Chapter 10 liquids and solids
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Chapter 10- Liquids and Solids. Bell Ringer: List and describe the 3 states of matter in molecular terms. Studying Solids, Liquids and Gases. The study of solids, liquids, and gases require an understanding of: The kinetic molecular theory The arrangement of atoms within a molecule.

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Chapter 10- Liquids and Solids

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Chapter 10 liquids and solids

Chapter 10- Liquids and Solids

Bell Ringer:

List and describe the 3 states of matter in molecular terms.


Studying solids liquids and gases

Studying Solids, Liquids and Gases

  • The study of solids, liquids, and gases require an understanding of:

    • The kinetic molecular theory

    • The arrangement of atoms within a molecule.

    • The intermolecular forces of attraction between particles.


Arrangement of atoms within molecules

Arrangement of Atoms Within Molecules

  • Important Factors to Consider:

    • Electronegativity differences

    • Molecular Symmetry


Intermolecular forces dipole dipole attraction

Intermolecular Forces- Dipole-Dipole Attraction

  • When the positive end of one polar molecule is attracted to the negative end of another polar molecule.


Chapter 10 liquids and solids

Intermolecular Forces- Hydrogen Bonding

  • When the positive hydrogen side of one polar molecule is attracted to the negative end of another molecule.


Intermolecular forces london dispersion

Intermolecular Forces- London Dispersion

  • Instantaneous Dipoles that are created by constantly moving electrons.


Comparisons of the three states of matter

Comparisons of the Three States of Matter


Three states of matter shape

Three States of MatterShape

  • Gases have no shape because of little attractive forces and independent movement. Liquids take the shape of their container but do not expand readily because of attractive forces. Solid molecules have definite shape and are held in fixed position.


States of matter density

Section 4 Changes of State

States of MatterDensity

  • Density is mass per unit volume and indicates the closeness of particles in a sample of matter.

  • Gas Liquid Solid

    Low High High


Three states of matter particle energy

Three States of MatterParticle Energy

  • Differences in attractive forces slow down particle movement.

  • Gases- high kinetic energy because of low attraction between particles.

  • Liquids- moderate kinetic energy and attraction

  • Solids- low kinetic energy and high attractive forces.


Chapter 10 liquids and solids

EQ: What does it mean for something to have heat, or be hot?

Temperature is related to the average kinetic energy of the particles in a substance.

KE = mv2

2


Chapter 10 liquids and solids

2. SI unit for temp. is the Kelvin

a. K = C + 273 (10C = 283K)

b. C = K – 273 (10K = -263C)

3. Thermal Energy – the total of all the kinetic and potential energy of all the particles in a substance.


Chapter 10 liquids and solids

Physical Standard for Temperature

“Kelvin”

Triple Point of Water

The temperature and pressure at which water, water vapor, and ice can coexist in equilibrium.

Temperature = 0.01C

Pressure = 4.58 mm Hg


Absolute zero the kelvin scale

Absolute Zero & the Kelvin Scale

The Kelvin scale is setup so that its zero point is the coldest possible temperature--absolute zero, at which point a substance would have zero internal energy. This is -273.15 °C, or -459.69 °F.

Absolute zero can never be reached, but there is no limit to how close we can get to it. Scientists have cooled substances to within 10-5 kelvins of absolute zero. How do we know how cold absolute zero is, if nothing has ever been at that temperature? The answer is by graphing Pressure vs. Temperature for a variety of gases and extrapolating.

P

gas A

A gas exerts no pressure when at absolute zero.

gas B

gas C

T (°C)

0°C

-273.15°C


Chapter 10 liquids and solids

4. Thermal energy relationships

a. As temperature increases, so does thermal energy (because the kinetic energy of the particles increased).

b. Even if the temperature doesn’t change, the thermal energy in a more massive substance is higher (because it is a total measure of energy).


Fahrenheit formula

Fahrenheit Formula

  • On the Fahrenheit scale, there are 180°F between the freezing and boiling points and on the Celsius scale, there are 100°C.

    180°F = 9°F =1.8°F

    100°C 5°C 1°C

  • In the formula for the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0°C to 32°F.

    TF = 9/5 TC + 32

    or

    TF = 1.8 TC + 32


Celsius formula

Celsius Formula

  • TC is obtained by rearranging the equation for TF.

    TF = 1.8TC + 32

  • Subtract 32 from both sides.

    TF - 32 = 1.8TC ( +32 - 32)

    TF - 32 = 1.8TC

  • Divide by 1.8 =°F - 32 = 1.8 TC

    1.8 1.8

    TF - 32 = TC

    1.8


Chapter 10 liquids and solids

Cup gets cooler while hand gets warmer

5. Heat

a. The flow of thermal energy from one object to another.

b. Heat always flows from warmer to cooler objects.

Ice gets warmer while hand gets cooler


Thermal equilibrium

Thermal Equilibrium

Two bodies are said to be at thermal equilibrium if they are at the same temperature. This means there is no net exchange of thermal energy between the two bodies. The top pair of objects are in contact, but since they are at different temps, they are not in thermal equilibrium, and energy is flowing from the hot side to the cold side.

hot

cold

heat

26°C

26°C

No net heat flow

The two purple objects are at the same temp and, therefore are in thermal equilibrium. There is no net flow of heat energy here.


Three states of matter compressibility

Three States of MatterCompressibility

  • Compressibility- the ability to move particles closer together.

  • Gases- Highly compressible

  • Liquids- Slightly compressible

  • Solids- Non-compressible


Three states of matter volume

Three States of MatterVolume


State of matter attractive forces between particles

State of MatterAttractive Forces Between Particles

  • Kinetic energy between solids and liquids is low allowing for dipole-dipole attraction, London dispersion or a crystalline lattice to occur. Attractive forces among gases are almost non-existent.


Three states of matter diffusion

Three States of MatterDiffusion

  • Diffusion- the spontaneous mixing of the particles of two substances caused by their random motion.

  • Gases Liquids Solids

    Quick Slow None


States of matter fluidity

States of MatterFluidity

A fluid is a substance that’s atoms or molecules are free to move past each other and therefore can take the shape of the container.

Gases Liquids Solids

Fluid Fluid Not Fluid


Three states of matter orderliness of particles

Three States of MatterOrderliness of Particles

  • Gases Liquids Solids

    Random Some Order Crystalline Lattice


Change of state terms

Change of State Terms


Examples of change in state

Examples of Change in State


Liquid terms condensation

Liquid TermsCondensation

  • Term that refers to a gas changing to a liquid.


Liquid terms evaporation

Liquid TermsEvaporation

  • Term that refers to the changing of a liquid into a gas.


Liquid terms melting freezing

Liquid TermsMelting/Freezing

  • Melting- the change of a solid into a liquid

  • Freezing- the change of a liquid to a solid


Chapter 10 liquids and solids

Liquid TermsViscosity

Resistance to flow (molecules with large intermolecular forces).


Liquid terms volatile

Liquid TermsVolatile

VOLATILE LIQUID

  • Liquids that evaporate readily and have very weak forces of attraction between particles.


Chapter 10 liquids and solids

Phase changes by Name


Characteristics of liquids surface tension

Characteristics of LiquidsSurface Tension

  • The force that tends to pull adjacent parts of a liquid’s surface together thereby decreasing surface area to smallest possible size. It causes liquid droplets to take a spherical shape.


Characteristics of liquids capillary action

Characteristics of LiquidsCapillary Action

  • The attraction of the surface of a liquid to the surface of a solid.


Bell ringer

Bell Ringer:

  • What is the term for the physical state change for each of the following

    Solid to a gas-Liquid to a gas-

    Gas to a solid-

    Gas to a liquid-

    Liquid to a solid-

Sublimation

Vaporization

Condensation

Deposition

Melting


Solid terms hygroscopic deliquescent

Solid TermsHygroscopic/Deliquescent

  • Hygroscopic- a substance that will capture water molecules from the air and hold them.

  • Deliquescent- a substance that is so hygroscopic that they take up enough water molecules from the air to dissolve and form a liquid.


Solid terms hydrated ions

Solid TermsHydrated Ions

HYDRATED IONS

  • Ions that are chemically bonded to the water molecule.


Solid terms anhydrous compound

Solid TermsAnhydrous Compound

ANHYDROUS COMPOUND

  • A compound in which the water of hydration has been removed.


Solid terms sublimation

Solid TermsSublimation

  • Sublimation

  • The change of state from a solid directly to a gas.


Solid terms deposition

Solid TermsDeposition

  • The change of state from a gas directly to a solid. The formation of frost is a popular example.


Types of solids amorphous solids

Types of SolidsAmorphous Solids

AMORPHOUS SOLID

  • One in which the particle arrangement is random.

  • Examples: glass, plastic


Types of solids crystalline solids

Types of SolidsCrystalline Solids

  • Consists of crystals with an orderly geometric repeating pattern. They have a highly regular arrangement of their components [table salt (NaCl), pyrite (FeS2)].


Ionic crystals electrons transferred

Ionic Crystals Electrons Transferred


Molecular crystals electrons are shared

Molecular CrystalsElectrons are Shared

  • Ammonium persulfate crystals are used as an alternative to traditional ferric chloride solutions for copper etching.


Covalent network diamond each atom is covalently bonded to the nearest atoms

Covalent Network- DiamondEach atom is covalently bonded to the nearest atoms


Metallic bonding electron sea formed by mobile valence electrons

Metallic BondingElectron sea formed by mobile valence electrons


Hydrogen bonding ice crystals

Hydrogen Bonding Ice Crystals


Dynamic equilibrium

Dynamic Equilibrium

  • Dynamic Equilibrium deals with the conditions under which liquid and vapor can coexist. When a system is in equilibrium, two opposing physical changes (ie. Evaporation /condensation will

    occur at equal rates


Liquid vapor equilibrium system

Liquid-Vapor Equilibrium System


Equilibrium vapor pressure

Equilibrium Vapor Pressure

  • The pressure exerted by a vapor that is in equilibrium with its corresponding liquid at a given temperature


Chapter 10 liquids and solids

The graph that shows the pressure and temperature conditions under which the phases of a substance exist is called a Phase Diagram

The line from the triple point to the critical point to the right is the vapor-liquid equilibrium line.


Triple point

Triple Point

  • Indicates the temperature and pressure conditions at which the solid, liquid and vapor of the substance can coexist at equilibrium.


Critical temperature

Critical Temperature

  • The temperature above which the substance cannot exist in the liquid state.


Critical pressure

Critical Pressure

  • Lowest pressure at which the substance can exist as a liquid at the critical temperature


Critical point

Critical Point

  • Indicates the critical temperature and pressure.


Le chatelier s principle

Le Chatelier’s Principle

  • When a system at equilibrium is disturbed by the application of a stress, it attains a new equilibrium position that minimizes the stress.


Based on the graph what phases of water are present for any corresponding pressure reading on

Based on the graph, what phases of water are present for any corresponding pressure reading on:

  • On curve AB?

  • On curve AD?

  • On curve AC?

  • At point A?


Chapter 10 liquids and solids

Given a sample of water on curve AC, what effect would each of the following changes have on that sample?

  • Increasing temperature at constant pressure.

  • Increasing pressure at constant temperature.

  • Decreasing temperature at constant pressure.

  • Decreasing pressure at constant temperature.


Chapter 10 liquids and solids

Given a sample of water on curve AD, what effect would each of the following changes have on that sample?

  • Increasing temperature at constant pressure.

  • Increasing pressure at constant temperature.

  • Decreasing temperature at constant pressure.

  • Decreasing pressure at constant temperature.


Molar heat of fusion molar heat of vaporization

Molar Heat of FusionMolar Heat of Vaporization

  • Molar Heat of Fusion- the amount of heat energy required to melt one mole of a solid at its melting point.

  • Molar Heat of Vaporization- the amount of heat energy needed to vaporize one mole of a liquid at its boiling point.


Heat of fusion heat of vaporization

Heat of Fusion/Heat of Vaporization


Latent heat phase change

Latent Heat & Phase Change

  • A transfer of heat energy from one substance to another does not always result in a temperature change.

  • For instance, ice water on a hot summer day will remain 0˚ C until all of the ice has melted before the temperature begins to rise

  • i.e. energy is lost due to phase change


Latent heat phase change1

Latent Heat & Phase Change

  • The energy needed to change the phase of a given pure substance is called Latent heat.

  • It is dependant on the nature of the substance and the phase change.

    solid  liquid liquid  gas

    333,000J/Kg2,263,300J/Kg

Q = mL

Q = heat (J)

m = mass (Kg)

L = latent heat (J/Kg)


Latent heat phase change2

Latent Heat & Phase Change

Latent heat of Fusion: melting or freezing

Q = mLf

Latent heat of Vaporization: boiling or condensation

Q = mLv

  • Latent heat is an experimental value

    • Dependent on inter & intermolecular forces


Chapter 10 liquids and solids

Latent Heat of Water

Pause for a Cause:

How much heat must be added to a 25g ice cube at 0ºC to change it to water at 0ºC if the latent heat of fusion for water is

3.33 X 10-5J/Kg?

Q = mLf

Q = 25g * 1Kg * 3.33 X 105J

1000g * Kg

= 8.3 X 103J


Chapter 10 liquids and solids

Specific Heat

a. Some things heat up or cool down faster than others.

Land heats up and cools down faster than water


Chapter 10 liquids and solids

b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K).

1) C water = 4184 J / kg C

2) C sand = 664 J / kg C

This is why land heats up quickly during the day and cools quickly at night and why water takes longer.


Chapter 10 liquids and solids

Why does water have such a high specific heat?

water metal

Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.


Chapter 10 liquids and solids

How to calculate changes in thermal energy

Q = mCpΔT

Q = m x T x Cp

Q = change in thermal energy

m = mass of substance

T = change in temperature (Tf – Ti)

Cp = specific heat of substance


Chapter 10 liquids and solids

c. A calorimeter is used to help measure the specific heat of a substance.

First, mass and temperature of water are measured

Knowing its Q value, its mass, and its T, its Cp can be calculated

This gives the heat lost by the substance

T is measured for water to help get its heat gain

Then heated sample is put inside and heat flows into water


Chapter 10 liquids and solids

Pause for a Cause:

Determine the specific heat of water if

8.8 x 105 Joules are lost when 3.00 Kg of water is cooled from 80.0C to 10.0C.

Q = mCpΔT

Cp = __Q__

mΔT

Answer: 4190J

Kg ºC


Chapter 10 liquids and solids

Pause for a Cause:

A metal bolt with a mass of 8.50 x 10-2 kg and a temperature of 85.0 ˚C is placed in a container of water. The mass of the water is 0.150 kg, and its temperature is 25.0˚C. What is the specific heat capacity of the bolt if the final temperature of the bolt and water is 28.4˚C? (c = 4186 J/kg*˚C)

Water

mw= 0.150 kg

Ti = 25.0 x ˚C

Tf = 28.4˚C

Cp = 4186 J/kg*˚C

Q = mCpΔT

Q = 0.15Kg * 4186 J (28.4 ˚C – 25 ˚C)

Kg ˚C

= 2134.86 J

Cp = __Q__

mΔT

Bolt

mb= 8.50 x 10-2 kg

Ti = 85.0 x ˚C

Tf = 28.4˚C

Cp = ?

Cp = ______2134.86 J__________

8.5x10-2 Kg (28.4 ˚C – 85.0 ˚C)

= 443 J

Kg ˚C


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