Gases, Liquids, and Solids - Saddleback College are some molecular compounds solid while others are gaseous and others are liquid at room temperature? Competing Powers • Intermolecular

Gases, Liquids, and Solids

Chapter 5

Educational Goals 1.  Define, compare, contrast the terms specific heat, heat of

fusion, and heat of vaporization. Know the equations that involve these concepts and be able to use them in calculations.

2.  Describe the meaning of the terms enthalpy change, entropy change, and free energy change. Explain if a process is spontaneous or not based on the free energy change.

3.  Convert between pressure units of atm, torr, and psi.

4.  List the variables that describe the condition of a gas and give the equations for the various gas laws.

Educational Goals (cont.)

5.  Explain Dalton’s law of partial pressures.

6.  Define the terms density and specific gravity. Given the density, and either the mass or volume, be able to determine the volume or mass (respectively).

7.  Know that a liquid will boil when its vapor pressure is equal to the atmospheric pressure.

8.  Describe, compare, and contrast amorphous solids and crystalline solids.

9.  Describe the makeup of the four classes of crystalline solids.

Phases of Matter and Energy Example: Three phases of water solid liquid

gas

ice water

steam

H2O(s) H2O(l) H2O(g)

Why are some molecular compounds solid while others are gaseous and others are

liquid at room temperature?

Competing Powers •  Intermolecular forces working to hold

particles together as liquids or solids • Kinetic Energy = Motion = Temperature,

work to separate particles

Kinetic Energy = Temperature

One major factor that is responsible for the varied behavior of solids, liquids, and gases is the nature of the interaction that attracts one particle (atom, ion, or molecule) to another.

What forces hold matter together to make liquids and solids?

The attractive forces that hold molecules together are called Intermolecular forces.

What forces hold matter together to make liquids and solids?

3 Types of Intermolecular Forces

1) Dipole-Dipole Forces 2) Hydrogen Bonding

3) London Forces –  also called Induced Dipole Forces

Other Noncovalent Interactions Noncovalent interactions are interactions that do not involve the sharing of valence electrons (covalent bonding).

Other noncovalent interactions due to the attraction of permanent charges.

•  1) Salt bridges •  2) Ion-dipole interactions

A salt bridge is another name for ionic bond. Ion-dipole interactions occur between ions with

a full charge and atoms with a partial charge.

Energy meets Matter Adding energy to liquids will overcome the forces holding the molecules together– boiling

Adding energy to solids will overcome the forces holding the molecules together– melting

Phase Changes Language

To reduce a fever, rubbing alcohol (2-propanol) can be applied to the skin. As the alcohol evaporates (liquid becomes a gas), the skin cools. Explain the changes in heat energy as this process takes place. Note: 2-propanol vapors are flammable, so care must be taken when using this technique.

Units of Energy •  One calorie is the amount of energy needed

to raise the temperature of one gram of water by 1°C

•  joule –  4.184 J = 1 cal

•  In nutrition, calories are capitalized –  1 Cal = 1 kcal = 1000 cal

60.1 cal = 4.184 J

1 cal 251 J

Converting Between Calories and Joules Example: Convert 60.1 cal to joules

Equivalence statement: 1 cal = 4.184 J

4.184 J 1 cal

1 cal 4.184 J Conversion

Factors

Calculations Involving Heat Energy

•  One of two things will happen if energy is added or removed from matter (assuming no chemical change takes place). –  1) Change the phase of the substance

–  Example: melt, freeze, vaporize (boil)

–  2) Change the temperature of the substance

•  You can only do one of these at a time!!! – See graph on the next slide

Energy calculations for phase changes may be carried out using the tabulated values for: •  heat of fusion (symbol = Hfus) for a substance (Table 5.2).

• Energy required to melt one gram of a solid • Change sign to negative for freezing (liquid to solid)

1) Phase Changes

Energy calculations for phase changes may be carried out using the tabulated values for:

1) Phase Changes

§  heat of vaporization (symbol Hvap) of a substance • Energy required to vaporize one gram of a liquid • Change sign to negative for gas going to liquid

Delta (∆) means “change in ___”

Calculations Involving Changing the Phase

Energy Change = (mass) x (heat of fusion or vaporization)

ΔE = m x (Hfus or vap)

Get from a table

ΔE = (mass) x (heat of fusion)

Calculations Involving Heat Energy

Example: Determine the amount of energy needed to melt 155 g of ice at 0°C, we use the heat of fusion of water (79.7 cal/g) as a conversion factor.

155 g = 1.24 x 104 cal 79.7 cal g

Note: No Temperature Change!

Ice (0oC) → Water (0oC)

x

Group Work A patient with a fever is sponged with 50.0 g of 2-propanol. How much energy is drawn from the patient when 2-propanol vaporizes?

(heat of vaporization for 2-propanol is 159 cal/g)

•  The amount the temperature of an object increases depends on the amount of energy added (Q). –  If you double the added heat energy the temperature

will increase twice as much.

•  The amount the temperature of an object increases depends on its mass. –  If you double the mass it will take twice as much heat

energy to raise the temperature the same amount.

2) Changing Temperature of Matter

Calculations Involving Changing the Temperature

•  Energy calculations may be carried out using the values for the specific heat of a substance.

•  Specific heat is the amount of energy required to raise the temperature of one gram of a substance by one Celsius degree.

C gJ 4.184 is water ofheat specific the,definitionBy °

Energy required = Specific Heat x Mass x Temperature Change

Q = S x m x (ΔT)

∆ is always: (final) – (initial)

(∆T) = Tfinal-Tinitial

Example: Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C

Mass = 7.40 g

Temperature Change (ΔT) = 46.0°C – 29.0°C = 17.0°C

Q = S x m x ΔT

Specific heat of water: 4.184 J g oC

4.184 J g oC

Q = (7.40 g) (17.0 oC) =526 J

Group Work How much energy needs to be removed from 175 g of water to lower the temperature from 23.0oC to 15.0oC ?

How much energy is required to convert 25g of ice at -7.0oC to water at 50.0oC

Ice -7.0oC

Water 50.0oC

Ice 0.0oC

Water 0.0oC

Temperature Change Q1=Sicexmx(∆T)

Step 1 Step 2

Step 3

Phase Change ∆E2=mxHfus

Temperature Change Q3=Swaterxmx(∆T)

∆Energy Total = Q1 + ∆E2 + Q3

Spontaneous vs. Nonspontaneous Changes

•  An important question to ask is why some changes are: –  spontaneous (continue to occur once they are started) OR –  nonspontaneous (will not run by themselves unless

something keeps them going). •  Energy is the key factor in determining this.

New Topic: Will a change occur?

Spontaneous vs. Nonspontaneous

Spontaneous vs. Nonspontaneous Changes

Energy vs. Free Energy The energy (E) of matter depends on the position (potential energy) and velocity (kinetic energy) of every molecule in the system.

E = Epotential + Ekinetic This is not practical to measure in the lab or to model in calculations!

When working at constant temperature and pressure, it is mathematically convenient and experimentally practical to look at the: Free Energy (G)

Energy vs. Free Energy

Just like the energy (E), in nature, given the chance, everything proceeds to the lowest possible free energy (G)!

Free Energy (G)

The “free energy” (ΔG) of a process can be thought of as the potential for change…. ∆G = Gf - Gi

A spontaneous process has a negative ∆G and a nonspontaneous process has a positive ∆G.

How to visualize a gas: Properties of Gases

Gases and Pressures

Gas molecules or atoms are very far apart from one another.

• different from liquids and solids!!

• Gas particles move in a straight line until they collide with another particle or the container wall.

Gases Have Low Density Properties of Gases

Because of the relatively large distances between gas particles, most of the volume occupied by a gas is empty space.

Gases completely fill their container.

Properties of Gases

Except for a few very heavy gases, most gasses will completely fill their container.

Gases Are Highly Compressible Properties of Gases

Gases are compressible

Liquids and Solids are not

Gases Are Highly Compressible Properties of Gases

Compressibility is the ability to make the space a substance takes up become smaller.

Gases can diffuse. Properties of Gases

• Gaseous molecules travel at high speeds in all directions and mix quickly with molecules of gases in the air in a process called diffusion.

• Diffusion is the movement of one substance within another substance until it is evenly distributed.

Examples of diffusion. Properties of Gases

• Ammonia

• Skunk in da house

Gas Pressure

•  Pressure = total force applied to a certain area –  larger force = larger pressure

•  Gas pressure is caused by gas molecules colliding with container walls or surfaces.

Air Pressure •  Constantly present when air present •  Decreases with altitude

–  less air

Air Pressure •  Measured using a barometer

–  Column of mercury supported by air pressure –  Force of the air on the surface of the mercury balanced by the

pull of gravity on the column of mercury

Various Units for Gas Pressure •  1) atmosphere (atm) •  2) height of a column of mercury (mm Hg, in Hg) •  3) Torr •  4) Pascal (Pa) •  6) pounds per square inch (psi, lbs./in2)

Units we will use for pressure: •  Atmospheres (atm) •  Pounds per square inch (psi) •  Millimeters of mercury (mm Hg)

–  also called torr (1mm Hg = 1 Torr) Relationships:

1 atm = 760. mmHg

1 atm = 760. Torr

1 atm = 14.7 psi

760. mm Hg 1 atm

1 atm 760. mm Hg

760. Torr 1 atm

1 atm 760. Torr

14.7 psi 1 atm

1 atm 14.7 psi

Pressure Unit Conversions A pressure of 690. Torr is how many atmospheres?

690. Torr =

1 atm 760 Torr

.908 atm

1 atm = 760 Torr

Group Work A pressure of 35.0 psi is how many atm?

A pressure of 812 mm Hg is how many atmospheres?

1 atm = 14.7 psi

1 atm = 760. mm Hg

Gas Laws

Instructional Goals Understand and be able to use the following gas laws in calculations: •  Boyles Law (relationship between pressure and volume) •  Charles’ Law (relationship between volume and

temperature) •  Gay-Lussac’s Law (relationship between pressure and

temperature) •  Avogadro’s Law (relationship between moles and volume) •  Combined Gas Law (relationship between pressure,

volume and temperature) •  Ideal Gas Law (relationship between pressure, volume,

number of moles, and temperature)

•  The gas laws are the mathematical equations that show the relationship between volume, temperature, pressure, and amount of gas.

The Gas Laws

• As with all laws, they were discovered by experiments.

Boyle’s Law •  Boyle studied the relationship between volume and

pressure.

• The inverse relationship between pressure and volume is known as Boyle’s law.

Boyle’s Law

• When the volume decreases, the pressure increases

low high

Pressure Gage

Boyle’s Law

•  When the volume increases, the pressure decreases

low high

Pressure Gage

Boyle’s Law •  Boyle also noticed that when the pressure and/

or volume of a gas is changed the product of the pressure and volume remains the same.

• PxV = Constant

P1 V1

Initial pressure

Initial volume

= P2 V2

Final pressure

Final volume

Boyle’s Law

• Remember that when using Boyle’s Law, that the temperature is never changing.

• Only the pressure and volume change.

Example

low high

Pressure Gage

The initial volume of the gas in the piston below is 3.00 liters and the initial pressure is 1.00 atm. The piston compressed (at constant temperature) to a new final volume of 1.00 L. What is the final pressure?

(1.00 L)

(3.00 L)

P1 = 1.00 atm P2 = ?

V1 = 3.00 L V2 = 1.00 L

Solution

= 3.00 atm V2

P2 P1

V2

V1 =

V2 P2 P1 V1 =

V2

= (1.00 atm)

P2 P1 V1 = V2

If the syringe shown has an initial volume of 0.50 mL and the gas in the syringe is at a pressure of 1.0 atm, what is the pressure inside the syringe if your finger is placed over the opening and the plunger is pulled back to give a final volume of 3.0 mL?

Group Work

Charles’ Law •  Charles observed that as the temperature increases, the

volume increases and vice versa.

Jacques Charles (1746-1823 )

• The direct relationship between temperature and volume is known as Charles’ law.

Charles’ Law When the temperature increases, the volume increases

Charles’ Law When the temperature decreases, the volume decreases

Charles’ Law •  Charles also noticed that ratio of volume to

temperature of a gas is always the same.

V T = Constant

Charles’ Law

V1

Initial volume

Initial temperature

T2

Final volume

Final temperature

T1

V2 =

Charles’ Law

•  Remember that when using Charles’ Law, that the pressure is never changing. – Only the temperature and volume change.

•  Temperature must be Kelvin (K). – Kelvin temperature scale is always positive – K = oC + 273.15

Example The initial volume of the gas in the piston below is 1.35 liters and the initial pressure is 1.00 atm. The temperature is lowered from 373 K to 250. K (at constant pressure). What is the final volume?

low high

Pressure Gage

(373K)

(1.35 L)

V1 = 1.35 L V2 = ?

T1 = 373 K T2 = 250. K

Solution

= 0.905 L T1

V2 T2

T2

V1 =

V2

T2 T1

V1 = T2

= (250.K)

V2

T2 T1

V1 =

A balloon is inflated to 665 mL volume at 27°C. It is immersed in a dry-ice bath at −79°C. What is its volume, assuming the pressure remains constant?

Group Work

Remember to convert to Kelvin (K)

K = oC + 273.15

V2

T2 T1

V1 =

Gay-Lussac’s Law Gay-Lussac’s observed that as the temperature increases, the pressure increases and vice versa.

Joseph Gay-Lussac (1778–1850)

• The direct relationship between temperature and pressure is known as Gay-Lussac’s Law.

Gay-Lussac’s Law

low high

Pressure Gage

•  When the temperature decreases, the pressure decreases.

low high

Pressure Gage

Gay-Lussac’s Law When the temperature increases, the pressure increases.

Gay-Lussac’s Law •  Gay-Lussac also noticed that ratio of pressure

to temperature of a gas is always the same.

P T = Constant

P1

Initial pressure

Initial temperature

T2

Final pressure

Final temperature

T1

P2 =

Gay-Lussac’s Law

•  Remember that when using Gay-Lussac’s Law, that the volume is never changing. – Only the temperature and pressure change.

•  Temperature must be Kelvin (K).

Example The initial pressure of the gas in the container below is .870 torr and the initial temperature is. 300.K. The temperature is raised from 300. K to 1250. K (at constant volume). What is the final pressure?

low high

Pressure Gage

(300.K)

(0.870 tor)

P1 = 0.870 torr P2 = ?

T1 = 300. K T2 = 1250. K

Solution

= 3.63 torr T1

P2 T2

T2

P1 =

P2

T2 T1

P1 = T2

= (1250. K)

P2

T2 T1

P1 =

An aerosol can containing gas at 25 atm and 22°C is heated to 55°C. Calculate the pressure in the heated can.

Group Work

Remember to convert to Kelvin (K)

K = oC + 273.15

P2

T2 T1

P1 =

The Combined Gas Law

•  Boyles’s, Charles’s, and Gay-Lussac’s Laws can be combined mathematically.

• The relationship between temperature, volume, and pressure is known as the Combined Gas Law.

P1

T2 T1

P2 =

The Combined Gas Law

V1 V2

At an ocean depth of 33 ft, where the pressure is 2.0 atm and the temperature is 285K, a scuba diver releases a bubble of air with a volume of 6.0 mL. What is the volume of the air bubble when it reaches the surface, where the pressure is 1.0 atm and the temperature is 298 K ?

Example

(1.0 atm)

(2.0 atm)

P1 = 2.0 atm P2 = 1.0 atm

V1 = 6.0 mL V2 = ?

T1 = 285 K T2 = 298 K

Solution

= 13 mL P2T1

V2 T2P1V1

=

P2

T2 T1

P1 =

= (298 K)

V2 V1

(6.0 mL) (285K)

Avogadro’s Law Avogadro’s observed that the volume of a gas is directly proportional to the number of gas molecules.

Amedeo Avogadro (1776–1856)

• The direct relationship between moles of gas molecules and volume is known as Avogadro’s Law.

Avogadro’s Law When the number of moles of gas decreases, the volume decreases.

low high

Pressure Gage

Avogadro’s Law When the number of moles of gas increases, the volume increases.

low high

Pressure Gage

Avogadro’s Law •  Avogadro noticed that ratio of volume to

the number of moles of a gas is always the same.

V n = Constant

V1

Initial volume

Initial # moles n2

Final volume

Final # moles n1

V2 =

Avogadro’s Law

• Remember that when using Avogadro’s Law, that the pressure and temperature are never changing. – Only the number of particles and

volume change.

Example The initial volume of the 3.5 moles of gas in a container is 1.5 L.

Amadeo adds 2.0 moles of gas. (at constant temperature and pressure). What is the final volume?

(5.5 mol)

(3.5 mol)

(1.5 L)

V1 = 1.5 L V2 = ?

n1 = 3.5 mol n2 = 5.5 mol

Solution

= 2.4 L n1

V2 n2

n2

V1 =

V2

n2 n1

V1 = n2

=

V2

n2 n1

V1 =

A balloon has a volume of 2.4 L and contains 0.12 moles of air. A child blows more air into the balloon until it has a final volume of 3.5 L. How many moles of gas are in the balloon?

Group Work

Gas Law Summary

V1

n2 n1

V2 =

P2

T2 T1

P1 = V2 V1 Combined Gas Law

The Ideal Gas Law No gas perfectly obeys all four of these laws under all conditions.

These assumptions work well for most gases and most conditions.

One way to model a gas’s behavior is to assume that the gas is an ideal gas that perfectly follows these laws.

The Ideal Gas Law

P x V = Cb V T = Cc

P T = Cg

V n = Ca

If we combine all these equations, we get the Ideal Gas Law.

PV nT

= R Gas Constant

PV nT

= R

The Ideal Gas Law The gas constant (R) is a mathematical combination of all the individual gas law constants (Cb, Cc, Cg, Ca)

The Ideal Gas Law is more commonly written as:

PV = nRT

The Ideal Gas Law The previous gas laws we studied involved a change in either P, V, T, or n.

V2

n2 n1

V1 =

P2

T2 T1

P1 =

V2

T2 T1

V1 =

V2 P2 P1 V1 =

The Ideal Gas Law

• The ideal gas law is used for any gas system, any time.

• No changes are involved in the equation

PV = nRT

The Ideal Gas Law The value of R is:

• When using this equation you must have the following units:

• Pressure = atm

• Volume = liters

• Temperature = K

PV = nRT

.0821 L.atm K.mol

The Ideal Gas Law There are 4 variables in this equation:

Pressure

PV = nRT Volume # Moles Temperature

In problems, we will always be given 3 of the 4 variables, then solve for the unknown variable.

Example: The Ideal Gas Law How many moles of gas are contained in 11.2 liters at 1.00 atm and 0.0°C?

PV = n P = _____

V = _____

n = _____

T = _____

1.00 atm

11.2 L

???

273.2 K RT

Partial Pressure

•  Dalton’s law of partial pressure states that the total pressure of a mixture of gases is the sum of the partial pressures of its components. – The partial pressure of a gas in a mixture is

the pressure that the gas would exert if alone.

PT = PA + PB + PC Total

pressure Partial pressure

of gas A Partial pressure

of gas C

Partial pressure of gas B

When two gases are present, the total pressure is the sum of the partial pressures of the gases.

Partial Pressures

VT x R x n P P P

n n nsame theare mixture in the

everything of volumeand re temperatutheV

T x R x n P V

T x R x n P

mixture ain B andA gasesfor

totalBAtotal

BAtotal

BB

AA

=+=

+=

==

The partial pressure of each gas in a mixture can be calculated using the Ideal Gas Law

A 1.00L flask contains 5.00 x 10-2 mol of neon and 5.00 x 10-3 mol of argon. At 30.0 °C, what is the partial pressure of each gas in atmospheres and what is the total pressure?

Let’s Try It!

Liquids Viscosity is the resistance to flow.

-  It is related to the strength of the noncovalent interactions between the molecules that make up the liquid - the stronger the attractions, the thicker the liquid.

-  Temperature has an effect on viscosity.

-  As temperature rises, the increase in the kinetic energy of the molecules in the liquid helps the molecules pull away from one another - higher temperature produces lower viscosity.

Glycerol is able to form more hydrogen bonds than 2-propanol. That is why glycerol is thicker (more viscous) than 2-propanol.

Density of a liquid (or any other substance) is the amount of mass contained in a given volume.

d = m V

Density is the relationship between the mass of a substance and its volume.

Due to collisions that take place between particles (atoms or molecules) that make up a liquid, particles at the surface are continually evaporating - being “bounced” off into the gas phase. At the same time gas phase molecules are being trapped and converted to liquid.

Vapor Pressure

•  The boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the atmospheric pressure.

•  Liquids boil when their vapor pressure equals the pressure of the air above them.

Vapor Pressure

•  The atoms, ions, or molecules that make up a solid are held close to one another and have a limited ability to move around.

•  Solids can be classified based on whether

or not the arrangement of these particles is ordered (in crystalline solids) or not (in amorphous solids).

Solids

Crystalline Solids •  Ionic

–  consist of oppositely charged ions held to one another by ionic bonds

•  Molecular –  consist of an ordered arrangement of molecules

attracted to one another by noncovalent interactions •  Covalent Networks

–  atoms are held to one another by an arrangement of covalent bonds that extends through the solids.

•  Metallic – An array of metal cations immersed in a cloud of

electrons that spans the entire crystalline structure.

Metallic Bonding

•  The valence electrons in metals are free to move about the entire crystal of metal nuclei and core electrons.

Metallic Solids

•  We can imagine it like a “sea of electrons” that are bonding the positive nuclei together

Properties of Metallic Substances Metallic substances are solid at room temperature. Except for :_______________________

Metallic substances are malleable (they can be hammered or beaten in thin sheets)

Metallic substances are ductile (they can be drawn, pulled, or extruded through a small opening to produce wire.

Metallic substances are good conductors of electricity.

A few substances exist as: Covalent Networks

•  Atoms are covalently bonded as if it was a huge molecule

•  Not too many covalent network substances exist •  Examples: Diamond (carbon)

Amorphous Solids- no regular repeating pattern of ions or molecules.

Example: rubber

Solids: Summary a)  Ionic b)  Covalent c)  Molecular d)  Metallic e)  Amorphous