CHM 201: Introduction to Physical Chemistry II

CHM 201: Introduction to Physical Chemistry II COURSEDEVELOPMENT Course Developer DrT. Busari University of Lagos Akoka, Lagos Unit Writer DrT. Busari University of Lagos Akoka, Lagos Course Coordinators Dr F. Peters Ahmadu Bello University Zaria Prof Aiyepeku University of lbadan Ibadan Alkpe National Open University of Nigeria Lagos National Open University of Nigeria National Open University of Nigeria Headquarters National Open University of Nigeria 14/16Ahmadu BeUo Way Victoria Island Lagos Ahuja Annaeoffice 245 Samuel Adesujo Ademulegun Street Central Business District Opposite Arewa Suites Ahuja E.mail: URL: © National Open University of Nigeria 2004 First published 2004 ISBN Published by Longman Nigeria Pic for National Open University of Nigeria 'l• _:.- I CHM 201: Introduction to physical chemistry II Course guide 1.0 Introduction CHM 122 is a second semester course in introductory physical chemistry. It is a continuation of CHM 115.In this course, there are several theoretical concepts and their practical applications.In fact, one expects that most of the fundamentals in the introduction to physical chemistry would have been covered.Some relevant concepts are however not covered in this course, especially in chemical thermodynamics,equilibrium,electrochemistry,photochemistry, and nuclear chemistry. These will be addressed later in other physical chemistry courses, as you proceed in your programme. Meanwhile, note that for you to effectively and successfully study CHM 122, you require such fundamental courses as Atoms and Molecules, Pure Mathematics, Mechanics (in Physics), and so on. This implies that when you are selecting electives, consideration should be given to some of those suggested, or more relevant courses that will facilitate your understanding of physical chemistry courses. As pointed out in CHM 115, physical chemistry is one of the fundamental courses for a Bach- elor 's degree programme in Science (B Sc) and Science Ed ucation (B Sc.Ed) in any university in the world. This course is compulsory for students studying for degrees in Chemistry and Chemistry Educa tion. It is a suitable elective for anyone who is undergoing a science or science-related pro- fessional programme. Just like CHM 115, it is a two-credit unit course and contains four modules, consisting of 20 units in all. The first module covers the terminology and first law of thermody- namics. The second and third laws of thermodynamics will be ta ught later in the programme. The second module focuses on thermochemistry. Here, enthalpy changes in different processes are dealt with.The third modules examines the introductory aspects of the concepts of kinetics and equ ilibrium. In the fourth mod ule, the concepts of ionic equilibria and some minor aspects of elect rochemistry a re discussed. This course, though adapted from IGNOU course materia ls went through serious surgery to fit '• into the systems we a re familia r with. You will be req uired to possess a scien tific calcula tor, instru- mental and problem-solving skills acquired thruu t:;h consistent practice, for you to successfully learn this course. Definitely, there are some implied practical works that you will be required to carry out in the labora tory, though they are not in some cases specified. This Course Guide provides you with additional information on the course aims, objectives and learning procedures, to enable you to derive maximum benefit from the four modules pre- pared for the course. 2.0 Course aims The aims of this course are to create that in-depth awareness that will enable students to have full understanding of the concepof energy, as the capacity for change in any chemical reaction;and to appreciate that strong relationship between energy and structure of matter, as the foundation of physical chemistry. Thus, the aims of this physical chemistry course are to: 1. discuss the mathematical viabilities and standard procedures required in effective study of ... physical chemistry; 2 • xplain the basic theories that elucidate various properties of matter; 3. discuss some aspects of physical phase equlibria; 4. build cases for the feasibilities of chemical reactions; 5. demonstrate the various energy relationships in chemical reactions; 6. give you an appreciation of systems at equilibrium and factors affecting such systems. 3.0 Course objectives Inorde1 to achieve the course aims, there are some overall objectives set for the course. Moreover, each module and each unit has their respective objectives, which you and your course tutor must constantly refer to, so that no objective is skipped or unachieved. It is, therefore, important that you outline what you have been able to achieve after completing a unit and compare the list with the unit objectives.This will help you to ascer tain whether you have accomplished what is required vfyou. All the module and unit objectives are specifics of the Cotll::>t:! ubjectl t!S. The course objectives are stated as follows: At the end of the course, you should be able to: i) familiarise yourself with basic instruments for physical chemistry labora tory work; ii) write and apply correctly 51 units; iii). measure in and convert to other units using 51 units; iv) explain and compare the main features of gas, liquid and solids states; v) appreciate the reasons why real gases deviate from gas laws; vi) analyse the properties of completely, partially and immiscible liquids; vii) discuss the essence of types of solutions; viii) analyse phase changes in matter; ix) describe the conditions regulating the solubility of liquid solutions. 4.0 Working through this course This course contains some packages that you will be given at the beginning of the semester, having satisfied the conditions for admission and registration.One of the packages is the course materials. The details of constitute course materials will be discussed nex•.. There are two areas you are expected to fulfi' at the end of this course. These are your full participation in the continuous assessment and the final written examinations. Though there is a practical course developed for this level, however it is advisable that some of the sections can be practicalised, with the assistance of the course tutor. Your tutor will be readily accessible to issue out the exercise you are to do and submit according to specified rules and regulations from the university authority. For you to have been counted as successful in this course, you will be expected to pass at an average level, both in continuous assessment and the final written examination. Thus, certain materials have been packaged for you, to enable you to prepare adequately for both continuous assessment and the final written examinations. First and foremost is the study unit material, which consists of 20 units in all. However, the units have been packaged for you in modules, as shown below: I Study units Module number Module Unit Title 1 Introduction 1 2 3 Units and dimensions Sl prefixes Separations techniques 2 Gaseous state of matter 4 Gases 5 Ideal gas 6 Kinetic theory of gases 7 Real gases 8 Liquefaction of gases 3 Solid and liquid states 9 Solids states and solid types of matter 10 Nature of bonds in solids 11 Structure of bonds in solids 12 Liquids 13 Other properties of liquids 4 Solutions and phase equilibria 14 Types of solutions 15 Solutions of solids and gases in liquids 16 Binary liquidssolutions-1 17 Binary liquids solutions-2 18 Partially miscible liquids 19 Some colligative properties 20 Phase rule By and large, you should be able to complete this two-<:redit unit course for about 30-35 weeks in a semester. Well spread out in each of the units are the introduction to the unit, specific objectives, reading materials on sub-topics,self-assessment questions (SAQ}, exercises,a conclusion, summary, tutor-marked assignments (TMA} and references. 5.0 Course materials Major course materials of the course are as follows: 1. Course guide:This looks like a blueprint that spells out what constitutes the course itself. It also extends to the basic information you require on how best to study this course. 2. Study units: These present an overview of what will be covered in the course and in how many units. The contents are spelt out under seven major headings: the introduction,objectives to guide, what to focus on, detailed content, conclusion, summary of the content, references (and other materials that will facilitate understanding} and tutor-marked assignments. In the body of content are exercises that should always attract your attention after every reading. :i:"• 3. Assignment files: These files contain challenging and tutorial questions known as Tutor Marked Assignments (TMAs}, which will enable you to assess yourself at the end of every assignment given by your tutor.Since we are operating a course unit system, your scores in these assignments will be added up, recorded and used for computing your cumulative grade score at the end of the course. I 4. Presentation schedule: This will feature information on a recommended study time-table, suggested number of hours of study per unit/ module, and assignment-submission procedure for effective self-teaching and progress-monitoring. 6.0 Study units Details of the study units have earlier been presented.It is spelt out in modules with corresponding units and titles. You will be expected to spend 3-4 hours studying a unit. The exercise items are provided to enable you to test your understanding of the areas covered and to help you monitor the progress made per sub-topic in the unit. There is also the TMA packages for you,but these will be given as assignments. 7.0 References and other resources Apart from this study unit, some reference materials are provided as additional reading materials to support your stud y.It is not the responsibility of NOUN to provide them, but you may be lucky to spot them in the designated library. 8.0 Instructional media As a distance learning university, several and relevant multimedia that can make learning more meaningful are available. Other instructional media will be disclosed and made accessible to you from time to time. 9.0 Assignment file This has been discussed earlier. It is mandatory to always tum in your assignments to the tutor assigned. 10.0 Assessment You will be expected to have completed at least ten assignments by the end of the course. Some of these will be in the form of a project, continuous assessment or written tests. Each of these ten assignments will carry ten marks, making a total of 100 marks. You will also be expected to write a final examination at the end of the course. Your overall (percentage) score at the end of the course will be the sum of your scores on continuous assessment and the written examination.The minimum pass score for the course will be 50% in continous assessment and 50% in the written examination. I Module 1: Unit 1: Energy relationships in chemical reactions 1.1 Introduction There is no chemical reaction in which energy is not involved.In the process of reacting substances, you either supply energy to the system or the reaction occurs on its own. But then, why and how do chemical reactions occur? What exactly is thermodynamics? Why do some chemical reactions occur without the supply of external energy? All these questions will be answered in this module. You will also be introduced to some of the terms used in thermodynamics. 1.2 Objectives By the end of this unit, you should be able to: i) define the term thermodynamics; ii) explain the terms- system, surroundings and thermodynamic variables; iii) state the zeroth law of thermodynamics; iv) differentiate between extensive and intensive variables, providing two examplr:s each. 1.3 Thermodynamics terminology In this section, a number of commonly used terms in thermodynamics are defined and explained. These terms should be understood clearly before you proceed further. Thermodynamics In respect of chemical reactions, we are not on1y concerned about how and why reactions take place but also why some substances are even more reactive than others. There is no doubt that before a chemical reaction can occur, energy is required. The energy required may be related to the substances undergoing the reaction. However, where the quantities of energy possessed by the reacting species are not adequate to initia te the reaction or take the reaction to compiPtion, energy may be supplied. In shor t, chemical reaction in a system requires some level of energy transforma tion. We have learnt before that heat is a form of energy. Thermodynamics is a compound word. Tlrermo means •of-heat' or:.concerned with heat and dynamics means •of physical power and forces producing motion'. Thermodynamics can thus be -L- 1 defined as the science of heat motion. In fact, you should consider it as the science of heat flow or transfer or disappearance of work attending Chemical and physical processes. Let us consider someexamples of thermodynamics,as follows: 1 Natural process:Water flowing down from a hilltop.An example is Erin - ljesha water fall in OsunState. 2 Controlled chemical reactions:You can determine the dissociation constant (pKa),e.g. for acetic acid. 3 Performance of engines:Any engine can be considered here, for instance, when you are deter- mining the efficiency of a blender engine. We should draw our attention to one fact that ther- modynamics cannot answer all questions due to its wide applicability. For instance, it carmot: i) answer how fast a reaction occurs; ii) give the character of the process of change. However, it can tell us whether a reaction will occur or not. In brief, the term thermodynamics is defined as the study of energy transformation in a system. In fact, thermodynamics deals with systems. The system under consideration in this unit is a chemical one, hence, our consideration of the concept of chemical thermodynamics. What then is a system? System Any part of the universe that is under study is called a system. Systems can be in different states. A system can be as simple as a gas contained in a closed vessel, or as complicated as a rocket shooting towardsthe moon. A system may be homogenous or heterogenous, depending on its contents and conditions. A system is homogenous if physical properties and chemical composition are identical throughout the system.Such a system is also called a single phase system.A heterogenous system consists of two or more phases separated by mechanical boundaries. Now, let us consider the surroundings of the reaction. Surroundings The rest of the universe around the system is considered as its surroundings. A system and its surroundings are always separated by boundaries across which matter and energy may be exchanged.The boundaries can be real (fixed or movable) or imaginary. Based on the exchange of matter and energy between the system and thesUFToundings,a system can be classified into the following three types: i) •isolated; ii) closed; iii) open. An isolated system is one that exchanges neither energy nor matter with its surroundings. There is no perfectly isolated system, but, a system that is thermally well-insulated (i.e. does not allow heat flow) and is sealed to inflow or outflow of matter can be considered as an isolated system. A sealed thermos flask having some matter may be regarded as an isolated system. A closed system allows for exchange of energy (heat or work) with the surroundings, but matter is not allowed to enter or leave it. A properly sealed system (to prevent the passage of matter across its boundary) can be considered as a closed system. An open system allows exchange of both matter and energy with its surroundings.This is the most common system encountered in our daily life. All living things are examples of an open system, since they are capable of exchanging energy and matter freely with their surroundings. Also, reaction vessels with permeable membranes are open systems. 2 B State variables Any thermodynamic system must be macroscopic, i.e. must have sufficiently large size. This facilitates the measurement of its properties such as pressure, volume, temperature, composition and derisity.Such properties are, therefore,called macroscopic or bulk properties.These are also called state or thermodynamic variables.1llese do not depend on the past history of the system. A state variable which depends on other variables is called a dependant variable. Others, on which it is dependent are called independent variables.For example, if you write the ideal gas equation as: v -p=nRT Then, V is the dependent variable, whereas n, T and Pareindependent variables. We know that R is the gas constant. On the other hand, if you write this equation as, nRT P=- V then, Pis the dependent variable, whereas n, T and V are independenvariables.The choice of dependent and independent variables is a malter of convenience. State of a system The state of a system can be defined in thermodynamics once you establish a small set of measurable parameters. For example,when you have a gasconfined in a container, the measurable parameters there will include pressure, volume, temperature and composition. In essence, the state of a system is defined when the state variables have definite values. It is not necessary to specify all the state variables since these are interdependent. For example, if the system is an ideal gas, then its pressure, volume, temperature and amount of the gas (number of moles) are related by the gru. equation. Thus, if we specify three of these, the fourth variable is automatically fixed.Similarly, many of its other properties, such as density, heat capacity, etc.,are also fixed, although through more complicated relations. We can change the state of a system by altering either the pressure or the volume. Exercise 1.1 Identify the type of system in each of the foUowing cases: i) a beaker covered with a lid. ii) a closed thermos flask. iii) a beaker without lid. 1.4 The zeroth law of thermodynamics The zeroth law of thermodynamics is based on the concept of thermal equilibrium.It helps us in defining temperature.If twoclosed systems are brought together so that they are in thermal contact, changes take place in the properties of both systems. But,eventually a state is reached when there is no further change in either of the systems. This is the state of thermal equilibrium. Both systems are at the same temperature. In order to find if two systems are at the same temperature, the two can be brought into thermal contact, and then the changes in their properties observed. If no changes occur, they are at the same temperature. The zeroth law of thermodynamics states that if a system A is in thermal equilibrium with system C, aud sy!>tem B is 'llso in thennal equilibrium with C, then A and B are also in thermal equilibrium with each other This is an experimental fact, which may be illustrated by assuming that systems A and Bare two vessels containing different liquids,and C is an ordinary mercury thermometer. If A is in thermal equilibrium with C, then the mercury level in the thermometer willshow a constant reading. This indicates the temperature of system A as well as that of C. Now, if A is also in thermal equilibrium with B, then the height of mercury level in the thermometer (in contact with B) is the same as before; B also has the same temperature as A.There is thermal equilibrium in both A and 3 .. B,or these are at the same temperature. Here we have only explained the concept of temperature, the temperature scale will be discussed later in thls course. 1.5 Extensive and intensive variables We have defined homogenous and hetergenous systems in Section 1.3. Let us now discuss the difference between the two, wi th respect to the value of some variables. The parameters mentioned earlier are also called variables. There are two types of variables, namely extensive and intensive variables. Extensive'variables An extensive property of a homogenous system is one which is dependent on the amount of a phase in the system or the mass of the system. For a heterogenous system made up of several phases, the total value of an extensive property is equal to the sum of the contributions from its various phases. Mass, volume, and energy are examples of extensive properties.Thus, if a system at equilibrium consists of 0.100 kg of ice and 0.100 kg of liquid water at 273.15K, the total volume of the system is the sum of the two volumes, each of which is directly proportional to its mass. Mass of ice Volume of 0.100 kg of ice = ----- Density of ice 0.100 kg = 917kgm-3 =1.09 x 10-4 m3 Mass of water Similarh•, the volume of 0.100 kg of water= -- Density of water 0.100 kg =-----'---- 1.00 x 103 kg m•3 = 1.00 x10-" m• Total volume= (1.09 + 1.00} x 10-•l m3 = 2.09 x 10•4 m3 Intemve variables An int. '1Sive propert) of a r:tase l mdepcr..• 11f of the aluvunt of the phaSE Thus, refracti 0 index, c ensity and pres::,ure are intens1•'e properttes. However, if astern consists •f several phases, then s0:ne of the mtenstve properties may be d1Herent. For example, c nsity is an ir.,ensive property, but its value is different for ice and liqUtd watt?r in equilibrium a t 273.15K. For thermal equilibrium, the ;ntensive property, i.e. temperature has to be tl1e same throughout the system Otherwise, heat will flow from one puint of thesystem toanother.Similarly, for mechanical eauilibriwn, the il nt ,sivP property, i.e. pressure, has to be the same throughout the J t. An exten::,iv ::ptvpe-tty .vnen divided by mass or molar mass of the system becomes an intensive property. Exercise 1.2 Identify the extensive or intensive variables from among those indicated below: i) energy required to cook your meals. ii) volume per unit mass of milk. iii) your body temperature. 4 I 1.6 Conclusion We are now familiar with various terms used in thermodynamics. The zeroth law has been stated. •••:l'he extensive and intensive variables, otherwise known as parameters are explained. In the next unit,we shall explain the terms:work, heat and heat capacity. We shall attempt to determine work done, using appropriate formulae derived in the course of explaining these concepts. 1.7 Summary In this unit, you ha ve learnt that: i) a system may be homogenous or heterogenous depending on its contents and conditions; ii) suttoundings refers to all other things in the universe which may interact with the system; iii) intensive varnble is independent of the quantity of matter present in a system. 1.8 Tutor-marked assignment 1. Which of the following are true or false: a) Thermodynamicscan explain how fast a reaction is. b) Flooding of Nigerian roads is an example of thermodynamics. c) Ina closed system,heat neither leaves nor enters. 2. Define thermodynamics. 1.9 References 1. IGNOU (199:1), Chemical Tltennodynamics: physical chemistry CHE-04, New Delhi. 2. Laidler, K J, Jnd Meiser, J. H, (1982), Physical Chemistry, California: The Benjamin Cummmgs Publ ishing Company, Inc. Unit 2: Work, heat and heat capacity 2.1 Introduction In Unit 1of thiscourse,we examined some terminology used in thermod ynamics and the concept of thermodynamics itself in matter.In this un]t, we will describe the various types of process.In addition, the definitions of work, heat and heat capaoty will be statednd some -ah:ulations made. 2.2 Objectives t By the end of this unit, you should be able to: i) describe isothermaJ, adiabatic and cyclic processes; ii) differentiate between these types of processes; iii) explain the terms- work, heat and heat capacity; iv) pPrform some calculations on work, heat and heat capc!Oty. '2.3 Types of ptocess When th St te of a system chanf es, it is said I) havt: tmdergrme a process Thus, a procest:. means a d1aJ !Je ir.1t least one of the s.a.e variabJe:.of the system. Th process may be accompanied oy an exchange o r.-.atter an-1 cnerg_, betVePn the system and the surroundmgs. There :u- • certain pwceoos in whiCh a p<l1Licubr state';, :ack (thermocl j'Ilam;c proper'v cf the sy' •err) •em<'in un, hanged.Such processes are or the followg types•isotherm.•l, adiabatic, sobaric o:nr:l isud10ric- proce5se . In an isothermal proce"s, thE tempt>r:=ttunof the S) ...tern emams :onstWanhte.n a system undergoe:; 1, c;u Prmal p;... ess it is r. i •'.'hermal contact with a large constant temp ratu•c ua.. l<..r>N" n • s •l1ermostat. The yt:.tem rna •1tains its temper!ure by exchar ge of heat w itthe thermostat. 1n an adiabatic process, no heat is allowed to enter or .e... the system. yst ms ;vr.;.mch Si.H.J a process occurs are thermally insulated from the surroundir bS An ad1abatic process may im olve an ncrease or a decrease in the temperature of the system. V e shall d1scuss these two processec; ir. detail later m t1us course. An isobanc process is one in which the pressure of the system remains unchanged. A .tuJuP ta kmg place in an open beaker 1s aJways at ahnospheric pressure, and the process IS, therefore isobaric. In an isochoric process, the volume of the system remains constant. Thus, a chemical reactior in a sealed flask of constant volume is an isochoric process. A process is cyclic if the system (after any number of intermediate changes) returns to 1t• s I original state.1he initial and final values of each thermodynamic variable are identical-after the completion of a cyclic process.The summation of energy in a cyclic process will be zero. Hence, for any cyclic process, the integral of energy is zero. This simply means that if there is a cyclic process, the quantity of work we do must be equal to the total amount of energy put in, that is,the energy expended.From this discussion on cyclic process,we can conclude that any engine process or any device operating in cycles for the purpose of converting heat to work can never produce more than the heat added. With this conclusion, we can then rule out the construction of the so- called perpetual motion machine of the first kind. Based on the value of the driving force applied, we can classify the processes into two types: reversible and irreversible. A reversible process is one in which at any time, the driving force exceeds the opposing force only very slightly. Hence, the direction of the process can be reversed by merely effecting a small change in a variable, such as temperature and pressure. The idea of a reversible process will become clear by consjdering the following example. Consider a gasat pressure pin a cylinder fitted with an air-tight piston.If theexternal pressure on the gas is equal to the pressure of the gas, then there is neither expansion nor compression and the piston remains at its position. However,on increasing theextemal pressure (Pext) infinitesimally, the gas can be compressed. On the other hand, by slightly decreasing the external pressure, the gas may be expanded. ...--- Fig.2.1 Static system (since piston does not move) ...--- ............ ,..... Thus, if Pext = p, the system is static and the piston does not move: Put = p + dp, the gas is compressed and the piston moves downwards,infi.t ltesimally slowly; Pext= p-dp, the gas expands and the piston moves outwards, infinitesim,illy sitJ>•:• Thus, for a reversible process, the direction is changed by changing the magnih1de of the driving force by a small amount. Any process that is not reversible is termed irreversible.All natural processes are irreversible. The flow of heat from a high temperature body to a low temperature body isa natural process and hence, irreversible. So is the expansion of a gas against vacuum, known as free eYransion. Irreversible processes are also called spontaneous processes. We will be studying revers1ble and irreversible processes in detail later in this course. I 7 2.4 Work, heat and heat capacity Work, heat and energy have thesame unit,called the joule (J).Energy is a thermodynamic property of a system, whereas work and heat are not.Work and heat are meaningful only when a process takes place. Heat By now, we all recognise heat as a form of energy. Heat is not the property of a system but it is exchanged between a system and its surroundings during a process, when there is a temperature difference between the two. Heat can be defined based on the ice-calorimeter. If we place ice in an ice-calorimeter and the stopper is pressed, there will be a rise in the level of Hp in the capillary and one can read the calibrated capillary.If the reaction mixture is placed in the reaction chamber of the ice-calorimeter and corked, two things may happen. The reaction may absorb heat from the surrounding jar, or the reaction mixture may produce heat. Now, what happens if heat is absorbed? If your answer is that ice will be formed, then you are right. If the reaction mixture produces heat, more ice will ( melt. Exercise 2.1 Complete the following: Assuming that more ice melts, a) lg of ice will occupy ..................... space than lg of f0 (less/ more) b) the volume of the system will (increase/decrease). Your answer to (a) above will reflect your understanding of the relationship between heat and volume change in a system. If the reaction absorbs heat from the surroundings, there will be a volume increase in the system as a result of ice-formation. There is a change in volume as a result of the quantity of heat that has been transferred between thecalorimeter and the reaction chamber. Heat is a measurable quantity. Work Let us now explain the term, work, and describe its different kinds. The amount of work which attends a thermodynamic state is very important. The simplest concept of work (W) is defined as the product of the force applied (F) and tht> distance (x} moved along the direction of the force. W = Force x distance (x) W = F.x 2.1 Forces have different physical origin, and work can be done in a variety of ways. 1. Gravitational work: When a body of mass m is moved through a height h against gravity, then force is equal to mg and the gravitational work done is mgh. 2. Electrical work:If an electric potential E is applied across a resistance R so that current I flows through it, then work done per second is EI and in t seconds it is equal to Elt. 3. Pressure-volume work:This is a type of mechanical work performed when a system changes its volume against an opposing pressure. This is also known as work of expansion or compression. We will study this in detail in later sections. Work will be very constantly referred to as pressure-volume work in this unit. The energy gained or lost during heat exchange between the system and its surroundings can be stated in terms of heat capacity values. Heat capacity Heat capacity is the heat required to raise the temperature of a body by lK. If,during the process, the volume of the system remains constant, then it is called heat capacity nt constant volume (C,). If the pressure remains unchanged, it is called heat atpacity at constant tempertJture (C/ For one mole of a pure substance, these are called molar heat capacity at constant pressure, C and molar heat capacity at censtant volume (C.). Heat capacity per unit ma55 is called specif fc heat. The heat s I capacities change with temperature.This means that the heat required to change the temperature by 1K is different at different temperatures.However,over small ranges of temperature, these are usually takenasconstant.The molar heat capacity and specific heat are intensive properties,whereas heat capacity is an extensive property. By changing the temperature of a particular system by dT if the heat required is dqo (at constant volume) or dq,(at constant pressure), we have dqy C..:::;nCv=- . dT Cp =nCp = dqp dT ..22 ..2.3 where n is the amount, i.e. number of moles of the substance constituting the system. From these equations, it is possible to deterinine the heat required for a process,by integration over the temperature range T1and T2• Hence, Clv = J-; C..dT= I-; nC:vdT 2.4 1i 1i CJp = CpdT= I-; nCpdT 2.5 Tt Tl In later sections, we will study the use of CP and Cv in the calculation of energy changes. Let us give an example here for the calculation ofif n, Cp' T 1 and T2 are given. Example2.1 The equation for the molar heatcapacityofbutaneis C = (19.41+0.233T) J mol•1 K"1 Calculate the heat necessary to raise the temperature of 3.00 mol J'f butane from 298 K to 573 Kat constant pressure. Solution We have to calculate qP•as per Equation 2.5. CJp Tl nCpdT where, 1i = 298 K 12 =573K n =3.00mol Cp = (19.41+ 0.233 T) J mor1K'1 573 CJp = 298 3.00 (19.41+ 0.233T) dT = 3.00 X 19.41 573dT + 3.00 X 0.233 J573 TdT 298 298 3.00 X 0.233 2 2 ] = [3.00 X 19.41(573-298) + = 9.97 X 104 J =99.7 kJ (573 2 - 298 ) J Hence, heat required to raise the temperature of 3.00 mol of butane from 298 K to 573.1< is 99.7 kJ. We will now give two general formulae for htegration. These two formulae. will be useful throughout this course in working out numericals. In this unit, Formula 1is used in Example 2.1 and Formula 2 in Example 2.2. I 9 ! "•t Formula 1 If m is not equal to -1. f x2 aXmdX=a Jx2 XmdX= _a_[l xm•t]"2 x, x, m+l x, where a is a constant. = a -(x2m+l -Xtm+l) (m+l) f f»rmu l a 2 If m i5:qual to -1. Formu Ia 2 finds use throughout our course (although not in this example). x 2 "t dx X2 a-=a In- X X1 Again, a is a constant. Note that 'In' stands for logarithm to the base e. Since we use natural logarithm, i.e. logarithm to the base 10 in our calculations, it is better to modify formula 2 as follows: "2 a dx = 2.303 a log (x2) Xt X X1 Note that In x = 2.303 log x Exercise 2.2 The molar heat capacity of ethane at constant pressure is 2.6 J K-1 moi-1. Calculate the energy required to heat 3.00 mol of ethane from 305 K to 405 K at constant pressure. Hints: i) Use Equation 2.5 ii) Integration is to be done as per formula 1and I term in Example 2.1. At constant pressure, the change in volume of Van der Waals gas, is accompanied by temperature change. Therefor.e, we can have W = P llV = nR {T - T ) 2.6 = nR llT (does not apply to Vander Waals equation) 2.7 When you compress a gas, applying work on that gaseous system, the sign of work done will be negative. When a gas expands, the volume of the gas will be smaller than the final volume.The change in volume is positive.Therefore, work done is positive. When a system is doing work on the surroundings, the work done will be positive. When work is done by the surroundings on a system, the work done is negative. Assuming you heat a known amount of water in a beaker, the amount of current passed is known, as well as'the work done on it. If the water is heated to boil, as in Fig.2.2 (b), there is the same quantity of heat in the beaker when the water starts boiling.That is,the amount of heat does not increase the content of the beaker. to 1 Fig. 2.2 Example 2.2 One mole of an ideal gas is heated at a constant pressure of 101.3K Nm•2, from 273.2 K to 373 K. a) Calculate, in Joules, the work involved. b) If the gas expanded isothermally at 273 K from the pressure of 101.3I<N m•2 to some pressure, P, what must be the pressure of the isothermal work that is equal to the work in (a) above? Solution W=nR(T2 -Td Joules =8314 -- (373- 273) K (1 mole) Kmol = 831.4 J mor1 b) W= fv2 dv v, v = nRTIn v2 (1 mole) vl Since n = 1 mole, then we have v2 =RTln- Vt It is known that: P1V1= P2V2 VW!can now have W = Rl' ln P.JP2 . 1 v2 t.e. -=v-. 8314 Joules 273 KIn (1013 x 10 3 831.4}/mole= x K-mol P 8314Joules 273 KIn (101.3 x to') 831.4}ImolD 4 Kmol )( ------- p P a S.3l Joules )( 273 KIn (101.3 )( lo')m)(ol Kmol m- 83tAJ P a 2731n 101.3 )( to' rd P a 273 )( 2.3031cJs 101.3 )(1o';i -6.37 )(107 Nm ' 11