Physics 123 Unit #2 Review I. Definitions & Facts thermal equilibrium ideal gas thermal energy internal energy heat flow heat capacity specific heat heat of fusion heat of vaporization phase change expansion coefficient thermal conduction convection radiation pascal (unit) state variables reversible process irreversible process adiabatic process free expansion degrees of freedom equipartition theorem efficiency of an engine engines and refrigerators mole Avagadro s number II. Mathematics & Tools Calorimetry is just bookkeeping with heat flow. If you re not sure if a phase change has occurred, try the bookkeeping without a phase change (because its simpler) and see if the results make sense. Thermodynamical processes can be represented on p-v diagrams. Work is the area under a curve on a p-v diagram. Entropy is only calculable from a reversible cycle, but since it is a state variable, we can choose any reversible path and get the same result. The probability of a given outcome is the number of ways that outcome can occur out of all possible outcomes. For uncorrelated events: and, or +. At least one 1 probability of none III. Basic Concepts Thermal energy is the translational kinetic energy of random motion (as opposed to bulk motion). Temperature is a measurement of the average thermal energy of a system. Internal energy is the sum of the random translational and rotational kinetic energy and potential energy associated with the molecules in a system. Heat flow is the transfer of thermal energy. When a phase change occurs, heat flowing into or out of the system rearranges the fundamental structure of the material without changing its temperature. The First Law of Thermodynamics results from energy conservation. Internal energy, temperature, pressure, volume are state variables.. The Second Law of Thermodynamics results from probability considerations. The universe moves toward its most probable configuration. The entropy of the universe cannot decrease in a process. Internal energy, temperature, pressure, volume, entropy are state variables. There is no such thing as a perfect engine or a perfect refrigerator. The Carnot engine is the most efficient. By using a molecular model, we can express macroscopic variables (T,U,C v ) in terms of microscopic variables (<v x2 >, <K>).
IV. Equations to Memorize Ideal Gas Law: pv = nrt = NkT heat flow: Q = C T = mc T, Q = ml f, Q=mL v Thermal expansion: L = L 0 T Thermal conduction: HkA dt dx Work W pdv 1st Law: Q = W + U Adiabatic process: pv = constant Heat: Q = n c V T, Q = n c P T Internal energy: 2nd Law: ds dq T U = n c v T (reversible), S0 Carnot efficiency: e W Q h 1 T c T h
Physics 123 Unit 2 Reading and Homework Assignments 19.1 Know the Zeroth law of Thermodynamics and the meaning of thermal equilibrium. Understand the difference between heat and temperature. 19.2 To define and measure temperature is not trivial. The details of this section are not important to the course, however. 19.3 Know the K, F, and C scales. It s useful to know the conversions for everyday life, but I won t test you on them. Gas thermometers are introduced only to make absolute zero make sense, but most of you are comfortable with the concept anyway. 19.4 Eq. 19.4 is useful. You may find the section on water interesting, but you won t be tested on it. In two dimensions, think of expansion as photographic enlargement. Look at Quick Quiz 19.2 and Ex. 19.3 (a). Don t worry about (b) as you may not be familiar with the definitions. 19.5 The Ideal Gas Law, Eq. 19.8, is crucial. It s primary difficulty lies in the fact that it s a relationship between three (four, if you count n) variables. Note that Eq. 19.10 can also be useful. Work Ex. 19.5 and 19.6. 20.1 Note the differences between internal energy and heat. This section may seem a little abstract, but the ideas will become clear. Look through Ex. 20.1. (There s not much physics, but it puts units in perspective.) 20.2 This and the following section make the foundation of calorimetry. We re mostly dealing with solids and liquids here. Eq. 20.4 is the key to this section. As the book mentions, c isn t really constant, but for solids and liquids we usually can treat it as if it were. Note that heat flow in is positive, heat flow out is negative. Be careful about the book s statement, when the temperature increases, Q and T are taken to be positive. With gasses, T can be positive or negative when Q is zero, for example. Do not think of Q as a rise in temperature. Work Ex. 20.2. 20.7 Qualitatively understand conduction, convection, and radiation. We will not use Eq. 20.17, but we will use 20.14. Look through Ex. 20.9. If you buy insulation, you ll find the definition of R = L / k useful, but we won t use it. 20.3 Eq. 20.6 is a summary, but you must understand how heat flows and temperature changes during phase transitions, Fig. 20.2. Also remember that Eq. 20.6 does not supply the sign of Q you must do that by hand. Calorimetry problems are then just bookkeeping. Eq. 20.5 is confusing. Just remember Q = 0, where Q is the total heat flow in the closed system. Work Ex. 20.5. 20.4 Now we turn our attention to gasses. The work done in a constant pressure process is p V. But since we can break up any process into a very large number of small constant pressure processes, we can calculate work as pdv, the area under a curve in a p-v diagram. p-v diagrams are crucial for the next few chapters. Note 1) work is path dependent and 2)work done by a volume of gas is positive, work done on the gas is negative. 20.5 The First Law is very important. It s just energy conservation - heat flowing into a gas either does work or increases the internal energy. For a monatomic ideal gas, E int is just the random part of the translational kinetic energy. For a diatomic gas, it also includes rotational kinetic energy and the potential energy of vibration. If we know what point a gas occupies on a p-v diagram we know T and E int as well. Any such quantity that depends only on the state (p,v,t) of a gas is called a state variable.
20.6 We will concentrate on process which are isothermal, constant pressure, constant volume, or adiabatic. Note that adiabatic means Q = 0 which is not equivalent to T = 0. Know p-v paths, W, Q, E int for each. Work Ex. 20.6. Look over Ex. 20.7 and 20.8, as these show how the First Law applies to systems other than gasses. 21.1 The basic idea is that there are so many molecules, we must use statistics to obtain averages. Carefully review the derivation of pressure this is central to the whole unit. Remember F = dp/dt is just Newton s Second Law. Look at one molecule. Equation 21.2 is an important result, but you don t need to memorize it. What it shows is that we can express a macroscopic (large scale) quantity, P, in terms of microscopic quantities. Now we use the ideal gas law to relate 21.2 to temperature. The equipartition of energy is introduced here, but we ll see more of it later. Work Ex. 21.1. It s easy. 21.2 Now we proceed to express everything else in sight in terms of microscopic quantities. We start with E int. Then we proceed with specific heats. Note that the book uses molar specific heats here. I introduced the same symbols as heat capacity in the last unit. The constant is important. Try Ex. 21.2. It s easy, too. 21.3 The important result here is that pv = constant for reversible (not free expansion!) adiabatic processes. Now we can calculate work for adiabatic transitions. Try it. Work Ex. 21.3. 21.4. Now for the equipartition theorem. The hard part is counting degrees of freedom. Technically, a degree of freedom is one term in the total energy of a molecule (counting ½mv x 2 as one term, etc.). Note that rotation counts as a degree of freedom only if the rotational inertia is non-zero. Atoms are taken to be points with no rotational inertia. The rotational inertia of a monatomic molecule is zero. The rotational inertia of a diatomic molecule around the axis joining the atoms is zero. Note how quantum mechanics causes rotational and vibration degrees of freedom to kick in only at higher temperatures. Ignore the specific heats of solids. 21.5 21.7. These sections might provide you with some interesting reading for optional work. 22.1 Remember that a heat engine runs between two heat reservoirs, even if the p-v diagram is an irregular shape. Q h is the heat coming from the hot reservoir and Q c is the heat passing to the cold reservoir. Note that these are both considered positive. Know the Kelvin-Plank form of the 2nd Law (colored box on p.672). For now, the 2nd Law may not seem to be a mathematical law; that comes later. 22.2 Read this section carefully. It just clarifies definitions. 22.3 The Carnot engine is always taken as the typical engine. Understand this engine thoroughly. Other engines are quite similar in most details. Eq. 22.4 is important. Note that no engine can be more efficient than this. You should understand Ex. 22.2 well. Try 22.3 on your own. 22.4 is optional. It contains some good examples, but we won t test you on these specific engines. 22.5 This isn t too important, it just defines something like efficiency for a refrigerator. 22.6. The concept of entropy is fairly familiar in words, but not so familiar mathematically. The connection between the words and Eq. 22.8 is not very easy. Remember that entropy is state variable. If we know the initial and final state of a system, we can measure the change in entropy by calculating it for the easiest path between the points. Increase in entropy simply means that systems move toward the configuration of greatest probability. Work through Ex. 22.6. 22.7 Work through this section and all the examples carefully. This will show you how we can use entropy, even if the concept remains a bit fuzzy. 22.8 is optional. It is interesting as it relates entropy to microscopic variables.
Homework 2-1 Sections 19.1-19.4 Questions: Questions 19-3,19-4,19-14 Essential Problems: Problems 19-11,19-15 Application Problems: Problems 19-12 Computational Problems: A. A steel bar is heated from 0C to 600C. Calculate its change in length using Eq. 19.4. Then do the same thing, but this time calculate its change in length in one degree steps using L from one step as L 0 for the next step. Repeat the same process for a material with = 0.01 C 1. Is there a difference? Why? Can you write and solve a differential equation which will give you the second answer? Contextual Problems: B. One liter of water is placed in a shallow container. The water is then frozen. Atmospheric pressure is 1.00 bar. What is the total work done by the water in freezing? What is the heat flowing out of the water and the change in its internal energy? Homework 2-2 Sections 19.5, 20.1, & 20.7 Questions: Questions 20-7 A. What temperature scale must you use in the ideal gas law? Why? B. Give some examples of heat transfer by conduction, convection, and radiation. Essential Problems: Problems 19-27,19-28,20-43 Application Problems: Problems 19-29,19-40,19-66,20-45 Computational Problems: C. Real gasses are better described by the van der Waals equation of state: pa n 2 V 2 (Vnb)nRT. For propane, a = 8.664 liters 2 / atm mole 2 and b = 0.08445 liter / mole. One mole of propane is initially at 100C and occupies a volume of one liter. Sketch a p-v diagram for an isothermal compression to 0.05 liter. Compare that with the result which would be obtained for the ideal gas case. Try the same thing for oxygen, a =1.360 liters 2 / atm mole 2 and b = 0.03183 liter / mole. Note R = 0.08207 atm liter / K mole. Contextual Problems: Problem 20-58, 20-66, 20-68 Homework 2-3 Sections 20.2-20.3 Questions: Questions 20-5,20-6,20-12 Essential Problems: Problems 20-3,20-11 Application Problems: Problems 20-5,20-15,20-17 Computational Problems: A. Calculate the answer to problem 20-39 by breaking the p-v curve into many small constant pressure and constant volume processes, and add up the work in each small process to get the total work. Note that this is the same as evaluating the integral using the midpoint rule. Contextual Problems: None Homework 2-4 Section 20.4-20.6 Questions: Questions 20-30,20-32 Essential Problems: Problems 20-27,20-29 Application Problems: Problems 20-32,20-35,20-40 Computational Problems: A. If the specific heat of a material is not constant, the temperature change and heat flow are not linearly related. Plot the temperature as a function of total heat flow into a sample for c = T 2 where is a constant. Context-rich Problems: B. In the Civil War, many battles were so violently fought that bullets would often collide head-on in the air and fuse together. Estimate a minimum muzzle velocity for Civil War rifles.
Homework 2-5 Sections 21.1-21.2 Questions: Questions 21.1, 21.3, 21.6 Essential Problems: Problems 21.11, 21.13 Application Problems: Problems 21.10, 21.15, 21.17 Computational Problems: A. Eq. 21.26 gives the probability of finding a gas molecule with a given speed as a function of temperature. Graph this for several temperatures with the different averages of Eq. 21.27-21.29 indicated on the diagrams. Your result should be similar to Figure 21.12. Context-rich Problems: Problem 21.4 Homework 2-6 Sections 21.3-21.4 Questions: Question 21.12 Essential Problems: Problems 21.25, 21.32 Application Problems: Problems 21.24, 21.30, 21.35 Computational Problems: None. Context-rich Problems: Problem 21.31 Homework 2-7 Sections 22.1-22.3,22.5 Questions: Questions 22.3, 22.5 Essential Problems: Problems 22.2, 22.4, 22.29 Application Problems: Problems 22.5, 22.7, 22.25 Computational Problems: None Context-rich Problems: Problem 22.15, 22.16 Homework 2-8 Section 22.6-22.7 Questions: Questions 22.9, 22.17 Essential Problems: Problems 22.39 A. What is the change in entropy when 1 mole of silver (108 g) is melted at 961C? Application Problems: Problems 22.33, 22.35, 22.42, 22:63 Computational Problems: B. Determine two points A and B on a p-v diagram. Find a rather complicated mathematical function which joins these points. Approximate the path with constant pressure and constant volume processes. Find the change in entropy from A to B by summing ds over all these processes. Compare the result to a single constant pressure process plus a single constant volume process between the points. Context-rich Problems: Problem 22.36
Physics 123 Section 2 Sample Exam #2 You are allowed to use your textbook, calculator, unit handouts, and anything you have personally written. There are no time limits. Be sure to show your work, as we can only give credit for what is on your paper. Be sure your copy of the test has fourteen problems. For the multiple choice problems, mark the one best answer. Assume all gasses are ideal. Possibly useful information: Terminology 1. (5 points) Define thermal energy. R = 8. 3145 J / mole K Q = ncv T, constant volume CP = CV + R = CP CV CV = Rf, γ /, / 2 2. (5 points) Define convection. 3. (5 points) Give two examples of quantities that are not state variables. 4. (5 points) Define adiabatic.
Conceptual Applications 5. (5 points) 37.2 J of thermal energy is put into each of two objects made of the same material but having different masses. The object which has the greatest temperature rise is the object with: A. greater heat capacity. B. smaller heat capacity. C. greater specific heat. D. smaller specific heat. E. None of the above. 6. (5 points) Which of the following statements is not true regarding the freezing of water: A. The water molecules slow down as the water begins to freeze. B. During the freezing process the temperature remains constant. C. Heat leaves the water as the water freezes. D. Q = m L f. 7. (5 points) A metal sphere is filled with helium gas at low pressure and sealed. A pressure gauge reads the pressure of the helium inside the sphere. The device is then used as a thermometer. Which of the following statements about the thermometer is not true? A. Ignoring small corrections, pressure is directly proportional to temperature. B. As the temperature increases, the thermal expansion of the metal causes the pressure to be slightly larger than if the metal did not expand. C. The thermometer ceases to function normally at the boiling point of helium. D. The thermometer must be in thermal equilibrium with its surroundings in order to read properly.
Equations and Tools For each of the following equations: A) Tell what each symbol means B) Write a short problem which can be solved with this equation The problem must be a word problem, not just a=5, b=4, what is c? 8. (5 points) pv = NkT 9. (5 points) Q = ml f 10. (5 points) E int = n C v T
Problems We can usually evaluate how well you understand the physics of a problem by the way you attempt to solve the problem. Occasionally, however, you may be unable to work a problem or a part of a problem even if you do understand the physics. If you cannot work a problem, please do the following so that you may receive partial credit: 1. Discuss the basic physics concepts involved. Include enough detail that the grader will know if you understand the physics. 2. Write down the equation(s) you need to solve the problem. Explain the meaning of each symbol in the equations. 3. Explain as much of the problem as you understand. 4. Explain why you feel you have difficulties with the problem. 11. (25 points) 11. (25 points) In a constant pressure process, 1.00 mole of an ideal gas is heated from 100 K to 300 K. The initial volume of the gas is 0.010 m 3. The molar specific heat at constant volume, C V, is 12.5 J/mole K and the molar specific heat at constant pressure, C P, is 20.8 J/mole K (a) What is the pressure of the gas? (b) What is the final volume of the gas? (c) How much heat flows into the gas in this process? (d) How much heat would flow would there be if the gas goes from the same initial state to the same final temperature, but via a constant volume process?
12. (25 points) A bimetallic strip of metal is often used as a thermometer. The thin strip is composed of two layers, each made of a different metal. (The thickness is greatly exaggerated in the drawing.) The linear coefficients of expansion are : upper material: 11 10 6 (C) 1 lower material: 19 10 6 (C) 1 (a) Describe how the bimetallic strip can be used as a thermometer. (6 points) (b) What is the ratio of the lengths of the upper and lower surfaces if the strip is at 100C and was made (with equal length strips) at 20C? (14 points)
13. (25 points) An engine runs through the cycle shown at the right. A B is constant pressure, B C is constant volume, and C A is adiabatic. The following information is given: 1 7 n = moles, C P = R, TA = 200K, TB = 400K, VA = 0. 001m 3 8. 31 2 (a) What is the heat which flows into the engine? (b) Show that V B = 0.002 m 3. (c) Show that the ratio of specific heats is = 1.4 (d) Show that p C = p 1 A 2 γ
14. (25 points) A refrigerator runs through the cycle ABCDA as shown. One side is constant volume, one is constant pressure, one is isothermal, and one is adiabatic. n = 2.000 moles. C P = 29.10 J/ mole K. C V = 20.79 J/ mole K. R = 8.3145 J/ mole K. The following are known: A B C D p 300,000 Pa 200,000 Pa 200,000 Pa 225,000 Pa V 0.01109 m 3 (a) Identify the type of process for each side of the cycle. (b) Find the temperatures at each vertex. (Assume all numbers are accurate to four significant digits.) (c) What is the net work put into the refrigerator during each cycle?
Answers Terminology 1. The energy associated with the random (as opposed to bulk) motion of molecules in a material. (It includes contributions from both kinetic and potential energy.) 2. The process of heat flow characterized by fluids moving because of thermally induced density changes. 3. Work and heat. 4. A process with no heat flow. Conceptual Applications 5. B 6. A 7. B Equations and Tools 8. p pressure V volume N number of molecules k Boltzmann s constant T temperature A 3.00 liter volume of gas has a pressure of 1.2 10 5 Pa and a temperature of 150C. How many molecules are present? 9. Q heat (flowing in or out) m mass L f latent heat of fusion How much heat is needed to melt 6.00 g of ice? 10. E int is the internal energy of a volume of gas n is the number of moles of gas C v is the molar specific heat at constant volume T is the temperature What is the internal energy of 0.01 m 3 of hydrogen gas at 420C and 1 bar pressure? Problems 11. (a) (b) p nr T V p nrt V 1.0mole 8.31J/mole K (100K) 83.1 kpa 0.010m 3 p nr T 1 T 2 V V 1 V 2 V 1 T 2 0.010m 3 300K 0.030m 3 2 T 1 100K (c) Q = n C P T = 1 mole 20.8 J / mole K 200 K = 4160 J (d) Q = n C V T = 1 mole 12.5 J / mole K 200 K = 2500 J 12. (a) Since the two metals expand differently, the strip cannot remain flat as the temperature changes. The curvature of the strip will be an indicator of temperature. (b) up down up up 0 0 up T down down 0 0 down T 1 up T 1 down T 111 106 80 119 10 6 80 0.99936
13. (a) For constant temperature and constant volume processes, the heat flow is given by the specific heat equations. It is readily seen from these equations that heat flows into the system when the temperature rises, so heat flows in during the constant pressure leg: QnC P P nr T A V A T B V B T 7nR 2 T 7 J/K 200 K 700J 2 (b) By the Ideal Gas Law: V B T V B A 400K 0.001m 3 0.002m 3 T A 200K (c) (d) P C P A C P C V PV constant P A V A P C V C V A V C 7/2R 7/2RR 7 5 1.4 P A V A V B P A 1 2 14. (a) AB is constant volume, BC is constant pressure, CD is adiabatic, and DA is isothermal. (b) At A we need the Ideal Gas Law: T A p V A A nr 200.1 K The temperature at D must be the same as the temperature at A: T D T A 200.1K This implies the volume at D is: V D P A V A P D 300000Pa 0.01109m3 225000 Pa 0.01479m 3 For the adiabatic process, The volume at C is then given by: C P C V 29.10 20.79 1.400 P C V C P D V D V C V D P D P C 1 0.01479m 3 225000 Pa 200000 Pa 1 1.400 0.01609 m 3 The temperature at C is then: T C P C V C nr 200000Pa 0.01609m 3 2.000mole 8.3145J/moleK 193.5K We know the volumes at A and B are the same: T B P B V B nr 200000Pa 0.01109m 3 2.000mole 8.3145J/moleK 133.4K ( c) Since Q = W, we can either calculate work or heat. We ll use the latter: QnC V T BA nc P T CB 0nR T D ln( V A /V D )234.1J
Physics 123 Unit Summary Unit #2 ID Number: Unit Score: (sum all boxes below): 1. Homework problems completed on time 10 = Homework problems completed on time 8 = Sum of previous lines 61 problems =, average score per problem Homework score: Average score per problem 2.5 = ( Maximum = 25 ) 2 Hours Item Hours Item Hours Total Hours (expected = 21, maximum = 36): Standard Work Score: Total Standard Hours 1.94 = ( Maximum = 70 ) 4. Reading Checks Reading score: 10 # correct = ( Maximum = 10 ) 8 4. Quizzes Quiz score: 10 # correct = ( Maximum = 10 ) 8 5. Walk-in Labs Lab score: = 15 #completed = ( Maximum = 15 )