Physics B Chapter 16 Notes Spring 018 These notes are thirteen pages. While some derivations and explanations are a bit lengthy, important results are boxed. A very brief summary: Temperature must always be expressed in Kelvins, although T in Kelvin and O C is the same value. We will use the ideal gas law throughout Chapters 16, 17 and 18. Thermal energy is associated with temperature and is actually the kinetic energy of molecules. Temperature It was in the 170 s that Gabriel Fahrenheit had enough with people asking what s the weather like outside? and the only available responses were nice or a little chilly or something similarly vague. Fahrenheit created a temperature scale that is very useful for measuring outdoor temperatures: 100 on the Fahrenheit scale is a hot day and zero is a cold day. Also in the 170 s, a Swedish fellow by the name of Anders Celsius got interested in doing scientific experiments measuring heat transfer from hot objects to cold objects. Since water was a very useful substance in these experiments, Celsius created a temperature scale specifically designed to work with the properties of water: he set 100 at the temperature at which water boils, and zero at the temperature at which water freezes. It is important to note that the Fahrenheit and Celsius scales were created for two different purposes, and they are each very useful for their intended purposes: Fahrenheit for measuring ambient temperatures, Celsius for measuring temperature of water in scientific experiments. How do these two scales compare, side by side? You probably addressed this question in elementary school, so it is more than a little silly to do it here. But what the heck, we can use a little silly every once in a while: If we imagine each temperature scale as steps on a staircase, then the Fahrenheit scale has 180 steps between the freezing and boiling points of water, while the Celsius scale has 100 steps. This tells us that the steps on the Celsius scale are bigger, and that an increase (or decrease) of 180 on the Fahrenheit scale is equivalent to a corresponding increase (or decrease) of 100 steps on the Celsius scale. This also means that an increase (or decrease) of 9 steps on the Fahrenheit scale corresponds to an increase (or decrease) of 5 steps on the Celsius scale. Page 1 of 13
Physics B Chapter 16 Notes Spring 018 The Celsius scale was used by scientists through the 1700 s, and a quirk of history established it as the common temperature scale (even for measuring ambient temperatures) for most of the world. In the aftermath of the French Revolution, around the year 1800, the new and exciting metric system was introduced to most of Europe by Napoleon. The metric system included a new standard way to measure mass and length (i.e. the kilogram and the meter), and somehow the Celsius scale tagged along and became the standard way to measure temperature. There was no good reason for the adoption as the Celsius scale as the only way to measure temperature, as there is nothing metric system about it. But history is littered with examples of illogical decisions people have made that others follow without questioning. Near the end of the 18 th century, James Watt started the Industrial Revolution, and changed the course of human history, with his invention of a new and vastly improved steam engine. The steam engine converted thermal energy into mechanical energy... that is, it used heat to do work. Watt became a very wealthy man, of which people took notice... and the study of thermodynamics took on a new level of importance. Into this exciting world of scientific development, William Thomson was born in 184. Young Billy s father was a professor of science and math at Glasgow University in Scotland. Billy was homeschooled by his father in his younger years, but at age 10 he entered a formal academy and moved on to Cambridge when he was 17. Billy was interested in math and science, and would become a professor himself at the age of. He would also become one of the more important scientists of the 19 th century, making contributions in many areas of physics and engineering, including writing the definitive book on the subject of thermodynamics (much as Newton had done for the study of mechanics.) In recognition of his accomplishments, he was knighted by Queen Victoria (Just as Newton was by Queen Anne... and Elton John, Paul McCartney and Mick Jagger have been by the current Queen Elizabeth. Back then it was scientists, now it s rock stars.) Later he was invited to join the British House of Lords (which, apparently, is better than being knighted.) They had to make up a lord-ish title for him (all the British lords get fake lord names) and since there was a river near the university where he taught called River Kelvin, the people that make up names for lords decided he should be known as Lord Kelvin (I am not making this up, I swear.) Lord Kelvin is best known today for the idea that the amount of thermal energy an object possesses must be finite, and so if the temperature of an object decreases as it loses this energy, there must be an absolute zero temperature for all objects. Kelvin was not the first person to propose this idea, but he developed a rigorous treatment of the theory and so created the defining work on the subject. Through careful experimentation, Kelvin established a value of absolute zero as -73 O C. It s worthy of note that while the technical definition of absolute zero has been refined over the past 150 years, and experimental techniques have improved tremendously, Kelvin s value has remained remarkably accurate. Page of 13
Physics B Chapter 16 Notes Spring 018 The definition of absolute zero effectively introduced another temperature scale: the Kelvin scale. While Fahrenheit and Celsius defined two points to create their scale, the same criteria necessary to define a straight line on a graph, the Kelvin scale was essentially created by defining only one point... and a slope. That is, the only point defined for the Kelvin scale is zero. The steps of the Kelvin scale are the same as that of the Celsius scale. This means: Zero on the Kelvin scale is equivalent to -73 on the Celsius scale. Every temperature on the Kelvin scale is equivalent to the corresponding temperature on the Celsius scale plus 73. A difference in temperature on the Kelvin scale has the same value as the corresponding difference on the Celsius scale. The study of thermodynamics is about the role of heat as a form of energy. We will use temperature as a way to measure this energy, in the same way that we used speed to determine an object s kinetic energy or height to determine an object s potential energy. For this reason, in our study of thermodynamics we must use the Kelvin scale to express temperatures. If temperatures are given in Celsius, convert them to Kelvin. If an expression includes a T, Celsius temperatures can be used because the value of T in Celsius is the same as the corresponding value in Kelvin. Thermal Expansion When things get hot they expand. Why this happens has much to do with the nature of thermal energy, i.e. the energy associated with temperature. As we ll discover at the end of these notes, thermal energy is simply kinetic energy on a molecular level. So things expand as they get hotter because their molecules move more. Molecules of a solid have a greater range of motion (although the motion is simple back and forth), as do molecules of a liquid. The greater range of motion relates to a greater volume of the object. How much does an object expand? To answer this question, we start with the simplest case: that of a solid object expanding in one dimension. We can refer to the length of this dimension and the expansion as L. L for any solid object depends on three things: the initial length the material the temperature increase Page 3 of 13
Physics B Chapter 16 Notes Spring 018 We can define a coefficient of thermal expansion for a solid material, which is a way to express how much a given material expands when it is heated. We will use α for this coefficient and write a simple expression to answer our question: L L i α T linear thermal expansion of a solid While the behavior of solid materials can be complicated, we will assume that the value of alpha remains constant within a reasonable temperature range. That is, for a given solid object with L i and α as constants, the increase in length is directly proportional to the increase in temperature. A few notes: L and L i have the same units, since they are both lengths. The units of α T combined must be unitless. Since you are expected to use Celsius for T, the unit of a is 1/ O C or O C -1 Thermal expansion of common materials is very small... it is not noticeable, but it is measurable. For example, the value of α for steel is 1. x 10-5 O C -1. This means that if the temperature of a one-meter long rod of steel is increased by 40 O C (from freezing to about 100 O F... i.e. very cold day to very hot day), it would expand by half a millimeter (i.e. the width of a mechanical pencil lead.) Another way to look at this: your car is about or 3 mm longer on a hot day than on a cold day. We can now consider what happens when the height and width (which are really just lengths...) of a rectangular solid increase. The area of the rectangle will increase, and we can write the increase in the area using the expression above for the increase in each of the height and width: The expansion of the area is represented by the shaded areas. We can write these as: A top area + right side + upper right corner A w h + h w + h w A w (h α T) + h (w α T) + (h α T) (w α T) Page 4 of 13
Physics B Chapter 16 Notes Spring 018 Notice that every term includes hw α T, so we can factor this out and: A A i α T ( 1 + 1 + α T) We now make an approximation. We know that T cannot be very large: if we heat solid materials more than several hundred degrees Celsius, they tend to melt. For most thermal expansion problems we are concerned about solids increasing or decreasing their temperature by around 100 degrees Celsius at most. And since the value of alpha for most solids is on the order of 10-5, this means that in most situations, α T will be on the order of 10-3. So ( 1 + 1 + α T) is approximately equal to simply for most situations (because α T is negligible.) This is equivalent to suggesting that since the individual length expansions, i.e. h and w, are tiny, the area in the upper right corner of the expansion is extremely tiny and can be ignored. The result? A A i α T area thermal expansion of a solid We can use similar logic to create the expression for the volume expansion of a solid material: V V i 3α T volume thermal expansion of a solid What about liquids? We do not define a length for a liquid, nor do we define an area. Liquids are simple: they have only volume. The volume of a liquid expands as the temperature of the liquid is increased, and we make the same simplification for liquids that we did for solids (i.e. that the expansion is proportional to the temperature change.) We can express the thermal expansion of a liquid: V V i β T volume thermal expansion of a liquid Note that we use β for a liquid in the place of α in the expressions for a solid. This is to help us clarify the difference in behavior between solids and liquids: α is the coefficient used for the expansion of the length of a solid, while β is the coefficient used for the volume expansion of a liquid. It is worth noting that values of β for typical liquids are on the order of 10-4 O C -1, which means that the volume of a typical liquid expands at a rate 10 to 100 times greater (for a given temperature increase) than that of a typical solid. This is important if a solid container is filled with liquid and both are subjected to the same temperature increase: while the solid container will expand, the liquid will experience a much greater expansion and will overflow the container. Page 5 of 13
Physics B Chapter 16 Notes Spring 018 The Ideal Gas Law While the volume of a liquid or solid will expand with temperature, the behavior of gases is a little different. This is because a given amount of a liquid or solid has a well defined volume, but a gas will always fill whatever volume it is allowed. If a gas is contained in a sealed, rigid container, the volume of the gas is the same as the interior of the container. If the gas is heated, its volume is still that of the interior of the container. Which means thermal expansion for gases cannot be addressed using the same equations we used for liquids and solids. Fortunately, the behavior of gases has been well documented for hundreds of years. In 166, Robert Boyle published his finding that if the temperature of a gas in a flexible container is held constant, then the pressure and the volume of the gas are inversely proportional. This is known as Boyle s Law. In the 1780 s, Jacque Charles determined that if the pressure of a gas is held constant, then the volume and temperature are directly proportional (much like the thermal expansion of a liquid or solid.) We know this today as Charles Law. We combine both of these laws to allow the pressure, volume and temperature of a given amount of gas to change and create the Ideal Gas Law : p V n R T Ideal Gas Law Where: p is the absolute pressure of the gas (Note: recall that gauge pressure is different!) V is the volume of the gas n is the number of moles of the gas R is a constant, which is usually referred to as the gas constant T is the temperature of the gas A few important notes regarding the Ideal Gas Law: An ideal gas is an approximation, but one that is appropriate for most situations. All of the problems we encounter in Chapters 16, 17 and 18 will assume that the gases involved qualify for this approximation. The pressure of the gas can be expressed in any units, but typically we will use either Pascals (which are SI units) or atmospheres (which are very useful non-standard units.) The volume of the gas can also be expressed in any units, but typically we will use either m 3 (which are SI units) or liters (which are very useful non-standard units.) Page 6 of 13
Physics B Chapter 16 Notes Spring 018 Watch your units! Either use SI units for both pressure and volume (i.e. Pascals and m 3 together) or use non-standard units (i.e. atmospheres and liters) for both. Avoid mixing standard units and non-standard units when using the Ideal Gas Law. Recall that there are 6.0 x 10 3 (i.e. Avagadro s number) molecules in one mole. Expressing the amount of gas in moles is just a more convenient way of expressing the number of molecules. Temperature must always be expressed in Kelvin! There are two versions of the gas constant, one for SI units and one for non-standard units: R 8.314 J / mol-k and R 0.0806 atm-l / mol-k Watch your units!! Always use the correct R for the units of your pressure and volume! Note that everything in the IGL (pressure, volume, number of moles, gas constant, temperature) must always be positive. None of these values can ever be negative. This is why temperatures must always be expressed in Kelvin! The IGL is the most important topic in our study of thermodynamics. It is an essential part of all three chapters. You will deal with two types of IGL problems: Single State : a state of a gas is a unique combination of p, V, n, and T. These parameters are related through the IGL. A single state problem is one in which three of these four parameters are provided and you are asked to find the value of the missing parameter. In other words, one equation (i.e. the IGL) and one unknown. Multiple State : in a multiple state problem, the gas will begin in some initial state and then some of the parameters will change. You will be given some information about each state, and you will be able to write the IGL for each state. The number of equations you will have will be equal to the number of states described in the problem, because you can write the IGL once for each state. Whether a single state or multiple state problem, always draw a proper picture and label relevant information. If you have multiple states, label them 1,, 3, etc and use these as subscripts for the parameters (i.e. p 1 V 1 n 1 and T 1 for State 1, p V n T for State, etc.) There are two more versions of the IGL which can be useful in certain situations. The first we can derive by multiplying both sides of the IGL by the molar mass of the gas. (Recall that molar mass is a measure of the mass per mole of a gas, which is a constant for a given gas.) M p V M n R T Page 7 of 13
Physics B Chapter 16 Notes Spring 018 Note that on the right side of the equation, M is the mass per mole and n is the number of moles... so their product is simply the mass of the gas! M p V m R T We can now divide both sides of this expression by the volume of the gas and we get: M p ρ R T Ideal Gas Law, density version This new version of the ideal gas law will be useful whenever the density of the gas is mentioned. Notice that the new version is actually simpler than the original: the new version has only three parameters that can change (pressure, density, temperature) compared to the four of the original. The third version of the IGL is one that we will not use much, although I will use it as part of the derivation in the following section, Kinetic Theory. We start with the original IGL and convert the number of moles to number of molecules by multiplying n by Avagadro s number: p V (n α) (R / α) T Or: pv N k T Ideal Gas Law, N version where N is the number of molecules and k is a new constant. Note that N is a huge number and that k must be a correspondingly small number. In fact: 8.314 J / mol-k k ---------------------------------------- 1.381 x 10-3 J / K 6.0 x 10 3 molecules/mol This new constant is called Boltzmann s Constant. (Ludwig Boltzmann contributed to the development of thermodynamics and modern physics in the 1870 s.) It is convenient to think of this constant as a tiny version of the gas constant. Note that the only difference between this new version and the original is the way we express the amount of gas: in moles or in molecules. If for some reason your problem provides you information on the number of molecules, you should consider using this version of the IGL. Page 8 of 13
Physics B Chapter 16 Notes Spring 018 Kinetic Theory Note: what follows is a lengthy derivation using simple principles of motion (kinetic energy, force, momentum) from last semester. While the steps of the derivation are useful in understanding how the concepts fit together, the results, in three boxes at the end of this section, are what you will use in homework. A central concern in thermodynamics is the connection between the macroscopic parameters of a gas (those that can be measured directly, like pressure, volume and temperature) and the microscopic parameters of the gas (which cannot be measured directly, such as the speed at which gas molecules move.) We will invoke some basic mechanics, from A, and a few logical arguments to make a simple, but important, connection between the kinetic energy of molecules and their temperature. We start with a box filled with gas molecules. The box can have any shape, but for simplicity we will assume it is rectangular with dimensions a, b and c. The box contains billions of billions of gas molecules, all flying in random directions at varying speeds. We will define: Coordinate axes: x, y and z; A molecule that has a kinetic energy equal to the average kinetic energy of the molecules. We assume this gas molecule has velocity in the x, y and z dimensions. But since it has the average kinetic energy, we will assume that its velocity in each dimension is identical. That is, its motion in each dimension separately represents the average motion of all the molecules in the dimension. We will consider the velocity in the x dimension of this molecule, which we can label v x : The molecule travels in the -x direction, hits the left wall of the container, and rebounds at a speed of v x in the +x direction (i.e. now moving to the right.) The change in momentum for the molecule is: p mv mv mv f i x We can now imagine that the molecule will rebound from the right wall, and return to the left wall to repeat the process. The time it takes for this cycle, i.e. between rebounds from the left wall, is found simply by considering the distance the molecule travels divided by the speed: Page 9 of 13
Physics B Chapter 16 Notes Spring 018 t c v x If there are N molecules in the container, and they all behave as if they have the same average motion in the x direction, then they will all follow the same behavior. That is, during the time it takes our one molecule to make a round trip across the box, all N molecules will do the same and rebound once from the left wall. The total change in momentum for all of these molecules is: p Nmv x We know that the total force the molecules exert on the left wall must be the same as the force the wall exerts on the molecules. And the force exerted on the molecules is responsible for the change in momentum of the molecules. The relationship between force and p is simply: p F avg t So in this case, we can determine the average force the molecules exert on the left wall: c Nmvx Favg or vx F avg Nmv c x Recall that we want to be able to connect the macroscopic parameters of the gas to the microscopic. In this last expression we can see that the force exerted on the wall (something we can measure directly) is related to the mass and velocity of the gas molecules (both microscopic parameters.) But we can do better. The area of the wall is simply ab. We can divide both sides of the last equation by the area of the wall to get an expression for the pressure exerted on the wall by the gas: p x x or Nmv abc Nmv V pv Nmv x where V is the volume of the box and therefore the volume of the gas. The left side of this last expression is the left side of the ideal gas law. So if our gas can be treated as an ideal gas, we can replace pv with NkT. We then get: NkT Nmv x or kt mv x or 1 kt 1 mv x Page 10 of 13
Physics B Chapter 16 Notes Spring 018 where k is Boltzmann s constant. Now we have a very simple expression with the temperature of the gas on the left side, and the kinetic energy of one molecule s one-dimensional (i.e. x dimension) motion. We could repeat the entire derivation, using the same logical arguments, for the other two dimensions. We would get: 1 1 kt mv x and 1 1 kt mv y and 1 kt 1 mv z If we add these three equations, we get: 3 1 kt m( vx + vy + vz ) or 3 kt 1 mv The final result is an expression for the average translational kinetic energy of a gas molecule in an ideal gas. Note that translational kinetic energy refers to the energy associated with the linear motion of the molecule; we will consider the additional energy of rotation in Chapter 19. 3 K avg kt average translational kinetic energy of a gas molecule where T is the temperature in Kelvins and k is Boltzmann s constant, 1.38 x 10-3 J/K. Note: this is the average translational kinetic energy of the molecule, i.e. the kinetic energy associated with its linear motion. We make this distinction because in Chapter 19 we will find that some molecules also have rotational kinetic energy, and to find the total kinetic energy of the molecule we will have to consider translational and rotational kinetic energies. We can now make a couple of simple observations. If there are N molecules of gas, and they have an average translantional kinetic energy described by the expression above, then we can determine the total translational kinetic energy of all the molecules of the gas by taking the expression above times N: 3 K total NkT total translational kinetic energy of a gas with N molecules Page 11 of 13
Physics B Chapter 16 Notes Spring 018 We can also determine the speed associated with the average kinetic energy. It would be logical to assume that this speed is the average speed of the molecules, but we have to be careful. Technically we are able to determine the average value of not the speed itself, but of the speed squared : 1 mv 1 m ( ) 3 kt avg 3 ( v ) kt avg 3kT v avg m It s tempting to think that we can just take the square root of both sides and we ll have the average speed, but that s not true. Consider this simple example: Individual speed of five particles: 1,, 3, 4, 5 Average? 3 Individual speed squared of particles: 1, 4, 9, 16, 5 Average? 11 And now... square root of the average speed squared? 3.3 Note that the square root of the average of the speed squared is not the average speed! It is actually a little higher than the average speed. We call this the root mean square (because it is the square root of the average, or mean, of the squares) speed, or rms speed. We make two observations: 1. The rms speed of a molecule is the speed associated with the average kinetic energy.. The rms speed is slightly higher than the average speed. We really are not too concerned about the average speed; the rms speed is much more practical because it is associated with the average translational kinetic energy of the molecules, and the study of thermodynamics is all about energy. With this in mind, we can take the square root of the expression above and we get: Page 1 of 13
Physics B Chapter 16 Notes Spring 018 3kT v rms m rms speed of a molecule in a gas with temperature T We can make a modification to this expression if we realize that k, which is Boltzmann s constant, is a very small number and is related to R, the gas constant, by a factor of Avagadro s number. And m, the mass of one molecule, is a very small number that is related to the molar mass of the gas by a factor of Avagadro s number. That is: R k Avagadro s number and M m Avagadro s number Where M is the molar mass of the gas. If we multiply the numerator and denominator of the fraction on the right side of the expression by Avagardo s number, we get: 3RT v rms M rms speed of a molecule in a gas with temperature T and molar mass M This expression is generally easier to use because while k and m typically include obnoxious powers of 10 (e.g. x 10-3 ), R and M are more reasonable numerically (i.e. can usually be written without using scientific notation.) Note that all values should be in standard units, so for this expression you should use R 8.314 J / mol-k and M should be in units of kg / mol. If all values are in standard units, your result for v rms will be in standard units of m/s. Page 13 of 13