Knight Chapter 12 Thermal Properties of Matter OCR PDF

Squid Propulsion BIO Squid usually move by using their tins, but they can utilize a form of jet propulsion, ejecting water at high speed to rocket them backward, as shown in Figure II. I. A 4.0 kg squid can slowly draw in and then quickly eject 0.30 kg of water. The water is ejected in 0.10 s at a...

Squid Propulsion BIO Squid usually move by using their tins, but they can utilize a form of jet propulsion, ejecting water at high speed to rocket them backward, as shown in Figure II. I. A 4.0 kg squid can slowly draw in and then quickly eject 0.30 kg of water. The water is ejected in 0.10 s at a speed of IO mis. This gives the squid a quick burst of speed to evade predators or catch prey. 17. What is the change in momentum of the ball during the collision? A. 1.4 kg· mis B. 1.8 kg· mis C. 2.9 kg· mis D. 5.1 kg· mis 18. What is the speed of the club head immediately after the collision? D. 11 mis A. 30 111/s B. 25 mis C. 19 mis 19. ls this a perfectly elastic collision? A. Yes B. No C. There is insufficient information to make this detemunation. 20. lf we define the kinetic energy of the club head before the collision as "what you had to pay" and the kinetic energy of the ball immediately after as "what you get," what is the efficiency of this energy transfer? A. 0.54 B. 0.46 C. 0.38 D. 0.27 Additional Integrated Problems FIGURE 11.1 13. What is the speed of the squid immediately after the water is ejected? A. 10 mis B. 7.5 mis C. 1.3 mis D. 0.75 mis 14. What is the squid's approximate acceleration, in g's? A. IOg B. 7.5g C. 1.0g D. 0.75g 15. What is the average force on the water during the jet? A. 100 N B. 30 N C. 10 N D. 3.0 N 16. This form of locomotion is speedy, but is it efficient? The energy that the squid expends goes two places: the kinetic energy of the squid and the kinetic energy of the water. Think about how to define "what you get" and "what you had to pay"; then calculate an efficiency for this particular form of locomotion. (You can ignore biomechanical efficiency for this problem.) Teeing Off A golf club has a lightweight flexible shaft with a heavy block of wood or metal (called the head of the club) at the encl. A golfer making a long shot off the tee uses a driver, a club whose 300 g head is much more massive than the 46 g ball it will hit. The golfer swings the driver so that the club head is moving at 40 mis just before it collides with the ball. The collision is so rapid that it can be treated as the collision of a moving 300 g mass (the club head) with a stationary 46 g mass (the ball); the shaft of the club and the golfer can be ignored. The collision takes 5.0 ms, and the ball leaves the tee with a speed of 63 mis. 21. Football players measure their acceleration by seeing how fast they can sprint 40 yards (37 m). A zippy player can, from a standing start, run 40 yards in 4.1 s, reaching a top speed of about 11 m/s. For an 80 kg player, what is the average power output for this sprint? A. 300W B. 600W C. 900W D. 1200W 22. The unit of horsepower was defined by considering the power output of a typical horse. Working-horse guidelines in the 1900s called for them to pull with a force equal to 10% of their body weight at a speed of 3.0 mph. For a typical working horse of 1200 lb, what power does this represent in W and in hp? 23. A 100 kg football player is moving at 6.0 mis to the east; a 130 kg player is moving at 5.0 111/sto the west. They meet, each jumping into the air and grabbing the other player. While they ,u-estill in the air, wl1ich way is the pair moving, and how fast? 24. A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed into elastic potential energy of the bending board; if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the force. Ultimately the board breaks at an applied force of 800 N and a deflection of 1.2 cm. a. To break the bmu-d with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand. b. If the hand is moving this fast and comes to rest in a distance of 1.2 cm, what is the average force on the hand? 25. A child's sled has rails that slide with little friction across the snow. Logan has an old wooden sled with heavy iron rails that has a mass of IO kg-quite a bit for a 30 kg child! Logan runs at 4.0 m/s and leaps onto the stationary sled and holds on tight as it slides forward. The impact time with the sled is 0.25 s. a. Immediately after Logan jumps on the sled, how fast is it moving? b. What was the force on the sled during the impact? c. How much energy was "lost" in the impact? Where did this energy go? 393 PART 111 Properties of Matter Male emperor penguins overwinter on the Antarctic ice, going long periods without eating in bitterly cold surroundings. Minimizing heat loss to the environment is crucial because replacing lost energy draws down precious reserves. In this section, we'll consider properties of matter, including thermal energy and its transfer as heat. We'll learn to understand and appreciate some of the adaptations these remarkable birds use to keep warm during the winter. Beyond the Particle Model The first 11 chapters of this text have made extensive use of the particle model in which we represent objects as point masses. The particle model is especially useful for describing how discrete objects move through space and how they interact with each other. Whether a ball is made of metal or wood is irrelevant to calculating its trajectory. But there are many situations where the distinction between metal and wood is crucial. If you toss a metal ball and a wood ball into a pond, one sinks and the other floats. If you stir a pan on the stove with a metal spoon, it can quickly get too hot to hold unless it has a wooden handle. Wood and metal have different physical properties. So do air and water. Our goal in Part Ill is to describe and understand the similarities and differences of different materials. To do so, we must go beyond the particle model and dig deeper into the nature of matter. Macroscopic Physics In Part Ill, we will be concerned with systems that are solids, liquids, or gases. Properties such as pressure, temperature, specific heat, and viscosity are characteristics of the system as a whole, not of the individual particles. Solids, liquids, and gases are often called macroscopic systems-the prefix "macro" (the opposite of "micro") meaning "large:· We'll make sense of the behavior of these macroscopic properties by considering a microscopic view in which we think of these systems as collections of particle-like atoms. This "micro-tomacro" development will be a key piece of the following chapters. In the coming chapters, we'll consider a wide range of practical questions, such as: ■ How do the temperature and pressure of a system change when you heat it? Why do some materials respond quickly, others slowly? ■ What are the mechanisms by which a system exchanges heat energy with its environment? Why does blowing on a cup of hot coffee cause it to cool off? Why are there three phases of matter-solids, liquids, gases? What happens during a phase change? ■ Why do some objects float while others, with the same mass, sink? What keeps a massive steel ship afloat? ■ What are the laws of motion of a flowing liquid? How do they differ from the laws governing the motion of a particle? Both Newton's laws and the law of conservation of energy will remain important tools-they are, after all, the basic laws of physics-but we'll have to learn how they apply to macroscopic systems. It should come as no surprise that an understanding of macroscopic systems and their properties is essential for understanding the world around us. Biological systems, from cells to ecosystems, are macroscopic systems exchanging energy with their environment. On a larger scale, energy transport on earth gives us weather, and the exchange of energy between the earth and space determines our climate. 395 LOOKING AHEAD • The Ideal Gas Heat and Temperature Heat Transfer The high pressurein a car tire is due to the countless collisions between the air molecules inside and the tire·s walls. Adding ice cools your drink as heat is transferred from the warm drink to the cold ice: even more heat is used to melt the ice. [ts large ears keep an elephant cool. Blood You'll learn how gas properties are related to the microscopic motion of the gas molecules. You'll learn how to compute the temperature changes that occur when heat is transferred or during a phase change such as melting. flowing through large vessels loses heat to the environment, returning Lo the body at a lower temperature. You'll learn about the heat-transfer mechanisms of conduction, convection, and radiation. IB•di To use the atomic model of matter to explain many properties of matter associated with heat and temperature. LOOKING BACK • 141-iii,iiiiMI Heat In Section 11.4 you learned about heat and the fin;t law of thermodynamics. In this chapter we will explore some of the consequences of transferring heat to or from a system. and doing work on the system. A blender does 5000 1 of work on the food in its bowl. During the time the blender runs. 2000 1 of heat is transferred from the warm food to the cooler environment. What is the change in the thermal energy of the food? You learned that a system's energy can be changed by doing work on it or by transferring heat to it. A. +20001 B. +30001 C. +70001 E. -30001 D. -20001 12.1 The Atomic Model of Matter 397 12.1 The Atomic Model of Matter We began exploring the concepts of thermal energy, temperature, and heat in Chapter 11, but many unanswered questions remain. How do the properties of matter depend on temperature? When you add heat to a system, by how much does its temperature change? And how is heat transferred to or from a system? These are questions about the macroscopic state of systems, but we'll start our exploration by looking at a microscopic view, the atomic model that we've used to explain friction, elastic forces, and the nature of thermal energy. In this chapter, we'll use the atomic model to understand and explain the thermal properties of matter. As you know, each element and most compounds can exist as a gas, liquid, or solid. You've certainly learned about these three phases of matter earlier in your education. An atomic view of the three phases is shown in FIGURE 12.1. Our atomic model makes a simplification that is worth noting. The basic particles in Figure 12.1 are drawn as simple spheres; no mention is made of the nature of the particles. In a real gas, the basic particles rnjght be helium atoms or nitrogen molecules. In a solid, the basic particles might be the gold atoms that make up a bar of gold or the water molecules that make up ice. But, because many of the properties of gases, liquids, and solids do not depend on the exact nature of the particles that make them up, it's often a reasonable assumption to ignore these details and just consider the gas, liquid, or solid as being made up of simple spherical particles. FIGURE 12.1 Atomic models of the three phases of matter: gas, liquid, and solid. In a gas, the particles are in random motion and interact only through elastic collisions. In a liquid, the particles have weak bonds that keep them close together. The particles can slide around each other, so the liquid c/n flow. Atomic Mass and Atomic Mass Number Before we see how the atomic model explains the thermal properties of matter, we need to remind you of some "atomic accounting." Recall that atoms of different elements have different masses. The mass of an atom is determined primarily by its most massive constituents: the protons and neutrons in its nucleus. The sum of the number of protons and the number of neutrons is the atomic mass number A: A = number of protons + number In a rigid solid, the particles are connected by relatively stiff spring-like ;°nds. of neutrons A, which by definition is an integer, is written as a leading superscript on the atomic symbol. For example, the primary isotope of carbon, with six protons (which 12 is what makes it carbon) and six neutrons, has A = 12 and is written C. The radio14 active isotope C, used for carbon elating of archeological finds, contains six protons and eight neutrons. 12 The atomic mass scale is established by defining the mass of C to be exactly 12 12 u, where u is the symbol for the atomic mass unit. That is, m( C) = 12 u. In kg, tbe atomic mass unit is • I u = 1.66 X 10- 27 kg Atomic masses are all very nearly equal to the integer atomic mass number A. For example, the mass of IH, with A = 1, is m = 1.0078 u. For our present purposes, it will be sufficient to use the integer atomic mass numbers as the values of the atomic mass. That is, we'll use m('H) = 1 u, m( 4He) = 4 u, and 111(160) = 16 u. For molecules, the molecular mass is the sum of the atomic masses of the atoms that form the molecule. Thus the molecular mass of the molecule 0 2 , the constituent of oxygen gas, is m.(O2 ) = 2m( 16O) = 32 u. lmii) ► An element's atomic mass number is not the same as its atomic number. The awmic n.u.mbe,;which gives the element's position in the periodic table, is the number of protons. ◄ lists the atomjc mass numbers of some of the elements that we' 11use for examples and homework problems. A complete periodic table, including atomjc masses, is found in Appendix B. TABLE12.1 TABLE 12.1 Some atomic mass numbers Element Symbol A Hydrogen IH Helium 4 He 4 Carbon 12c 12 Nitrogen 14N 14 Oxygen 160 16 Neon 20Ne 20 Aluminum 21AI 27 Argon -I0Ar 40 Lead 201Pb 207 398 CHAPTER 12 Thermal Properties of Matter The Definition of the Mole TABLE 12.2 Monatomic gases and diatomic Monatomic Diatomic Helium (He) Hydrogen (H2) Neon (Ne) Nitrogen (N 2) Argon (Ar) Oxygen (0 2) One way to specify the amount of substance in a system is to give its mass. Another way, one connected to the number of atoms, is to measure the amount of substance in moles. 1 mole of substance, abbreviated 1 mol, is 6.02 x 1023 basic particles. The basic particle depends on the substance. Helium is a monatomic gas, meaning that the basic particle is the helium atom. Thus 6.02 X I 023 helium atoms are l mo) of helium. But oxygen gas is a diatomic gas because the basic particle is the two-atom diatomic molecule 0 2 . l mol of oxygen gas contains 6.02 X 1023 molecules of 0 2 and thus 2 X 6.02 X 1023 oxygen atoms. TABLE 12.2 lists the monatom.ic and diatomic gases that we will use for examples and problems. The number of basic particles per mole of substance is called Avogadro's number NA. The value of Avogadro's number is thus NA= 6.02 X 1023 mol- 1 The number II of moles in a substance containing N basic particles is 11=- N (12.1) NA Moles of a substance in terms of the number of basic particles The molar mass of a substance, MmoI, is the mass in grams of l mo! of substance. To a good approximation, the numerical value of the molar mass equals the numerical value of the atom.ic or molecular mass. That is, the molar mass of He, with m = 4 u, is Mmol (He)= 4 g/mol, and the molar mass of diatomic 0 2 is Mmol (02) = 32 g/mol. You can use the molar mass to determ.ine the number of moles. In one of the few instances where the proper units are grams rather than kilograms, the number of moles contained in a system of mass M consisting of atoms or molecules with molar mass M 11101 is 11 = M (in grams) (12.2) Mmol Moles of a substancein terms of its mass One mole of helium, sulfur, copper, and mercury. EXAMPLE 12.1 Detennining quantities of oxygen A system contains I 00 g of oxygen. How many moles does it contain? How many molecules? SOLVE The diatomic oxygen molecule 0 2 bas molar mass Mmol = 32 g/mol. From Equation 12.2. 100 g 11=----=3.I 32 g/mol mo! Each mole contains NA molecules. so the total number is N = 11NA= 1.9 X I 024 molecules. Volume The volume V of a macroscopic system is the amount of space the system occupies. The SI unit of volume is 1113. Nonetheless, both cm 3 and, to some extent, liters (L) are widely used metric units of volume. In most cases, you must convert these to 1113 12.2 before doing calculations. Some important conversions for volumes are listed here. Although it is true that I m = 100 cm, it is not true that I 1113 = l 00 cm 3 . FIGURE 12.2 shows that the volume conversion factor is 1 1113 = 106 cm3 . l 3 111 = 1000 liters = I 06 cm 3 l liter= 1000 cm 3 399 The Atomic Model of an Ideal Gas FIGURE 12.2 There I cm 3 = 1 mL are l0 6 crn3 in I 3 111 . Subdivide the l m X I m X I rn cube into little cubes I cm on a side. You will get 100 5,ubdivisions along each edge. \ ... I cm 3 ··~,c;,J-...,.-=-~---_-_-_-_ -_-.... ( Which system contains more atoms: 5 mol of hebum (A = 4) or 1 mo! of neon (A= 20)? STOP TO THINK 12.1 A. Hebum B. Neon I 1113 I 111 100 cm down C. They have the same number of atoms. '--'------ ••.L.::-...J,~ Im There are 100 X 100 X 100 = 106 little I cm 3 cubes in the big I m3 cube. 12.2 The Atomic Model of an Ideal Gas Sobds and liquids are nearly incompressible because the atomic particles are in close contact with each other. Gases, in contrast, are highly compressible because the atomic particles are far apart. When we work with gases, we' II assume a set of simplifications that we call the ideal-gas model, summed up in FIGURE 12.3. It turns out that most real gases at the temperatures and pressures we work with can be explained quite well using this model; the simpljfi.cations are not far from reality. We discussed thermal energy in the context of a gas in • SECTION 11.3. The thermal energy of a gas is simply the total kinetic energy of the atoms of the gas; the temperature is a measure of the average kinetic energy of the atoms. Adding heat to a gas makes its atoms move faster, increasing the thermal energy and the temperature. For an ideal gas, there is a straightforward relationship between the temperature and the average kinetic energy per atom Kavg: T=-- 2 Kavg FIGURE 12.3 The ideal-gas model. The gas is made up of a large number N of par\.1',,.i· ... c..les of mass 111, each moving randomly. The particles are quite far from each other and interact only rarely when they co~lide. \., Ill J (12.3) 3 ks .. where ks is a constant known as Boltzmann's constant. Its value is The collisions of the particles with each other (and with walls of the container) arc elastic; no energy is lost in these collisions. ks= 1.38 X 10-23 J/K We can rearrange Equation 12.3 to give the average kinetic energy in terms of the temperature: (12.4) The thermal energy of an ideal gas consisting of N atoms is the sum of the kinetic energies of the individual atoms: E,. (L2.5) Them1al energy of an ideal gas of N atoms E,. I /9 ~gT ~N PHOPORTIOtlAI. PROPORTIONAi. For an ideal gas, thermal energy is directly proportional to temperature. Consequently, a change in the thermal energy of an ideal gas is proportional to a change in temperature: (12.6) 400 c HA PTE R 12 Thermal Properties of Matter Finding the energy needed to warm up a room EXAMPLE 12.2 A large bedroom contains about I X I 027 molecules of air. [f we model the air as an ideal gas. how much energy is required 10 raise the temperature of the air in the room by 5°C. STRATEGIZE We can use Equa1ion 12.6 to detem1ine the change in thermal energy corresponding ro the noted change in 1emperarure. PREPARE The actual temperature of the gas doesn·t maner-only the cbange. The temperature increase is given as 5°C, implying a change in the absolute temperature by the same amount: t:i.T= 5 K. Is it cold in space? The International Space Station orbits in the upper thermosphere, about 400 km above the smface of the earth. There is still a trace of atmosphere left at this altitude, and il has quite a high temperature-over I 000°C. Although the average speed of the air molecules here is high, there are so few air moleculespresent that the thermal energy is extremely low. SOLVE We can use Equation 12.6 to calculate the amount by which the room ·s thermal energy must be increased: 6.£11,= INk 2 8 t:i.T= I(I x 1027)(1.38 x 10-13 J/K)(S K) =Ix 2 This is the energy we would have 10 upply-probably nace-to raise the temperature. 105 J = 100 kJ in the fom1 of beat from a fur- I00 kJ isn't that much energy. Table 11.2 showed it to be less than the food energy in a carrot! This seems reasonable because you know that your furnace can quickly warm up the air in a room. Heating up the walls and furnishings is another story. ASSESS Molecular Speeds and Temperature FIGURE 12.4 The distribution of molecular speeds in nitrogen gas at 20°c. The atomic model of an ideal gas is based on random motion, so it's no surprise that the individual atoms in a gas are moving at different speeds. FIGURE 12.4 shows data from an experiment to measure the molecular speeds in nitrogen gas at 20°C. The The height of a bar representsthe percentageof moleculeswith speeds results are presented as a histogram, a bar chart in which the height of the bar indiin the rangeon !,'._1·.,e horizontalaxis. cates what percentage of the molecules have a speed in the range of speeds shown Percentage below the bar. For example, 16% of the molecules have speeds in the range from of molecules ..····Therms speed ~-· is 510 m/s. 600 m/s to 700 111/s.The most probable speed, as judged from the tallest bar, is 20 ., -500 mis. This is quite fast: -1200 mph! N2 molecules 15 Because temperature is proportional to the average kinetic energy of the atoms, it at 20°C will be useful to calculate the average kinetic energy for this distribution. An indi10 vidual atom of mass m and velocity v has kinetic energy K = ½mv2 . Recall that the 5 average of a series of measurements is found by adding all the values and then dividing the total by the number of data points. Thus, we can find the average kinetic 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 N -st r-- 00 0 energy by adding up all the kinetic energies of all the atoms and then dividing the I "'I I "'I '°I 0 I I °'I I I I 0 0 0 0 0 0 0 0 total by the number of atoms: 0 0 0 0 0 0 0 0 0 0 "' "' st '° r- 00 0°' 00 2 2 V) Speedrange(m/s) Kavs = L½mv _ N 1 - 2m Lv _ 1 N - 2117( V2 ) avg (12.7) The quantity L v2 JN is the sum of the values of v2 for all the atoms divided by the number of atoms. By definition, this is the average of the squares of all the individual speeds, which we've written ( 1/)avg· The square root of this average is about how fast a typical atom in the gas is moving. Because we'll be taking the square root of the average, or mean, of the square of the speeds, we define the root-mean-square speed as v,-ms = ~ = speed of a typical atom ( 12.8) The root-mean-square speed is often referred to as the rms speed. The rms speed isn't the average speed of atoms in the gas; it's the speed of an atom with the average kinetic energy. But the average speed and the rms speed are very nearly equal, so we'll interpret an rms speed as telling us the speed of a typical atom in the gas. Rewriting Equation 12.7 in terms of v,ms gives the average kinetic energy per atom: 1 Kavg ? = 211111~ms ( 12.9) 12.2 The Atomic Model of an Ideal Gas 401 We can relate the temperature to the speeds of the atoms by substituting Equation 12.9 into Equation 12.4: (12.10) Solving Equation 12.10 for therms speed of the atoms, we find that (12.11) rms speed of an atom of mass lmii]I ► III in an ideal gas at temperature T You must use absolute temperature, in kelvin, to compute rms speeds. ◄ We have been considering ideal gases made of atoms, but our results are equally valid for real gases made of either atoms (such as helium, He) or molecules (such as oxygen, 0 2 ). At a given temperature, Equation 12.11 shows that the speed of atoms or molecules in a gas varies with the atomic or molecular mass. A gas with lighter atoms will have faster atoms, on average, than a gas with heavier atoms. The equation also shows that higher temperatures co1Tespond to faster atomic or molecular speeds. The rms speed is propottional to the square root of the temperature. Tl1is is a new mathematical form that we will see again, so we will take a look at its prope1ties. ~ Martian airsicles The atmosphere of Mars is mostly carbon dioxide. At night, the temperature may drop so low that the molecules in the atmosphere will slow down enough to stick together-the atmosphere actually freezes. The frost on the stni"acein this image from the Viking 2 lander is composed partially of frozen carbon dioxide. Square-root relationships Two quantities are said to have a square-root relationship if y is propor- tional to the square root of x. We write the mathematical relationship as Increasing .r by a factor of ➔ Y .-······· increases y by a factor of 2. 2A "" v=A-VX ") y=AVx ' y is proportional to the square root of x The graph of a square-root relationship is a parabola that has been rotated by 90°. ' A is the constant of proportionality.Ltis the value of y when x = I. 0+------------~x 0 2 3 SCALING If x has the initial value x 1, then y has the initial value Yt. Changing x from x 1 to x 2 changes y from y 1 to y 2 . The ratio of y 2 to y 1 is Y2 _ A\/x; _ {½ Yi- A¼J-'.J-;; which is the square root of the ratio of x 2 to x,. ■ If you increase x by a factor of 4, you increase y by a factor of V4 = 2. ■ If you decrease x by a factor of 9, you decrease y by a factor of v'9 = 3. These examples illustrate a general rule: Changing x by a factor of c changes y by a factor of Vc. Exercises 7, 8 ml EXAMPLE 12.3 Finding the speeds of air molecules Most of the ea,th's atmosphere is the gas nitrogen, which consists of molecules, N 2 . At the coldest temperature ever recorded on earth, -95°C, what is the root-mean-square speed of the nitrogen molecules? Does the temperature at the earth's surface ever get high enough that a typical molecule is moving at twice this speed'l (The nighest temperature ever recorded on earth was 54°C.) STRATEGIZE The first part of the problem is a straightforward calculation. The second part is best done by considering ratios. The rms speed varies as the square root of the temperalllre, so we will determine the factor by which we'd need to increase the temperature to double therms speed. Co11ti11ued 402 c HAP TE A 12 Thermal Properties of Matter PREPARE We can use the periodic table to determine that the mass of a nitrogen atom is 14 u. A molecule con ists of two atoms. so its mass is 28 u. Thus the molecular mass in SI units (i.e .. kg) is Ill I 66 = 28 u X • J0- 27 ko0 X lu = 4.6 X 10- 26 k!! - The problem statement gives two temperatures we·11call T1 and T2 : we need to express these in kelvin. The lowest temperature ever observed on earth is T1 = -95 + 273 = 178 K: the highest temperature is T2 = 54 + 273 = 327 K. SOLVE We use Equation 12.11 to find "rms for the nitrogen mol- ecules at T1: 3(1.38 X I0- 23 J/K)(I78 4.6X Because the rrns speed is proportional to the square root of the temperature. doubling the m1s speed would require increasing the temperature by a factor of -t The ratio of the highest temperature ever recorded to the lowest temperature ever recorded is less than this: 10-26 kg K) 400 mis T, 327 K -=-=--= T 1 178 K 1.8 The temperature at the earth's surface is never high enough that a typical nitrogen molecule would move at twice the computed speed. ASSESS We can use the square-root relationship to assess our computed result for the molecular speed. Figure 12.4 shows an rms speed of 510 mis for nitrogen molecules at 20°C. or 293 K. Temperature T1 is about 0.6 of this. so we·d expect to compute a speed that is lower by the square root of 0.6 about 0.8, which is what we found. Pressure Everyone has some sense of the concept of pressure. If you get a hole in your bicycle tire, the higher-pressure air inside comes squirting out. It's hard to get the lid off a vacuum-sealed jar because of the low pressure inside. But just what is pressure? Let's take an atomic-scale view of pressure, defining it in terms of the motion of particles of a gas. Suppose we have a sample of gas in a container with rigid walls. As particles in the gas move around, they sometimes collide with and bounce off the walls, creating a force on the walls, as illustrated in FIGURE12.sa. FIGURE12.5 The pressure in a gas is due to the force of the particles colliding with the walls of its container. (b) On a macroscopic scale, these microscopic collisions exert a force Fon an area A of the wall that is proportional to A. There are an enormous number of collisions of particles against the wall per second. Each collision e{~-rtsa force on the wall. The net force due lo all the collisions causes the gas to have a pressure. These countless microscopic collisions are what lead to the pressure in the gas. On a small patch of the container wall with surface area A, these colbsions result in a continuous macroscopic force of magnitude F directed perpendicular to the wall, as shown for the bicycle tire in FIGURE12.sb. If the size of the patch is doubled, then twice as many particles will hit it every second, leading to a doubling of the force. Thjs implies that the force is proportional to the area of the patch, and so the ratio Fl A is constant. We define this ratio to be the pressure p of the gas: p=- F A ( 12.12) Definition of pressure in a gas You can see from Equation 12.12 that a gas exerts a force of magnitude F=pA on a surface of area A. ( 12.13) 12.2 lmiDI ► We sometimes talk informally about "the force exerted by the pressure," but it's really the gas that exerts the force, not the pressure. ◄ From its definition, you can see that pressure has units of N/m 2 . The SI unit of pressure is the pascal, defined as J pascal 403 The Atomic Model of an Ideal Gas llll llfll A video to support a section's topic is embedded in the eText. Video Force and Pressure N = l Pa = I ~ Ill 2 A pascal is a very small pressure, so we usually see pressures in kilopascals, where l kPa = 1000 Pa. The pressure with the most significance for our daily lives is the pressure of the atmosphere, caused by the microscopic collisions of the air molecules that smTound us. The pressure of the atmosphere varies with altitude and the weather, but the global average pressure at sea level, called the standard atmosphere, is l standard atmosphere = l atm = 101,300 Pa = l 0 l. 3 kPa Just as we measured acceleration in units of g, we will often measure pressure in units of atm. In the United States, pressure is often expressed in pounds per square inch, or psi. When you measure pressure in the tires on your car or bike, you probably use a gauge that reads in psi. The conversion factor is 1 atm FIGURE 12.6 The net force depends the pressure difference. on (b) Area A ""l = 14.7 psi I The total force on the surface of your body due to the pressure of the atmosphere is over 40,000 pounds. Why doesn't this enormous force pushing in simply crush you? The key is that there is also a force pushing out. FIGURE 12.sa shows an empty plastic soda bottle. The bottle isn't a very sturdy structure, but the inward force due to the pressure of the atmosphere outside the bottle doesn't crush it because there is an equal pressure due to air inside the bottle that pushes out. The forces pushing on both sides are quite large, but they exactly balance and so there is no net force. A net pressure force is exerted only where there's a pressure difference between the two sides of a surface. FIGURE 12.sb shows a sutface of area A with a pressure difference b..p = p 2 - p 1 between the two sides. The net pressure force is Fnet = F2 - F1 = /J2 A - p I A = A ( /J2 - p I) I /Ji \. The pressure inside lhe bottle equals the pressure outside. so there is no nel force. ' I ~ : F2 /J2 > /Ji ./ There is a ditfe(~nce in pressure between the two sides. = A !:i.p This is the force that holds the lid on a vacuum-sealed jar, where the pressure inside is less than the pressure outside. Decreasing the number of molecules in a container decreases the pressure because there are fewer collisions with the walls. The pressure in a completely empty container would be p = 0 Pa. This is called a pe1j'ect vacuum. A pe1fect vacuum cannot be achieved because it's impossible to remove every molecule from a region of space. In practice, a vacuum is an enclosed space in wruch p << 1 atm. Using p = 0 Pa is then a good approximation. The basic principle behind pressure measurements is expressed by Equation 12.12. A pressure gauge measures the force exerted by the gas on a known area. The actual or absolute pressure p is directly proportional to this force. Because the effects of pressure depend on pressure differences, most gauges measure not the absolute pressure but what is called the gauge pressure /Jg, the difference between the absolute pressure and atmospheric pressure. Most gauges measure this difference directly because they measure a force on a swface. FIGURE 12.1 shows how this works for a tire gauge. The gauge measures the difference between the pressure inside the tire and the pressure outside the tire, which is just the pressure of the atmosphere, Patmos· The calculations we'll do later in the chapter involve absolute pressure, but tire pressures and other common pressures are repotted as gauge pressures, so we' II often need to convert between the two. The gauge pressure is, as in the case of the Too little pressure too fast 4miJ The rockfish is a popular game fish that can be caught at ocean depths up to 1000 ft. The great pressure at these depths is balanced by an equally large pressure inside the fish's gas-filled swim bladder. If the fish is hooked and rapidly raised to the lower pressure of the suiface, the pressure inside the swim bladder is suddenly much greater than the pressure outside, causing the swim bladder to expand dramatically. When reeled in from a great depth, these fish seldom survive. 404 c HA PTER 12 Thermal Properties FIGURE 12.7 A tire gauge measures the difference between the tire's pressure and atmospheric pressure. of Matter tire gauge, the difference between an absolute pressure and atmospheric pressure, so we can write /Jg = P - Patmos This fixed disk holds the end of the spring. The scale can slide through it. P If you are at sea level, you can assume that you'll need to use the local pressure. EXAMPLE 12.4 = Pg + Patmos Patmos= (12.14) 1.0 atm; at other elevations, Finding the force due to a pressure difference Patients suffering from decompression sickness may be treated in a hyperbaric oxygen chamber filled with oxygen at greater lhan atmospheric pressure. A cylindrical chamber with flat end plates of diameter 0.75 mis filled with oxygen to a gauge pressure of 27 kPa. What is the resulting force on the end plate of the cylinder? STRATEGIZEThe net force F0 e, = A /:j.p on the end plate depends on /:j.p, the difference between !he pressure inside and the pressure oulside the chamber. This is lhe gauge pressure, 27 kPa. The difference·········· PREPAREThe end plates are circular, with area A = r,,- 2 = r,(0.375 m) 2 = 0.442 m2 . between p,;,•eand pushes the piston forward against the spring, moving the scale outward. SOLVEThe pressure difference results in a net force Patmos Fnet= A {j.p = (0.442 m2 )(27,000 Pa)= 12 kN ASSESS The area of the end plate is large, so we expect a large force, which is exactly what we get-che force is equivalent to the weight of about 15 people. If you·ve seen chambers like this, you know that the end plate is fastened in place with stout bolts. It is remarkable to think that this large force results from the collisions of tiny air molecules with the plate! The Ideal-Gas Law We can use the fact that the pressure in a gas is due to the colfoions of particles with the walls to make some qualitative predictions. FIGURE12.8 presents a few such predictions. FIGURE 12.8 Relating gas pressure to other variables. 'o y " ;I ~ ~ y J s ... ,I _t Increasing the temperature of the g~s means the particles move at higher speeds. They hit the walls more often and with more force, so there is more pressure. Decreasing the volume of the c~ntainer means more frequent collisions with the walls of the container, and thus more pressure. Increasing the number of particles in the container means more frequent collisions with the walls of the container, and thus more pressure. Based on the reasoning in Figure 12.8, we expect the following proportionalities: ■ ■ ■ Pressure should be proportional to the temperature of the gas: p ex T. Pressure should be inversely proportional to the volume of the container: p ex 1/V. Pressure should be proportional to the number of gas particles: p ex N. In fact, careful experiments back up each of these predictions, leading to a single equation that expresses these proportionalities: p=CPumping up a bicycle tire adds more particles to the fixed volume of the tire, increasing the pressure. NT V The proportionality constant C turns out to be none other than Boltzmann's constant k 8 , which allows us to write 12.2 pV=Nk 8 The Atomic Model of an Ideal Gas 405 (12.15) T Ideal-gas law, version I Equation 12.15 is known as the ideal-gas law. Equation 12.15 is written in terms of the number N of particles in the gas, whereas the ideal-gas law is stated in chemistry in terms of the number n of moles. But the change is straightforward to make. The number of particles is N = nN;1, so we can rewrite Equation 12.15 as (12.16) Ideal-gas law, version 2 In this version of the equation, the proportionality constant-known constant-is R = NAk 8 = 8.31 J/rnol as the gas •K The units may seem unusual, but the product of Pa and 1113, the units of pV, is equivalent to J. Let's review the meanings and the units of the vaiious quantities in the ideal-gas law: Number of moles in the sample or container of gas Absolute pressure (Pa)··········· \ ._..•··········· Gas constant, 8.31 J/mol • K .. "pV= nRT,.. Volume of the sample or·········_..container of gas (m 3) EXAMPLE 12.5 ·············Temperature in kelvin (K) Finding the volume of a mole of a gas What volume is occupied by I mole of an ideal gas at a pressure of 1.00 arm and a temperature of 0°C? We recall from earlier in the chapter that 1.00 m3 = I 000 L, so we can write STRATEGfZE We are given the pressure. the temperature. and the number of moles. We will use version 2 of the ideal-gas law to find the volume. \I= 22.4 L The first step in ideal-gas law calculation is to convert all quantities to SI units: PREPARE At this temperature and pressure. we find that the volume of I mole of a gas is 22.4 L. a result you might recall from chemistry. When we do calculations using gases, it will be useful to keep this volume in mind to see if our answers make physical sense. ASSESS p = 1.00 atm = IOI .3 X !03 Pa T = 0 + 273 = 273 K We rearrange version 2 of the ideal-gas law equation to compute SOLVE nRT V = -- P ( 1.00 mo1)(8.3 I J/mol • K)(273 K) = -'----_:_.:.._ ____ 101.3Xl0 3 --'-'---- = 0.0224 m3 Pa STOP TO THINK 12.2 A sample of ideal gas is in a sealed container. The temperature of the gas and the volume of the container ai·e both increased. What other properties of the gas necessarily change? (More than one answer may be correct.) A. The rms speed of the gas atoms C. The pressure of the gas B. The thermal energy of the gas D. The number of molecules of gas 406 c HAP TE A 12 Thermal Properties of Matter 12.3 Ideal-Gas Processes Suppose you measure the pressure in the tires on your car on a cold morning. How much will the tire pressure increase after the tires warm up in the sun and the temperature of the air in the tires has increased? We will solve this problem later in the chapter, but, for now, note these properties of this process: FIGURE 12.9 The state of the gas and ideal-gas processes can be shown on a ■ pV diagram. ■ (a) Each state of an ideal gas is represented as a point on a p V diagram. ~O(kPa\ = I mol ···--•••........ ,; Processes with these properties are called ideal-gas processes. For gases in sealed containers, the number of moles (and the number of molecules) does not change. In that case, the ideal-gas law can be written as T2= 3600 K T1 ■ The quantity of gas is fixed. No air is added to or removed from the tire. There is a well-defined initial state. The initial values of pressure, volume, and temperature will be designated Pi, Vi, and T;. There is a well-defined final state in which the pressure, volume, and temperature have values /Jr, Vr, and Tr. • = 900 K pV 5 - T = nR = constant The values of the variables in the initial and final states are then related by 3 0 +------------- V (m ) 0 2 3 (b) A process that changes the gas from one state to another is represented /J (k.Pa) by a trajectory on a pV diagram. 2 Pr Vr P; Vi Tr T; (12.17) Initial and final states for an ideal gas in a sealed container 10 This before-and-after relationship between the two states, reminiscent of a conservation law, will be valuable for solving many problems. EmDI ► Because pressure and volume appear on both sides of the equation, 5 \I (m 3) 0 +---~---------.--0 2 3 FIGURE 12.10 A constant-volume (a) process. As the temperature increases, so does the pressure ...... '::, (b) p f /Jr ... •····"A constant-volume •····· process appears on a p V diagram i as a vertical line. f ~-------v V pressure and volume can be in any units; we don't necessarily need to convert to SI units. However, temperature must be in K. Unit-conversion factors for pressure and volume are multipl.icative factors, and the same factor on both sides of the equation cancels. But the conversion from K to °C is an additive factor, and additive factors in the denominator don't cancel. ◄ pVDiagrams It's useful to represent ideal-gas processes on a graph called a pV diagram. The important idea behind a pV diagram is that each point on the graph represents a single, unique state of the gas. This may seem surprising because a point on a graph specifies only the values of pressure and volume. But knowing p and V, and assuming that n is known for a sealed container, we can find the temperature from the ideal-gas law. Thus each point on a p V diagram actually represents a triplet of values (p, V, T) specifying the state of the gas. For example, FIGURE 12.sa is a p V diagram showing three states of a system consisting of 1 mol of gas. The values of p and V can be read from the axes, then the temperature at that point calculated from the ideal-gas law. An ideal-gas process-a process that changes the state of the gas by, for example, heating it or compressing it-can be represented as a "trajectory" in the pV diagram. FIGURE 12.sb shows one possible process by which the gas of Figure 12.9a is changed from state 1 to state 3. Constant-Volume Processes Suppose you have a gas in the closed, rigid container shown in FIGURE 12.10a. Warming the gas will raise its pressure without changing its volume. This is an example of a constant-volume process. Vr = Vi for a constant-volume process. Because the value of V doesn't change, this process is shown as the vertical line i - f on the pV diagram of FIGURE 12.,ob. A constant-volume process appears on a p V diagram as a vertical line. 12.3 EXAMPLE 12.6 the ratio of the two volumes. which is equal volume process: STRATEGIZEA tire is (to a good approximation) a sealed container with constant volume. so this is a con tant-volume process. pr= 44.7 p Tr = 30°C + 273 = 303 K SOLVE We can use Equation 12.17 10 solve for the final pressure. In thi equation. we divide both ides by Vr.and then cancel Processes Many gas processes take place at a constant, unchanging pressure. A constant-pressure process is also called an isobaric process. For a constant-pressure process Pr= Pi· One way to produce a constant-pressure process is shown in FIGURE 12.11a, where a gas is sealed in a cylinder by a Jjghtweight, tight-fitting cap-a piston-that is free to slide up and down. Tn fact, the piston will slide up or down, compressing or expanding the gas inside, until it reaches the position at which Pgas = /Jcxt· That's the equilibrium position for the piston, the position at which the upward force fgas = /JgasA, where A is the area of the face of the piston, exactly balances the downward force F0 , 1 = /Jcx,A due to the external pressure Pext· The gas pressure in thjs situation is thus equal to the external pressure. As long as the external pressure doesn't change, neither can the gas pressure inside the cylinder. (a) process. (b) In equilibrium, pressure must be the same on both ~ Pex1 /Jp., p t- Initial A constant-pressure process appears on a p V diagram as a horizontal line. .;. p Fexl /J\:.t ···........ . F,._, Final V; ~ V1 Movable piston IX 303 K K = 49.6 psi 273 ASSESS The temperature has changed by 30 K, which is a bit more than 10% of the initial temperature. so we expect a large change in pressure. Our re ult seems reasonable. and it has practical implications: If you check the pressure in your tires when they are at a panicular temperature, don't expect the pressure to be the same when conditions change! o·c+ 213 = 213 K FIGURE 12.11 A constant-pressure . (pg)r = Pr- 1.00 atm = 49.6 psi - 14.7 psi= 34.9 p i Temperatures must be in kelvin, o we conven: Constant-Pressure I for this constant- This is an ab olute pressure. but the problem asks for the measured pressure in the tire-a gauge pressure. Convening to gauge pressure gives 1.00 atm = 30.0 psi+ 1-U psi= 44.7 psi T; = 10 The units for Pr will be the same as tho e for p;, so we can keep the initial pressure in psi. The pressure at the higher temperature is PREPAREWe weren't told anything about elevation, so we assume that the atmospheric pressure is 1.0 atm. When we solve this problem. we don't need to conven units, but we do need to be sure that we are using the correct form of the pressure. The measured tire pressure is a gauge pressure, but the ideal-gas law requires an absolute pre sure. We can use Equation 12.14 10 conven to absolute pre ure. The initial pressure i = (pg);+ 407 Computing tire pressure on a hot day The pressure in a car tire measures 30.0 psi on a cool morning when the air temperature is 0°C. After the day warms up and bright sun shines on the black tire, the temperature of the air inside the tire reaches 30°C. What is the tire pressure at this temperature? p, Ideal-Gas Processes L---,----~-- V • ••••Because the external pressure doesn't change, the gas pressure remains (,~ constant as the gas expands. Suppose we heat the gas in the cylinder. The gas pressure doesn't change because the pressure is controlled by the unchanging external pressure, not by the temperature. But as the temperature rises, the faster-moving atoms cause the gas to expand, pushing the piston outward. Because the pressure is always the same, this is a constant-pressure process, with a trajectory as shown in FIGURE 12.11b. A constantpressure process appears on a pV diagram as a horizontal line. 408 CHAPTER EXAMPLE 12.7 12 Thermal Properties of Matter Finding the final volume of a compressed gas A gas in a cylinder with a movable piston occupies 50.0 cm3 at 50°C. The gas is cooled at constant pressure until the temperature is 10°C. What is the final volume? SOLVE STRATEGIZE The pressure of the gas doesn't change, so this is a constant-pressure process, with p/JJr = l. Our answer makes sense. The gas cools, so we expect a smaller volume. The cooling is-in absolute temperature termsrelatively modest, so we expect a modest change in volume. In this example and in Example 12.6, we have not converted pressure and volume units because these multipUcative factors cancel. But we did convert temperature to kelvin because this additive factor does not cancel. The temperatures must be in kelvin, so we convert: PREPARE Ti = 50°C + 273 = 323 K Tr= 10°C +273 = 283 K We can use Equation 12.17 to solve for Vr: Pi Tr Vr = Vi - 0 ~ A constant-temperature process. (a) Push - ,_ -~ V; Vr , Final Initial I ConstantT ConstantT (b) JJ .. /Jr _..-A constanHemperature process appears on a p V diagram as a hyperbola. 283 K 323K X 1 X -- = 43.8 cm 3 ASSESS Constant-Temperature FIGURE 12.12 = 50.0 cm3 Processes A constant-temperature process is also called an isothermal process. For a constant-temperature process Tr= Ti. One possible constant-temperature process is illustrated in FIGURE 12.12a. A piston is being pushed clown to compress a gas, but the gas cylinder is submerged in a large container of liquid that is held at a constant temperature. If the piston is pushed slowly, then heat-energy transfer through the walls of the cylinder will keep the gas at the same temperature as the surrounding liquid. This is an isothermal compression. The reverse process, with the piston slowly pulled out, is an isothermal expansion. Representing a constant-temperature process on the pV diagram is a little more complicated than the two preceding processes because both p and V change. As long as T remains fixed, we have the relationship p.116 nRT constant (12.18) P --- i V V ~\I INVERSE .: /Ji V V; Vr (c) JJ Different temperatures correspond to different isotherms. Because there is an inverse relationship between p and V, the graph of a constant-temperature process is a hyperbola. Figure l 2. l 2a shows an isothermal compression, which is represented graphically as the hyperbola i ~fin FIGURE 12.12b. An isothermal expansion would move in the opposite direction along the hyperbola. The graph of an isothermal process is known as an isotherm. The location of the hyperbola depends on the value of T. If we use a higher constant temperature for the process in Figure 12.12a, the isotherm will move farther from the origin of the pV diagram. FIGURE 12.12c shows three isotherms for this process at tlu·ee different temperatures. A gas undergoing a constant-temperature process will move along the isotherm for the appropriate temperature. EXAMPLE 12.8 .____________ v Why does the bag expand? Residents of Colorado, where it's easy to drive rapidly to different elevations, are familiar with this scenario: You buy a sealed bag of chips at a low elevation and then drive your car up into the mountains. As you climb, the atmospheric pressure drops, the air inside the bag expands, and the bag gets "puffy." Explain what is happening. The chip bag is sealed-no gas gets in or out-so we can treat this as an ideal-gas process. Folks in the car will use the heater or the air conditioner to keep the temperature in the car steady, so this is a constant-temperature process. Chip bags aren't very sturdy, so the pressure inside the bag is the same as the pressure outside the bag. As the pressure outside the bag drops, so does the pressure inside the bag. The volume of air in the bag is inversely proportional to the pressure, so as the pressure drops, the volume of the air increases and the bag expands. REASON The result makes sense. A change in pressure also occurs on an airplane, so you may have seen something similar during a fUght, or you may have had bottles with caps that pop off and contents pushed out as the gas inside expands. ASSESS 12.3 Ideal-Gas Processes EXAMPLE 12.9 409 Finding the volume of air in the lungs (iJ A snorkeler takes a deep breath at the surface, filling his lungs with 4.0 L of air. He then descends to a depth of 5.0 m, where the pressure is 0.50 atm hjgber than at the surface. At thjs depth, what is the volume of air in the snorkeler's lungs? Most of the snorkeler's air is nitrogen, whlch his body doesn't use. Oxygen is used by the body, but for every oxygen molecule taken from tbe lungs, a molecule of carbon dioxide is returned. The ideal-gas law doesn't care what the molecules are, just the total number. The total number of gas molecules stays approximately constant, and we can treat this as an ideal-gas process. The air stays at body temperature, so this is a constanttemperature process with Tr= T;, so Tr/Ti = 1. As the snorkeler descends, the pressure inside his lungs rises to match the pressure of the surrounding water because the body can't sustain large pressure differences between inside and out. STRATEGIZE Thermodynamics We need absolute pressures for our calculation. At the surface, the pressure of tbe air inside the snorkeler's lungs is 1.0 atm-it's the pressure at sea level. The additional pressure at the noted depth is 0.50 atm, so the absolute pressure at thjs depth is 1.5 atm. PREPARE SOLVE The ideal-gas law for a sealed container (the lungs) gives Vr= vi Pi Tr Pr Ti J.5 atm 1 =2.7 L Notice that we didn't need to convert pressure to SI uruts. As long as the units are the same in the numerator and the denorrunator, they cancel. ASSESS The air has a smaller volume at the higher pressure, as we would expect. The air inside your lungs does compresssignificantly 1-when you dive below the surface. of Ideal-Gas Processes Chapter 11 introduced the first law of thermodynamics, and we saw that heat and work are just two different ways to add energy to a system. We've been considering the changes when we heat gases, but now we want to consider the other form of energy transfer-work. When gases expand, they can do work by pushing against a piston. This is how the engine under the hood of your car works: When the spark plug foes in a cylinder in the engine, it ignites the gaseous fuel-air mixture inside. The hot gas expands, pushing the piston out and, through various mechanical linkages, turning the wheels of your car. Energy is transferred out of the gas as work; we say that the gas does work on the piston. Similarly, the gas in Figure 12.11 a does work as it pushes on and moves the piston. You learned in ~SECTION 10.2 that the work done by a constant force Fin pushing an object a distanced is W = Fcl. Let's apply this idea to a gas. FIGURE 12.13a shows a gas cylinder sealed at one end by a movabl~piston. Force Fgas is due to the gas pressure and has magnitude Fgas = pA. Force Fext, perhaps a force applied by a piston rod, is equal in magnitude and opposite in direction to Fgas. The gas pressure would blow the piston out if the external force weren't there! Suppose the gas expands at constant pressure, pushing the piston outward from x; to xr, a distanced= xr - x;, as shown in FIGURE 12.13b. As it does, the force due to the gas pressure does work Wgas 1.0 atrn --=4.0LX--X The expanding gas does work on the piston. FIGURE 12.13 The gas pu~hes on the piston with force Fgas· To keep the piston in place, (a) an external force must be equal an/ opposite to F'gns· Pressure p \ \ - " Fg.,---- = Fgasd = (pA)(xr - x;) = p(xrA - X;A) But x;A is the cylinder's initial volume V; (recall that the volume of a cylinder is the length times the area of the base) and xrA is the final volume Vr. Thus the work done is (b) (12.19) As the piston moves a distance d, the ga__ ..?loes work F'ga,d, Piston areaA Work done by a gas in a constant-pressure process where LlV is the change in volume. Equation 12.19 has a particularly simple interpretation on a pV diagram. As FIGURE 12.14a on the next page shows, p fl V is the "area under the p V graph" between V; and Vr, Although we've shown this result for only a constant-pressure process, it turns out to be true for all ideal-gas processes. That is, as FIGURE 12.14b shows, W gas = area under the p V graph between V; and Vr 0 / X; Xf The gas volume changes by ~V = dA. 410 c HAP TE R 12 Thermal Properties of Matter FIGURE 12.14 Calculating the work done There are a few things to clarify: in an ideal-gas process. ■ (a) In order for the gas to do work, its volume must change. No work is done in a constant-volume process. ■ The simple relationship of Equation 12.19 applies only to constant-pressure processes. For any other ideal-gas process, you must use the geometry of the p V diagram to calculate the area under the graph. ■ To calculate work, pressure must be in Pa and volume in m3 . The product of Pa (which is N/m 2) and m3 is N • m. But l N ·mis J J-the unit of work and energy . ■ Wgas is positive if the gas expands ( I:::. V > 0). The gas does work by pushing against the piston. In this case, the work done is energy transferred out of the system, and the energy of the gas decrea~s. Wgas is negative if the piston compresses the gas ( I:::. V < 0) because the force Fgas is opposite the displacement of the piston. Energy is transferred into the system as work, and the energy of the gas increases. We often say "work is done on the gas," but this just means that Wgas is negative. For a constant-pressureprocess, the area under the graph p!).V is the work done by the gas as it e7ands from V~to Vr, p p ... • I .w V c_-1------~ V, Vr (b) Tngeneral,the work done by a /J i In the first law of thermodynamics, l:::.E 111= Q + W, W is the work done by the environment-that is, by force F'ext acting on the system. But F'ext and F'gas are equal and opposite forces, as we noted previously, so the work done by the environment is the negative of the work done by the gas: W = -Wgas· Consequently, the first law of thermodynamics can be written as gas is th~t=~du~~-erthe graph f lVg.,, area (12.20) V c_-+-----l-- V; The thermal energy of an ideal gas depends only on its temperature as £ 1h = ~ Nk8 T. Comparing Equations l2.15 and 12.16 shows that Nk8 is equal to nR, so we can write the change in thermal energy of an ideal gas as 3 3 !:::.E = -nR l:::.T 1h = -NkB l:::.T 2 (12.2]) 2 Finding the heat for an ideal-gas process EXAMPLE 12.10 A cylinder with a movable piston contains 0.016 mol of helium. A researcher expands the gas via the process illustrated in FIGURE12.15. To achieve this, does she need to heat the gas? If so, bow much heat energy must be added or removed? FIGURE 12.15 pVdiagram PREPARE The graph tells us the pressure and the volume, so we can use the ideal-gas law to compute the temperature at the initial and final poinrs. To do this we need the volumes in SI units. Reading the initial and final volumes on the graph and convening. we find for Example 12.10. l m3 V; = 100 cm 3 X --- p (kPa) 106 cm 3 400 I m3 I06cm3 Vr = 300 cm 3 X --200 = 1.0 X 10-4 m3 = 3.0 X 10-4 m3 SOLVE The initial and final temperatures are found using the V (cm3 ) 0-+--~--~-~0 100 200 ideal-gas law: 300 T= As the gas expands, it does work on the piston. Its temperature may change as weU. implying a change in them1al energy. We will use die version of the first law of thennodynamics in Equation 12.20 to describe the energy changes in the gas. We'll first find the cbange in thermal energy by computing the temperature change, then calculate how much work is done by looking at the area under the graph. Once we know Wgasand tlE 111 , we will use the first law to detemune the sign and the magnitude of the heat-telling us whether heat energy goes in or out, and how mucb. ' P; V; (4.0 X 1()5 Pa)( 1.0 X 10-4 m3 ) = --------= 300 K 11R (0.016 mol)(8.31 J/mol • K) STRATEGIZE PrVr (2.0 X 105 Pa)(3.0 X 10-4 m 3 ) Tr= = -'---------11R (0.016 mol)(8.31 J/mol • K) = 450 K The temperature increases, and so must the thermal energy. We can use Equation 12.21 to compute this change: .lE111= 3 (0.016 2 - mol)(S.31 J/mol • K)(4:>0 K - 300 K) = 30J T