Chapter 15 Gases, Liquids, and Solids

www.cengage.com/chemistry/cracolice

Mark S. CracoliceEdward I. Peters

Mark S. Cracolice • The University of Montana

Chapter 15Gases, Liquids, and Solids

Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures

The partial pressure of a component in a gas mixture is the pressure that component would exert if it alone occupied the same volume at the same temperature.

The total pressure exerted by a mixture of gases is the sum of the partial pressures.

Ptotal = p1 + p2 + p3 + p4 + - - - -

Dalton’s Law of Partial PressuresExample:The partial pressures in a mixture of gases are:Helium, 344 torrNeon, 298 torrArgon, 109 torrWhat is the total pressure?

Solution:P = p1 + p2 + p3 = pHe + pNe + pAr

= (344 + 298 + 109) torr = 751 torr

Laboratory Collection of Oxygen• A mixture of oxygen and water vapor are collected at 220C, at

which water vapor pressure is 19.8 torr. If the total pressure of the mixture is 755 torr, what is the oxygen pressure?

Laboratory Collection of OxygenPtotal = Patm when the water levels are equalized.Poxygen = Ptotal – Pwater vapor = 755 torr – 19.8 torr = 735 torr

Properties of LiquidsComparison of Properties of Liquids

with Properties of Gases

1. Gases may be compressed; liquids cannot.2. Gases expand to fill their containers; liquids do not.3. Gases have low densities; liquids have relatively high

densities.4. Gases may be mixed in a fixed volume; liquids cannot.

Properties of Liquids

Gas particles are far apart.Forces between the particles are negligible.

In a liquid, the particles are very close to one another. The intermolecular attractions in a liquid are strong.

Properties of Liquids: Vapor Pressure

The partial pressure exerted by a vapor in equilibrium with its liquid phase is the vapor pressure.

The stronger the intermolecular forces, the lower the vapor pressure.

The vapor pressure is higher at higher temperature.

Properties of Liquids: Vapor Pressure

Properties of Liquids: Heat of Vaporization

The energy required to vaporize one gram of substance is called heat of vaporization (the unit is kJ/g)

The energy required to vaporized one mole is called molar heat of vaporization (the unit is kJ/mol).

The stronger the intermolecular attraction, the higher the molar heat of vaporization.

.

Properties of Liquids: Heat of Vaporization

Properties of Liquids: Boiling Point

Boiling point is the temperature at which the vapor pressure of a liquid is equal to the external pressure

above its surface.

The temperature at which the vapor pressure is equal to one atmosphere is called the normal boiling point.

Properties of Liquids: Boiling Point

Properties of Liquids: Boiling Point

In order for a stable bubble to form in a boiling liquid, the vapor pressure within the bubble must be high enough to push back the surrounding liquid and the atmosphere above the liquid

The boiling point of a substance increases with the surrounding pressure.

Properties of Liquids

Properties of Liquids: Viscosity

ViscosityA measure of the resistance of a liquid to flow.

The stronger the intermolecular attractive forces, the greater the resistance to flow, so the higher the

viscosity.

Properties of Liquids: Viscosity

A measure of the resistance of a liquid to flow.water<syrup<honey

Properties of Liquids: Surface tension

Surface TensionThe tendency for a liquid to form the

minimum possible surface area.

The stronger the intermolecular attractive forces, the stronger each particle is attracted in all directions by

the particles around it, so the higher the surface tension.

Properties of Liquids: Surface Tension

Molecules at the surface possess a higher potential energy than those within the body of the liquid. To minimize this energy, a liquid tends to have minimum surface.

Three Kind of Intermolecular Forces

Dipole forces

Induced dipole forces

Hydrogen bonds

Of the three kinds of intermolecular attractions, hydrogen bonds are the strongest. Dipole forces are

next, and dispersion forces are the weakest.

Types of Intermolecular Forces

Dipole ForcesPolar molecules have permanent dipole moments.

The dipole force is the force of attraction between the positive pole of one molecule and the negative pole of another molecule.

Types of Intermolecular Forces

Types of Intermolecular Forces

Types of Intermolecular Forces: Induced Dipole Forces

Induced Dipole: The presence of an external electric field may cause an induced, or temporarily, polarization of the molecule.

Molecules in which a dipole can be induced in this way are said to be polarizable.

Attractions between induced dipoles are also called dispersion forces.

Types of Intermolecular Forces: Induced Dipole Forces

Attractive forces between instantaneous electric dipoles on neigbboring molecules

Types of Intermolecular Forces Induced Dipole Forces

Large molecules are more easily polarized than small molecules. Induced dipole forces increase

with molar massBr2 molecules have 70 electrons.Bromine (left) is a liquid.Iodine, I2 , molecules have 106 eIodine (right) is a solid.

.

Types of Intermolecular Forces Hydrogen bonds

The covalent bond between hydrogen atom and atom of nitrogen, oxygen, or fluorine is strongly polar.

Types of Intermolecular Forces: Hydrogen bonds

Hydrogen bonds: The attractive forces between molecules that have a hydrogen atom covalently

bonded to nitrogen, oxygen, or fluorine atom. Hydrogen atom lies between two small strongly electronegative atoms with lone pairs of electrons.

A hydrogen bond is an intermolecular force, an attraction between hydrogen atom of one molecule

with the negative pole of another molecule.

Types of Intermolecular Forces: Hydrogen bonds between H2O molecules

A hydrogen bond is denoted by dots to distinguish it from a covalent bond.

The Solid State

Solid substances can be classified based onhow the particles are arranged in the solid:

Crystalline SolidParticles are arranged in a geometric pattern that

repeats itself over and over in three dimensions.

Amorphous SolidNo long-range ordering of the particles in the solid.

The Solid State

The Solid State

Crystalline Solid and amorphous solid

Types of Crystalline Solids

Crystalline solids can be divided into four classeson the basis of the type of forces that

hold particles together in the crystal lattice:

Ionic crystalsMolecular crystals

Covalent network solidsMetallic crystals

Types of Crystalline Solids: Ionic Crystals

Ionic CrystalsIonic crystals have ions at the points of the lattice that describes the structure of the solid.

Examples are sodium chloride, calcium carbonate.

Ionic Crystals: NaCl

Ionic Crystals: CaCO3

Molecular CrystalsMolecular Crystals

A molecular solid has covalently bonded molecules at each of its lattice points.

Examples are sugar, I2, S8

Molecular Crystal of Sulfur

Molecular Crystal of S8 and liquid form of sulfur

Covalent Network Solids

Covalent Network Solids

The atoms bond to each other with strong directional covalent bonds that lead to giant

molecules, or networks, of atoms.

Examples are diamond, graphite, quartz.

Diamond

Covalent Network Solids: Quartz

Covalent Network Solids: Quartz

Types of Crystalline Solids: Metallic Crystals

Metallic CrystalsA crystal of positive ions through which

valence electrons move freely.

Examples: Al, Cu, Fe

Metallic Crystal

Metal ions in an electron sea

Metallic Crystal: Aluminum Metal

Properties of Crystalline Solids

Energy and Change of State

Energy and Change of State: Vaporization

To boil a liquid (vaporization), heat must be added.

For a pure substance at its boiling point, the amount of heat added to vaporize the liquid is proportional to

the mass of the liquid:

q = ∆Hvap x m

∆Hvap is the enthalpy of vaporization or the heat of vaporization.

Energy and Change of State

Energy and Change of StateExample:How much energy is required to vaporize 19.6 g of water at

100°C?

Solution:Solve with algebra.

GIVEN: 19.6 g; ∆Hvap = 2.26 kJ/g (Table 15.4)WANTED: Energy (assume kJ)

kJ 44.3 = g

kJ 2.26 g 19.6 = H m = q vap

Energy and Change of State

To melt a solid (fusion), heat must be added.

For a pure substance at its melting point, the amount of heat added to melt the solid is proportional to the

mass of the solid:

q = ∆Hfus x m

∆Hfus is the enthalpy of fusion or the heat of fusion.

Energy and Change of StateExample:How much energy is required to melt 19.6 g of ice at 0°C?

Solution:Solve with algebra.

GIVEN: 19.6 g; ∆Hfus = 333 J/g (Table 15.4)WANTED: Energy (assume kJ)

J 10 6.53 = g

J 333 g 19.6 = H m = q 3fus

Energy and ∆T: Specific Heat

Experiments indicate that the heat flow, q, in heating orcooling a substance is proportional to the mass, m, and

its temperature change, ∆T

q = m x c x ∆T

C is the specific heat of the substance

Energy and ∆T: Specific Heat

Specific Heat, cThe heat flow required to change the temperature

of one gram of a substance by one degree Celsius.

A substance with a low specific heat gains little energy in warming through a temperature change, as compared with a

substance with a higher specific heat.

Energy and ∆T: Specific Heat

Energy and ∆T: Specific HeatExample:How much energy is required to change the temperature of

25.0 g of water from 10.0°C to 50.0°C?

Solution:Solve with algebra.

GIVEN: 25.0 g; c = 4.18 J/g • °C (Table 15.5)WANTED: Energy (assume J)

Changes in Temperature & StateIf you steadily apply heat to a pure substance

in the solid state, five things will happen:

1. The solid will warm to its melting point temperature.2. The solid will change to liquid at the melting point

temperature.3. The liquid will warm to its boiling point temperature.4. The liquid will change to gas at the boiling point temperature.5. The gas will become hotter.

Changes in Temperature & State

Changes in Temperature & StateExample:Calculate the energy that must be added to 19.6 g of ice at–12°C to change it to steam at 115°C. Answer in kJ.

Solution:Step 1 is to sketch a graph of temperature vs. heat flow.

Changes in Temperature & StateStep 2 is to calculate the heat flow for each step.

A to B: Solid at –12°C to solid at 0°C

GIVEN: 19.6 g; c = 2.06 J/g • °C (Table 15.5)WANTED: Energy in kJ

kJ 0.48 = J 1000

kJ 1 C(—12)]— [0 C gJ 2.06 g 19.6 B)Q(A

Changes in Temperature & StateB to C: Solid at 0°C to liquid at 0°C

GIVEN: 19.6 g; ∆Hfus = 333 J/g (Table 15.4)WANTED: Energy in kJ

kJ 6.53 = J 1000

kJ 1 g

J 333 g 19.6 C)Q(B

Changes in Temperature & StateC to D: Liquid at 0°C to liquid at 100°C

GIVEN: 16.9 g; c = 4.18 J/g • °C (Table 15.5)WANTED: Energy in kJ

kJ 8.19 = J 1000

kJ 1 C0)— (100 C gJ 4.18 g 16.9 D)Q(C

Changes in Temperature & StateD to E: Liquid at 100°C to gas at 100°C

GIVEN: 19.6 g; ∆Hvap = 2.26 kJ/g (Table 15.4)WANTED: Energy in kJ

kJ 44.3 = g

kJ 2.26 g 19.6 E)Q(D

Changes in Temperature & StateE to F: Gas at 100°C to gas at 115°C

GIVEN: 19.6 g; c = 2.00 J/g • °C (Table 15.5)WANTED: Energy in kJ

kJ 0.59 = J 1000

kJ 1 C100) - (115 C gJ 2.00 g 19.6 F)Q(E

Changes in Temperature & StateStep 3 is to add the heat flows.

qtotal = qA to B + qB to C + qC to D + qD to E + qE to F =

0.48 kJ + 6.53 kJ + 8.19 kJ + 44.3 kJ + 0.59 kJ =

60.1 kJ