Physical chemistry Thermodynamics Lecture notes

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Physical thermodynamics Hilary 2015 1 PHYSICAL THERMODYNAMICS HILARY TERM 2015 Lecturer Dr. Richard F. Katz; richard.katzearth.ox.ac.uk Demonstrator Helen Ashcroft; helen.ashcroftearth.ox.ac.uk Books We will frequently refer to two books:  Concepts in Thermal Physics by Blundell and Blundell, published by Oxford Uni- versity Press. (Hereafter BB)  Fundamentals of Physics by Halliday, Resnick, and Walker, published by John Wiley. (Hereafter HRW) You are encouraged to purchase these for use during the course, for consultation during future courses at Oxford, and as a reference throughout your career in science. Both books are available at University and college libraries. Lectures There will be about three lectures per week for the rst four weeks of term. Please see the lecture schedule for details. Lecture topics (Subject to modi cation) Lecture 1 Preliminaries. Motivation; key concepts; the Ideal Gas law; temperature and heat; the Zeroth law of thermodynamics; review of probability. Lecture 2 Kinetic theory I. Microstates vs. macrostates; a statistical de nition of temperature; the Boltzmann distribution; geophysical application: isothermal at- mosphere; the velocity distribution. Lecture 3 Kinetic theory II. The Maxwell-Boltzmann distribution; pressure and the Ideal Gas law; mean-free path. Lecture 4 Kinetic theory III. Di usion of heat; di usion of chemical species; the heat equation. Lectures 5 Classical thermodynamics I. Internal energy; First law of thermodynamics; adiabatic and isothermal processes; geophysical application: adiabatic atmosphere. Lectures 6 Classical thermodynamics II. Second law of thermodynamics; heat engines; the Carnot cycle. Lecture 7 Classical thermodynamics III. Entropy; internal energy revisited; geophys- ical application: adiabatic mantle and potential temperature. Lecture 8 Classical thermodynamics IV. Phase transitions; latent heat; Clausius- Clapyron; phase diagrams. Lecture 9 Spill-over from previous lectures. Questions and discussion. Practice sheets There will be three assigned practice sheets, to be distributed at the third lecture each week. Each sheet will consist of several quantitative problems, and several essay questions. The sheets are intended to help you develop, clarify, and assess your understanding of the material. You should complete them carefully and thoroughly,2 Physical thermodynamics Hilary 2015 taking the time neccessary to learn the requisite material. Do not expect that this will be easy At the end of this series of lectures, you should schedule a tutorial to discuss your results. The names of possible tutors will be made available in lecture. Note that these problems will not neccessarily cover the full scope of the taught materialthey are merely representative of the depth of understanding that is expected of you. Problem classes There are three scheduled problem classes, one per week, starting in week 2. Come to these sessions prepared: having read, understood, and planned how to solve the problems. Other important points Please read carefully:  These notes are not a complete description of the information that is to be understood as part of the course. The complete description is contained in the union of these notes, the lectures themselves, and the book-sections referred to below and in the lectures. The level of mathematics that will be required on an examination of this material is approximately equivalent to that of the assigned problems.  Questions in the notes are meant as a check on your understanding. Try to answer them; if a question confuses you, it is an indication that you should review the course materials and, perhaps, seek help from a peer, a tutor, or the lecturer.  You are responsible for developing and assessing your own understanding of the material by completing the problem sheets and discussing them with your tutors.Physical thermodynamics Hilary 2015 3 Contents 1 Preliminaries 4 1.1 Why study thermal physics? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Physics, mathematics, and this course . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Key concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4 The Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.5 Temperature and heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.6 The Zeroth law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 7 1.7 Basic probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2 Kinetic theory I 9 2.1 Microstates vs. macrostates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2 A statistical de nition of temperature . . . . . . . . . . . . . . . . . . . . . . 10 2.3 The Boltzmann distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.4 Geophysical application: isothermal atmosphere . . . . . . . . . . . . . . . . . 12 2.5 The probability distribution for velocity . . . . . . . . . . . . . . . . . . . . . 13 3 Kinetic theory II 15 3.1 The Maxwell-Boltzmann distribution . . . . . . . . . . . . . . . . . . . . . . . 15 3.2 The mean kinetic energy of a gas molecule . . . . . . . . . . . . . . . . . . . . 16 3.3 Pressure and the Ideal Gas law revisited . . . . . . . . . . . . . . . . . . . . . 17 3.4 The frequency of collisions and the mean free path . . . . . . . . . . . . . . . 18 4 Kinetic theory III 21 4.1 Aside: partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2 Thermal transport in an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . 23 4.3 Chemical di usion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.4 Thermal di usion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 5 Classical thermodynamics I 29 5.1 Internal energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 5.1.1 Functions of state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 5.2 The rst law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 30 5.3 Application of the rst law: heat capacity . . . . . . . . . . . . . . . . . . . . 314 Physical thermodynamics Hilary 2015 5.3.1 Heat capacity of an ideal gas . . . . . . . . . . . . . . . . . . . . . . . 32 5.4 Reversibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 5.4.1 Application of the rst law: reversible isothermal expansion of an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5.4.2 Application of the rst law: adiabatic expansion of an ideal gas . . . . 34 5.5 Geophysical application: adiabatic atmosphere . . . . . . . . . . . . . . . . . 36 6 Classical thermodynamics II 37 6.1 The Carnot engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 6.1.1 Eciency of the Carnot cycle . . . . . . . . . . . . . . . . . . . . . . . 39 6.2 The second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 39 6.3 Clausius' theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 7 Classical thermodynamics III 42 7.1 A de nition of Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 7.1.1 Entropy in the Carnot cycle . . . . . . . . . . . . . . . . . . . . . . . . 42 7.2 Irreversible changes and entropy . . . . . . . . . . . . . . . . . . . . . . . . . 42 7.3 Entropy in kinetic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 7.4 Entropy, internal energy, and the rst law . . . . . . . . . . . . . . . . . . . . 45 7.4.1 Entropy: kinetic & classical . . . . . . . . . . . . . . . . . . . . . . . . 46 7.5 Geophysical application: adiabatic mantle . . . . . . . . . . . . . . . . . . . . 46 8 Classical thermodynamics IV 49 8.1 Latent heat and phase transitions . . . . . . . . . . . . . . . . . . . . . . . . . 49 8.1.1 Latent heat of vaporisation for an ideal gas . . . . . . . . . . . . . . . 50 8.2 Phase diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8.3 The Clausius-Clapeyron equation . . . . . . . . . . . . . . . . . . . . . . . . . 51 8.3.1 Phase boundary for a liquid ideal gas transition . . . . . . . . . . . 52 8.3.2 Phase boundary for a solid liquid transition . . . . . . . . . . . . . 52Lecture 1: Preliminaries 5 1 Preliminaries In this lecture we consider the motivation for a study of thermodynamics. We learn basic concepts that form the basis for the rest of the material. We introduce the Ideal Gas law, heat capacity, the Zeroth law of thermodynamics. There is a brief review of probability. 1.1 Why study thermal physics? Material From the top of the atmosphere, to the centre of the inner core, to the rst forms life, Earth science is concerned with materials and how they behave under variable conditions. Energy Energy drives all terrestrial processes, and it links them together. From the nucleus of an atom, the metabolism of an organism, the climate, the production and eruption of molten rock, to the tectonic motions of the Earth's surface plates, energy is the key physical connection. Mathematics Mathematics is the language that can best describe the physical world. Words are a starting point, a means for communicating mathematical results, and a fall-back in cases where we are yet unable to develop mathematical models. 1.2 Physics, mathematics, and this course The purpose of this course of lectures is to introduce you to the physics of energy. We will describe this physics with words, diagrams, and mathematics. The equations we will write down are not recipes for obtaining a numerical answer to a problem; instead they encapsulate the physical concepts, giving them a concise expression that can be manipulated using mathe- matical rules. Hence the derivation of these equations is the process by which we go from a physical idea to a mathematical expression. It will not be required that you reproduce these derivations. However you should understand the physical concepts and assumptions that go into them, and the mathematical results that come out, and how the two are connected. 1.3 Key concepts BB 1.1, 1.2 Mole A mole is a quantity of discrete objects. To be speci c, a mole is equal to an Avogadro 23 number of objects: N = 6:022 10 . This is the number of atoms that are contained A 12 in 12 g of the isotope C. It is merely a reference number. The molar mass of an object is the mass of one object times the Avogadro number. For 23 example, the molar mass of British one-pound coins is 9:5 g N = 57:2 10 g. A Number density The number density n is the number of particles within a given volume. If you have N molecules of gas in a volume V then the number density of molecules is n =N=V .6 Physical thermodynamics Hilary 2015 Pressure A force applied over an area is called a pressure, and has units of Pascals: " 2 force kg-m/s pressure = ;  Pa : 2 area m Thermodynamic limit (See the raindrops-on-roof analogy, BB 1.2) A very large number of instances of a random variable, such that the di erences of that variable from its mean value average to nought. In nitessimal An in nitessimal is a mathematical representation of a vanishingly small quantity. For some variable x, an in nitessimal change in x is written as dx. We can take dx as being arbitrarily small: as small as we need it to be for the computation at hand. The ratio of in nitessimal quantities is a derivative, e.g. dy=dx. Extensive variable A quantity that scales with the size of the body it measures. Examples (think of a container of a gas): volume, mass, kinetic energy. If you cut the container in half, each new container will contain half the volume, mass, and kinetic energy. Intensive variable A quantity that does NOT scale with the size of the body it measures. Examples (again, think of a container of a gas): temperature, pressure, density. If you cut the container in half, each half will have the same temperature, pressure, and density as the original container of gas. Classical thermodynamics deals with macroscopic properties of a system such as its pres- sure, temperature, and volume. It consists of laws relating variations among these quantities. Kinetic theory of gasses considers the microscopic motions of molecules in a gas and uses statistical averages to derive, in the thermodynamic limit, the macroscopic properties of the gas. 1.4 The Ideal Gas Law BB 1.3 The following are known from experiments on con ned gasses 1 p/V Boyle's law V /T Charles' law p/T Gay-Lussac's law Putting these together gives pV /T . We might also expect the pressure and the volume to scale with the number of molecules N of the gas. Incorporating this and including a constant of proportionality gives pV =Nk T; (1) B 1 1 1 where k =R=N J K is Boltzmann's constant (R = 8:315 J mol K is the Universal B A Gas Constant). Verify that substitution for k gives the familiar form pV = n RT , where B m n =N=N is the number of moles of gas molecules. m A The Ideal Gas law is an equation of state, and is very useful. It tells us, for an ideal gas, how any one property of the gas changes as we change other properties. It raises several questions and comments:Lecture 1: Preliminaries 7  This law was stated based on empirical measurements. How does it follow from rst principles? We will address this question using Kinetic Theory, starting in Lecture 2.  What is an \ideal gas" anyway? Kinetic Theory will de ne this rigorously. For now, lets say it is a dilute gas (i.e. there is a lot of empty space between molecules) in which intermolecular forces and molecular diameter are both negligibly small. As it turns out, these assumptions apply (approximately) to many gasses of interest.  Not all gasses are well-described by this law. Not all materials that will concern us are gasses.  Consider the variables in equation (1): p,V ,N, andT . The rst three are rather clear in their meaning: p is pressure or force per unit area; V is volume; N is the number of molecules of gas. We can imagine simple ways to measure these. Temperature T , although intuitively obvious, is not so clear, when you consider it carefully. What is temperature? 1.5 Temperature and heat (A temporary de nition) Temperature is a measure of \hotness" or \coldness." But what are \hotness" and \coldness?" To answer this question, consider two identical BB 4.1 HRW blocks of material: block one possesses \hotness," and block two possesses \coldness" (both 18-7 are completely insulated from the rest of existence). Now put these two blocks into thermal contact with each other and leave them for a whilethe blocks become indistinguishable. The hot block (block one) has lost its hotness and the cold block (block two) has lost its coldness, and hence the two are at the same temperature. The blocks are then said to be in thermal equilibrium. How did the process of thermalisation between our two blocks of di erent temperature occur? Thermal energy owed from block one to block two. BB 2.1 Thermal energy in transit is called heat Q. The transfer of heat raised the temperature of block two and lowered the temperature of block one until they were in thermal equilibrium. How much heat was transferred? That depends on the di erence in temperature between the blocks, the mass of the blocks, as well as the speci c heat capacity of the material in the blocks. BB 2.2 HRW Speci c heat capacity is the amount of heat dQ required to change the temperature of a 18-8 unit mass of material by a speci ed, small amount dT: From this de nition, we can write that for a block of mass M kg, dQ =cM dT or 1 dQ c = ; M dT 1 1 where c is the speci c heat capacity in J K kg .8 Physical thermodynamics Hilary 2015 Now if blocks one and two were initially at temperatures T and T , respectively, with 1 2 T T , then we can calculate their nal temperatures, and the amount of heat transferred. 1 2 As stated above, the blocks are identical, so they have the same c and M, and we'll assume that c is independent of temperature. At thermal equilibrium, they both have the same temperature, T . Because they are in thermal contact and otherwise insulated, the heat f transferred into each block must sum to zero. Thus we can write Q +Q =cM(T T ) +cM(T T ) = 0 1 2 f 1 f 2 and solve to obtain T = (T +T )=2: The heat transferred out of block one is thus Q = f 1 2 1 cM(T T )=2 J; Q is negative because block one lost heat. 2 1 1 Question: Suppose blocks one and two start with di erent speci c heat capacities (c and 1 c ) and di erent masses (M andM ). What are their nal temperatures and how much heat 2 1 2  was transferred? There's just one small complication to be added to the discussion of heat capacity above: BB 2.2 we know that materials tend to expand or contract as they change temperature. We did not specify whether our two-block thought experiment was carried out at constant pressure or at constant volume (in a rigid container). In fact, there is a di erent speci c heat for each of y these cases . They are de ned as:   1 dQ c = ; (2a) p M dT p   1 dQ c = : (2b) V M dT V 1.6 The Zeroth law of thermodynamics th The concept of thermal equilibrium, introduced above, allows us to state the 0 law. Zeroth law of thermodynamics: Two systems, each separately in thermal equilibrium with a third, are in equilibrium with each other. This is exactly as simple as it sounds. Suppose the two systems are blocks, and the third is a thermometer. If both blocks are in thermal equilibrium with the same thermometer, then both blocks are at the same temperature, and hence they are in thermal equilibrium with each other. (N.B. the thermometer is a system with a temperature; in thermal equilibrium that temperature is the same as the system that it is used to measure.) Another way to state th the 0 law: thermometers work. 1.7 Basic probability Read: Chapter 3 of BB.  Answer: when the blocks reach thermal equilibrium, they have the same temperaturelet's call it T . As f before, the heat transferred into each block must sum to zero, since the system is isolated. Hence we can write: c M (T T ) +c M (T T ) = 0; 1 1 f 1 2 2 f 2 and we can solve this equation for T . f y This distinction is especially important if the blocks are actually containers of gas.Lecture 1: Preliminaries 9 A discrete probability distribution gives the likelihood that a discrete random variable will take on any given value. Consider a 6-sided die; let the discrete random variable x represent i the di erent results of rolling and P the likelihood of obtaining the value x (assume the die i i is not \crooked."). Then x =f1; 2; 3; 4; 5; 6g; i   1 1 1 1 1 1 P = ; ; ; ; ; : i 6 6 6 6 6 6 Note that the sum of the probabilities must be equal to unity X P = 1; (3) i i and that the mean (or average, or expected value) of x is X hxi = xP; (4) i i i and the mean of any function of x is given by X hf(x)i = f(x )P: (5) i i i 2 Question: What arehxi, hax +bi, and x for the die example above? (a and b are constants.) A continuous probability distribution gives the likelihood that a continuous random vari- able x will take on a value with a given range (say x xx ): 1 2 Z x 2 P (x)dx: x 1 Analogous to the discrete distribution, Z P (x)dx = 1; (6) Z hxi = xP (x)dx; (7) Z hf(x)i = f(x)P (x)dx: (8) Be sure to familiarise yourself with the properties of a Gaussian distribution, example 3.3 in BB. Also review the variance (BB 3.4) and the mean of the product of two independent random variables (BB 3.6).10 Physical thermodynamics Hilary 2015 2 Kinetic theory I The purpose of the next three lectures is to introduce the Kinetic Theory of gasses. This is a theory that aims to explain how macroscopic features of a gas system (pressure, temperature, volume) arise from the microscopic motions and forces of molecules. We will see that many macroscopic properties arise from the kinetic energy of moleculesthis is the same kinetic energy that you learned about in the lectures on Mechanics. In this lecture we derive a statistical de nition of temperature. We learn how, for a gas, temperature determines the probability distribution of molecular velocities. 2.1 Microstates vs. macrostates How can we describe the state of a collection of molecules that compose a closed box of gas? For kinetic theory, we will need two descriptions: one that accounts for each individual molecule, and another that accounts for the system as a whole. Before we talk about the gas, let's consider two simpler systems, dice and coins. A microstate is a description of the state of each individual element of a system at a given instant. In a system consisting of three individually labelled, six-sided dice on a table, the microstate would be given by: microstate s = Die A: 3, Die B: 1, Die C: 6, where the number corresponds to the face of each die that is pointing upward. A macrostate is a description of the state of a system as a whole. In the dice example, the macrostate could be de nied as the sum of the three dice. In that case, for the given microstate above, the macrostate would be macrostate S = 3 + 1 + 6 = 10: Next consider a system composed of ve randomised coins, numbered one through ve, BB 4.3 but otherwise indistinguishable. One microstate of the system is s = H;T;T;T;T , where a H indicates \heads" and a T indicates \tails." Clearly there are ve di erent microstates s that correspond to the macrostate S = 1, which indicates that one of the ve coins is heads. These are: Microstates corresponding to S = 1: s = H;T;T;T;T 1 s = T;H;T;T;T 2 s = T;T;H;T;T 3 s = T;T;T;H;T 4 s = T;T;T;T;H 5 5 In total, there are 2 = 32 distinct microstates of this 5-coin system. Key concept: each distinct microstate is equally probable In contrast, there are only six macrostates S = i f0; 1; 2; 3; 4; 5g; these have di erent probabilities.Lecture 2: Kinetic theory I 11 What is the probability of each macrostate? There is only one microstate corresponding to the macrostate S = 0 so it has probability P = 1=32. There are 5=(3 2) microstates 0 corresponding to three heads S = 3 (why?), so it has P = 10=32. The probabilities of all 3  1 5 10 10 5 1 macrostates are P = ; ; ; ; ; . i 32 32 32 32 32 32 Question: Consider a system composed of three tetrahedral (4-sided) dice. (a) How many microstates does this system have? (b) What is the probability that after rolling the dice, the system is in the microstate s = 4; 4; 4? (c) What is the probability that it is in the  macrostate S = 4? 2.2 A statistical de nition of temperature How can we understand temperature in terms of the motion of a collection of molecules? Consider a container of gas with N total molecules; this system has a microstate and a macrostate. The microstate might be de ned as the kinetic energy  of each molecule in the system at a given instant, s =  ;  ;  ;:::;  ; 1 2 3 N while the macrostate could be the temperature of the system at the same time. In general, BB 4.4 the microstate is unknowable, but we can assume that number of possible microstates depends on the total energy E of the system; we denote the number of possible microstates for a given total energy as (E). Now consider two containers of gas, each with N molecules. Suppose that initially, each container has a di erent quantity of energy, E and E , corresponding to a di erent number 1 2 of microstates, (E ) and (E ). The containers are brought into thermal contact but are 1 1 2 2 otherwise isolated so the total energy E = E +E is xed. The number of microstates of 1 2 the combined system is (E ) (E ). 1 1 2 2 How will the total energy be distributed in thermal equilibrium? E and E will take on 1 2 values such that the combined system obtains the macrostate that corresponds to the largest possible number of microstates. Why does this occur? 1. The system is rapidly changing from one microstate to another; 2. each of the microstates is equally likely; 1 2 3. over a long enough time interval, the system will spend an equal amount of time in each microstate. Its macrostate will be that which corresponds to the largest number of di erent microstates. So we seek the maximum of (E ) (E ) for all values of E and E . We know that 1 1 2 2 1 2 these energies are not independent, however; in fact, E +E =E and so dE =dE . To 1 2 1 2 nd the maximum, we can thus search over all values of E : 1 d ( (E ) (E )) = 0: (9) 1 1 2 2 dE 1  Answers: (a) 64 microstates, (b) P = 1=64 = 0:015625, (c) P = 3=64 = 0:046875.12 Physical thermodynamics Hilary 2015 Using the product rule and the relationship between dE and dE we can write 1 2 d ln d ln 1 2 = : (10) dE dE 1 2 This condition will hold when the blocks are in thermal equilibrium. We know that in thermal equilibrium the temperature of each block is equal to the other. This motivates us to de ne the temperature as 1 d ln = : (11) k T dE B You can verify that combining equations (10) and (11) gives T = T . The choice of the 1 2 constant k means that T has units of Kelvin and that our de nition will have a useful B physical interpretation, as we shall see later. 2.3 The Boltzmann distribution In a gas of temperature T , how much kinetic energy does any given molecule have? Since all the molecules are bouncing around in the gas with di erent speeds, we can only answer this in terms of a probability distribution, the Bolzmann distribution. The Boltzmann distribution describes the probability that a single molecule within our container of gas will have a given kinetic energy . Consider a microscopic system (one molecule, chosen at random) in thermal contact with a huge energy reservoir (the reservoir is huge because it contains many molecules, and much more energy than ). There are several BB 4.6 important points to note  The total energy of the reservoir plus microscopic system (our one molecule) is xed at E. The reservoir has energy E.  Since  E; uctuations of  have a minuscule e ect on the energy of the reservoir, and we can assume the reservoir is at a xed temperature, T .  The one molecule, as a system by itself, is so simple that for each allowed value of its energy, there is only one associated microstate: () = 1.  For a given energy , the total number of microstates available to the coupled reservoir and microscopic system is (E)1: The probability that the microsystem has energy  is therefore given by P ()/ (E). Since E, we can make a simple, linear approximation of the quantity ln (E) d ln ln (E) ln (E) + (); (12) d(E) which is actually quite accurate because  is very very small compared to E. Using our de nition of temperature from equation (11), we can write  ln (E) ln (E) ; (13) k T BLecture 2: Kinetic theory I 13 where T is the temperature of the reservoir. Exponentiating equation (13) and using P ()/ (E) we have =(k T) B P ()/ e : (14) When properly normalised, equation (14) is the Boltzmann distribution. A single molecule is coupled to a reservoir at temperature T . The probability P that the  single molecule has energy  is proportional to the Boltzmann factor: P ()/ e , where 1 = (k T ) . B The Boltzmann factor states that the probability of our single molecule having an energy  decreases as  gets larger, and that the rate at which probability decreases with increasing  is determined by the temperature of the gas reservoir. This is best understood with a graph, Figure 1. In this gure, we compare the probability density functions for a single molecule in two di erent containers: one at temperature T and one at temperature T . a b Figure 1: A plot of the Boltzmann factor for two di erent temperatures, T and T . (Axis labels and a b lines to be added by the student). The Boltzmann distribution is  P () =Ce ; (15) R 1 where C is a normalisation factor, which ensures that P ()d = 1. We will use this 0 important formula in the next lecture. 2.4 Geophysical application: isothermal atmosphere We can use the Boltzmann factor to construct a model for the density of the atmosphere as a function of height. To do so, we assume that the atmosphere is isothermal, with tempera- BB 4.7 ture T . Method 1 using the Boltzmann distribution The potential energy of a molecule with mass m at height z in a gravity eld with acceleration g is mgz. Hence we can write mgz=(k T) B P (z)/ e : The number density of molecules at height z in the atmosphere will be proportional to the probability P (z). This means that the number density (molecules per cubic meter) is mgz=(k T) B n(z) =n(0)e ; where n(0) is the number density at the surface of the Earth.14 Physical thermodynamics Hilary 2015 Method 2 using the Ideal Gas law We can check this result by trying to derive it with a di erent approach. Consider a layer of the atmosphere between z andz + dz that has number density of particles n. The downward pressure exerted by this layer is dp =nmg dz: Now recall equation (1), the ideal gas law. Dividing both sides by V gives p =nk T; where B n is, again, the number density of molecules. Di erentiating both sides gives dp = dnk T: B Combining this with our previous expression for dp gives dn mg = dz; n k T B which we can integrate to give, again, mgz=(k T) B n(z) =n(0)e ; which is consistent with our earlier result. We predict that the number density of the atmosphere decreases exponentially with height. This turns out to be wrong, because our assumption of an isothermal atmosphere is wrong. We will return to this problem later in the course. 2.5 The probability distribution for velocity We now apply the Boltzmann distribution to determine the probability distribution g that a molecule of gas will have a given velocity v = (v ;v ;v ). To do so, we make three important x y z BB 5.1 assumptions.  Molecular size is much smaller that intermolecular spacing, so collisions between molecules are rare and negligiable.  There are no intermolecular forces.  Each molecule behaves like a microsystem coupled to a thermal reservoir at temperature T , composed of all the other molecules in the gas. Recall that the kinetic energy of a molecule is given by 1 1 1 1 2 2 2 2 mv + mv + mv = mv ; (16) x y z 2 2 2 2 wherev =jvj. We can use the kinetic energy in each direction to de ne the Boltzmann factor; for example, in the x-direction we have 2 mv =(2k T) x B g(v )/ e : (17) xLecture 2: Kinetic theory I 15 Figure 2: A Gaussian distribution. (Axis labels and lines to be added by the student). To nd the normalisation factor for the Boltzmann distribution, we evaluate the integral r Z 1 2 2k T B mv =(2k T) x B e dv = : (18) x m 1 R 1 Then, since we require that g(v )dv = 1, we have x x 1 r m 2 mv =(2k T) B x g(v ) = e : (19) x 2k T B 2 2 This is a Gaussian distribution with a mean of zero and a variance of  = (v hvi) = x x x k T=m; it is shown in Figure 2. B Note that there is nothing special about the x-direction; the distributions for the y- and z-directions are exactly the same as equation (19). r m 2 mv =(2k T) y B g(v ) = e : (20) y 2k T B r m 2 mv =(2k T) B z g(v ) = e : (21) z 2k T B In the next lecture, we'll consider a slightly more complex quantity, the distribution of molecular speeds. The above results will be useful to keep in mind.16 Physical thermodynamics Hilary 2015 3 Kinetic theory II In this lecture we derive the probability distribution of molecular speeds in a gas, a result that will be very useful to us. We then examine how this distribution gives rise to a statistical de nition of gas pressure. We also introduce the concept of molecular collisions, and de ne the mean free path. In the last lecture we derived an expression for the distribution of velocity in each of the three Cartesian directions, r 2 m mv =(2k T) B j g(v ) = e ; (22) j 2k T B where j can be replaced with x, y, or z. We can use this result to determine the fraction of molecules with velocity between v = (v ;v ;v ) and v + dv = (v + dv ;v + dv ;v + dv ) x y z x x y y z z by simply multiplying them together: 2 2 2 mv =(2k T) mv =(2k T) mv =(2k T) x B y B z B g(v )dv g(v )dv g(v )dv / e dv e dv e dv ; x x y y z z x y z 2 2 2 m(v +v +v )=(2k T) B x y z / e dv dv dv ; (23) x y z 2 mv =(2k T) B / e dv dv dv : (24) x y z 3.1 The Maxwell-Boltzmann distribution BB 5.2 A more useful quantity, however, is the distribution of molecule speed v =jvj. In particular, we can ask: what is the fraction of molecules that is travelling with speed between v and v + dv? In velocity space, this corresponds to the spherical shell between radii v and v + dv, shown in Figure 3. Figure 3: Velocity space. (Axis labels and lines to be added by the student). What is the volume of velocity-space between the two shells? The volume of each spherical shell is 4 3 V (v) = v : 3 Therefore the volume between the two shells is dV 2 dV = dv = 4v dv: dvLecture 3: Kinetic theory II 17 Now we can de ne the probability distributionf such that the probability that a molecule will have speed between v and v + dv is given by 2 2 mv =(2k T) B f(v)dV /v e dv; (25) where the factor 4 has been dropped since this is a statement of proportionality, not equality. As with the velocity distribution, we need to normalise the function f such that R 1 f(v)dv = 1. (Note that we integrate from zero because speed is the absolute value of 0 velocity and thus cannot be negative.) Z r 1 2 1  2 mv =(2k T) B v e dv = ; (26) 3 4 m=(2k T ) B 0 and thus we can write an equation for f(v) as   3=2 4 m 2 2 mv =(2k T) B p f(v) = v e : (27)  2k T B This very important result is the Maxwell-Boltzmann speed distribution. A graph is shown in Figure 4. Figure 4: The Maxwell-Boltzmann probability distribution. (Axis labels and lines to be added by the student). We can calculate the expected values of equation (27) as r Z 1 8k T B hvi = vf(v) dv = ; (28) m 0 Z 1 3k T B 2 2 v = v f(v) dv = : (29) m 0 D E 2 2 2 2 2 Note that v + v + v = 3 v = 3k T=m = v , from equation (19). B x y z j p 2 Question: Calculate the mean hvi and root-mean-square hvi velocity of an oxygen molecule (O ) at room temperature (300 K). 2 3.2 The mean kinetic energy of a gas molecule 1 2 The mean kinetic energy of a molecule in the gas described above ishE i = m v . Using KE 2 the result in equation (29) we can calculate that 3 hE i = k T: (30) KE B 218 Physical thermodynamics Hilary 2015 This equation states the the average kinetic energy of a molecule in a gas depends only on the temperature of the gas. The mean kinetic energy of a gas molecule is independent of its mass. It is directly proportional to the temperature of the gas. 3.3 Pressure and the Ideal Gas law revisited 3 HRW Consider a cubic box of gas, shown in Figure 5, with volume V =L . The box is full of gas 19-4 molecules with number densityn (so there areN =nV molecules in the box). Each molecule has a mass m and velocity v, and the distribution of these velocities is given by equations (19), (20), and (21). L v z y x L L Figure 5: A cubic box lled with gas. One molecule of the gas is shown, with its velocity vector. The edges are of length L. We can assume that the particles don't collide with each other; they only hit the walls. Each time a molecule hits a wall, it undergoes a perfect elastic rebound. For example, if a molecule that has velocity v = (v ;v ;v ) hits the shaded wall in Figure 5, it rebounds before x y z with velocity v = (v ;v ;v ). In the process in imparts some momentum onto the wall. after x y z The momentum imparted ism v = 2mv . Since the molecule doesn't lose any speed in the x x collision with the wall, it will return to the same wall every t = 2L=v . x Since force is given by F = d(momentum)=d(time), the force of one molecule on the shaded wall is 2 2mv mv x x F = = : (31) 2L=v L x To nd the total force on the shaded wall, we must sum the forces of all the N molecules on that wall: N 2 X m(v ) i x F = ; (32) tot L i=1Lecture 3: Kinetic theory II 19 but we can replace the sum with the total number of molecules times the mean-squared velocity to give m 2 F = N v : (33) tot x L We can also use the following 2  Pressure equals force divided by area. In this case, the area is L .  Since N is large and all three directions are equivalent, the mean-squared speed in any 2 2 one direction is one third the mean squared speed: v = v =3. x Applying both these points gives mN mN 2 2 p = v = v ; (34) 3 3L 3V a statistically derived expression for the pressure of a gas. 2 Finally, using our result from equation (29), v = 3k T=m, and rearranging gives B pV =Nk T; (35) B the Ideal Gas law To review the path that we took to this point: 1. We assumed a dilute gas with no intermolecular forces and no intermolecular collisions. 2. We used the idea that each molecule is coupled to the thermal reservoir of gas at temperature T to derive a probability distribution for molecular energy. 3. We used this result to derive the probability distribution of molecular speed in the gas. 4. We calculated the mean-squared molecular speed. 5. We calculated the pressure exerted by the gas on the wall of a box in terms of the frequency and momentum change of molecular collisions with the wall. 6. We used our expression for the mean-squared molecular speed to write the pressure in terms of temperature. And hence we derived, from rst principles, the empirically known Ideal Gas equation. 3.4 The frequency of collisions and the mean free path Molecules in a gas are not points of zero diameter, they are objects with nite size and hence they sometimes collide. This has important consequences for molecular transport (di usion), HRW as we shall see in the next lecture. Now we quantify occurrence of collisions. We assume that 19-6 collisions between molecules are perfectly elastic, like collisions between molecules and walls.20 Physical thermodynamics Hilary 2015 2 1. Consider a molecule in a gas with speedv and cross-sectional area =d , whered is 0 the molecule radius. For now, assume that all the other molecules are not moving and have zero diameter. As our molecule bounces around within the gas, it sweeps out a (zigzag) cylindrical path behind it. 2. In a time t, the molecule goes a distance v t and thus sweeps through a volume of 0 v t. 0 3. The number of collisions per t is equal to the number of other molecules that are in the BB 8.1 swept volume. For a gas with particle density n, this number is, on average, nv t. 0 Thus the mean frequency of collisions for our molecule is nv ; the mean time between 0 1 collisions is  = (nv ) . 0 0 4. The average distance travelled between collisions by the molecule that we are watching is the mean free path  . For this one molecule the mean free path is its speed times 0 1 the time-interval between collisions:  =v  = (n) . 0 0 0 BB 8.3 5. However, we are interested in the mean free path for all molecules in the gas. In that case, we need the mean speed:  =hvi. 6. Furthermore, we must substitute the mean time between collisions for all molecules in the gas. In this case, all the molecules are moving, and the collision interval is determined by the mean relative speedhvi =hjvji. In this case r r v = v v ; r 1 2 2 2 2 v = v  v =v +v 2v  v : r r 1 2 r 1 2 2 When we take the mean of v we must handle the cross term carefully: r hv  vi = 0 because hcosi = 0 1 2 where is the angle between the two velocity vectors. Since the vectors are independent and can point in any direction,  ranges from zero to . The mean value of cos over this interval is zero, so the cross term drops out and we have 2 2 2 2 v = v + v = 2 v ; r 1 2 where the second equality is true because v and v are drawn from the same probability 1 2  distribution. We can then make two approximations : p p 2 hvi hvi 2hvi: r r p  1 From this we have  2nhvi , which we can substitute into  =hvi to give 1 p ; (36) 2n the (approximate) mean free path for all molecules in the gas.  recall that Z hvi = v f(v )dv r r r while s Z p 2 2 hvi = v f(v )dv: r r r

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