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Lecture Outlines Chapter 17
Physics, 3rd Edition James S. Walker
Chapter 17
Phases and Phase Changes
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
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.
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 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, the pressure increases:
17-1 Ideal Gases
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 are atoms in 12 g of carbon-12.
Example 17-1 Take a Deep Breath
Syre är 21% av luftmolekylerna. Hur många O2-molekyler finns i lungorna?
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 kPa). Hur många syremolekyler finns i lungorna? Luft kan behandlas som en ideal gas.
PV = NkT ger direkt totala antalet molekyler N = 101 kPa•6 10-3 m3/1,38 10-23 J/K•310 K =
= 1,42•1023 (stycken)
Antalet syremolekyler = 3,0•1022 (stycken)
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 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 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 luftmolekyler finns det i en
basketboll? Trycket är 171 kPa, T = 293 K och bollens diameter är 30,0 cm. Luft kan behandlas som en ideal gas.
PV = nRT ger direkt antalet mol eftersom V = 4πr3/3 = 0,01414 m3
n = 171 kPa•0,01414 m3/[8,3145 J/(mol•K)•293 K] =
= 0,993 (mol)
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 (O2). 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•16,00 g/mol /N = 5,314•10-23 g
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, 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
När trycket är 130 kPa är höjden i pistongen 25 cm. När man sedan placerar en större massa på locket stiger trycket till 170 kPa och höjden
ändras till h2, medan temperaturen (T = 290 K) är konstant. Beräkna den nya höjden.
PV = konstant och V = A•h ger h2 = h1•P1/P2 = 19 cm
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 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
När trycket är 130 kPa är höjden i pistongen 25 cm. När man sedan ökar temperaturen från 290 K till 330 K ändras höjden till h2 (medan trycket är konstant). Beräkna den nya höjden.
V/T = konstant och V = A•h ger h2 = h1•T2/T1 = 28 cm
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
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
När Δpx = pf,x – pi,x = 2 mvx Δt = 2L/vx
F = Δpx/Δt = mvx2/L
Nu erhålls 17-9 enkelt
P = F/A = (mvx2/L)/L2 = mvx2/L3 = mvx2/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 x- komponent och resultatet blir detsamma.
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 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(vx2 )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 (vx2)av = (vy2)av = (vz2)av
(vx2 )av = v2av /3
P = N/V • mv2av /3
P = 2/3 • (N/V) • (mv2)av/2 = 2/3 • (N/V) • Kav 17-11
17-2 Kinetic Theory
Comparing this expression with the ideal gas law (PV=NkT) allows us to relate average
kinetic energy and temperature:
The square root of is called the root mean square (rms) speed.
17-2 Kinetic Theory
Solving for the rms speed gives:
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
Luft består huvudsakligen av kvävemolekyler
(78%) och syremolekyler. Är rms-hastigheten för N2 (28,0 g/mol) >=< rms hastigheten för O2 (32,0 g/mol)? Beräkna hastigheterna vid T = 293 K.
vrms = (3kT/m)1/2 = [R = NAk, M= mNA] så att
vrms = (3RT/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 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 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 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•1010 N/m2 F = Y•A•(ΔL/L0)
ΔL = (F•L0 )/(Y•A) = 8,2 μm
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/L0)= (YA/L0)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 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 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
F = 1,2 N d = 13 cm
Δx = 2,5 cm L0 = 9,0 cm A = π•d2/4
F = S•A•(Δx/L0)
S= (F•L )/(A•Δx) = 330 (325,47) N/m
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 bulk modulus.
Active Example 17-4 A Gold Doubloon: Find the Change in Volume
t(jockleken) = 2,0 mm d(iametern) = 6,1 cm A = π•d2/4
B = 22 •1010 N/m2
Ekvation 15-7 ger direkt (gaugetrycket) ΔP = ρ•g•h (h = 770 m)
Ekvation 17-19 ger då direkt
ΔV = - V0•ΔP/B = - A•t•ΔP/B = - 2•10-10 m3
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)
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 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 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 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 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 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 i boken)
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 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 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 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 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 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 inte isen utan går direkt i gasfas (sublimerar) och vice versa.
Example 17-5 Steam Heat
Example 17-5 Steam Heat
För att få (vatten)ånga tillför man 560 kJ 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 kJ För att förånga 0,220 kg vattenånga åtgår Q = m•L = 0,220 kg•2,26 MJ/kg = 497 kJ
Den återstående värmemängden (17 kJ) används för att höja ångans temperatur (Tabell 16-2)
ΔT= Q/(mc )=17 kJ/(0,220 kg•2010(J/kg•K)= 38,4°C
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
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 kJ
För att smälta 0,0450 kg is åtgår värmemängden Q = mi•L = 0,0450 kg•335 kJ/kg = 15,1 kJ
Den återstående värmemängden bestämmer
sluttemperaturen för den ”nya” (större) mängden vatten ΔT= Q/(m+mi)c = (330,7 -15,1)kJ/(3,995 kg•4186 {J/kg•K}) = 18,9°C
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:
• 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 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 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:
• 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)
• 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 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
