Introductory Physics PHYS101
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1 Introductory Physics PHYS101
2 Dr Richard H. Cyburt Office Hours Assistant Professor of Physics My office: 402c in the Science Building My phone: (304) My TRF 9:30-11:00am F 12:30-2:00pm Meetings may also be arranged at other times, by appointment In person or is the best way to get a hold of me. Check my schedule on my office door. PHYS101
3 PHYS101: Introductory Physics 400 Lecture: 8:00-9:15am, TR Science Building Lab 1: 3:00-4:50pm, F Science Building 304 Lab 2: 1:30-3:20pm, M Science Building 304 Lab 3: 3:30-5:20pm, M Science Building 304 Lab 20: 6:00-7:50pm, M Science Building 304 PHYS101
4 Mastering Physics Online Go to HYPERLINK " Under Register Now, select Student. Confirm you have the information needed, then select OK! Register now. Enter your instructor s Course ID (RCYBURTPHYS101), and choose Continue. Enter your existing Pearson account username and password and select Sign in. You have an account if you have ever used a Pearson MyLab & Mastering product, such as MyMathLab, MyITLab, MySpanishLab, or MasteringChemistry. If you don t have an account, select Create and complete the required fields. Select an access option. Enter the access code that came with your textbook or was purchased separately from the bookstore. PHYS101
5 Bonus Homework for 3 Opens today! Adds points to Midterm 3 = 0.3x(Bonus-Exam) Due Wednesday, Nov 5:59pm Midterm 3 available for pick up after due date. PHYS101
6 Introductory Physics PHYS101
7 Douglas Adams Hitchhiker s Guide to the Galaxy PHYS101
8 You re already know physics! You just don t necessarily know the terminology and language we use!!! Physics of NASCAR Physics of Anger Birds PHYS101
9 Go Cubbies!!!! Go Cubs! World Series Champs!! PHYS101
10 In class!! PHYS101
11 This lecture will help you understand: Ideal-Gas Processes Thermal Expansion PHYS101
12 Section 12.3 Ideal-Gas Processes
13 Ideal-Gas Processes Ideal-gas processes have the following properties: The quantity of gas is fixed. There is a well-defined initial state. The initial values of pressure, volume, and temperature will be designated p i, V i, and T i. There is a well-defined final state in which the pressure, volume, and temperature have values p f, V f, and T f.
14 Ideal-Gas Processes The values of the variables in the initial and final states are related by
15 pv Diagrams In a pv diagram, each point on the graph represents a single, unique state of the gas. An ideal-gas process that changes the state of the gas can be represented as a trajectory.
16 pv Diagrams Text: p. 389
17 Constant-Volume Processes (Isochoric) Warming a gas in a closed container will raise the pressure without changing its volume. This is an example of a constant-volume process, where V f = V i. A constant-volume process appears on a pv diagram as a vertical line.
18 QuickCheck The temperature of a rigid (i.e., constant-volume), sealed container of gas increases from 100 C to 200 C. The gas pressure increases by a factor of (the pressure doesn t change)
19 QuickCheck The temperature of a rigid (i.e., constant-volume), sealed container of gas increases from 100 C to 200 C. The gas pressure increases by a factor of (the pressure doesn t change) Temperatures MUST be in K, not C, to use the ideal-gas law.
20 Example 12.6 Computing the tire pressure on a hot day The gauge pressure in a car tire is 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?
21 Example 12.6 Computing the tire pressure on a hot day (cont.) PREPARE A tire is (to a good approximation) a sealed container with constant volume, so this is a constant-volume process. The measured tire pressure is a gauge pressure, but the ideal-gas law requires an absolute pressure. We must correct for this. The initial pressure is Temperatures must be in kelvin, so we convert:
22 Example 12.6 Computing the tire pressure on a hot day (cont.) SOLVE The gas is in a sealed container, so we can use the ideal-gas law as given in Equation to solve for the final pressure. In this equation, we divide both sides by V f, and then cancel the ratio of the two volumes, which is equal to 1 for this constant-volume process:
23 Example 12.6 Computing the tire pressure on a hot day (cont.) The units for p f will be the same as those for p i, so we can keep the initial pressure in psi. The pressure at the higher temperature is
24 Example 12.6 Computing the tire pressure on a hot day (cont.) This is an absolute pressure, but the problem asks for the measured pressure in the tire a gauge pressure. Converting to gauge pressure gives
25 Example 12.6 Computing the tire pressure on a hot day (cont.) 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 result seems reasonable, and it has practical implications: If you check the pressure in your tires when they are at a particular temperature, don t expect the pressure to be the same when conditions change!
26 QuickCheck A sample of gas is in a cylinder with a moveable piston. The force on the piston can be varied, altering the pressure and volume. A sample of gas is taken from an initial state to a final state following a curve on the pv diagram shown. The final temperature is Higher than the initial temperature. The same as the initial temperature. Lower than the initial temperature. p (atm) End 6 2 Start n mol V (cm 3 )
27 QuickCheck A sample of gas is in a cylinder with a moveable piston. The force on the piston can be varied, altering the pressure and volume. A sample of gas is taken from an initial state to a final state following a curve on the pv diagram shown. The final temperature is Higher than the initial temperature. The same as the initial temperature. Lower than the initial temperature. p (atm) End 6 2 Start n mol V (cm 3 )
28 Constant-Pressure Processes An isobaric process is a constant-pressure process. A piston slides to the position where the external pressure and the gas pressure are equal. As long as the external pressure is constant, so must be the gas pressure. A constant-pressure process appears on a pv diagram as a horizontal line.
29 QuickCheck A sample of gas is in a cylinder with a moveable piston. The force on the piston can be varied, altering the pressure and volume. A sample of gas is taken from an initial state to a final state following a curve on the pv diagram shown. The final temperature is p (atm) Higher than the initial temperature. The same as the initial temperature. Lower than the initial temperature. 6 2 End n mol Start V (cm 3 )
30 QuickCheck A sample of gas is in a cylinder with a moveable piston. The force on the piston can be varied, altering the pressure and volume. A sample of gas is taken from an initial state to a final state following a curve on the pv diagram shown. The final temperature is p (atm) Higher than the initial temperature. The same as the initial temperature. Lower than the initial temperature. 6 2 End n mol Start V (cm 3 )
31 QuickCheck A steady force pushes in the piston of a well-insulated cylinder. In this process, the temperature of the gas Increases. Stays the same. Decreases. There s not enough information to tell.
32 QuickCheck A steady force pushes in the piston of a well-insulated cylinder. In this process, the temperature of the gas Increases. Stays the same. Decreases. There s not enough information to tell. No heat flows (well insulated) but work is done on the gas. First law: Q + W = ΔE th Work increases the gas s thermal energy and with it the temperature.
33 Example 12.7 A constant-pressure compression A gas in a cylinder with a movable piston occupies 50.0 cm 3 at 50 C. The gas is cooled at constant pressure until the temperature is 10 C. What is the final volume? PREPARE This is a sealed container, so we can use Equation The pressure of the gas doesn t change, so this is an isobaric process with p i /p f = 1. The temperatures must be in kelvin, so we convert:
34 Example 12.7 A constant-pressure compression (cont.) SOLVE We can use the ideal-gas law for a sealed container to solve for V f : ASSESS In this example and in Example 12.6, we have not converted pressure and volume units because these multiplicative factors cancel. But we did convert temperature to kelvin because this additive factor does not cancel.
35 Constant-Temperature Processes A constant-temperature process is an isothermal process. A gas cylinder can maintain a constant temperature if submerged in a larger container with a constant temperature. Pushing the piston down would be an isothermal compression; the reverse would be an isothermal expansion.
36 Constant-Temperature Processes Because there is an inverse relationship between p and V, the graph of an isothermal process is a hyperbola.
37 Constant-Temperature Processes An isotherm is a graph of an isothermal process. The location of the isotherm depends on temperature. The direction along the isotherm depends on the process.
38 QuickCheck A sample of gas is in a cylinder with a moveable piston. The force on the piston can be varied, altering the pressure and volume. A sample of gas is taken from an initial state to a final state following a curve on the pv diagram shown. The final temperature is Higher than the initial temperature. The same as the initial temperature. Lower than the initial temperature. p (atm) End 6 2 n mol Start V (cm 3 )
39 QuickCheck A sample of gas is in a cylinder with a moveable piston. The force on the piston can be varied, altering the pressure and volume. A sample of gas is taken from an initial state to a final state following a curve on the pv diagram shown. The final temperature is Higher than the initial temperature. The same as the initial temperature. Lower than the initial temperature. p (atm) End 6 2 n mol Start V (cm 3 )
40 Example 12.8 Compression air in the lungs 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 higher than at the surface. At this depth, what is the volume of air in the snorkeler s lungs? PREPARE At the surface, the pressure of the air inside the snorkeler s lungs is 1.0 atm it s atmospheric pressure at sea level. As he descends, the pressure inside his lungs must rise to match the pressure of the surrounding water, because the body can t sustain large pressure differences between inside and out. Further, the air stays at body temperature, making this an isothermal process with T f = T i.
41 Example 12.8 Compression air in the lungs (cont.) SOLVE The ideal-gas law for a sealed container (the lungs) gives Notice that we didn t need to convert pressure to SI units. As long as the units are the same in the numerator and the denominator, they cancel. ASSESS The air has a smaller volume at the higher pressure, as we would expect. The air inside your lungs does compress significantly! when you dive below the surface.
42 Thermodynamics of Ideal-Gas Processes When gases expand, they do work on the piston. If the gas expands under constant pressure, pushing the piston (with area A) from x i to x f a distance d = x f - x i, then the work done is This can be written as
43 Thermodynamics of Ideal-Gas Processes For all ideal-gas processes, the work is the area under the pv graph between V i and V f.
44 Thermodynamics of Ideal-Gas Processes 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 applies only to constant-pressure processes. For any other ideal-gas process, you must use the geometry of the pv diagram to calculate the area under the graph. To calculate work, pressure must be in Pa and volume in m 3. The product of Pa ( which is N/m 2 ) and m 3 is N m. But 1 N m is J the unit of work and energy.
45 Thermodynamics of Ideal-Gas Processes (cont.) W gas is positive if the gas expands (ΔV > 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 decreases. W gas is negative if the piston compresses the gas (ΔV < 0) because the force F gas 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 W gas is negative.
46 Thermodynamics of Ideal-Gas Processes The first law of thermodynamics can be written Because the thermal energy only depends on its temperature, we can also write the change in thermal energy of an ideal gas as
47 Adiabatic Processes Adiabatic processes are processes during which heat is not transferred. In these processes, Q = 0. An adiabatic expansion lowers the temperature of a gas. An adiabatic compression raises the temperature of a gas. Adiabatic processes allow you to use work, rather than heat, to change the temperature of the gas.
48 QuickCheck A cylinder of gas has a frictionless but tightly sealed piston of mass M. A small flame heats the cylinder, causing the piston to slowly move upward. For the gas inside the cylinder, what kind of process is this? Isochoric Isobaric Isothermal Adiabatic None of the above
49 QuickCheck A cylinder of gas has a frictionless but tightly sealed piston of mass M. A small flame heats the cylinder, causing the piston to slowly move upward. For the gas inside the cylinder, what kind of process is this? Isochoric Isobaric Isothermal Adiabatic None of the above
50 QuickCheck A cylinder of gas has a frictionless but tightly sealed piston of mass M. The gas temperature is increased from an initial 27 C to a final 127 C. What is the final-to-initial volume ratio V f /V i? Not enough information to tell
51 QuickCheck A cylinder of gas has a frictionless but tightly sealed piston of mass M. The gas temperature is increased from an initial 27 C to a final 127 C. What is the final-to-initial volume ratio V f /V i? Not enough information to tell Isobaric, so V f = T f = 400 K V f T f 300 K
52 QuickCheck A cylinder of gas floats in a large tank of water. It has a frictionless but tightly sealed piston of mass M. Small masses are slowly placed onto the top of the piston, causing it to slowly move downward. For the gas inside the cylinder, what kind of process is this? Isochoric Isobaric Isothermal D. Adibatic E. None of the above
53 QuickCheck A cylinder of gas floats in a large tank of water. It has a frictionless but tightly sealed piston of mass M. Small masses are slowly placed onto the top of the piston, causing it to slowly move downward. For the gas inside the cylinder, what kind of process is this? Isochoric Isobaric Isothermal D. Adibatic E. None of the above
54 Conceptual Example Adiabatic curves on a pv diagram FIGURE shows the pv diagram of a gas undergoing an isothermal compression from point 1 to point 2. Sketch how the pv diagram would look if the gas were compressed from point 1 to the same final pressure by a rapid adiabatic compression. REASON An adiabatic compression increases the temperature of the gas as the work done on the gas is transformed into thermal energy. Consequently, as seen in FIGURE 12.18, the curve of the adiabatic compression cuts across the isotherms to end on a higher-temperature isotherm when the gas pressure reaches p 2.
55 Conceptual Example Adiabatic curves on a pv diagram (cont.) ASSESS In an isothermal compression, heat energy is transferred out of the gas so that the gas temperature stays the same. This heat transfer doesn t happen in an adiabatic compression, so we d expect the gas to have a higher final temperature. In general, the temperature at the final point of an adiabatic compression is higher than at the starting point. Similarly, an adiabatic expansion ends on a lower-temperature isotherm.
56 QuickCheck What type of gas process is this? Isochoric Isobaric Isothermal Adiabatic None of the above
57 QuickCheck What type of gas process is this? Isochoric Isobaric Isothermal Adiabatic None of the above
58 QuickCheck A gas follows the process shown. What is the final-to-initial temperature ratio T f /T i? Not enough information to tell
59 QuickCheck A gas follows the process shown. What is the final-to-initial temperature ratio T f /T i? Not enough information to tell
60 QuickCheck 12.7 If the volume of a sealed container of gas is decreased, the gas temperature Increases. Stays the same. Decreases. Not enough information to tell.
61 QuickCheck 12.7 If the volume of a sealed container of gas is decreased, the gas temperature Increases. Stays the same. Decreases. Not enough information to tell.
62 Section 12.4 Thermal Expansion
63 Thermal Expansion Thermal expansion is the expansion of a material (including solids) when heated. Volume thermal expansion is The constant b is the coefficient of volume expansion. Its value depends on the material of the object and has units of K 1.
64 Thermal Expansion Linear thermal expansion is The constant α is the coefficient of linear expansion. Its value depends on the material of the object and has units of K 1. Expanding spans: A long steel bridge will slightly increase in length on a hot day and decrease on a cold day. Thermal expansion joints let the bridge s length change without causing the roadway to buckle.
65 Thermal Expansion
66 Thermal Expansion The expressions for thermal expansions are empirical formulas; they are a good fit to data, but not all materials follow the equations for volume and linear thermal expansions. Water at low temperatures is one example.
67 QuickCheck Two rods are precisely cm long. Rod 1 is made of aluminum, with a coefficient of linear expansion α = /K. Rod 2 is made of steel, and has α = /K. Rod 1 is heated by 10 C while rod 2 is heated by 20 C. After heating, which rod is longer? Rod 1 Rod 2 Both rods have the same length.
68 QuickCheck Two rods are precisely cm long. Rod 1 is made of aluminum, with a coefficient of linear expansion α = /K. Rod 2 is made of steel, and has α = /K. Rod 1 is heated by 10 C while rod 2 is heated by 20 C. After heating, which rod is longer? Rod 1 Rod 2 Both rods have the same length.
69 QuickCheck Aluminum rod 1 is cm long, while aluminum rod 2 is cm long. Both rods are heated by 20 C. After heating, which rod has undergone the greatest change in length? Rod 1 Rod 2 The lengths of both rods have changed by the same amount.
70 QuickCheck Aluminum rod 1 is cm long, while aluminum rod 2 is cm long. Both rods are heated by 20 C. After heating, which rod has undergone the greatest change in length? Rod 1 Rod 2 The lengths of both rods have changed by the same amount.
71 Example Problem In the United States, railroad cars ride on steel rails. Until the mid-1900s, most track consisted of 11.9 m lengths connected with expansion joints that allowed for the rails to expand and contract with temperature. If a section of rail is exactly m long on a hot, sunny day when it warms up to 50 C, how long will it be on a cold 10 C winter morning?
72 Try It Yourself: Thermal Expansion to the Rescue If you have a stubborn lid on a glass jar, try this: Put the lid under very hot water for a short time. Heating the lid and the jar makes them both expand, but, as you can see from the data in Table 12.3, the steel lid and the opening in the lid that fits over the glass jar expands by more than the glass jar. The jar lid is now looser and can be more easily removed.
73 Example How much closer to space? The height of the Space Needle, a steel observation tower in Seattle, is 180 meters on a 0 C winter day. How much taller is it on a hot summer day when the temperature is 30 C? PREPARE The steel expands because of an increase in temperature, which is T = T f - T i = 30 C - 0 C = 30 C = 30 K
74 Example How much closer to space? (cont.) SOLVE The coefficient of linear expansion is given in Table 12.3; we can use this value in Equation to compute the increase in height: L = al i T = ( K -1 )(180 m)(30 K) = m
75 Example How much closer to space? (cont.) ASSESS You don t notice buildings getting taller on hot days, so we expect the final answer to be small. The change is a small fraction of the height of the tower, as we expect. Our answer makes physical sense. Compared to 180 m, an expansion of 6.5 cm is not something you would easily notice but it isn t negligible. The thermal expansion of structural elements in towers and bridges must be accounted for in the design to avoid damaging stresses. When designers failed to properly account for thermal stresses in the marble panels cladding the Amoco Building in Chicago, all 43,000 panels had to be replaced, at great cost.
76 Special Properties of Water and Ice Water differs from other liquids at colder temperatures. Instead of decreasing volume with decreasing temperature, water increases in volume at temperatures below 4 C. This happens at lower temperatures because the water molecules form clusters that are more strongly bound but are located farther apart.
77 Special Properties of Water and Ice Freezing the water results in an even greater increase in volume. Because water expands as it freezes, ice is less dense than liquid water and floats.
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