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# Solucionario Fundamentos de Física Serway 9na Edición Capitulo 10

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Fundamentos de Física Serway Solucionario 9na edición Capitulo 10

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### Solucionario Fundamentos de Física Serway 9na Edición Capitulo 10

1. 1. 10 Thermal Physics CLICKER QUESTIONS Question J1.01 Description: Identifying preconceptions about and relating heat, temperature, and the perception of hotness. Question Body A has a higher temperature than body B. Which of the following statements is true? 1. Body A will feel hotter than body B. 2. Body A contains more energy than body B. 3. If placed in contact with each other, energy will fl ow from body A to body B. 4. If placed in contact with a third body having temperature greater than body A, body B will absorb more heat than A. 5. More than one statement is true. Commentary Purpose: To hone and relate the concepts of temperature, heat, and the perception of “hotness.” Description: If two bodies of different temperatures are placed in thermal contact, heat always fl ows from the higher-temperature one to the lower-temperature one. Period. This behavior defi nes the concept of temperature. Thus, answer (3) is always true. The thermal energy contained in a body depends on its temperature, its size, and the thermal properties of the material constituting it (it’s specifi c heat). For example, 1 cm3 of lead holds much more thermal energy than 1 cm3 of plastic at the same temperature, and 1 m3 of lead holds much more heat than 1 cm3 of lead at the same temperature. So, (2) is not necessarily true. For the same reason, (4) is not necessarily true: if body A were much smaller than body B or made of a different material, it might absorb less energy before equilibrating with the third body than B would. How “hot” an object feels when you touch it depends on the object’s temperature, its size and material, and how quickly heat can fl ow through the object and into your hand. So (1) is not necessarily true, either. Key Points: • The temperatures of two bodies, and nothing else, determines which direction heat will fl ow between them if they are placed in thermal contact. • The thermal energy contained by a body depends on its temperature, size (mass), and material (specifi c heat). • How “hot” a body feels—and how badly it might burn you—depends on more than just its temperature. 499
2. 2. 500 Chapter 10 For Instructors Only Connecting these ideas to students’ personal experience is easy, and can be interesting, motivating, and enlightening. For example, it is OK to lick snow at 10°C but you better not lick metal at that temperature. Relating this topic to latent heat is also useful. For example, steam can give a very nasty burn very quickly. Question J1.02 Description: Understanding temperature. Question Temperature is a measure of: 1. The total amount of heat possessed by something 2. The fl ow of energy to or from something 3. The total energy possessed by something 4. The total kinetic energy of the constituents of something 5. The amount of useful work that could be extracted from something 6. None of the above Commentary Purpose: To introduce the concept of temperature and distinguish it from the concept of heat. Discussion: All of the choices are tempting, but none is quite right, so none is a proper description of temperature. The best choice is (6), “None of the above.” Heat is work done at the microscopic level to exchange thermal energy (the kinetic energy of microscopic constituent particles) between one body and another. It is the amount of energy transferred due to a difference in temperature. Heat cannot be “possessed” by something, so choice (1) is false. Heat, not temperature, describes the fl ow of thermal energy to or from something, so choice (2) is also false. When two objects at different temperatures are in contact, the object at the lower temperature absorbs energy from the other, but that does not mean that the object at higher temperature has more energy. What if a very small hot object touches the ocean? The ocean has lots more energy than the hot object, yet it is the ocean that gains energy, and the small object that loses energy. The total energy and total kinetic energy will depend on temperature, but they also depend on the quantity and nature of the material. (3) and (4) are false. Temperature is associated with the average kinetic energy per molecule in something. That is why a small hot object can give energy to a large cool object. Equilibrium is reached only when all of the molecules in contact with each other have (on average) the same kinetic energy. Like (3) and (4), choice (5) is false. The amount of useful work depends on the temperature, but it also depends on the quantity and nature of the material. So, temperature is certainly associated with many quantities and processes, but it is not a “measure” of these. It would be like saying mass is a measure of momentum. Mass and momentum are related, but momentum depends on more than just the mass; it also depends on the velocity.
3. 3. Thermal Physics 501 Our bodies sense temperature by the fl ow of energy (heat). When we touch something hotter than our skin, we absorb energy and interpret the sensation as a “warm” or “hot” temperature. The hotter it is, the larger the rate of energy fl ow. Likewise when we touch something “cold,” energy fl ows out of our bodies to the “cold” object. Our sense of temperature is inconsistent, however. Metal objects at room temperature feel cold. If our hands are cold, warm things feel hot. Key Points: • Temperature is associated with the average kinetic energy of a object’s constituent particles. • Heat describes the fl ow of this “internal energy” from one substance or object to another. • Temperature is related to heat, total internal energy, and our sensation of “hotness,” but it is not a direct measure of any of these. For Instructors Only As described above, there are many quantities associated with temperature, but none is accurately a “measure” of temperature. The closest is the average kinetic energy per particle, which also depends on the number of degrees of freedom. An even better, though cumbersome, answer is that the temperature is a measure of the average translational kinetic energy per particle, since this is what is equilibrates when two bodies at different temperatures come into contact. Basically, the problem is that temperature is not a description of any other quantity; it is a quantity unto itself, justifi ed because it is useful for predicting how heat will fl ow between two objects placed in thermal contact. Some students might choose (6) without necessarily understanding why the fi rst fi ve choices are not valid. They might simply have their own defi nitions of temperature that does not fi t any of the others. Question J6.01 Description: Sensitizing to assumptions when working with the ideal gas law. Question Consider the two systems below, labeled A and B. F F A B Which gas has the higher pressure? 1. A 2. B 3. Neither; the pressures are the same. 4. Impossible to determine
4. 4. 502 Chapter 10 Commentary Purpose: To develop your awareness of the role different assumptions play in answering an ideal gas law question. Discussion: The ice-water mixture keeps the gas in the container at 0°C. Thus, we know the temperatures of the two systems are the same, and the volume of system A is larger than that of system B. If both systems contain the same amount (number of moles) of gas, then the ideal gas tells us that system B must have the higher pressure. However, we don’t know anything about the amounts of gas, so answer (2) is unjustifi ed. If the applied force F is the same for both systems, and if both pistons have the same size, shape, and mass, then the pressure applied to the gas must be the same in both cases. The question statement doesn’t say that these conditions are true, but we might assume that the label “F” is a variable that represents the same value for both cases, and the picture seems to show two identical pistons. So answer (3) is defensible, but requires some assumptions. The only thing we can say for sure is that the volumes are different and the temperatures are the same, which means that either the pressure or the amount of gas must differ. The most conservative answer is therefore (4). Key Points: • The ideal gas law relates the pressure, volume, temperature, and amount of a gas. Don’t overlook the amount (number of moles) as a variable. • Be aware of the assumptions you make, and choose them carefully. For Instructors Only Answer (2), that system B has a higher pressure, tends to be common; this is generally because students implicitly assume both systems contain the same amount of gas. As always, we recommend focusing on the consequences and reasonableness of various possible assump-tions rather than on the correctness of answers. Additional Questions: 1. Which system has the higher temperature? 2. Which system has the higher number of moles of gas? 3. Assume now that a thermodynamic process takes system A and turns it into system B. How would you accomplish this?
5. 5. Thermal Physics 503 QUICK QUIZZES 1. (c). When two objects having different temperatures are in thermal contact, energy is transferred from the higher temperature object to the lower temperature object. As a result, the temperature of the hotter object decreases and that of the cooler object increases until thermal equilibrium is reached at some intermediate temperature. 2. (b). The glass surrounding the mercury expands before the mercury does, causing the level of the mer-cury to drop slightly. The mercury rises after it begins to get warmer and approach the temperature of the hot water, because its coeffi cient of expansion is greater than that for glass. 3. (c). Gasoline has the highest coeffi cient of expansion so it undergoes the greatest change in volume per degree change in temperature. 4. (c). A cavity in a material expands in exactly the same way as if the cavity were fi lled with material. Thus, both spheres will expand by the same amount. 5. (b). Since the two containers are at the same temperature, the average kinetic energy per molecule is the same for the argon and helium gases. However, helium has a lower molar mass than does argon, so the rms speed of the helium atoms must be higher than that of the argon atoms. ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. TF = 9 TC + = (− ) + = − 5 9 32 5 25° 32° 13° F, and the correct response is choice (e). 2. T T C F = 5 ( − ) = ( − ) = 9 5 32 9 162 32 72.2°C, then T T K C = + 273 = 72.2 + 273 = 345 K, so choice (c) is the correct answer. 3. Δ Δ L L T = ( ) = × ( ) ⎡⎣ α − − ( )( ) Cu °C m °C 0 ⎤⎦ 6 1 17 10 93 10 = 1.6 × 10−2 m = 1.6 cm and choice (c) is the correct order of magnitude. 4. The correct choice is (b). When a solid, containing a cavity, is heated, the cavity expands in the same way as it would if fi lled with the material making up the rest of the object. 5. From the ideal gas law, with the mass of the gas constant, PV T PV T 2 2 2 1 2 1 = . Thus, V P P T T = 1 4 1 0 50 2 1 V 2 2 2 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( )( )( . m3 ) = .0 m3 and (c) is the correct choice. 6. From the ideal gas law, with the mass of the gas constant, PV T PV T 2 2 2 1 2 1 = . Thus, P V V T T = 1 4 2 1 2 P P P 2 2 2 1 1 1 1 ⎛ ⎞ ⎟ ⎠ ⎜⎝ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =⎛⎝ ⎞⎠ ( ) = and (d) is the correct choice.
6. 6. 504 Chapter 10 7. Remember that one must use absolute temperatures and pressures in the ideal gas law. Thus, the original temperature is TK = TC + 273.15 = 25 + 273.15 = 298 K, and with the mass of the gas constant, the ideal gas law gives T P P V V 2 T 2 1 2 1 1 1 07 106 5 00 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × . . Pa 0 3 00 298 191 6 Pa K K ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( . )( ) = and (d) is the best choice. 8. The internal energy of n moles of a monatomic ideal gas is U nRT = 32 , where R is the universal gas constant and T is the absolute temperature of the gas. For the given neon sample, T = T + = ( + ) = C 273.15 152 273.15 K 425 K, and n= = 26 0 1 29 . . g 20.18 g mol mol Thus, U = ( ) ⋅ ( )( ) = × 32 1.29 mol 8.31 J mol K 425 K 6.83 103 J, and (b) is the correct answer. 9. The root-mean-square speed of molecules in a gas with molar mass M (expressed in kilograms per mole) and absolute temperature T is vrms = 3RT M. The molar mass of methane (CH ) 4 is M = [ + ( )] ⎛ g mol kg 3 g 12 0 4 1 00 = × − ⎝ ⎜ ⎞ ⎠ ⎟ 1 10 . . 1.6 10 2 kg mol and its absolute temperature is T = T + = ( + ) = C 273.15 25.0 273.15 K 298 K. Therefore, vrms 3 8 31 298 J mol K K 1.60 kg mol = ( ⋅ )( ) × = − 10 6 2 . 81 m s making (b) the correct response. 10. The kinetic theory of gases does assume that the molecules in a pure substance obey Newton’s laws and undergo elastic collisions, and the average distance between molecules is very large in comparison to molecular sizes. However, it also assumes that the number of molecules in the sample is large so that statistical averages are meaningful. The untrue statement included in the list of choices is (a). 11. In a head-on, elastic collision with a wall, the change in momentum of a gas molecule is Δp m m m f = (v − v ) = (−v − v ) = − v 0 0 0 0 2 . If the molecule should stick to the wall instead of rebounding, the change in the molecule’s momentum would be Δ p = m(0 − ) = −m 0 0 v v, which is half that in the elastic collision. Since a gas exerts a pressure on its container by molecules imparting impulses to the walls during collisions, and the impulse imparted equals the magnitude of the change in the molecular momentum, decreasing the change in momentum during the collisions by a factor of 2 would halve the pressure exerted. Thus, the correct response is choice (b). 12. The rms speed of molecules in the gas is vrms = 3RT M. Thus, the ratio of the fi nal speed to the original speed would be ( v ) rms RT M )= f f = f K ( = = v 200 K rms T RT M T 0 0 0 3 3 600 3 Therefore, the correct answer to this question is choice (d).
7. 7. Thermal Physics 505 ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS 2. The pressure inside the balloon is greater than the ambient atmospheric pressure because the pressure inside must not only resist the external pressure, but also the force exerted by the elastic material of the balloon. 4. The lower temperature will make the power line decrease in length. This increases the tension in the line to the point that it is near breaking. 6. At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Strong latches hold them together, but they would explode apart if you tried to open the hot cooker. 8. The measurements are too short. At 22°C the tape would read the width of the object accurately, but an increase in temperature causes the divisions ruled on the tape to be farther apart than they should be. This “too long” ruler will, then, measure objects to be shorter than they really are. 10. The existence of an atmosphere on a planet is due to the gravitational force holding the gas of the atmosphere to the planet. On a small planet, the gravitational force is very small, and the escape speed is correspondingly small. If a small planet starts its existence with atmosphere, the molecules of the gas will have a distribution of speeds, according to kinetic theory. Some of these molecules will have speeds higher than the escape speed of the planet and will leave the atmosphere. As the remaining atmosphere is warmed by radiation from the Sun, more molecules will attain speeds high enough to escape. As a result, the atmosphere bleeds off into space. 12. Doubling the volume while reducing the pressure by half results in no change in the quantity PV that appears in the ideal gas law equation. Consequently, the temperature and hence the internal energy remain the same. PROBLEM SOLUTIONS 10.1 (a) TF = TC + = (− ) + = − 9 5 32 9 5 273.15 32 460°F (b) T T C F = ( − ) = ( − ) = 5 9 32 5 9 98.6 32 37.0°C (c) T T T F C = + = ( − ) + = (− ) + = − 9 5 32 9 5 273 15 32 9 5 . 173.15 32 280°F
8. 8. 506 Chapter 10 10.2 When the volume of a low density gas is held constant, pressure and temperature are related by a linear equation P = AT + B, where A and B are constants to be determined. For the given constant-volume gas thermometer, P = 0.700 atm when T = 100°C ⇒ 0.700 atm = A(100°°C) + B [1] P = 0.512 atm when T = 0°C ⇒ 0.512 atm = A(0) + B [2] From Equation [2], B = 0.512 atm. Substituting this result into Equation [1] yields A = − = × − 0 700 0 512 . atm . atm 1 . 88 10 3 100 °C atm °C so, the linear equation for this thermometer is: P = ( × )T + 1.88 10−3 atm °C 0.512 atm. (a) If P = 0.0400 atm, then T = − = − P B A = 0.0400 atm 0.512 atm 1.88 10-3 atm °C − 251°C × (b) If T = 450°C, then P = (1.88 × 10−3 atm °C)(450°C) + 0.512 atm = 1.36 atm 10.3 (a) T T F C = + = (− )+ = − °F 9 5 32 9 5 196 32 321 T T K C = + 273.15 = (−196 + 273.15) K = 77 K (b) T T F C = + = ( )+ = °F 9 5 32 9 5 37.0 32 98.6 T T K C = + 273.15 = (37.0 + 273.15) K = 310 K 10.4 (a) T T C F = ( − ) = ( − ) = 5 9 32 5 9 134 32 56 7 . °C and TC= (− − ) = − 5 9 79.8 32 62.1°C (b) T TK C = + 273.15 = (56.7 + 273.15) K = 330 K and T TK C = + 273.15 = (−62.1°C + 273.15) K = 211 K 10.5 Start with TF = −40°F and convert to Celsius. T T C F = ( − ) = (− − ) = − 5 9 32 5 9 40 32 40°C Since Celsius and Fahrenheit degrees of temperature change are different sizes, this is the only temperature with the same numeric value on both scales.
9. 9. Thermal Physics 507 10.6 Since we have a linear graph, we know that the pressure is related to the temperature as = + , where A and B are constants. To fi nd A and B, we use the given data: P A BTC 0.900 atm = A + B(−80.0°C) [1] and 1.635 atm = A + B(78.0°C) [2] Solving Equations [1] and [2] simultaneously, we fi nd: A = 1.27 atm and B = 4.65 × 10−3 atm °C Therefore, P TC = 1.27 atm + (4.65 × 10−3 atm °C) (a) At absolute zero the gas exerts zero pressure (P = 0), so TC = − 1 27 × = − − 4 65 10 273 3 . . atm atm °C °C (b) At the freezing point of water, TC = 0 and P = 1.27 atm + 0 = 1.27 atm At the boiling point of water, TC = 100°C, so P = 1.27 atm + (4.65 × 10−3 atm °C)(100°C) = 1.74 atm 10.7 Apply T T F C = + 95 32 to two different Celsius temperatures, T T C C ( ) ( ) 1 2 and , to obtain 9 5 T T F C ( ) = ( ) + 1 1 32 [1] and 9 5 T T F C ( ) = ( ) + 2 2 32 [2] Subtracting Equation [1] from [2] yields 9 5 T T T T F F C C ( ) − ( ) = ( ) − ( ) ⎡⎣ ⎤⎦ 2 1 2 1 or ΔT ΔT F C = (9 5) 10.8 (a) Using the result of Problem 10.7 above, ΔT ΔT C F = 5 ( ) = ( ) = 9 5 9 57.0 °C 31.7°C . (b) ΔT T T T T K C,out C,in C,out = ( + 273.15)− ( + 273.15) = − C,in C ( ) = ΔT = 31.7 K
10. 10. 508 Chapter 10 10.9 T T F C = + = ( )+ = + ( ) = 95 32 43 32 77 32 °F 109°F . 95 Yes . The normal body temperature is 98 6 . °F, so this patient has a high fever and needs immediate attention. 10.10 (a) Since temperature differences on the Rankine and Fahrenheit scales are identical, the temperature readings on the two thermometers must differ by no more than an additive constant (i.e., T T constant R F = + ). To evaluate this constant, consider the temperature readings on the two scales at absolute zero. We have TR = 0°R at absolute zero, and T T F C = + = (− )+ = − 9 5 32 9 5 273.15 32 459.67°F Substituting these temperatures in our Fahrenheit to Rankine conversion gives 0° = −459.67° + constant or constant = 459.67° giving T T R F = + 459.67 . (b) We start with the Kelvin temperature and convert to the Rankine temperature in several stages, using the Fahrenheit to Rankine conversion from part (a) above. 5 9 T = T + 273 15 = ( T − 32 )+ 273 15 = ( T − ) K C F R 5 9 459 67 . . . − ⎡⎣ ⎤⎦ + = ( − )+ = − 32 273 15 5 9 491 67 273 15 5 9 5 . T . . T R R9 491 67 273 15 5 9 273 15 273 15 5 9 ( . )+ . = T − . + . = T R R 10.11 The increase in temperature is ΔT = 35°C − (−20°C) = 55°C Thus, Δ Δ L L T = ( ) = × ( ) ⎡⎣ α − − ( )( ) = 0 ⎤⎦ 6 1 11 10 °C 518 m 55°C 0.31 m = 31 cm 10.12 (a) As the temperature drops by 20°C, the length of the pendulum changes by ΔL =α L (ΔT ) 0 = × ( ) ⎡⎣ 19 10−6 °C −1 (1.3 m)(−20°C) = −4.9 × 10−4 m = − 0.49 mm ⎤⎦ Thus, the fi nal length of the rod is L = 1.3 m − 0.49 mm . (b) From the expression for the period, T = 2π L g , we see that as the length decreases the period decreases. Thus, the pendulum will swing too rapidly and the clock will run fast . 10.13 We choose the radius as our linear dimension. Then, from ΔL =α L (ΔT ) 0 , ΔT T − L L L C = − = = − 2 21 2 20 0 20 . 0 0 × − . . °C cm cm ( ) = −1 2 20 α 130 10 6 °C cm °C ( ) ⎡⎣ ⎤⎦ 35 0 . . or TC = 55 0 . °C
11. 11. Thermal Physics 509 10.14 (a) The diameter is a linear dimension, so we consider the linear expansion of steel: d = d [ + ( T )] = ( ) + ( × − ( ) − ) 0 6 1 1 2 540 1 11 10 1 α Δ . cm °C 00 25 2 542 °C °C cm − ( ) ⎡⎣ ⎤⎦ = . 0. Then, using ΔV = βV (ΔT ) 0 , (b) If the volume increases by 1%, then ΔV = (1.0 ×10−2)V where β = 3α is the volume expansion coeffi cient, we fi nd Δ Δ T V V = = × 1 0 10 × ( ) ⎡⎣ ⎤⎦ = × − − − 0 2 6 1 3 11 10 3 0 10 β . . °C 2 °C 10.15 From ΔL = L − L = L (ΔT ) 0 0 α , the fi nal value of the linear dimension is L = L + L ( T ) 0 0 α Δ . To remove the ring from the rod, the diameter of the ring must be at least as large as the diameter of the rod. Thus, we require that L L Brass Al = , or L L T L L 0 0 0 0 ( ) + ( ) ( ) = ( ) + ( Brass Brass Brass Al Al α Δ α ) ( ) Al ΔT This gives ΔT = ( ) − ( ) L L 0 Al 0 Brass L L ( ) − ( ) α α Brass 0 Brass Al 0 A l (a) If L0 ( ) = 10 01 Al . cm, ΔT = − . . . 10 °C -6 × ( ) ⎡⎣ × ( ) ⎡⎣ ( )− − ⎤⎦ 10 01 10 00 19 10 00 24 1 = − − − 10 °C ⎤⎦ ( ) °C 6 1 10 01 199 . so T = T + T = − = − 0 Δ 20.0° C 199° C 179°C which is attainable (b) If L0 ( ) = 10 02 Al . cm ΔT = − . . . 10 °C 6 × ( ) ⎡⎣ × ( ) ⎡⎣ − − ( )− ⎤⎦ 10 02 10 00 19 10 00 24 1 = − − − 10 °C ⎤⎦ ( ) °C 6 1 10 02 396 . and T = T + T = − 0 Δ 376°C, which is below absolute zero and unattainable . 10.16 ρ ρ β 0 0 Δ β Δ 1 β Δ 1 Δ ) = = + = + ( ) = + ( ) = + ( m V m V V m V V T m V T T 0 0 0 10.17 (a) Using the result of problem 10.16, with β = 3α , gives ρ ρ β = + ( ) = × 11 3 10 . kgm ⎡ × − ( )− 0 3 ΔT 10 ⎣ ⎤⎦ 1 1 1+3 29 °C 3 6 ( − ) = × 90 0 11 2 103 °C °C . kg m3 (b) No. Although the density of gold would be less on a warm day, the mass of the bar would be the same, regardless of its temperature, and that is what you are paying for. (Note that the volume of the bar increases with increasing temperature, whereas its density decreases. Its mass, however, remains constant.)
12. 12. 510 Chapter 10 10.18 When the temperature drops by 10.0°C, the wire will attempt to contract by ΔL =α L ΔT = ( × − ( )− )( )( ) Cu °C m °C 0 6 1 17 10 10.0 10.0 = 1.70 × 10−3 m If the ends of the wire are held stationary, and the wire is not allowed to contract, it will develop an additional tension force ΔF suffi cient to keep it stretched by an additional 1.70 × 10−3 m beyond its natural length. The required additional tension force is ΔF = Y A ΔL L Cu ( )0 , where YCu is Young’s modulus for copper and A is the cross-sectional area of the wire. Thus, ΔF = ( × )( × − ) × − 10 5 1 7 10 11 10 2 40 10 3 Pa m m 10.0 . 2 . m N ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 449 The total tension in the wire, after the decrease in temperature, is then F = F + F = + = 0 Δ 75.0 N 449 N 524 N 10.19 The difference in Celsius temperature in the underground tank and the tanker truck is ΔT ΔT C F = ( ) = ( − ) = °C 5 9 5 9 95.0 52.0 23.9 If V 52°F is the volume of gasoline that fi lls the tank at 52.0°F, the volume this quantity of gas would occupy on the tanker truck at 95.0°F is V V V V V T V T 95 F 52 F 52 F 52 F 52 F 1 ° ° ° ° ° = + = + ( ) = + ( Δ Δ Δ β β ) ⎡⎣ ⎤⎦ = (1 00 × 103 ) 1+ (9 6 × 10−4 ( )−1 ) 23 9 . . . m °C ° 3 C m3 ( ) ⎡⎣ ⎤⎦ = 1.02 × 103 10.20 Consider a regular solid with initial volume given by V AL 0 0 0 = at temperature T0 . Here, A is the cross-sectional area and L is the length of the regular solid. As the temperature undergoes a change ΔT = T − T0, the change in the cross-sectional area is ΔA = A − A = A (ΔT ) = A (ΔT ) 0 0 0 γ 2α , giving A = A + A ( T ) 0 0 2α Δ . Similarly, the new length will be L = L + L ( T ) 0 0 α Δ , so the new volume is V A A T L L T A L A L = + ( ) ⎡⎣ 0 2⎤⎦0 Δ ⎤⎦ 0 α 0 Δ α = 0 0 + 30 0 (+ 2 α ΔΔT ) 2 A L (ΔT ) + ( ) ⎡⎣ 2 α 0 0 The term involving α 2 is negligibly small in comparison to the other terms, so V = A L + A L ( T ) = V + V ( T ) 0 0 0 0 0 0 3α Δ 3α Δ . This is of the form ΔV = V −V = V ΔT 0 0 β ( ), where β = 3α . 10.21 [Note that some rules concerning signifi cant fi gures are deliberately violated in this solution to better illustrate the method of solution.] When the temperature of a material is raised, the linear dimensions of any cavity in that material expands as if it were fi lled with the surrounding material. Thus, the fi nal value of the inner diam-eter of the ring will be given by L = L + L ( T ) 0 0 α Δ as Dinner cm °C c = + × ( ) ⎡⎣ − − 2 168 1 42 10 5 1 2 168 . . ( . m)(100°C−15.0°C) = 2.171 cm ⎤⎦
13. 13. Thermal Physics 511 10.22 [Note that some rules concerning signifi cant fi gures are deliberately violated in this solution to better illustrate the method of solution.] Let L be the fi nal length of the aluminum column. This will also be the fi nal length of the quantity of tape now stretching from one end of the column to the other. In order to determine what the scale reading now is, we need to fi nd the initial length this quantity of tape had at 21.2°C (when the scale markings were presumably painted on the tape). Thus, we let this initial length of tape be (L ) 0 tape and require that L = (L ) [ + ( T )] = (L ) + ( T 0 0 1 1 tape steel column Al α Δ [ α Δ )] which gives ( ) ( L ) [ 1 + ( T )] L = 0 column Al 1 tape + ( ) T 0 steel α α Δ Δ or L0 6 1 18 700 1 24 10 29 4 ( ) = ( ) + ( × − ( )− ) tape m °C °C . . − ( ) ⎡⎣ ⎤⎦ + ( × − ( )− ) − 21 2 1 11 10 6 1 29 4 21 2 . . . °C °C °C °°C m ( ) = 18.702 = 2π = 2π (5.0 × 10−3 m) = 3.1 × 10−2 m. The amount this 10.23 The initial length of the band is L r 0 0 length would contract, if allowed to do so, as the band cools to 37°C is α − − ( × − ) 0 Δ Δ L L T = = × ( ) ⎡⎣ 17.3 10 6 °C 1 3.1 10 2 m (80°°C − 37°C) = 2.3 × 10−5 m ⎤⎦ Since the band is not allowed to contract, it will develop a tensile stress given by Stress Y = ( × ) × Δ − L L = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 10 5 18 10 2 3 10 3 Pa . m .1 10 = × − m 1 3 10 2 8 × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . Pa and the tension in the band will be F A Stress = ( ) = × ( ) × ( ) ⎡⎣ 4.0 10−3 m 0.50 10−3 m (1.3 × 108 Pa) = 2.7 × 102 N ⎤⎦ 10.24 The expansion of the pipeline will be ΔL =α L (ΔT ) 0 , or 11 10−6 °C −1 (1300 × 103 m)[35°C − (−73°C)] = 1.5 × 103 m = 1.5 km ΔL = × ( ) ⎡⎣ ⎤⎦ This is accommodated by accordion-like expansion joints placed in the pipeline at periodic intervals. 10.25 Both the drum and the carbon tetrachloride expand as the temperature rises by ΔT = 20 0 . °C. Since the drum was completely fi lled when the temperature was 10.0°C, the initial volume for the drum and the carbon tetrachloride are the same. From ΔV = βV (ΔT ) 0 , where β = 3α is the coef-fi cient of volume expansion, we obtain β α ( ) carbon = Δ − Δ = − V V V spillage carbon tetrachloride steel drum ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ tetrachloride steel 3 0 V T Δ or Vspillage = × − (°C)− − ( × − (°C)− 5 81 10 4 1 3 11 10 6 1 . ) ⎡⎣ ⎤⎦ (50.0 gal)(20.0°C) = 0.548 gal
14. 14. 512 Chapter 10 = = ( × )( ) 3 80 × 10− 3 m ρ . . 10.26 (a) m V 0 0 2 3 7 30 10 10 0 . kg m gal 3 gal kg 1 27 7 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . (b) V = V + V = V + V ( T ) = V [ + ( T )] 0 0 0 0 Δ β Δ 1 β Δ or V = ( ) + × ( ) ( )( ) ⎡⎣ 1 000 1 9 60 10−4 −1 20 0 ⎤ . m3 . °C . °C ⎦ = 1.02 m3 (c) Gasoline having a mass of m = 7.30 × 102 kg occupies a volume of V0 = 1.000 m3 at 0°C and a volume of V = 1.02 m3 at 20.0°C. The density of gasoline at 20.0°C is then ρ20 = = 716 × = m V 7 . 30 102 1 . 02 kg m kg m 3 3 3 80 10 = = ( × )( ) × − kg m3 gal m ρ . . (d) m V 20 20 2 3 7 16 10 10 0 . gal kg 3 1 27 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = . (e) Δm = m − m = − = 0 20 27.7 kg 27.2 kg 0.5 kg 10.27 (a) The gap width is a linear dimension, so it increases in “thermal enlargement” as the temperature goes up. (b) At 190°C, the length of the piece of steel that is missing, or has been removed to create the gap, is L = L + L = L [ + ( T )] 0 0Δ 1 α Δ . This gives 1 600 1 11 10−6 −1 190 − 30 0 . cm ( °C ( °C . °C)) = 1.603 cm L = ( ) + × ( ) ⎡⎣ ⎤⎦ 10.28 If allowed to do so, the concrete would expand by ΔL =α L (ΔT ) 0 . (a) Since it is not permitted to expand, the concrete experiences a compressive stress of Stress Y Δ = (Δ ) = ( × ) × L L ⎛ = ⋅ Y T ⎝ ⎜ ⎞ ⎠ ⎟ 0 9 7 00 10 12 α . Pa 10 30 0 6 1 − − ( ) ⎡⎣ °C ( . °C) ⎤⎦ or Stress = 2.5 × 106 Pa . (b) Since this stress is less than the compressive strength of concrete, the sidewalk will not fracture . 10.29 (a) From the ideal gas law, PV = nRT, we fi nd P T = nR V. Thus, if both n and V are constant as the gas is heated, the ratio P T is constant giving P T P T f f = i or T T i P P f P f i P i 300 i = = i = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( )⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 K 900 K 627°C (b) If both pressure and volume double as n is held constant, the ideal gas law gives P V PV f i T T P V PV T f f i i i i i i i = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ( )( ) ⎝ ⎜ ⎞ ⎠ ⎟ = 2 2 4T 4 300 1 200 927 i = ( K) = K = °C
15. 15. Thermal Physics 513 10.30 (a) Ti = TC + 273.15 = (19.0 + 273.15) K = 292 K PV RT i (b) n = i i = i × ( ) ( )( 9 50 10 20 0 10 1 5 3 . . Pa L cm L 3 )( ) ⎡⎣ ⎤⎦ 1 10 ( ⋅ )( ) = 292 7 8 m 6 cm 8.31 J mol K K 3 3 . 3 mol (c) MCO2 g mol = [12.0 + 2(16.0)] = 44.0 g mol (d) m nM i i = =( )( ) = CO2 7.83 mol 44.0 g mol 345 g (e) T T T f i = + Δ = 292 K + 224 K = 516 K n m M Δ 345 82 0 = f = m m i f M − = − = CO2 CO2 g g 44.0 g mol 5 . .98 mol (f) Neglecting any change in volume of the tank, V V f i ≈ , and we have P V P V n RT n RT P n T n T f f i i f f i i f f f i i = ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Pi (g) P n n T T f i = P f i f i ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ 5.98 mol 7.83 mol ⎝ ⎜ ⎞⎠ ⎟ ⎛⎝ ⎜ ( × ) = × 516 9 50 105 1 28 10 K ⎞⎠ ⎟ 292 K . Pa . 6 Pa 10.31 (a) n PV RT = = (1 . 013 × 10 5 )(1 . 0 × 10− 6 ) 8 . 31 Pa atm m J 3 mol K K mol ( ⋅ )( ) = × − 293 4.2 10 5 Thus, N = n ⋅N = ( .2 × 10 − 5 mol ) × A molecules mol 4 6.02 1023 ⎛⎝ ⎜ ⎞⎠ ⎟ = 2.5 ×1019 molecules (b) Since both V and T are constant, n n P V RT P V RT P P 2 1 2 2 2 1 1 1 2 1 = = or n 1 0 10 11 = P P 2 n 2 1 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × × ⎛ ⎝ ⎜ . − Pa 1.013 105 Pa ⎞ ⎠ ⎟ (4.2 × 10−5 mol) = 4.1 × 10−21 mol 10.32 The volume of helium in each balloon is V r b = 4π 3 3. The total volume of the helium at P2 = 1.20 atm will be V P P 1 ⎛ V 2 2 1 150 1 20 = 0 100 ⎝ ⎜ ⎞ ⎠ ⎟ =⎛⎝ ⎜ ⎞⎠ ⎟ atm . atm ( . m3 ) = 12.5 m3 Thus, the number of balloons that can be fi lled is N V Vb = 2 =( )( ) = 3 12 5 4 3 0 150 884 . . m m balloons 3 π
16. 16. 514 Chapter 10 10.33 The initial and fi nal absolute temperatures are Ti = T i + = ( + ) = C, 273 25.0 273 K 298 K and T T f f = + = ( + ) = C, 273 75.0 273 K 348 K The volume of the tank is assumed to be unchanged, or V V f i = . Also, since two-thirds of the gas is withdrawn, n n f i = 3 . Thus, from the ideal gas law, P V P V n RT n RT P n n T T f f i i f f i i f f i f i = ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =⎛⎝ ⎜ ⎞⎠ ⎟ ⎛⎝ ⎜ ⎞⎠ ⎟ 1 3 348 P ( i 11 0 K 298 K . atm) = 4.28 atm 10.34 If the volume and the temperature are both constant, the ideal gas law gives P V P V n RT n RT f f i i f f = or n i i P P f i = n f i ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =⎛⎝ ⎜ ⎞⎠ ⎟ 5 00 1 50 . . atm 25.0 atm ( mol) = 0.300 mol so the amount of gas to be withdrawn is Δn n n i f = − = 1.50 mol − 0.300 mol = 1.20 mol . 10.35 With n held constant, the ideal gas law gives V V 0 030 = P P T T 1 2 2 1 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ . atm 1.0 atm ⎝ ⎜ ⎞⎠ ⎟ ⎛⎝ ⎜ 300 = × − 4 5 10 2 K ⎞⎠ ⎟ 200 K . 3 = ( ) Since the volume of a sphere is V = (4π 3)r3, V V r r 1 2 1 2 Thus, r V V = 1 4 5 10 2 1 3 20 7 1 r 1 2 1 3 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( . × − ) ( m) = . m 10.36 The mass of the gas in the balloon does not change as the temperature increases. Thus, ρ f ρ i ( m V ) ⎛ V = f V i or ρ = ρ i ( ) = f i m V V V i f f ⎝ ⎜ ⎞ ⎠ ⎟ From the ideal gas law with both n and P constant, we fi nd V V T T i f i f = and now have T T ρ ρ i f i f = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( 0 ⎞⎠ 179 )⎜⎛⎝ 273 . kgm K 3 ⎟ = 0.131 kg m3 373 K 10.37 The pressure 100 m below the surface is found, using P P gh 1= + atm ρ , to be P1 = 1.013 × 105 Pa + (103 kg m3 )(9.80 m s2 )(100 m) = 1.08 × 106 Pa The ideal gas law, with T constant, gives the volume at the surface as V P P V P P 1 V 2 2 1 1 1 08 106 1 = ⎛ ⎞ ⎟ ⎠ ⎜⎝ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × atm . Pa .013 10 1 50 16 0 × 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( ) = Pa . cm3 . cm3
17. 17. Thermal Physics 515 10.38 (a) We assume the density ρ can be written as ρ = naVbMc where n is the number of moles, V is the volume, and M is the molecular weight in kilograms per mole, while a, b, and c are constants to be determined by dimensional analysis. In terms of mass (M), length (L), time (T), and number of moles (N), the fundamental units of density are [ρ ] = [mass / volume ] = ML−3 , those of n are [n] = N, for volume [V ] = L−3, and molecular weight [M] = [kg mol ] = MN−1. In terms of basic units, our assumed equation for density becomes [ρ ] = [n] [V ] [M] a b c or M1L−3 = Na (L3 )b (MN−1 )c = Na−cL3bMc and equating the powers of each of the basic units on the two sides of the equation gives: 1 = c ⇒ c = 1; −3 = 3b ⇒ b = −1; 0 = a − c ⇒ a = c = 1 so our expression for density, derived by dimensional analysis, is ρ = n1V−1M1, or ρ = nM V where M is in kilograms per mole. (b) From the ideal gas law, PV = nRT, or P = (n V ) RT . But, from the result of part (a), we may write n V = ρ M, so the ideal gas law may be written in terms of the density of the gas as P ρ = RT M where M is in kilograms per mole. (c) For carbon dioxide, M = 44 g mol = 44 × 10−3 kg mol. Then, if the pressure is P = (90.0 atm)(1.013 × 105 Pa 1 atm) = 9.12 × 106 Pa, and T = 7.00 × 102 K, the density of the atmosphere on Venus is ( × )( × − ) PM RT ρ= = 9.12 106 Pa 44 10 3 kg mol 8.31 J mol K K kg m3 ( ⋅ )( × ) = 69 0 . 2 7 00 10 . (d) The density of the evacuated steel shell would be ρ = = × shell . = 47 7 . kg m3 shell shell π kg ( ) 4 1.00 m M V 2 00 10 3 2 3 Since , this shell would shell atmosphere ρ ρ rise in the atmosphere on Venus . 10.39 The average kinetic energy of the molecules of any ideal gas at 300 K is KE m k T = = = × ⎛⎝ ⎜ ⎞⎠ ⎟ J K 1 3 3 − ( ) 2 2 2 . K = 6.21×10−21 J v2 1 38 10 23 300 B
18. 18. 516 Chapter 10 10.40 Since the sample contains three times Avogadro’s number of molecules, there must be 3 moles of gas present. The ideal gas law then gives P nRT V = = ( 3 mol )( 8 31 J mol ⋅ K )( 293 K ) ( 0 200 m ) . . 3 = 9.13 × 105 Pa The force this gas will exert on one face of the cubical container is F = PA = (9.13 × 105 Pa)(0.200 m)2 = 3.65 × 104 N = 36.5 kN 10.41 One mole of any substance contains Avogadro’s number of molecules and has a mass equal to the molar mass, M. Thus, the mass of a single molecule is m = M NA. For helium, M = 4.00 g mol = 4.00 × 10−3 kg mol, and the mass of a helium molecule is m = × × = 4 00 10− 6 02 10 6 64 3 23 . . . kg mol molecule mol × 10−27 kg molecule Since a helium molecule contains a single helium atom, the mass of a helium atom is matom = 6.64 ×10−27 kg 10.42 The rms speed of molecules in a gas of molecular weight M and absolute temperature T is vrms = 3RT M . Thus, if vrms = 625 m s for molecules in oxygen O2 ( ), for which M = 32.0 g mol = 32.0 × 10−3 kg mol, the temperature of the gas is T 3 2 . 2 kg mol m s 3 M ( × − )( ) vrms = = R 32 0 10 625 . J mol K 3 8 31 K ( ⋅ ) = 501 10.43 The average translational kinetic energy per molecule in an ideal gas at absolute temperature T is KE k T molecule = 3 B 2, where kB= 1.38 ×10−23 J K is Boltzmann’s constant. Thus, if the absolute temperature is T = T + = ( + ) = C 273.15 77.0 273.15 K 350 K, we have KEmolecule = ( × − J K)( K) = × 3 2 1.38 10 23 350 7.25 10−21 J 10.44 (a) The volume occupied by this gas is V = 7.00 L(103 cm3 1 L)(1 m3 106 cm3 ) = 7.00 ×10−3 m3 Then, the ideal gas law gives T PV nR = = ( 1.60 × 106 Pa )( 7.00 × 10−3 m 3 ) ( 3.50 mol )( ⋅ ) = 8 31 385 . JmolK K (b) The average kinetic energy per molecule in this gas is KE k T molecule = B = ( × J K)( K) = 3 − 2 3 2 1.38 10 23 385 7.97 ×10−21 J (c) You would need to know the mass of the gas molecule to fi nd its average speed, which in turn requires knowledge of the molecular weight of the gas .
19. 19. Thermal Physics 517 10.45 Consider a time interval of 1.0 min = 60 s, during which 150 bullets bounce off Superman’s chest. From the impulse–momentum theorem, the magnitude of the average force exerted on Superman is F I t p t m ( − ) ⎡⎣ = = bullet = t av ⎤⎦ = Δ Δ Δ Δ 150 150 150 8 0 v v .0 10 400 400 60 16 3 × ( ) ( ) − − ( ) ⎡⎣ ⎤⎦ = − kg m s m s s N 10.46 From the impulse–momentum theorem, the average force exerted on the wall is F I t N p = = molecule = t − ( ) ⎡⎣ N m t av ⎤⎦ Δ Δ Δ Δ v v0 , or Fav kg m s = (5.0 ×1023 )(4.68 ×10−26 ) (300 ) − −300 m s s N ( ) ⎡⎣ ⎤⎦ = 1 0 14 . The pressure on the wall is then P F A av = × = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 2 2 N 8.0 cm cm m 14 10 1 1 8 10 4 . 4 N m2 = 18 kPa 10.47 As the pipe undergoes a temperature change ΔT = 46.5°C − 18.0°C = 28.5°C, the expansion of the horizontal segment is ΔL L ΔT x x = ( ) = × ( ) ⎡⎣ − − ( ) ⎤⎦ α 0 6 1 17 10 °C 28.0 cm (28.5°C) = 1.36 × 10−2 cm = 0.136 mm The expansion of the vertical section is ΔL L ΔT y y = ( ) = × ( ) ⎡⎣ α − − ( ) 0 6 1 17 10 °C 134 cm (28.5°°C) = 0.649 mm ⎤⎦ The total displacement of the pipe elbow is ΔL ΔL ΔL x y = 2 + 2 = ( )2 + ( )2 = 0.136 mm 0.649 mm 0.663 mm at θ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛⎝ tan− tan− . . 1 1 0 649 0 136 Δ Δ L L y x mm mm ⎞⎠ = 78.2° or L Δ = 0.663 mm at 78.2° below the horizontal 10.48 (a) Δ Δ L L T = ( ) = × ( ) ⎡⎣ α − − ( )( 0 6 1 9.0 10 °C 20 cm 75 °C) = 1.4 × 10−2 cm ⎤⎦ (b) Δ Δ D D T = ( ) = × ( ) ⎡⎣ α − − ( ) 0 6 1 9.0 10 °C 1.0 cm (75 °C) = 6.8 × 10−4 cm ⎤⎦ continued on next page
20. 20. 518 Chapter 10 (c) The initial volume is V π D 2 π = 0 1 0 20 16 L 0 0 2 4 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( ) ( ) = . cm cm cm3. = ( ) Δ Δ V V T 0 6 1 3 3 9.0 10 °C (16 cm3 )(75 °C) = 3.2 × 10−2 cm3 ⎡⎣ = V ( Δ T ) = × ( ) ⎤⎦ − − β α 0 10.49 The number of moles of CO2 present is n= = 6 50 0 148 . . g 44.0 g mol mol Thus, at the given temperature and pressure, the volume will be V = =(0.148 mol)(8.31 J mol ⋅K)(293 K) nRT P 1.013 × = × − = 10 m L 5 10 Pa 3.55 3 3 3.55 10.50 (a) The sketch at the right shows a thermometer marked with the Fahrenheit scale on one side and Réaumer’s scale on the other. Note that 80 RE scale divisions span the range from the freezing to the boiling of water, but 180 F divisions are required to do this. Then, note that the current temperature is TRE RE divisions above the freezing point of water, but is TF ( − 32) F divisions above this same level. TRE 212 F 80 RE divisions 180 F divisions TRE RE divisions (TF 32)F divisions Since the range between the current temperature and the freezing of water is the same fraction of the span between freezing and boiling points of water on both sides of the thermometer, we must have = − T T RE F 80 32 180 or T T T RE F F = 80 ( − ) = ( − ) 180 32 4 9 32 (b) Normal body temperature TF ( = 98.6°F) on the Réaumer scale is = 4 9 T T RE F °RE ° = − = ( ) − ⎡⎣⎢ ⎤⎦⎥ 14 2 4 9 . 98.6 14.2 29.6 RE 32 F divisions TF Boiling of water Freezing of water 80 RE 0 RE 32 F 0 F
21. 21. Thermal Physics 519 10.51 The ideal gas law will be used to fi nd the pressure in the tire at the higher temperature. However, one must always be careful to use absolute temperatures and absolute pressures in all ideal gas law calculations. The initial absolute pressure is Pi = Pi,gauge + Patm = 2.5 atm +1.0 atm = 3.5 atm. The initial absolute temperature is T T i i = + = ( + ) = , . . C 273 15 15 273 15 K 288 K. and the fi nal absolute temperature is T T f f = + = ( + ) = , . . C 273 15 45 273 15 K 318 K. The ideal gas law, with volume and quantity of gas constant, gives the fi nal absolute pressure as P V P V n RT n RT = ⇒ P = i T T P f f i i f f i i f f i ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 318 288 3 5 3 9 K K atm atm ⎛⎝ ⎞⎠ ( . ) = . The fi nal gauge pressure in the tire is P P P f , f . . . gauge atm = − = 3 9 atm −1 0 atm = 2 9 atm . 10.52 When air trapped in the tube is compressed, at constant temperature, into a cylindrical volume 0.40-m long, the ideal gas law gives its pressure as P 1 5 = V V P L L 1 P 2 2 1 1 2 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =⎛⎝ . m ⎞ 0.40 m⎠ ( × )1.013 105 Pa = 3.8 ×105 Pa This is also the water pressure at the bottom of the lake. Thus, P = P + gh atm ρ gives the depth of the lake as h ( − ) × ( ) 2 = P − P = g 5 3 8 1 013 10 10 3 9 8 atm 3 Pa . . ρ kg m ( = 28 m . 0 m s 2 ) 10.53 The mass of CO2 produced by three astronauts in 7.00 days is m = 3(1.09 kg d)(7.00 d) = 22.9 kg, and the number of moles of CO2 available is n m M = = 22 9 × . kg = − 520 3 44.0 10 kg mol mol The recycling process will generate 520 moles of methane to be stored. In a volume of V = 150 L = 0.150 m3 and at temperature T = −45.0°C = 228 K, the pressure of the stored methane is P = =(520 )(8 31 ⋅ )(228 ) nRT V mol J mol K K . = 6.57 ×106 Pa = 6.57 MPa 0 . 150 m 3
22. 22. 520 Chapter 10 10.54 (a) The piston in this vertical cylinder has three forces acting on it. These are: (1) a downward gravitational force, mg, the piston’s own weight; (2) a downward pressure force, F PA d = 0 , due to the atmospheric pressure above the piston; and (3) an upward pressure force, F PA u = , due to the absolute pressure of the gas trapped inside the cylinder. Since the piston is in equilibrium, Newton’s second law requires ΣFy = 0 ⇒ Fu − mg − Fd = 0 or PA = mg + P A 0 [1] From the ideal gas law, the absolute pressure of the trapped gas is P nRT V nRT Ah = = [2] Substituting Equation [2] into [1] yields A mg PA ⎛ nRT Ah ⎝ ⎜ ⎞ ⎠ ⎟ = + 0 or h nRT mg P A = + 0 (b) From Equation [1] above, the absolute pressure inside the cylinder is P = mg A + P0 where P0 is atmospheric pressure. This is greater than atmospheric pressure because mg A 0. (c) Observe from the result of part (a) above, if the absolute temperature T increases, the equilibrium value of h also increases . 10.55 (a) As the acetone undergoes a change in temperature ΔT = (20.0 − 35.0) °C = −15.0 °C, the fi nal volume will be V V V V V T V T f= + = + ( ) = [ + ( )] = ( ) + Δ β Δ β Δ mL .50 10 4 1 15.0 99.8 × ( ) ( ) − ( ) ⎡⎣ 0 0 0 0 1 100 1 1 ⎤⎦ − °C − °C = mL (b) When acetone at 35°C is poured into the Pyrex fl ask that was calibrated at 20°C, the volume of the fl ask temporarily expands to be larger than its calibration markings indicate. However, the coeffi cient of volume expansion for Pyrex [β = 3α = 9.6 ×10−6 (°C)−1] is much smaller than that of acetone [β = 1.5 ×10−4 (°C)−1]. Hence, the temporary increase in the volume of the fl ask will be much smaller than the change in volume of the acetone as the materials cool back to 20°C, and this change in volume of the fl ask has negligible effect on the answer .
23. 23. Thermal Physics 521 10.56 If Pi is the initial gauge pressure of the gas in the cylinder, the initial absolute pressure is Pi,abs Pi Patm = + , where Patm is atmospheric pressure. Likewise, the fi nal absolute pressure in the cylinder is P P P f ,abs f atm = + , where Pf is the fi nal gauge pressure. The initial and fi nal masses of gas in the cylinder are m nM i i = and m nM f f = , where n is the number of moles of gas present and M is the molecular weight of this gas. Thus, m m n n f i f i = . We assume the cylinder is a rigid container whose volume does not vary with internal pressure. Also, since the temperature of the cylinder is constant, its volume does not expand nor contract. Then, the ideal gas law (using absolute pressures) with both temperature and volume constant gives P V P V n RT n RT m m f i f i f i , , abs abs = = or m P P f i = m f i ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , , abs abs and in terms of gauge pressures, m P P f P P i = m f i + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ atm atm 10.57 (a) The volume of the liquid expands by ΔV V ΔT liquid = β ( ) 0 and the volume of the glass fl ask expands by ΔV V ΔT flask = (3 ) ( ) 0 α . The amount of liquid that must overfl ow into the capil-lary is V V V V T overflow liquid flask = Δ − Δ = ( − )(Δ ) 0 β 3α . The distance the liquid will rise into the capillary is then ⎜ ⎛⎝ = V = T Δh Δ A V A ⎞⎠ ⎟ overflow 0 (β − 3α )( ) (b) For a mercury thermometer, βHg = 1 82 ×10−4 (°C)−1 . and (assuming Pyrex glass), 3 3 3 2 10 6 1 9 6 10 6 1 α glass = × − °C − = × − (°C)− ( . ( ) ) . . Thus, the expansion of the mercury is almost 20 times the expansion of the flask , making it a rather good approximation to neglect the expansion of the fl ask. 10.58 (a) The initial absolute pressure in the tire is P P P 1 1 = + ( ) = 1 00 +1 80 = 2 80 atm gauge . atm . atm . atm and the fi nal absolute pressure is P2 = 3.20 atm. The ideal gas law, with volume constant, gives T P P 2 ⎛ T 2 1 1 3 20 2 80 = 300 ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎞ ⎟ ⎠ ⎜⎝ . . atm atm ( K) = 343 K (b) When the quantity of gas varies, while volume and temperature are constant, the ideal gas law gives n n P P 3 2 3 2 = . Thus, when air is released to lower the absolute pressure back to 2.80 atm, we have n n 3 2 = = 0 875 . 2 80 3 . 20 . atm atm At the end, we have 87.5% of the original mass of air remaining, or 12.5% of the original mass was released.
24. 24. 522 Chapter 10 10.59 After expansion, the increase in the length of one span is ΔL = L (ΔT ) = × ( ) ⎡⎣ − − ( ) ⎤⎦ α . . 0 6 1 12 10 °C 125 m (20 0°C) = 0.0300 m giving a fi nal length of L = L + L = + 0 Δ 125 m 0.0300 m From the Pythagorean theorem, y = L2 − L = ( + ) − ( ) = 2 2 2 125 0.0300 m 125 m 2.74 m 0 10.60 (a) From the ideal gas law, PV T PV T P P V V T T 2 2 2 1 1 1 , or 2 = 1 2 1 2 1 = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The initial conditions are = 1 atm, = 5.00 L = 5.00 × 10− 3 m3 , and = 20.0°C = 293 K 1 P V T 1 1 The fi nal conditions are V V A h T 2 2 1 2 1 1 2 = + + ⋅ P F A k h A atm = atm , = + ⋅ , and = 50 °C = 523 K Thus, = k h A atm K 1 1 1 523 293 1 + ⋅ ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⋅ ⎛ ⎝ ⎜ ⎞ A h V K ⎠ ⎟ ⎛⎝ ⎞⎠ or ( 2 . 00 × 10 3 N m )⋅ 5 + m2 N m2 1 ( . )( . × 0 0100 1 013 10 h ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ( )⋅ ( × ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . . m 0 0100 5 00 10 3 − 1 m 2 h 3 =⎛⎝ ⎞⎠ 523 293 Simplifying and using the quadratic formula yields h = 0.169 m = 16.9 cm k h A 2 (b) P 5 3 1 1 013 10 ( ) 2 00 10 0 1 = + ⋅ = × + ( × ) atm Pa N m . . .69 1 35 105 m 0.0100 m = . × Pa 2 L y L0125 m
25. 25. Thermal Physics 523 10.61 (a) The two metallic strips have the same length L0 at the initial temperature T0. After the tem-perature has changed by ΔT = T − T0, the lengths of the two strips are L L T 1 0 1 = [1 +α (Δ )] and L L T 2 0 2 = [1 +α (Δ )] The lengths of the circular arcs are related to their radii by L r 1 1 = θ and L r 2 2 = θ , where θ is measured in radians. Thus, Δ Δ r r r α α L L L T = − = − = ( − ) ( ) 2 1 2 1 2 1 0 θ θ θ or θ ( − ) ( ) 2 1 0 L T α α = r Δ Δ (b) As seen in the above result, θ = 0 if either = or α =α 1 ΔT 0 2 . (c) If ΔT 0, then θ is negative so the bar bends in the opposite direction . 10.62 (a) If the bridge were free to expand as the temperature increased by ΔT = 20°C, the increase in length would be ΔL =α L (ΔT ) = ( × − − )( )( ) = × 0 12 10 6 °C 1 250 m 20°C 6.0 10−2 m = 6.0 cm (b) Combining the defi ning equation for Young’s modulus, Y = Stress Strain = Stress (ΔL L) with the expression, ΔL =α L (ΔT ), for the linear expansion when the temperature changes by ΔT yields Stress Y L L Y L T L Y T = ⎛⎝ ⎜ ⎞⎠ ⎟ = ( ) ⎛⎝ ⎜ ⎞⎠ ⎟= ( ) Δ Δ Δ α α (c) When ΔT = 20°C, the stress in the specifi ed bridge would be Stress =αY (ΔT ) = (12 × 10−6 °C−1 )(2.0 × 1010 Pa)(20°°C) = 4.8 × 106 Pa Since this considerably less than the maximum stress, 2.0 × 107 Pa, that concrete can withstand, the bridge will not crumble . 10.63 (a) As the temperature of the pipe increases, the original 5.0-m length between the water heater and the fl oor above will expand by ΔL =α L (ΔT ) = ( × − )( )( − ) = 0 17 10 6 °C 5.0 m 46°C 20°C 2.21 × 10−3 m If this expansion occurs in a series of 18 “ticks,” the expansion per tick is movement per tick = ΔL 18 = (2.21 × 10−3 m) 18 = 1.23 × 10−4 m = 0.12 mm continued on next page
26. 26. 524 Chapter 10 (b) When the pipe is stuck in the hole, the fl oor exerts a friction force on the pipe preventing it from expanding. Just before a “tick” occurs, the pipe is compressed a distance of 0.123 mm. The force required to produce this compression is given by the equation defi ning Young’s modulus, Y = (F A) (ΔL L), as F YA Δ = ( × )( × − ) × L L = ⎛⎝ ⎞⎠ 10 5 1 23 Pa . m2 11 10 3 55 10 . 10 96 ⎛ −4 ⎝ ⎜ ⎞ = m ⎠ ⎟ 5.0 m N 10.64 Let container 1 be maintained at T T 1 0 = = 0°C = 273 K, while the temperature of container 2 is raised to T2 = 100°C = 373 K. Both containers have the same constant volume, V, and the same initial pressures, P P P 0 2 0 1 0 ( ) = ( ) = . As the temperature of container 2 is raised, gas fl ows from one container to the other until the fi nal pressures are again equal, P P P 2 1 = = . The total mass of gas is constant, so 0 1 + =( ) + ( ) [1] n n n n 2 1 02 From the ideal gas law, n = PV RT, so Equation [1] becomes PV RT 1 2 PV RT PV RT PV RT + = 0 + , or P 0 0 0 1 1 2 T T P T 1 2 0 0 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = Thus, P = 2 2(1 00 ) ⋅ 273 37 0 P T T T T T = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 273 0 1 2 1 2 . atm 3 273 373 1 15 + ⎛⎝ ⎜ ⎞⎠ ⎟ = . atm