initial systems, and therefore entropy is an expression of disorder or randomness. Fig. 4-1. Ice melting in a warm room is a common example of increasing entropy since it represents an increase in the desegregation of ice molecules. Whereas the entropy of ice increases, that of the room reduces For example, when mixing ink with water, the system starts from low entropy state (each fluid is separate and distinct hence orderly with dS = 0) to a disordered state of high entropy (when particles are thoroughly mixed). No spontaneous unmixing (net decrease of entropy) is ever observed. Likewise, a perfect crystal, free of impurities and defects, would have zero entropy at 0K. As the temperature of the crystal is increased, its atoms absorb energy by thermal motion, some disordering occurs, and as a consequence, the entropy of the crystal increases to a value characteristic of that temperature and degree of randomness. We can now define Entropy using a macroscopic approach by considering heat transfer in a system. Using he above example, the entropy change (dS) of a system such as the above crystal at some temperature (T) that has occurred from the initial state (i) to final state (f), assuming a reversible process, is given mathematically by 84 SPH 302: Thermodynamics f dQ Heat added / Re moved Change in Entropy (dS ) R Absolute Temperatur e i T i.e. dS dQR T ………………..…….(4.1) Where QR = the heat absorbed reversibly. Entropy as measure of disorder We have seen that entropy determines the direction of flow of thermal energy and such a direction should always be such as to reduce the state of order of the initial systems. Therefore entropy is an expression of disorder or randomness. For example, when heat is added to any system, there is always a change in volume. A gas system expands when heated and will be in a more disorderly state since the gas molecules now move in a larger volume and have more randomness of position. From 1ST law of thermodynamics, we have that dQ PdV nRT dV V We see that the fractional increase in the volume is given by dV dQ V nRT dQ dV C dS T V dV is a measure of the increase in disorder and is dQ T V or the Entropy (S). We can therefore see that entropy is a qualitative measure of disorder or randomness of a system since any heat input into a system creates disorder and increases the entropy. The fractional increase Definition of Entropy We can therefore define Entropy as a measure of randomness or disorder and it is a thermal property of a system that increases in every irreversible process but remains constant in a reversible (adiabatic) process (just as temperature remains constant in an isothermal process). 4.4 Entropy and 2ND Law of Thermodynamics The principle of increasing Entropy states categorically that “The Entropy of an irreversible processes always increases; It never decreases”, since in such processes, there is an increase in disorder. This statement can be written as 85 SPH 302: Thermodynamics dSirrev >0. On the other hand, for a reversible (adiabatic process, dQ = 0), the Total entropy change is always zero i.e., dSRev = 0. From these expressions, we can obtain the general statement for the definition of entropy as: Irreversible dS 0 [for any cycle] Reversible dQ T But since dS TdS dQ 0 ………………………(4.2) Equation (4.2) gives the direction of transformation from one state to another and we see that any such direction should be such as to increase the Entropy. This is the general form of the Second Law of Thermodynamics. Thus the second law can be stated quantitatively in terms of entropy as follows: “No natural process is possible in which the total entropy decreases when all systems taking part in the process are included”. This is equivalent to the “Engine” and/or the “Refrigerator” statement of the 2 ND law. The above statement implies that heat will not flow from a colder body to a hotter body without the application of work (the imposition of order) to the colder body. Secondly, it is impossible for any device operating on a cycle to produce net work from a single temperature reservoir without ejecting any heat to the colder reservoir. As a result, there is no possibility of a perpetual motion system. Finally, it follows from the second law of thermodynamics that the entropy of a system that is not isolated may decrease. An air conditioner, for example, may cool the air in a room, thus reducing the entropy of the air of that system. The heat expelled from the room (the system), involved in the operation of the air conditioner, will always make a bigger contribution to the entropy of the environment than will the decrease of the entropy of the air of that system. Thus, the total of entropy of the room plus the entropy of the environment increases, in agreement with the second law of thermodynamics. Take Note; Significance of Entropy Since Entropy is a measure of randomness or disorder in a system, it thus provides us with a criterion of reversibility and thereby enable us determine the direction of naturally occurring thermodynamic processes. 86 SPH 302: Thermodynamics Take Note; Significance of 2nd Law We see from Equation (4.2) that the 2 ND law limits the availability of energy and the way in which it can be used and converted. It is the law of nature describing the direction of natural thermodynamic processes. Take Note: Entropy of the universe Since all natural processes taking place in the universe are irreversible i.e., processes involving friction, free expansion and even life itself is irreversible, this essentially means that the entropy of the universe is increasing i.e., dS[Universe] > 0. The universe is therefore running down and we are approaching the “heat death” of the universe. Is this true? 4.4.1 The central Equation of Thermodynamics. From the first law of thermodynamics, we have that dQ = dU + PdV Combining the 1ST and 2ND law of thermodynamics, we get TdS dU + PdV [For any process] ……………..…(4.3) Equation (4.3) holds for all processes (both reversible and irreversible) since all the functions in the equation are state function. This is called the central Equation of Thermodynamics. 4.4.2 Entropy and degradation of energy From the Central Equation of thermodynamics (TdS dU + PdV), we have that, dW TdS dU This expression implies that the system does less work in an irreversible process (since TdS > 0). Hence, of the available energy, some must go into increasing the Entropy (S) and hence the energy is degraded in that it is less useful for doing work. The energy involved in the entropy change is tied up in the random arrangement and thermal motion of the atoms, it is often referred to as unavailable energy. For a reversible process, dSsystem = -dS surrounding i.e., total entropy change in system + surrounding is zero. More work is done in this case. 87 SPH 302: Thermodynamics Worked Examples 1. One kilogram of ice at 0 oC is melted and converted to water at 0 oC. Compute the change in Entropy. Take the specific latent heat of fusion of ice (SLHF) as = 333,624.2 JKg-1. Solution Since the temperature remains constant, S S 2 S1 1 Q 333.624.2 dQ 1222.1JK 1 T T 273 2. 1 kg of water at 0oC is heated to 100oC. Compute the change in entropy Solution From dQ mcdT T2 T2 1 1 T dS dT dS mc mc ln 2 1.3110 3 JK 1 T T1 T T T 3. Compute the entropy change that occurs when 1kg of water at 100oC is mixed with 1 kg of water at 0oC. Solution We shall assume the entropy of water is zero when it is in the liquid state at 0oC. From the last example, the entropy of water at 100 oC is 1310.85 JK-1 and the entropy of 1kg of water at 0oC is zero. Entropy of the system before mixing = 1310.85JK-1 After the hot and cold water have been mixed, we have 2 Kg of water at 50oC. The entropy of the system is now: mc p ln 323 323 2 x4200 x ln 1412.72 JK 1 273 273 Total increase in entropy = 1412.72 1310.85 101.87 JK 1 88 SPH 302: Thermodynamics 4.5 Entropy and the performance of heat Engines: T-S Diagrams We saw in lecture 1 that thermodynamic changes in the state of a substance can be represented by plotting P-V diagrams. Such changes can also be represented on Temperature – Entropy or (T – S) diagrams. The T – S diagram can be obtained from P – V diagram as shown below. For example, Figure 4.1 shows a Carnot’s reversible cycle abcd on a P-V diagram while Fig 4.2 shows the same cycles on a T-S diagram whereby the resulting T-S diagram is a rectangle abcd. The isothermal process ab in fig 4-1 is represented by a straight line on a T-S diagram in Fig 4-2. For the adiabatic process (dQ = 0) such that dS = 0. These processes are represented by isentropic lines (dS = 0) as shown in Figure 4-2 below. P V1, P1 a Isothermal expansion, Th Qh b Adiabatic compression V4, P4 V2, P2 Adiabatic expansion d c QC Fig. 4-1. The Carnot Cycle on a P-V diagram V Isothermal expansion, Th T Tc V3 , P 3 Qh a b Isentropic dS2 = 0 Isentropic dS1 = 0 Th d QC S1 c S2 S Fig. 4-2. The T-S Diagram of a Carnot cycle 89 SPH 302: Thermodynamics Along the isothermal expansion ab, the gain in entropy of a working substance is dS = S 2 S1 Qh Th ………………………….(a) Along bc and along da, there in no change in Entropy, dS = 0 Along cd, loss in the entropy of the working substance is dS = S 2 S1 QC TC …………………………(b) Combining Equations (a) and (b) we get Qh QC Th TC S 2 S1 Where Qh – QC = external work done in the reversible cycle and Th TC S 2 S1 = Area of rectangle on T-S diagram. Thus, the net heat absorbed is area adcd = external work done in a reversible Carnot’s cycle. 4.5.1 Efficiency of Carnot’s Engine on T-S diagram The efficiency of a Carnot engine can be obtained using a T-S diagram as follows Efficiency, External work in each cycle Heat absorbed from source Qh QC Th TC S 2 S1 Qh Th S 2 S1 Or 1 T h TC Th TC Th …..………………………………..(4.5) 90 SPH 302: Thermodynamics Worked Examples 1. A fridge working on the reversed Carnot cycle has a power requirement of 5 KW. If the maximum and minimum temperatures in the cycle are 40 oC and 10 oC respectively. Draw a T-S diagram and determine (i) The coefficient of performance (ii) The rate of heat extracted from the cold space (iii) The change in entropy for each of the processes in the cycle, given a refrigerant mass flow rate of 0.32 Kg/s Solution Given W = - 5 Kw, TC = 273.15 + 40 = 313.15 K, TC = 273.15 – 10 = 263.15K T Q1 4 3 T2 1 Q2 T1 2 S Reversed Carnot Cycle T2 5.263 T1 T2 Q (b) From COP Q1 COP WNET 26.315KW WNET (a) COP (c) From 2ND law, dS dQ T Along Process 1 2 2 Q Q dQ 1 1 dS dQ 1 1 0.3125 KJ Kg K T1 1 T1 mT1 1 T Along Process 2 3 and 41, it is isentropic S.T dS = 0 Along Process 32 dS Q2 mT2 dS Q2 0.3125 KJKg 1 K 1 mT2 but Q2 dW Q1 31.315kW 91 SPH 302: Thermodynamics 4.6 Third Law of Thermodynamics The third law of thermodynamics is concerned with the limiting behaviour of systems (or Entropy) as temperatures approach absolute zero. All Statements of the 3RD law of thermodynamics (i.e., Plank’s statement, Simon’s statement and The Nernst statement) point out that the entropy of a system tends to zero at absolute zero of temperature. The Nernst Statement It states that : “The Entropy change in a process associated with a change in the external parameters (e.g., P and T) tends to zero as temperatures approach absolute zero” i.e.,dS 0 as T 0. This means that as T 0, a strange behaviour is expected in materials i.e., an orderly arrangement of molecules should occurs in materials. For example, ice is more orderly than water. The reason for this orderly arrangement can be explained from a microscopic viewpoint (Quantum mechanics) where is assumed that particles of a system can exist in different quantum states with different discrete energies. However, as T 0, all particles pack in the lowest energy levels (ground state) i.e., assuming they are bosons. Since the ground state in nondegenerate, g = 1, and the number of ways the particles can be arranged is 1. Hence, there is complete order and S = 0. In real cases however, the decrease in S depends infact on the number of states per unit energy range (density of states) with energy rather than just on the occupancy of the ground level as our simple theorem suggest. In this case, S may not fall to zero as T 0 for both fermions and bosons (refer to your statistical mechanics for details). 4.6.1 Some Consequences of the 3RD Law As a consequence of dS 0 as T 0, certain measurable parameters vanish at the absolute zero of temperature. Typical examples are: (a) The Heat Capacity (CV) vanishes as T 0 Recall the heat capacity at constant volume is given by dQ S S CV T dT V T V ln T V Thus, as T 0, dS 0 but d(lnT) 0 since lnT . Hence, CV should 0 as T 0. Similar arguments hold for other heat capacities. 92 SPH 302: Thermodynamics (b) The Thermal expansion Coefficient vanishes as T 0 The volume cubic expansivity (or isobaric expansivity), is defined as the fractional increases in the volume per unit temperature rise i.e., 1 dV V dT P By suing Maxwell’s relation, this becomes 1 S V P T Thus, as T 0, dS 0 and 0. (c) Slope of phase boundary in 2ND order phase transition vanishes as T 0 Recall: The Slope of phase boundary in 2ND order phase transition is given by: P S T V Thus, as T 0, dS 0 and P or the slope 0. This is observed in T 4 He as shown in the figure below. P Solid Helium Slope = 0 Liquid Helium II Liquid Helium I 2.2K K Fig. 4-6. The phase boundary of liquid He as T 0. Questions Can you now list some of the consequences of the 3rd law of thermodynamics 93 SPH 302: Thermodynamics 4.6.2 The Unattainability of Absolute Zero The other statements of 3RD law says that “It is impossible to reach absolute zero even on using a finite number of processes”. Although much progress has been made in recent years to reach extremely very low temperatures e.g., the super fluid of 3He, the attainability of absolute zero is still elusive 4.7 Summary In this lecture, we have learned that: 1. Entropy is a measure of randomness or disorder and is a thermal property of a system that increases in every irreversible process 2. The change in Entropy (dS) of a system is given by dS dQR dT 3. The Principle of Increasing Entropy states that:- “The entropy of a thermally isolated system increases in every irreversible process and is unaltered in a reversible process” i.e., dS 0 4. The significance of Entropy is that it provides us with criteria for reversibility and enables us determine the direction of naturally occurring processes 5. The 2nd law describes the direction of natural thermodynamic processes and states that “No natural process is possible in which the total entropy of the system decreases” 6. The central Equation of Thermodynamics is given by TdS = dU + PdV 7. The Third Law of thermodynamics says that “The Entropy change of a system tends to zero as temperatures approach absolute zero” i.e.,dS 0 as T 0. 8. Some of the consequences of the 3RD law are (a). The Heat Capacity (CV) vanishes as T 0 (b) The Thermal expansion Coefficient vanishes as T 0 (c) Slope of phase boundary in 2ND order phase transition vanishes as T 0 9. Attainability of absolute zero is still elusive 94 SPH 302: Thermodynamics Questions for Discussion 1. Discuss the change in entropy of a gas that expands (a) at constant temperature and (b) adiabatically 2. Discuss the entropy change that occurs when you (a) bake a loaf of bread and (b) consume the bread. 3. Discuss three common examples of natural processes that involve an increase in entropy. Be sure to account for all parts of each system under consideration. 4. If a supersaturated sugar solution is allowed to evaporate slowly, sugar crystals form in the container. Hence, sugar molecules go from a disordered form (in solution) to a highly ordered crystalline form. Does this process violet the second law of thermodynamics?. Explain. 5. Suppose your roommate is “Mr. Clean” and tidies up your messy room after a big party. Because your roommate is creating more order, does this represent a violation of the second law of thermodynamics? 6. “Energy is the mistress of the universe and Entropy is her shadow”. Writing for an audience of general readers, argue for this statement with examples. Alternatively, argue for the view that “Entropy is like a decisive hands-on executive instantly determining what will happen, while energy is like a wretched back-office bookkeeper telling us how little we can afford”. 7. When we put cards together in a deck or put bricks together to build a house, for example, we increase the order in the physical world. Does this violet the second law of thermodynamics? Explain. 8. “The first law of thermodynamics says that you can’t really win, and the second law says you can’t even break even.” Explain how this statement applies to a particular device or process; alternatively, argue against the statement. Memorable Quotes ABILITY is what we are capable of doing, MOTIVATION determines what we do, and ATTITUDE determines how well we do it. -Lou Holtz 95 SPH 302: Thermodynamics PROBLEM SET 4 1. 2.0 Kg liquid water at 363 K is mixed adiabatically and at constant pressure with 3.0 Kg liquid water at 283 K. Determine the total entropy change resulting from this process. 2. 1 mole of an ideal gas for which CV = 25.12 and CP = 33.44 J mol-1 K, expands adiabatically from an initial state at 340 K and 500 KPa to a final state where its volume has doubled. Find the final temperature of the gas, the work done and the entropy change of the gas for (ii) a reversible expansion (iii) a free expansion of the gas into an evacuated space (Joule's expansion). 1. The specific heat capacity of a solid at low temperature is given by C aT bT 2 where a and b are constants Determine the Entropy of the solid as a function of temperature if the entropy is zero at T = 0oC. 2. 10 Kg of water at 20 oC is converted to superheated steam at 250 oC at constant atmospheric pressure. Determine the Entropy change of water. [C P (liquid) = 4180 J/Kg oC; CP (vapour) = 1670 + 0.4904 T + 1.86 106 T-2; Latent heat of vapourization (100oC) = 22.6 105 J/Kg]. 3. One gram of water when converted to steam at atmospheric pressure occupies a volume of 1671 cm3. The latent heat of vapourization at this temperature is 539 cal/gm. (i) Compare the volume of steam with the volume that would be occupied at this temperature and pressure if the water vapour were an ideal gas (ii) Compute the increase in the internal energy U, and the Entropy S when one gram of water is evaporated at this temperature and pressure. 4. In a refrigerator, the cooling chamber is maintained at 290 K while the outside temperature is 305 K. The motor (located outside) has compression cylinders operating at 320 K and the expansion coils inside the chamber operating at 280 K. If the motor operates reversibly (i) Give a schematic diagram illustrating the sequence of events (ii) Calculate the efficiency of the refrigerator (iii) How much work must be done for each transfer of 500 J of heat from the chamber? (iv) What entropy change occurs inside and outside the chamber for this amount of refrigeration? 96 SPH 302: Thermodynamics 4.8 References 1. Finn C. B. J., Thermal Physics, 1986. 2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics., S. Chand & Company Ltd, New Delhi. 3. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons, Inc., NY., Pg. 517-561. 97 SPH 302: Thermodynamics Lecture 5 THERMODYNAMIC POTENTIALS AND MAXWELL’S RELATIONS Outline 5.1 5.2 5.3 5.4 5.5 5.6 Introduction Objectives Maxwell’s Relations Maxwell’s Relations using Cyclic rule Applications of Maxwell’s Relations Summary 5.1 Introduction A mechanical system is said to be in a stale equilibrium when the potential energy of the system is minimum. This means that the system must proceed in such a direction so as to acquire minimum potential energy. This is what we observe in nature, viz. water flows from a higher to lower level, electric current flows from higher to lower potential, heat flows from higher to lower temperature, a body falls from higher to lower potential due to gravitational field and so on. Likewise, in thermodynamics, the behaviour of the state functions namely; internal energy (U), Helmholtz Free Energy (F), Enthalpy (H), and Gibbs Free Energy (G) is similar to potential energy in mechanics. To achieve thermodynamic equilibrium, the direction of thermodynamical processes must be in such a direction as to minimize the respective thermodynamical function (potential). Since the four functions U, F, H and G play in thermodynamics the same role as played by potential energy in mechanics, they are called thermodynamic potentials and all have the dimensions of energy. 98 SPH 302: Thermodynamics The four thermodynamic potentials are all functions of their natural thermal state variables (temperature T or entropy S ) and their mechanical state variable (pressure P or volume V ). For example, the internal energy (U) is given by U = TS Thermal Variable – PV Mechanical Variable Therefore, the thermodynamic variables associated with a state function can be derived from the thermodynamic functions by their differentiations with respect to the independent variables associated with them. The resulting set of equations are called Maxwell’s Equations In this lecture, we shall derive some (four) useful general thermodynamic relations between the four thermodynamic variables viz: P, V, T, and S referred to as the four Maxwell’s equations and discuss the significance of each of the thermodynamic potentials. 5.2 Objectives At the end of this lecture, you should be able to: 1. Define the four thermodynamic potential (U, F, H and G) and state their significance 2. Derive Maxwell’s equations from the thermodynamic potentials and also by using the cyclic rule 3. Apply Maxwell’s equations to solve problems related the mechanical behaviour of materials 5.3 Maxwell’s Relations In lecture 2, we introduced the thermodynamic potential (U) i.e., internal energy of a system and in lecture 4, we introduced the concept of Entropy (S). By combining the first and the second laws of thermodynamics, we came up with the central equation of thermodynamics given by TdS dU PdV Though this equation is quite powerful, it is not well suited in analysis of some certain thermodynamic processes. It is therefore convenient to introduce three additional thermodynamic potentials (also called state functions), all closely related to (U) and all having dimensions of energy. These are, the Enthalpy (H), the Helmholtz Free Energy (F), and the Gibb’s function (G). There is also a fifth 99 SPH 302: Thermodynamics function, the chemical potential ( ), which is useful in discussing the thermodynamics of open systems where the mass of the system is not constant. The four thermodynamic potentials are defined in terms of their natural thermal state variables (T and S) and mechanical state variables (P and T). (for a system with TWO degrees of freedom) as follows: Internal Energy; U = TS – PV Enthalpy; H = U + PV Helmholtz Free Energy; F = U – TS Gibbs Free Energy; G = H – TS = U – TS + PV We see that the four potentials U(S,V), F(T,V) H(S,P) and G(P,T) are functions of thermodynamic variables S, T, P and V. Thus, these thermodynamic variables can be derived from the thermodynamical functions by their differentiations with respect to the independent variables associated with them. The resulting is the equalities of each of the four thermodynamic potentials w.r.t their natural variables (T,S,P and V) and is called Maxwell’s Equations. Let us now derive the Maxwell’s equation from the thermodynamical potentials: Take Note Since the state of a system can be specified by any pair of the four quantities, viz, pressure (P), volume (V), temperature (T), and entropy (S). Taking two of the four state variables at a time, we see that there are six possible pairs, i.e., (P,V), (P,T), (P,S), (V,T), (V,S) and (T,S). However, the pair (P,V) is connected with composite and is exact differential quantity dW as dW = PdV and the pair (T,S) with dQ as dQ = TdS. Hence these two pairs can be eliminated. Thus, we are left with four pairs of the thermodynamical variables (P,T), (P,S), (V,T), and (V,S) and corresponding to each pair, we can write a thermodynamical relations known as Maxwell’s Thermodynamic relations Questions 1. What is a thermodynamic potential 2. Write down the four Thermodynamical potentials in terms of their thermodynamic variables 100 SPH 302: Thermodynamics 5.3.1 Internal Energy and The First Maxwell’s Equation Consider the central Equation of Thermodynamics given by dU TdS PdV (5.1) This suggests that U is a function of Entropy (S) and volume (V) and we write U in terms of its independent pair of variables (its natural variables) S and V i.e., U U ( S ,V ) Thus, the change in U is thus given by U U dU dS dV S V V S (5.2) By comparing the two expressions (5.1) and (5.2) we see that; U T S V and U P V S (5.3) Thus, if U is known in terms of its natural function S and V, then T and P can be found from the natural functions. Now, since U is a state function, dU is an exact differential. By using the condition that the differential to be exact (see condition below) we get T P V S S V (5.4) This is the First Maxwell’s Equation. Notice that the natural variables of U, are S and V which are the qualities appearing outside the partial differentials. Activity 5.1 Beginning from the central equation of thermodynamics, derive the first Maxwell’s relation 101 SPH 302: Thermodynamics Take Note: Condition for a differential to be Exact. Consider a mathematical function G x, y of x and y that takes unique values for each pair of x and y. when x and y change by dx and dy, the infinitesimal change in G is G G dy dG dx x y y x Xdx Ydy where X and Y are the general functions of both x and y The condition for to be exact is that the differential of the coefficient of dx with respect to y while holding x constant should be equal to the differential of the coefficient of dy when differentiated with respect to x while holding y constant i.e., X y Y 2G x x y y x We see that this two partial differentials are equal, the order of the differential being in material. Thus the necessary condition for dG to be exact differential is that X y Y x x y This condition will be used in derivation of Maxwell’s Relations eeeeeeeeeeeeeeeeeeee Activity 5.2 Consider a differential given by dG 2 xy dx 4 x y dy . Show that dG is an exact differential 4 Solution X 2 xy 4 X y and 2 3 y 4x 2 y 3 Y 8 xy 3 x y x Hence dG is an exact differential 102 SPH 302: Thermodynamics Questions Check whether the following differential is exact? dF xy 4 dx 4 x 2 y 3 dy Heat capacity at constant volume, CV Let’s now write down an expression for CV in terms of S and T. Now from the central equation of Thermodynamics, we have that for an isochoric reversible process (dV =O) TdS dU CV dT (5.5) Where Cv is the molar heat capacity It follows that dU dS CV T dT V dT V (5.6) 5.3.2 Enthalpy and The Second Maxwell’s Equation We have defined enthalpy (H) as; U PV (5.7) Differentiating we obtain; dH dU PdV VdP Or dH TdS VdP (5.8) This suggests that H is a function of Entropy (S) and pressure (P) and we write H in terms of its independent pair of variables (its natural variables) S and P i.e., H H ( S , P) Thus, the change in H is thus given by d dS dP S P P S (5.9) By comparing the two expressions (5.8) and (5.9) we see that; 103 SPH 302: Thermodynamics T S P and H V P S (5.10) Which means that if we know H in terms of its natural variables, S and P we can find both the temperature T and the volume V. Using the condition that dH is an exact differential we have? T V p S S P (5.11) This is The Second Maxwell’s Equation. Take Note Significance of Enthalpy From dH = TdS + VdP We see that for a reversible isobaric process (dP = 0), then dH = dQP The Enthalpy change is the heat evolved in a reversible isobaric process such as a chemical reaction. If dH is +ve, heat is absorbed during the reaction and the reaction is called endothermic. If dH is –VE, heat is evolved during the reaction and the reaction is called exothermic. Heat capacity at constant pressure, Cp Likewise, we can also write the expression for CP in terms of S and T. Now from the central equation of Thermodynamics, we have that for an isochoric reversible process (dP =O) TdS dH CP dT (5.12) Where CP is the molar heat capacity Thus, if the temperature dependency of the heat capacity is known, the change in heat content which occurs during the heating of a given material can be calculated from the following expression T2 dH C P dT T1 It follows that 104 SPH 302: Thermodynamics dH dS CP T dT P dT P (5.13) 5.3.3 Helmholtz Free Energy and The Third Maxwell’s Equation For processes where temperature and volume are the important variables, we introduce the state function: The Helmholtz Free Energy (F) defined by F U TS (5.14) For an infinitesimal change (differentiating), we have; dF dU TdS SdT But dU TdS PdV dF PdV SdT (5.15) From which we see that the natural variables of F are V and T. Which suggests that? F = F(V,T) (5.16) Thus, the change in F is thus given by F F dF dV dT V T T V (5.17) Comparing the coefficients of the two expressions for dF we see that; F P v T and F S T V (5.18) Hence if we know F in terms of V and T we can calculate P and S. As F is a function of state, dF is an exact differential and using the conditions for an exact differential, we obtain; P S T V V T (5.19) This is the Third Maxwell’s Equation, with the natural variables V and T appearing outside the differentials. Take Note Significance of Helmholtz Free Energy From dF = -PdV - SdT We see that for a reversible isothermal process (dT = 0), then 105 SPH 302: Thermodynamics Activity 5.3 The Helmholtz Free Energy (F) defined by . F U TS . Derive the related Maxwell’s Equation and state the significance of Helmholtz Free Energy (H) 5.3.4 Gibbs function (G) and The Fourth Maxwell’s Equation This state function is designed for use in problems where pressure and temperature are the important variables. It finds enormous application in the study of equilibrium conditions in systems that are mixtures of two phases such as the ice and water mixture. The Gibbs function is defined as; G H TS (5.20) For an infinitesimal change; dG dH TdS SdT But dH dU PdV VdP dG dU PdV VdP TdS SdT But also TdS dU PdV dG VdP SdT (5.21) From which we see that the natural variables of G are thus P and T. Thus, we can write G in terms of its independent pair of variables (its natural variables) P and T as; 106 SPH 302: Thermodynamics G = G(P,T) (5.22) From which we obtain the differential; G G dG dP dT P T T P (5.23) Comparing the two equations (5.21 and 5.23) for dG it shows that G V P T and G S T P (5.24) Hence if we know G as a function of P and T we can find the volume and the entropy. Since G is a state function, dG is an exact differential, thus; V S T P P T (5.25) This is the Fourth Maxwell’s Equation with the natural variables P and T appearing outside the partial differentials. Take Note: Significance of Gibbs Function (G) From dG dH TdS SdT We see that in a chemical reaction, P and T remain constant such that dG dH –TdS or dG = dH - dQ Now, a chemical reaction requires heat (dQ) in order to take place. Hence: If dQ > dH, then dG < 0 and the reaction is spontaneous. If dQ = dH, then dG = 0, and the reaction is at equilibrium If dQ < dH, then dG > 0, and the is no reaction Thus, the Gibbs free energy, G is a measure of the available energy. It represents the driving force for a reaction. For a reaction to occur spontaneously, the free energy change must be negative (the free energy of the system decreases). At equilibrium, the Gibbs function is at its minimum (dG = 0). “The condition for thermodynamic equilibrium in a system in thermal and mechanical contact is that the Gibbs function is a minimum”. Example 5.1 107 SPH 302: Thermodynamics Consider the reaction: ice water . For ice to melt, dS has to increase (dS >0) as the system moves from order to disorder. Also, dH is +ve since ice requires latent heat in order to melt. dH goes to increase the potential energy of the molecules as they break free from their bound positions in solid. Now When heat is supplied to ice at very low temperatures ( < 0 oC), then, dH > TdS, hence dG > 0 and ice does not melt. At high temperatures, ( > 0 oC), TdS > dH and dG < 0. Hence, ice melts into water. At ( = 0 oC), dH =TdS, and dG = 0. At this temperature, water and ice coexist. Activity 5.4 1. Show that the enthalpy change in an isobaric chemical reaction is equal to the heat of reaction. 2. State the significance of G The above four are by no means the only Maxwell relationships. When other work terms involving other natural variables besides the volume work are considered or when the number of particles is included as a natural variable, other Maxwell relations become apparent. Take Note: Significance of Thermodynamic Potentials As mentioned earlier, a mechanical system is said to be in state of equilibrium when the potential energy of the system is minimum. This means that for thermodynamic equilibrium, the behaviour of the internal energy (U), Helmholtz free energy (F), enthalpy (H), and Gibbs free energy (G) is similar should proceed in a such a direction such as to minimize these potentials. As we have seen, the direction of isothermal-isochoric process is to make Helmholtz free energy (F) minimum. In isothermal-isobaric process, Gibbs free energy (G) tends to be minimum while in an isobaric-adiabatic process, the Enthalpy (H) tends to be minimum. Questions State the significance of thermodynamic potential. Why are they important in thermodynamics 108 SPH 302: Thermodynamics 5.4 Derivation of Maxwell’s Relationships using cyclic rule Recall the four thermodynamic potentials as functions of their natural thermal and mechanical variables (for a system with TWO degrees of freedom) are given as follows: Internal Energy; U = TS – PV Enthalpy; H = U + PV Helmholtz Free Energy; F = U – TS Gibbs Free Energy; G = H – TS = U – TS + PV Their differential forms are dU = TdS – PdV dH = TdS + VdP dF = -SdT – PdV dG = -SdT + VdP where the two terms on the right hand side correspond to the 2 degrees of freedom The corresponding Maxwell’s equations are From the internal energy, U; 2U T P V S S V VS From enthalpy, H; T V P S S P From the Helmholtz function, F; P S T V V T From the Gibb’s function, G; V S T P P T These equations relating the partial differentials of the thermodynamic variables P, V, T and S can be easily derived using the Mnemonic as follows STEP 1 Write P, V, T and S as a mnemonic as follows 109 SPH 302: Thermodynamics S P V T The phrase for remembering the sequence is “Let’s join the Society for the Prevention of Teaching of Vectors”. STEP 2 To derive the Maxwell’s equation, all one has to do is to go round this array cyclically in opposite directions. For example the First Maxwell’s equation would follow from starting at T as; S P S V P T V T T P V S S V STEP 3 The sign is positive if T appears with P in the differential e.g., S P S V P T V T T V P S S P Activity 5.5 Using the cyclic rule, derive the four Maxwell’s equations 110 SPH 302: Thermodynamics 5.5 Application of Maxwell’s Relationships to simple thermodynamic systems The state of a system can be specified by any pair of quantities, viz, pressure (P), volume (V), temperature (T), and entropy (S). In solving any thermodynamical problem, the most suitable pair of variables as given by Maxwell’s relations is chosen. Worked Example 5.2 The central equation of thermodynamics is given by TdS dU PdV . If the Entropy S is a function of T and V i.e., S = S(T,V), using Maxwell’s relationships, show that the above equation can be written as dP TdS CV dT T dV dT where symbols have their usual meanings. Solution If S S T ,V S S dS dT dV T V V T By using the mnemonic S S P P V T V S P V T T V T S P TdS T dT T dV T V T V S CV T V But T dP TdS CV dT T dV dT 111 SPH 302: Thermodynamics 5.6 Summary The four thermodynamic potential are Internal Energy; U = TS – PV Enthalpy; H = U + PV Helmholtz Free Energy; F = U – TS Gibbs Free Energy; G = H – TS = U – TS + PV The respective Maxwell’s equations are 2U T P V S S V VS From U; T V P S S P From H; From F; P S T V V T V S T P P T From G; The Enthalpy change is the heat evolved in a reversible isobaric process such as a chemical reaction. The dF represents the maximum amount of work done by the system in an isothermal process and is equal to the decrease in the Helmholtz function The condition for thermodynamic equilibrium in a system in thermal and mechanical contact with a heat and pressure reservoir, is that the Gibbs function is a minimum. Maxwell’s equations can be applied to solve problems in any mechanical system such as magnetic systems, elastic rods etc. 5.7 References 8. Finn C. B. J., Thermal Physics, 1986. 9. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics., S. Chand & Company Ltd, New Delhi. 10. Resnick R., Halliday D., Krane K. S., PHYSICS Vol 1. John Wiley and Sons, Inc., NY., Pg. 545-561. 112 SPH 302: Thermodynamics PROBLEM SET 5 1. The central equation of thermodynamics is given by TdS dU PdV . If the Entropy S is a function of T and V i.e., S = S(T,V), using Maxwell’s relationships, show that the above equation can be written as dV TdS C P dT T dP dT P where symbols have their usual meanings. 2. One gram of water when converted to steam at atmospheric pressure occupies a volume of 1671 cm3. The latent heat of vapourization at this temperature is 539 cal/gm. (iii) Compare the volume of steam with the volume that would be occupied at this temperature and pressure if the water vapour were an ideal gas (iv) Compute the increase in the internal energy U, the Entropy S, the Enthalpy H, and Gibbs function G when one gram of water is evaporated at this temperature and pressure. (v) Using your answer for G above, what can you say about the nature of this process 3. The Gibbs function of 1 mole of a certain gas is given by G RT ln P A BP C P2 P3 D . 2 3 where A, B, C and D are constants. Find the equation of state of the gas. 4. Show that the difference between the isothermal and the adiabatic compressibility can be given by V 2 K T K S T CP 5. A wire undergoes an infinitesimal change from one equilibrium state to another. Obtain expressions for the infinitesimal changes in length and in tension in terms of the Young’s modulus and the thermal expansivity. Memorable Quotes “Ignorant people make life so difficult for some of us” -Albert Einstein 113 SPH 302: Thermodynamics Lecture 6 PHASE CHANGES AND PHASE EQUILIBRIA Outline 6.1 6.2 6.3 6.4 6.5 6.6 6.7 Introduction Objectives Systems, phases and components Phase transitions First order phase changes Second order phase changes Summary (a) a glass of water showing the three phases of water: ice, water and vapour (b) ice, water and steam coexisting naturally 114 SPH 302: Thermodynamics 6.1 Introduction The chemical and physical properties of most materials are related to the number, composition, and distribution of the phases present. Temperature, pressure, and composition are the principle variables which determine the kinds and amounts of the phases present under equilibrium conditions. For example, water changes to vapour when heated to a temperature of 100 oC at a pressure of 760 mmHg. On the contrary, at low temperatures of 0 oC, water turns to ice. Ice, water and steam are the three phase of a single component called water. An understanding of phase equilibrium in material systems is central to the utilization and development of materials particularly in refractories, glass and other high-temperature technologies. Phase equilibrium address significant questions related to the flexibility and constraints dictated by forces of nature, on the evolution of phase assemblages in materials. Phase boundaries also assist in the evaluation of the service stability of materials specially ceramics, both in the long and short time frames. Thus, knowledge of the stability of a material (e.g., ceramic or glass) component in high-temperature or high-pressure environments can often be obtained from an appropriate stable or metastable phase diagram. The phase rule developed by J. Willard Gibbs was derived from the first and second laws of thermodynamics and phase relationships can be presented in temperature-composition-pressure diagrams. In a system with more than one phase, each phase may be considered as a separate system within the whole. Thermodynamics parameters of the whole system may then be applied to the component phases to determine the equilibrium conditions for the two phases. In this lecture, we shall study the change of phase that occur in a substance as a result of change in temperature and pressure as well as the effect of Pressure on the melting and boiling point of a substance. We shall study the equilibrium conditions for existence of two phases and the characteristics of first order and second order phase changes respectively. We shall restrict ourselves to a single component system such as water since 2 or 3 component systems are more complex to analyze their phase changes. 6.2 Objectives At the end of this lecture, you should be able to 1. Explain what is a phase and what causes phase changes 2. 3. Explain the importance of phases changes in thermodynamic systems State the equilibrium conditions for existence of two phases 4. Explain the effect of pressure on the melting and boiling point of a substance 5. State the characteristics of first order and second order phase transitions. 115 SPH 302: Thermodynamics 6.3 Systems, Phases, and components Before proceeding with the development of the concept of equilibrium between phases and the derivation of the phase rule, it will be useful to define certain terms commonly used in a treatment of phase equilibrium. 6.3.1 Definitions Phase:- Any portion including the whole of a system which is physically homogeneous within itself and bounded by a surface so that it is mechanically separable from any other potions. Example, Ice, water and steam are the three phases of water. Components:- Are the smallest number of independently variable chemical constituents necessary and sufficient to express the composition of each phase present in any state of equilibrium. For example, in the water system, H2O is the component of the system while in the alumina-silica system, Al2O3 and SiO2 are the components of the system since all phases and reactions can be described by using only these two materials. Equilibrium:- In lecturer 1, we defined thermodynamic equilibrium as a state in which a system experiences thermal, mechanical and chemical equilibrium. A more comprehensive thermodynamic definition of equilibrium states that a system at equilibrium has a minimum free energy as defined by the Gibbs free energy (G). 6.3.2 Phase diagrams of One-Component System Since the change in phase of a substance is dependent mainly on P and T, phase diagrams are best illustrated on P-T diagrams. For pressure and temperature changes, typical P- T projections are as shown in Fig. 6-1. P (Pa) B C Melting freezing SOLID LIQUID Vapourization condensation Sublimation TP VAPOUR deposition TTP TC T (K) Fig. 6-1. Hypothetical P-T projections of one-component system 116 SPH 302: Thermodynamics TP is the triple point i.e., the temperature (TTP) at which all the three phases (liquid, solid and vapour) coexist. Point C is the critical point i.e., where liquid and vapour are indistinguishable. Below pt C, the system exists in two phases (liquid and vapour), while beyond C, (or temperature TC), the system is gaseous. At the triple point (TP), the solid and liquid are in equilibrium. If the volume were held constant as heat is added to the system, the solid would melt. The increased volume of the melt increases the pressure and temperature and as such, the equilibrium between the solid and liquid shifts to a position of higher pressure and temperature i.e., towards pt B on curve IB (Fig. 6-1). Similar projections can be worked out for solid-vapour and for liquid-vapour systems at equilibrium. 6.4 Phase Transitions 6.4.1 Equilibrium Conditions for Two Phases In the example above, we saw that a single component such as water can exist in 3 phases (solid, liquid and vapour) depending on the conditions of Pressure and temperature. We need to determine the conditions under which any 2 phases e.g. solid and liquid can exist in equilibrium. The conditions for thermodynamic equilibrium of any two phases of a system, is that the Gibbs function for both phases (phases 1 and phase 2) should be minimum. Thus by Applying Gibbs potential (G), it is possible to investigate the equilibrium between a liquid and its vapor or between any two phases of a substance. For example, consider a closed system containing a liquid in equilibrium with its saturated vapour. For equilibrium conditions, the temperature and pressure are equal in both the phases and must remain constant throughout the phases. The thermodynamic coordinate V, S, U and G will be equal to the product of the specific value and the mass of the substance in that phases. Suppose the phases have masses M1 and M2, where M1 + M2 = M (total mass) and Gl and G2 are Gibbs functions for the two phases respectively. Then, for the whole system, G m1G1 m2G2 …………….…….(6.1) (NB: G is extensive variable i.e., dependent on system size. If a small quantity of the liquid changes into vapour, then the change in G is obtained by differentiating equation (6.1) giving dG m1G1 m2G2 ………….……….(6.2) 117 SPH 302: Thermodynamics But for equilibrium condition, dG = 0. m1G1 m2G2 0 But since the system is closed, the total mass is constant i.e., m1 + m2 = M (total mass) or m1 m2 G1 G2 …………………..……..(6.3) This shows that the equilibrium condition for existence of 2 phases is that their specific Gibbs functions (G) must be equal. Equation (6.3) is applicable to the processes of evaporation, fusion and sublimation. This equality of (G) leads to Clausius-Clapeyron Heat Equation which characterizes phase transition between two phases of matter. Activity 6.1 1. Explain what is a phase 1. Show that the condition for thermodynamic equilibrium of 2 phases in a system in thermal and mechanical contact with a heat and pressure reservoir, is that their Gibb’s function must be equal. 6.4.2 The Clausius-Clapeyron Equation The Clausius–Clapeyron relation is a way of characterizing a discontinuous phase transition between two phases of matter. On a pressure-temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of this curve and as such, this relation can be used to (numerically) find the relationships between pressure and temperature for the phase change boundaries. Entropy and volume changes (due to phase change) are orthogonal to the plane of the P-T diagram. To derive the Clausius–Clapeyron relation, let us consider a system in equilibrium such as a liquid and its saturated vapour, such that G1 G2 . If p and T are changed (by dT and dP respectively) as shown by a portion of a phase boundary in Fig. 6-2 below, but in such a way that the system is still kept in equilibrium, then dG1 dG2 as the equilibrium condition 118 SPH 302: Thermodynamics P B PHASE 1 (T+dT, P+dP) PHASE 2 A (T,P) T (K) Fig. 6.2. Portion of a phase boundary between two phases Remembering that (i.e., from Gibbs Relation) dG d VdP SdT Then But V1dP S1dT V2 dP S 2 dT V dP S dT V S dQ Q L T T T 1 V2 dP S 2 S1 dT ………………………..………(a) ……..…………..…….(b) where L = the latent heat of vapourization Substituting (b) into (a) we get dP L dT T (V2 V1 ) where dP dT ………………………….………(6.4) is the slope of the coexistence curve, and ΔV is the volume change of the phase transition. Equation (6.4) is the Clausius-Clapeyron equation for the slope of the phase boundary and is used to characterize first order phase transitions. 119 SPH 302: Thermodynamics Activity 6.2 Can you now deduce and state the first latent heat equation, dP L . dT T (V2 V1 ) 6.4.3 Applications of Clausius-Clapeyron Equation (a) Effect of Pressure on the melting point of a substance When a solid is converted into a liquid, there is a change in volume. Usually, the substance may expand on melting in which case the volume of the liquid (V L = V2) > the volume of the solid (VS = V1) or it may contract on melting such that VL < VS. We wish to look at the effect of pressure on these two cases. (i) if VL > VS i.e., for substances that expands on melting Then, dP > 0 i.e., the slope is +ve the melting point of the substance dT will increase with increase in pressure and vice versa as shown in Fig 6-3 below by the graph IB. P (Pa) D Under Normal P B dP 0 dT dP 0 dT CP LIQUID SOLID I VAPOUR TP TTP TC T (K) Fig. 6-3. Substances that expand and contract on melting 120 SPH 302: Thermodynamics (ii) if VL < VS i.e., for substances that contract on melting such as water and antimony, the volume of water formed is less than the volume of ice taken: In this case, dP < 0 or the slope is –ve the melting point of the dT substance will decrease with increase in pressure and vice versa as shown by graph ID. Thus, at high pressure, ice melts at a temperature lower than 0oC. NB. Ice melts at 0 oC only at a normal pressure of 76 cm of Hg. Hence, increase in pressure lowers the melting point of ice. (b) Effect of Pressure on the boiling point Let us now look at the effect of pressure on the boiling point. When a liquid is converted into a gaseous state, the volume V g (Vg = V2) of the gas is always greater than the corresponding volume VL (VL = V1) of the liquid i.e., Vg > VL. Therefore, dP is a +ve quantity. dT Thus, increase in pressure increases the boiling point of a substance and vice versa. For example, water boils at 100 oC only at 76 cm of Hg pressure. In the labs, while preparing steam, the boiling point is less than 100 oC because the atmospheric pressure is less than 76 cm of Hg. In pressure cookers, the liquid boils at a higher temperature because the pressure inside is more than the atmospheric pressure and this makes food cook faster. Questions Can you now explain the following 1. Will water on top of mount Kenya boil at a lower or higher temperature compared to the same water at the coast 2. Meat cocks faster in a pressure cooker than in a sufuria Worked Example 6.1 Calculate the change in the boiling point of water when the pressure is increased by 1 atmosphere. Boiling point of water is 373 K. Specific volume of steam = 1.671 m3 kg-1 and latent heat of steam is 2.268 106 J Kg-1. Solution Given V2 = 1.671 m3Kg-1, V1 = 110-3m3 Kg-1, T = 373K, dP = 1atm = 105 Nm-2. V2 – V1 = 1.670 m3Kg-1, Steam = 2.268 J/Kg. 121 SPH 302: Thermodynamics From dP L dT T (V2 V1 ) dT T (V2 V1 )dP 27.47 K L Worked Example 6.2 Find the increase in the boiling point of water at 100 oC when the pressure is increased by one atmosphere. Latent heat of vaporization of steam is 540 cal/gram and 1 gram of steam occupies a volume of 1677 cm3. Solution Given V1 = 1.00 cm3, V2 = 1,677 cm3, T = 373K, dP = 7613.6980 dynes cm-2. From dP L dT T (V2 V1 ) T (V2 V1 )dP 27.92 K L dT Increase in boiling point = 27.92K or 27.97 oC 6.5 First Order Phase Changes In the modern classification scheme, phase transitions are divided into two broad categories, namely First-order and Second order phase transitions. First-order phase transitions are those that take place at constant temperature and pressure and involve a latent heat. During such a transition, a system either absorbs or releases a fixed (and typically large) amount of energy as the temperature remains constant. The system is in a "mixed-phase regime" in which some parts of the system have completed the transition and others have not. Familiar examples are the melting of ice or the boiling of water (the water does not instantly turn into vapour, but forms a turbulent mixture of water and vapor bubbles). Definition Apart from the transference of heat, a more concise classification of phases transitions is based on Ehrenfest classification. According to Ehrenfest classification, phase transitions are classified based on the behavior of the thermodynamic free energy (G) as a function of other thermodynamic variables. Under this scheme, A first order phase transition is defined as one in which the Gibbs function (G) remains the same in both the phases at equilibrium. However, the first order derivative of the Gibbs function with respect to pressure and temperature change discontinuously at the transition point. 122 SPH 302: Thermodynamics The various solid/liquid/gas transitions are classified as first-order transitions because they involve a discontinuous change in density, which is the first derivative of the free energy (G) with respect to chemical potential. 6.5.1 Characteristics of 1St order phase transitions First order phase transition is characterized by (i) The Gibbs function remains the same in both the phases at equilibrium i.e., dG = 0 (since GS = GL). (ii) A transference of heat and a subsequent change in specific volume (V) and entropy (S) leading to discontinuity in the first order derivative of the Gibbs function with respect to pressure and temperature respectively at the G G and V are discontinuous are T P T P transition point i.e., S shown in the G-T and G-P curves of Figs 6-5b and 6-7b respectively. Activity 6.3 1. List the characteristics of a First order phase change 1. Give an example of a First order phase change 6.5.2 Variation of G, S and V in First Order Phase Transitions Since a first order phase is characterized by a transfer of latent heat, we thus expect a change in volume (V) and entropy (S). These changes are characterized by the Clausius-Clapeyron heat equation. (a) Change in S The entropy S can be obtained from the expression dG = VdP – SdT where we see that G S = slope of G vs. T curve T P The change in entropy (S) can be determined by considering the variation of the Gibb’s function (G), along a constant pressure section (XY) on the P-T diagram shown in Figure 6-4a below From the graph, the corresponding G-T plots for both the liquid and solid phases i.e., GL and GS are shown in Fig. 6-4b. Note that the G vs. T plots has a –ve slope as S is always +ve. At the phase boundary, (at temperature To), the Gibbs functions for the solid and liquid phases are equal (i.e., Gs = GL). For T < To, the solid phase is more stable, and the Gs curve is much lower in this section in order to minimize G. For T > To, the liquid phase is more stable, and the GL curve is much lower in this section compared to the Gs curve. 123 SPH 302: Thermodynamics G P X Y L S TP GL GS CP V To S L T (K) To Fig 6-4 (a) P-T curve of water T (K) (b) G-T curve along X-Y in Fig (a) The G-T curves of Figure 6-4(b) can be combined to give a general variation of G (= Gs + GL) shown in Fig. 6-5a. Note the discontinuity in the slope of this graph at To. The corresponding variation in Entropy (S) vs. T curve is shown in Fig 6-5b. Higher temperature phase has a higher entropy (i.e., SL> Ss). Note the discontinuity in S at To. S Discontinuity in Slope FIRST ORDER G S G T S L L To T (K) 6-5(a) General G-T curve To T (K) (b) S-T curve from Fig (a) (b) Change in V The volume of the phases (V) can be obtained from the expression dG = VdP – SdT G 0 = slope of G vs. P curve P T V Thus, the change in volume (V) can be determined from the variation of the Gibb’s function (G), along a constant temperature section (X’Y’) on the P-T diagram shown in Figure 6-6a below 124 SPH 302: Thermodynamics Figure 6.5(b) shows the G-T plots for both the liquid and solid phases i.e., GL and GS as function of P as obtained from Fig 6(a). Note that the G vs. P plots have a +ve slope since the volume of a substance should increase upon decrease in pressure. For P < Po, the liquid phase is more stable, and the GL curve is much lower in this section in order to minimize G. For P > Po, the solid phase is more stable, and hence GL curve is much lower in this section as compared to the G s curve. X’ P G Po L S CP TP GS V GL Y’ S L T (K) Po Fig 6-6 (a) P-T curve of water P(Mpa) (b) G-T curve along X’-Y’ in Fig (a) The general variation of the above G-T curves (G = Gs + GL) is shown in Fig. 67a. Note the discontinuity in the slope of this graph at Po. The slope of this graph gives the variation in specific volume (V) which is plotted as a function of P in Fig 6-7b. Note that higher pressure phase (solid phase) has a smaller specific volume (i.e., VL> Vs) since substances suffer a decreases in volume upon increase in pressure (obvious). The discontinuity in V at Po is a result of the discontinuity in Fig 6-7a. V L Discontinuity in V at Po G dP L S Po VL > V S V Discontinuity in Slope FIRST ORDER G P(Mpa) Fig. 6-7(a) General G-P curve along X’Y’ S Po P(Mpa) (b) V-P curve 125 SPH 302: Thermodynamics St Example of 1 order phase transitions When water is heated at 100 oC and 1 atmosphere, it changes from liquid to vapour. The density of water is 1000 Kg cm -3, while that of vapour is 0.6 kg m-3 representing an increase in volume and consequently an increase in the Entropy (S). Therefore, the transformation of water into vapour at constant temperature and pressure is a first order phase transition. Similarly, the transformation of ice into water at 0 oC and at 1 atmospheric pressure is a first order phase transition since the volume of water is less than the volume of ice due to increase in density. 6.6 Second Order Phase Changes Recent investigations have revealed that during second order phase transitions, there is no transfer of heat and hence, there is no change in volume. Second order phase transitions are defined as the phenomena that take place with no change in volume and entropy at constant pressure and temperature respectively. As such, second-order phase transitions are also called continuous phase transitions 6.6.1 Characteristics of 2ND order phase transitions Second order phase transition are characterized by (i) No transfer of heat and hence no change in specific volume (V) and entropy (S) (ii) No discontinuity in first order derivatives of the Gibbs function with respect to G = 0 T P pressure and temperature at the transition point i.e., S and V G = 0 S and V are continuous at the P T phase boundary as shown in Fig 6-8(b) and 6-9(b) respectively. (iii) 2G C S P 0 However, second order derivatives of G i.e., 2 T P T P T 2G V and 2 P T P T change discontinuously at the phase boundary as shown in Fig 6-8(c). 126 SPH 302: Thermodynamics S S CP T P Discontinuity in slope No Discontinuity in Slope SECOND ORDER G G T S S Superfluid Liquid 4He II L Normal Liquid 4He I L To T (K) To Fig 6-8(a) General G-T curve in 2ND order phase transition 2.2K G V P T No Discontinuity in Slope L L K (c) The anomaly in CP of 4He. (b) S-T curve from Fig (a) showing continuity in S SECOND ORDER G T (K) S S Po P(Mpa) Fig 6-9(a) General G-P curve Po P(Mpa) G vs. P curve showing P T (b) continuity in slope 6.6.2 Implications of second order phase changes The discontinuity in the second order derivatives of G implies that certain material properties such as CP, β and K that are derivatives of Entropy (S) and Volume (V) are discontinuous. For example, the heat capacity is defined by S CP T T P is discontinuous for liquid helium at 2.2K Likewise The volume cubic expansivity () and the bulk modulus defined by 127 SPH 302: Thermodynamics 1 dV 1 S V dT P V P T K 1 S V P T is discontinuous and is discontinuous This classification of phase changes (Ehrenfest Classification) can also be extended to higher order phase changes. Examples of 2ND Order Phase Changes 1. Transition in helium from normal liquid helium I to superfluid (liquid helium II) at very low temperature (λ = 2.2K). In a superfluid, there is complete absence of internal friction (viscosity) and a rotating fluid would go on rotating forever. 4He has two phases and its heat capacity has a discontinuity at 2.2K (see Fig. 6-10) 2. Transition of ferromagnetic material to a paramagnetic material at the curie point 3. Transition of superconducting metal into an ordinary conductor in the absence of a magnetic field S CP T P Superfluid Liquid Helium II Normal Liquid Helium I 2.2K K Fig. 6-10. The anomaly in CP of 4He. Activity 6.4 1. List the characteristics of a second order phase change 2. Give examples of second order phase transitions 128 SPH 302: Thermodynamics Take Note The superfluidity of 4He can be explained with aid of Quantum mechanics. 4He are bosons with zero integral spins (2P with opposing spins and and 2N with opposing spins and ). At low temperatures, all the 4He atoms pack in lowest same quantum states and their effect on each other is minimal-hence the superfluidity. On the other hand 3He are fermions (i.e., with ½ integral spins: for protons and for neutrons). As such, they do not pack in the same quantum state. 1.7 Summary In this lecture, we have learned that 1. A Phase is any portion including the whole of a system which is physically homogeneous within itself and bounded by a surface so that it is mechanically separable from any other potions. 2. The equilibrium condition for the existence of two phases is the minimum in the Gibbs function G. 3. The slope of the phase boundary of a substance is defined by the dP L Clausius-Clapeyron equation given by . dT T (V2 V1 ) 4. Increase in pressure increases the melting point and the boiling point of a substance and visa versa. 5. First order phase transition is characterized by (i) the Gibbs function remains the same in both the phases at equilibrium i.e., dG = 0 (since GS = GL). (ii) A transference of heat and hence a change in specific volume (V) and entropy (S) leading to discontinuity in the first order derivative of the Gibbs at the transition point. 6. Second order phase transition is characterized by (i) No transference of heat and hence no change in volume (V) and entropy (S). Therefore no discontinuity in the first order derivatives of the Gibbs function at the transition point. S S V (ii) Second order derivatives of G i.e., , and , T P P T T P V and change discontinuously at the transition point. P T 129 SPH 302: Thermodynamics PROBLEM SET 6 1. Calculate under what pressure ice freezes at 272K if the change in specific volume when I kg of water freezes is 91 10-6 m3. Given latent heat of ice = 3.36 105 JKg-1. 2. Calculate the depression of melting point of ice produced by one atmosphere increase of pressure. Given that latent heat of ice = 80 cal/gm and specific volume of ice and water at 0oC are 1.091 cm3 and 1.0 cm3 respectively. 3. Calculate the change in the melting point o ice when it is subjected to a pressure of 100 atmospheres. 4. Calculate the change in temperature of boiling water when the pressure is increased by 27.12 mm Hg. The normal boiling point of water at atmospheric pressure is 100 oC. Explain why the boiling temperature of a liquid increase with pressure? 5. Discuss the effect of increase of pressure on ice and wax under isothermal conditions. 6. 7. Show that for a single component system, the necessary conditions for any of its two phases to be at thermodynamic equilibrium is the equivalence of their specific Gibbs functions. Hence derive the Clausius – Clapeylon equation for a First – Order phase change. dP L Establish and state the Clausius-Clapeyron’s equation, and explain the dT T (V2 V1 ) effect of pressure on (a) boiling point of a liquid, and (b) melting point of a solid. 8. The Figure below shows a P-T diagram of a substance which expands on melting. P (Pa) C.P Liquid Solid T.P Vapour T(K) (a) (b) (c) 9. Make a clearly labeled P-T diagram of a real substance which contracts on melting. Briefly explain the effect of increasing pressure on the melting point of the substance whose diagram you have. Suggest an example of such a substance. The phase diagram for water is given in the figure below where TP is the triple point. P Po X Y S L CP 130 SPH 302: Thermodynamics (a) (b) 10. Discuss with the aid of G–T diagrams the variation of Gibbs function along the section XY for both FIRST ORDER and for SECOND ORDER phase changes. Sketch the corresponding Entropy-Temperature behaviour for both FIRST ORDER and SECOND ORDER phase changes and compare the entropy of the solid phase with the entropy of the liquid phase in both the two cases The phase diagram for water is given in the figure below where TP is the triple point. X’ P Po LIQUID CP SOLID TP Y’ To (a) (b) VAPOUR T (K) Discuss with the aid of G–T diagrams the variation of Gibbs function along the section X’Y’ for both FIRST ORDER and for SECOND ORDER phase changes. Sketch the corresponding Volume-Pressure behaviour for both FIRST ORDER and SECOND ORDER phase changes and compare the volume of the solid phase with the volume of the liquid phase in both the two cases. References 1. Finn C. B. J., Thermal Physics, 1986. 2. Brij L., Subrahmanyam N., Hemne P. S., (2008), Heat Thermodynamics and Statistical Physics., S. Chand & Company Ltd, New Delhi. 3. Bergeron C. G., et al, (1984), Introduction t Phase Equilibrium in Ceramics, American ceramic society, Inc, Columbus, Ohio, pp 1- 19. 131 SPH 302: Thermodynamics APPENDIX A SOLUTIONS TO PROBLEM SETS - EVEN QUESTIONS PROBLEM SET 2 2. Given n = 2, T = 300K, dT = 0 and V2 = 2V1 V2 = 3.46 103 J V 1 (i) dW nRT ln (ii) From 1ST law, dQ = dU + dW. But dU = 0J for a reversible process dQ = dW = 3.46 103 J. (iii) dU = 0J 4. P W = 30 J c b dQ= 80J 10 J a dQ = ? d From 1ST law: dQ = dU + dW Along a c b, dQ = 80J and dW = 30J dU = (80 – 30)J = 50J. V NOTE: dU depends only on temperature or the final states and not the path taken. Thus dUacb = dUadb. (i) Along a d b, dW = 10J and dU = Ub – Ua = 50J dQ = dU + dW = (50 + 10) J dQ = 60 J absorbed (ii) For the curved path b a, dW = -20J and dU = Ua – Ub = -50J dQ = dU + dW = (-50 - 20) J dQ = -70 liberated (iii) dUad = 40 J. dQ = dU + dW = (40 + 10) J dQ = 50J For d b, dUdb = (50 – 40) J = 10 J and dW = 0J dQ = dU = 10 J. 132 SPH 302: Thermodynamics 10. 18. Since pressure remains constant, then (i) For isothermal process, W PV2 V1 (vi) V 23 3 For adiabatic process, W P1V1 1 1 2 V2 (iii) For PV = const, W Given m = 16g, Mr =32 n 1 PV P V 1 1 1 2 2 m 16 0.5 M r 32 (a) For isovolumetric cooling, dV = 0 1. dQ = dU = nCvdT = 0.5 (21)(200K) = 2100 J (b) For process 2, dP = 0 dU = dQ – nRdT = nCpdT – nRdT But Cp = R + Cv = 29.314 J 1. dU = 2100 J (c) dW = PdV = nRdT = 831.45 J (d) dQ = dW + dU = nCpdT = 2931 J (e) W = PdV = 831 J PROBLEM SET 3 2. Given Th = 1,000K, Tc = 250K, Qh = 1.5 KJ (i) 1 (ii) From Tc 250 1 0.75 =0.75 or 75% Th 1000 T Qh Th 250 Qc Qh c 1,500 J 375 J Qc Tc 1000 Th 8. Given Qh = ?, Qc = 2100J, Th = 300K, Tc = 260K (i) From Qh Th Qc Tc Th 300 2,100 J 2,432 J 260 Tc Qh Qc (v) W = Qh – Qc = 323 J 133 SPH 302: Thermodynamics PROBLEM SET 4 2. Given dQ =0 with n = 1, Cv = 25.12, Cp = 33.44, T1= 340K, P1 = 500Kpa, V2 = 2 V1 1 (i) (ii) C V From the adiabatic expression; T2 T1 1 with P 1.33 C V2 V 0.33 T2 340 0.5 270.3K By 1ST law, W = -dU = -CPdT = 1,351 J dS = 0 since process is reversible In a free, dW = 0, dQ = 0 (adiabatic) dU = 0 = CvdT But Cv ≠ 0, therefore dT = 0 T2 = T1 Entropy must change if process is irreversible. Hence from central equation of thermodynamics TdS – PdV = dU = CvdT 3. CV dT PdV R dV for n = 1 T T V V dS R ln 2 5.76 JK 1 V1 dS Given m = 10 Kg, T1 = 20 oC = 293K, T2 = 100 oC = 373K, T3 = 250 oC = 523K, dP = 0, Cp (Liquid) = 4180 J/Kg oC, Cp (vapour) = 1670 + 0.4904T + 1.86 106 T-2, L (100 oC) = 22.6 105 J/Kg. Now, for heating water to 373 K, we have dQ , but dQ C P dT T 373 T dT C P ln 2 10,090 J dS1 C P T1 293 T dS1 When water is converted to steam, Q = mL where L = latent heat of vapourization dS 2 mLCP (vapour) 60,590 J T When water is converted to superheated steam, 523 dT 1670 0.49T 1.86 10 6 T 2 m dS 3 mC P dt T T T 273 T3 2 = 1.25 1012 J Thus dSTOTAL dS1 dS 2 dS 3 7.7110 J 4 134 SPH 302: Thermodynamics 135 View publication stats
