Lecture Outlines Chapter 17 Physics 3 rd Edition

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Lecture Outlines Chapter 17 Physics, 3 rd Edition James S. Walker © 2007 Pearson

Lecture Outlines Chapter 17 Physics, 3 rd Edition James S. Walker © 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Chapter 17 Phases and Phase Changes

Chapter 17 Phases and Phase Changes

Units of Chapter 17 • Ideal Gases • Kinetic Theory • Solids and Elastic

Units of Chapter 17 • Ideal Gases • Kinetic Theory • Solids and Elastic Deformation • Phase Equilibrium (fasjämvikt) and Evaporation (förångning) • Latent Heats (smält/ångbildningsvärme) • Phase Changes and Energy Conservation

17 -1 Ideal Gases are the easiest state of matter to describe, as all

17 -1 Ideal Gases are the easiest state of matter to describe, as all ideal gases exhibit similar behavior. An ideal gas is one that is thin enough, and far away enough from condensing, that the interactions (växelverkan) between molecules can be ignored.

17 -1 Ideal Gases If the volume of an ideal gas is held constant,

17 -1 Ideal Gases If the volume of an ideal gas is held constant, we find that the pressure increases with temperature:

17 -1 Ideal Gases If the volume and temperature are kept constant, but more

17 -1 Ideal Gases If the volume and temperature are kept constant, but more gas is added (such as in inflating a tire or basketball), the pressure will increase:

17 -1 Ideal Gases Finally, if the temperature is constant and the volume decreases,

17 -1 Ideal Gases Finally, if the temperature is constant and the volume decreases, the pressure increases:

17 -1 Ideal Gases Rearranging gives us the equation of state for an ideal

17 -1 Ideal Gases Rearranging gives us the equation of state for an ideal gas: Instead of counting molecules, we can count moles. A mole is the amount of a substance that contains as many elementary entities as there atoms in 12 g of carbon-12.

Example 17 -1 Take a Deep Breath Syre är 21% av luftmolekylerna. Hur många

Example 17 -1 Take a Deep Breath Syre är 21% av luftmolekylerna. Hur många O 2 -molekyler finns i lungorna?

Example 17 -1 Take a Deep Breath En persons lungor kan innehålla 6, 0

Example 17 -1 Take a Deep Breath En persons lungor kan innehålla 6, 0 liter vid normal kroppstemperatur (T = 273+37=310 K) och normalt lufttryck (101 k. Pa). Hur många syremolekyler finns i lungorna? Luft kan behandlas som en ideal gas. PV = Nk. T ger direkt totala antalet molekyler N = 101 k. Pa • 6 10 -3 m 3/1, 38 10 -23 J/K • 310 K = = 1, 42 • 1023 (stycken) Antalet syremolekyler = 3, 0 • 1022 (stycken) [Antal mol = 0, 0494]

Figure 17 -4 Moles of various substances (H, Cu, Hg, S)

Figure 17 -4 Moles of various substances (H, Cu, Hg, S)

17 -1 Ideal Gases Experimentally, the number of entities (atoms or molecules) in a

17 -1 Ideal Gases Experimentally, the number of entities (atoms or molecules) in a mole is given by Avogadro’s number: (Avogadro 1778 -1856) Therefore, n moles of gas will contain molecules.

17 -1 Ideal Gases Avogadro’s number and the Boltzmann constant can be combined to

17 -1 Ideal Gases Avogadro’s number and the Boltzmann constant can be combined to form the universal gas constant and an alternative equation of state:

Active Example 17 -1 The Amount of Air in a Basketball Hur många mol

Active Example 17 -1 The Amount of Air in a Basketball Hur många mol luftmolekyler finns det i en basketboll? Trycket är 171 k. Pa, T = 293 K och bollens diameter är 30, 0 cm. Luft kan behandlas som en ideal gas. PV = n. RT ger direkt antalet mol eftersom V = 4πr 3/3 = 0, 01414 m 3 n = 171 k. Pa • 0, 01414 m 3/[8, 3145 J/(mol • K) • 293 K] = = 0, 993 (mol)

17 -1 Ideal Gases The atomic or molecular mass of a substance is the

17 -1 Ideal Gases The atomic or molecular mass of a substance is the mass, in grams, of one mole of that substance. For example, Helium: Copper: Furthermore, the mass of an individual atom is given by the atomic mass divided by Avogadro’s number: (17 -6)

Exercise 17 -1 Bestäm massan för en Cu-atom och för en syremolekyl (O 2).

Exercise 17 -1 Bestäm massan för en Cu-atom och för en syremolekyl (O 2). Atommassor är tabellerade i appendix E. Eftersom (massan av en atom) • NA = atomvikten NA = 6, 022 • 1023 /mol och atommassorna för Cu och O är 63, 546 g/mol och 16, 00 g/mol respektive fås m(Cu) = 63, 546 g/mol/6, 022 1023/mol =1, 055 • 10 -22 g m(O 2) = 2 • 16, 00 g/mol /NA = 5, 314 • 10 -23 g Observera att massorna undantagsvis är givna i gram.

Conceptual Checkpoint 17 -1 Air Pressure Om man fryser litet och drar upp sin

Conceptual Checkpoint 17 -1 Air Pressure Om man fryser litet och drar upp sin termostat, så känns luften efter ett tag varmare. Om rummet kan anses “isolerat” har då trycket > = <? (Vad är det som gör detta? )

17 -1 Ideal Gases Boyle’s law, which is consistent with the ideal gas law,

17 -1 Ideal Gases Boyle’s law, which is consistent with the ideal gas law, says that the pressure varies inversely with volume. These curves of constant temperature (hyperbel) are called isotherms.

Example 17 -2 Under Pressure

Example 17 -2 Under Pressure

Example 17 -2 Under Pressure När trycket är 130 k. Pa är höjden i

Example 17 -2 Under Pressure När trycket är 130 k. Pa är höjden i pistongen 25 cm. När man sedan placerar en större massa på locket stiger trycket till 170 k. Pa och höjden ändras till h 2, medan temperaturen (T = 290 K) är konstant. Beräkna den nya höjden. PV = konstant och V = A • h ger h 2 = h 1 • P 1/P 2 = 19 cm

17 -1 Ideal Gases Charles’s law, also consistent with the ideal gas law, says

17 -1 Ideal Gases Charles’s law, also consistent with the ideal gas law, says that the volume of a gas increases with temperature if the pressure is constant (isobar process).

17 -1 Ideal Gases In this photograph, the balloon was inflated at room temperature

17 -1 Ideal Gases In this photograph, the balloon was inflated at room temperature and cooled with liquid nitrogen (kokpunkt 77 K). The decrease in volume of the air in the balloon is obvious.

Active Example 17 -2 Find the Piston Height

Active Example 17 -2 Find the Piston Height

Active Example 17 -2 Find the Piston Height När trycket är 130 k. Pa

Active Example 17 -2 Find the Piston Height När trycket är 130 k. Pa är höjden i pistongen 25 cm. När man sedan ökar temperaturen från 290 K till 330 K ändras höjden till h 2 (medan trycket är konstant). Beräkna den nya höjden. V/T = konstant och V = A • h ger h 2 = h 1 • T 2/T 1 = 28 cm

17 -2 Kinetic Theory The kinetic theory relates microscopic quantities (läge och hastighet hos

17 -2 Kinetic Theory The kinetic theory relates microscopic quantities (läge och hastighet hos enskilda gasatomer/molekyler) to macroscopic ones (tryck, temperatur). Assumptions: • N identical molecules of mass m are inside a container of volume V; each acts as a point particle. • Molecules move randomly and always obey Newton’s laws. • Collisions with other molecules and with the walls are elastic (dvs både p och K bevaras).

17 -2 Kinetic Theory Pressure is the result of collisions between the gas molecules

17 -2 Kinetic Theory Pressure is the result of collisions between the gas molecules and the walls of the container. It depends on the mass and speed of the molecules, and on the container size:

Figure 17 -7 Force exerted by a molecule on the wall of a container

Figure 17 -7 Force exerted by a molecule on the wall of a container När Δpx = pf, x – pi, x = 2 mvx Δt = 2 L/vx F = Δpx/Δt = mvx 2/L Nu erhålls 17 -9 enkelt P = F/A = (mvx 2/L)/L 2 = mvx 2/L 3 = mvx 2/V För att förenkla deriveringen antogs att molekylen rörde sig i x-led. Om den rör sig med en vinkel relativt x-axeln, så gäller beräkningen rörelsens xkomponent och resultatet blir detsamma.

17 -2 Kinetic Theory Not all molecules in a gas will have the same

17 -2 Kinetic Theory Not all molecules in a gas will have the same speed; their speeds are represented by the Maxwell distribution, and depend on the temperature and mass of the molecules.

17 -2 Kinetic Theory We replace the speed in the previous expression for pressure

17 -2 Kinetic Theory We replace the speed in the previous expression for pressure with the average speed: Including the other two directions, Therefore, the pressure in a gas is proportional to the average kinetic energy of its molecules.

17 -2 Speed distibution of Molecules För en molekyl där P = m(vx 2

17 -2 Speed distibution of Molecules För en molekyl där P = m(vx 2 )av / V Eftersom alla molekyler (”ensemblen”) följer samma fördelning får man för hela gasen med N molekyler 17 -10 P = Nm(vx 2)av / V x-riktningen är godtyckligt vald så det gäller att (vx 2)av = (vy 2)av = (vz 2)av (vx 2 )av = v 2 av /3 P = N/V • mv 2 av /3 P = 2/3 • (N/V) • (mv 2)av/2 = 2/3 • (N/V) • Kav Gastrycket är direkt proportionellt mot tätheten av partiklar och medelvärdet av den kinetiska energin 17 -11

17 -2 Kinetic Theory Comparing this expression with the ideal gas law (PV=Nk. T)

17 -2 Kinetic Theory Comparing this expression with the ideal gas law (PV=Nk. T) allows us to relate average kinetic energy and temperature: The square root of mean square (rms) speed. is called the root

17 -2 Kinetic Theory Solving for the rms speed gives:

17 -2 Kinetic Theory Solving for the rms speed gives:

17 -2 Kinetic Theory The rms speed is slightly greater than the most probable

17 -2 Kinetic Theory The rms speed is slightly greater than the most probable speed and the average speed.

Example 17 -3 Fresh Air

Example 17 -3 Fresh Air

Example 17 -3 Fresh Air Luft består huvudsakligen av kvävemolekyler (78%) och syremolekyler. Är

Example 17 -3 Fresh Air Luft består huvudsakligen av kvävemolekyler (78%) och syremolekyler. Är rms-hastigheten för N 2 (28, 0 g/mol) >=< rms hastigheten för O 2 (32, 0 g/mol)? Beräkna hastigheterna vid T = 293 K. vrms = (3 k. T/m)1/2 = [R = NAk, M= m. NA] så att vrms = (3 RT/M )1/2 = (3 • 8, 31 J/[mol K])293 K/28, 0 g/mol)1/2 = (tänk på dimensionerna!!) = 511 m/s vrms(M=32, 0 g/mol) = 478 m/s

17 -2 Kinetic Theory The internal energy of an ideal gas is the sum

17 -2 Kinetic Theory The internal energy of an ideal gas is the sum of the kinetic energies of all its molecules. In the case where each molecule consists of a single atom (som ädelgaserna Ar, Ne, Xe), this may be written:

17 -3 Solids and Elastic Deformation Solids have definite shapes (unlike fluids), but they

17 -3 Solids and Elastic Deformation Solids have definite shapes (unlike fluids), but they can be deformed. Pulling on opposite ends of a rod can cause it to stretch:

17 -3 Solids and Elastic Deformation The amount of stretching will depend on the

17 -3 Solids and Elastic Deformation The amount of stretching will depend on the force; Y is Young’s modulus and is a property of the material: (17 -17)

Example 17 -4 Stretching a Bone m = 21 kg Y = 1, 6

Example 17 -4 Stretching a Bone m = 21 kg Y = 1, 6 • 1010 N/m 2 F = Y • A • (ΔL/L 0) ΔL = (F • L 0 )/(Y • A) = 8, 2 μm

Conceptual Checkpoint 17 -3 Compare Force Constants Det finns en direkt koppling mellan 17

Conceptual Checkpoint 17 -3 Compare Force Constants Det finns en direkt koppling mellan 17 -17 och Hooke’s lag (6 -4) för en fjäder. F = Y • A • (ΔL/L 0)= (YA/L 0)ΔL = k • x så om två identiska fjädrar sätts ihop blir då fjäderkonstanten k a) > b) < c) = med fjäderkonstanten för en fjäder?

17 -3 Solids and Elastic Deformation Another type of deformation is called a shear

17 -3 Solids and Elastic Deformation Another type of deformation is called a shear deformation (skjuvning), where opposite sides of the object are pulled laterally in opposite directions.

17 -3 Solids and Elastic Deformation As expected, the deformation is proportional to the

17 -3 Solids and Elastic Deformation As expected, the deformation is proportional to the force. S is the shear modulus (skjuvmodulen).

Active Example 17 -3 Deforming a Stack of Pancakes: Find the Shear Modulus

Active Example 17 -3 Deforming a Stack of Pancakes: Find the Shear Modulus

Active Example 17 -3 Deforming a Stack of Pancakes: Find the Shear Modulus F

Active Example 17 -3 Deforming a Stack of Pancakes: Find the Shear Modulus F = 1, 2 N d = 13 cm Δx = 2, 5 cm L 0 = 9, 0 cm A = π • d 2/4 F = S • A • (Δx/L 0) S= (F • L 0)/(A • Δx) = 330 (325, 47) N/m 2

17 -3 Solids and Elastic Deformation Finally, if a solid is uniformly compressed, it

17 -3 Solids and Elastic Deformation Finally, if a solid is uniformly compressed, it will shrink.

17 -3 Solids and Elastic Deformation Here, the proportionality constant, B, is called the

17 -3 Solids and Elastic Deformation Here, the proportionality constant, B, is called the bulk modulus.

Active Example 17 -4 A Gold Doubloon: Find the Change in Volume t(jockleken) =

Active Example 17 -4 A Gold Doubloon: Find the Change in Volume t(jockleken) = 2, 0 mm d(iametern) = 6, 1 cm A = π • d 2/4 B = 22 • 1010 N/m 2 Ekvation 15 -7 ger direkt (gaugetrycket) ΔP = ρ • g • h (h = 770 m) Ekvation 17 -19 ger då direkt ΔV = - V 0 • ΔP/B = - A • t • ΔP/B = - 2 • 10 -10 m 3

17 -3 Solids and Elastic Deformation The applied force per unit area is called

17 -3 Solids and Elastic Deformation The applied force per unit area is called the stress (skjuvkraft), and the resulting deformation is the strain (skjuvning). They are proportional to each other until the stress becomes too large; permanent (plastisk) deformation will then occur.

17 -4 Phase Equilibrium and Evaporation If a liquid is put into a sealed

17 -4 Phase Equilibrium and Evaporation If a liquid is put into a sealed container so that there is a vacuum above it, some of the molecules in the liquid will vaporize. Once a sufficient number have done so, some will begin to condense back into the liquid. Equilibrium is reached when the numbers remain constant.

17 -4 Phase Equilibrium and Evaporation The pressure of the gas when it is

17 -4 Phase Equilibrium and Evaporation The pressure of the gas when it is in equilibrium with the liquid is called the equilibrium vapor (ång-) pressure, and will depend on the temperature.

17 -4 Phase Equilibrium and Evaporation and Conceptual Checkpoint 17 -4 (p. 567) The

17 -4 Phase Equilibrium and Evaporation and Conceptual Checkpoint 17 -4 (p. 567) The vaporization curve determines the boiling point of a liquid: A liquid boils at the temperature at which its vapor pressure equals the external pressure. This explains why water boils at a lower temperature at lower pressure – and why you should never insist on a “ 3 -minute egg” in Denver (1610 m över havet)! Conceptual checkpoint 17 -4 När vattnet kokar på en bergstopp, är då temperaturen > < = 100° C?

17 -4 Phase Equilibrium and Evaporation This curve can be expanded. When the liquid

17 -4 Phase Equilibrium and Evaporation This curve can be expanded. When the liquid reaches the critical point, there is no longer a distinction between liquid and gas; there is only a “fluid” phase.

17 -4 Phase Equilibrium and Evaporation The fusion curve is the boundary between the

17 -4 Phase Equilibrium and Evaporation The fusion curve is the boundary between the solid and liquid phases; along that curve they exist in equilibrium with each other. Almost all materials have a fusion curve that resembles (a); dvs när trycket ökar så kan man få vätskan att bli “solid”. Water, due to its unusual properties near the freezing point, follows (b).

17 -4 Phase Equilibrium and Evaporation Finally, the sublimation curve marks the boundary between

17 -4 Phase Equilibrium and Evaporation Finally, the sublimation curve marks the boundary between the solid and gas phases. The triple point is where all three phases are in equilibrium. This is shown on the phase diagram below.

Photo 17 -4 Patterned ground on Mars (Jämför denna bild med motsvarande från Arktis

Photo 17 -4 Patterned ground on Mars (Jämför denna bild med motsvarande från Arktis i boken)

17 -4 Phase Equilibrium and Evaporation A liquid in a closed container will come

17 -4 Phase Equilibrium and Evaporation A liquid in a closed container will come to equilibrium with its vapor. However, an open liquid will not, as its vapor keeps escaping – it will continue to vaporize without reaching equilibrium. As the molecules that escape from the liquid are the higher-energy ones, this has the effect of cooling the liquid. This is why sweating cools us off. (Svettdroppe på huden. )

17 -4 Phase Equilibrium and Evaporation If we look at the Maxwell speed distributions

17 -4 Phase Equilibrium and Evaporation If we look at the Maxwell speed distributions for water at different temperatures, we see that there is not much difference between the 30° C (648 m/s) curve and the 100° C (719 m/s) curve. This means that, if 100° C water molecules can escape, many 30° C molecules will also.

17 -4 Phase Equilibrium and Evaporation This same evaporation process can cause a planet

17 -4 Phase Equilibrium and Evaporation This same evaporation process can cause a planet to lose its atmosphere – some molecules will have speeds exceeding the escape velocity. The evaporation process will be faster for lighter molecules and for less massive planets.

17 -5 Latent Heats When two phases coexist, the temperature remains the same even

17 -5 Latent Heats When two phases coexist, the temperature remains the same even if a small amount of heat is added. Instead of raising the temperature, the heat goes into changing the phase of the material – melting ice, for example.

17 -5 Latent Heats The heat required to convert from one phase to another

17 -5 Latent Heats The heat required to convert from one phase to another is called the latent heat. The latent heat, L, is the heat that must be added to or removed from one kilogram of a substance to convert it from one phase to another. During the conversion process, the temperature of the system remains constant.

17 -5 Latent Heats The latent heat of fusion is the heat needed to

17 -5 Latent Heats The latent heat of fusion is the heat needed to go from solid to liquid; the latent heat of vaporization from liquid to gas.

Photo 17 -7 Ice lake on Mars På grund av det låga trycket smälter

Photo 17 -7 Ice lake on Mars På grund av det låga trycket smälter inte isen utan går direkt i gasfas (sublimerar) och vice versa.

Example 17 -5 Steam Heat

Example 17 -5 Steam Heat

Example 17 -5 Steam Heat För att få (vatten)ånga tillför man 560 k. J

Example 17 -5 Steam Heat För att få (vatten)ånga tillför man 560 k. J till 0, 220 kg vatten med ursprungstemperaturen 50, 0° C. Bestäm sluttemperaturen. För att höja vattnets temperatur till 100° C åtgår Q = m • c • ΔT = 0, 220 kg • 4186 J/(kg • K) • 50 K= 46, 0 k. J För att förånga 0, 220 kg vattenånga åtgår = m • L = 0, 220 kg • 2, 26 MJ/kg = 497 k. J Q Den återstående värmemängden (17 k. J) används för att höja ångans temperatur (Tabell 16 -2) ΔT= Q/(mcå)=17 k. J/(0, 220 kg • 2010(J/kg • K)= 38, 4°C T = 100° C + ΔT = 138° C

17 -6 Phase Changes and Energy Conservation Solving problems involving phase changes is similar

17 -6 Phase Changes and Energy Conservation Solving problems involving phase changes is similar to solving problems involving heat transfer, except that the latent heat must be included as well.

Example 17 -6 Warm Punch

Example 17 -6 Warm Punch

Example 17 -6 Warm Punch En ”punchbål” innehåller (i princip) 3, 95 kg vatten

Example 17 -6 Warm Punch En ”punchbål” innehåller (i princip) 3, 95 kg vatten vid 20, 0° C. En iskub med vikten 0, 0450 kg och temperaturen 0° C läggs i vattnet. Vad blir sluttemperaturen? Återstår då någon is? Bortse från omgivningens inverkan. För att sänka vattnets temperatur till 0° C kan vi ”disponera” värmemängden Q = m • c • ΔT = 3, 95 kg • 4186 J/(kg • K) • 20 K= 330, 7 k. J För att smälta 0, 0450 kg is åtgår värmemängden Q = mi • L = 0, 0450 kg • 335 k. J/kg = 15, 1 k. J Den återstående värmemängden bestämmer sluttemperaturen för den ”nya” (större) mängden vatten ΔT= Q/(m+mi)c = (330, 7 -15, 1)k. J/(3, 995 kg • 4186 {J/kg • K}) = 18, 9°C

Summary of Chapter 17 • An ideal gas is one in which interactions between

Summary of Chapter 17 • An ideal gas is one in which interactions between molecules are ignored. • Equation of state for an ideal gas: • Boltzmann’s constant: • Universal gas constant: • Equation of state again: • Number of molecules in a mole is Avogadro’s number:

Summary of Chapter 17 • Molecular mass: • Boyle’s law: • Charles’s law: •

Summary of Chapter 17 • Molecular mass: • Boyle’s law: • Charles’s law: • Kinetic theory: gas consists of large number of pointlike molecules • Pressure is a result of molecular collisions with container walls

Summary of Chapter 17 • Molecules have a range of speeds, given by the

Summary of Chapter 17 • Molecules have a range of speeds, given by the Maxwell distribution • Relation of kinetic energy to temperature: • Relation of rms speed to temperature:

Summary of Chapter 17 • Internal energy of monatomic gas: • Force (“normalkraft”) required

Summary of Chapter 17 • Internal energy of monatomic gas: • Force (“normalkraft”) required to change the length of a solid: • Force (“skjuvkraft”) required to deform a solid:

Summary of Chapter 17 • Pressure required to change the volume of a solid:

Summary of Chapter 17 • Pressure required to change the volume of a solid: • Applied force per area: stress • Resulting deformation: strain • Deformation is elastic if object returns to its original size and shape when stress is removed

Summary of Chapter 17 • Most common phases of matter: solid, liquid, gas (plasma)

Summary of Chapter 17 • Most common phases of matter: solid, liquid, gas (plasma) • When phases are in equilibrium, the number of molecules in each is constant • Evaporation occurs when molecules in liquid move fast enough to escape into gas phase • Latent heat: amount of heat required to transform from one phase to another • Latent heat of fusion: melting or freezing

Summary of Chapter 17 • Latent heat of vaporization: vaporizing or condensing • Latent

Summary of Chapter 17 • Latent heat of vaporization: vaporizing or condensing • Latent heat of sublimation: sublimation or condensation directly between gas and solid phases • When heat is exchanged within a system isolated from its surroundings, the energy of the system is conserved