Ch10.4 Attractive Forces

• Intermolecular Forces are the forces holding molecules to each other.

• Solids have strong forces

• Gases (vapor) have weak forces

Intermolecular forcesdetermine the phase of matter.

Intermolecular Forces

• Types:

1. Dipole – Dipole Attraction are strong. (ex. Hydrogen bonding)

2. Dispersion forces are weaker (nonpolar)

STRONG INTERMOLECULAR FORCES

• High melting and boiling points– Higher molar masses

• Low vapor pressure

• Nonvolatile Substances

• High viscosity

• High surface tension

WEAK INTERMOLECULAR FORCES

• Low melting and boiling points

–Lower molar masses

• High vapor pressure

• Volatile Substances

• Low viscosity

• Low surface tension

Solids

• Definite volume

• Definite shape

• Orderly arrangement of particles

• Particles constantly vibrating

Liquids

• Definite volume

• no definite shape

(takes shape of container)

• Difficult to compress

• disorderly arrangement of particles

• Flowing motion of particles

Liquid Properties

• Viscosity- the resistance of a fluid to flow

•Thick fluids have high viscosity

•Ex. Syrup

Liquid Properties

• Surface Tension- Ability of liquid molecules to hold on to each other.

Ch. 10.5Changing States

liquid Gas/vaporsolid

Changes of State

Exothermic Process

Endothermic Process

Melting and Freezing Points

• Temperature which substances freeze and melt is the same.

• Each substance has an unique Melting/freezing point

Evaporation

• Conversion of a liquid to a vapor below its boiling point.

• It occurs only at the surface.

• Remember the difference between a vapor and gas.

–Vapor is normally a liquid or solid at room temperature

Ch 11.5 Vapor Pressure

• Vapor pressure measures how easily a liquid changes into vapor

• Liquids with high vapor pressures turn into vapors very easily. (Volatile liquids)

Ex. Gasoline, perfume

Once equilibrium is reached, the vapor particles will begin

to condense back to a liquid at the same rate they change into a vapor.

Dynamic Equilibrium

Vapor Equilibrium reached

Ch. 10.5, 11.5 Boiling Point

• The temperature at which the vapor pressure of the liquid equals the atmospheric pressure

• The entire liquid is changing state, not just the surface.

• Liquids with low boiling points are considered volatile

Difference between Evaporation and Boiling

Super Heated Water

Ch. 10.5 Condensation

• The changing of a gas/vapor to a liquid

Distillation

• A method of separating the substances of a mixture with different boiling points.

• Used in desalinating sea water.

Heating/Cooling Curve

• Used to show how much enthalpy energy (Heat transfer) is needed to change phase.

• Enthalpy (heat) of Fusion- energy required to change from solid to liquid

• Enthalpy (heat) of Vaporization- energy required to change from liquid to vapor/gas

Enthalpy of Fusion

Enthalpy of Vaporization

Heating Curve

Sublimation• Process where solid goes directly to a

gas (vapor), because the vapor pressure is so high, liquid phase does not exist.

• Ex. Iodine, Dry Ice

• Deposition is a gas back to solid

Ch 11.1 Gases

• Definite volume

• Definite shape

(takes shape and volume of container)

• Easy to compress

• Particles are far apart

Ch. 11.1 Kinetic Theory of Gases

•Describe Ideal Gas Behavior

1.Gases consist of tiny particles that move randomly at high velocities.

2.All collisions between gas molecules are perfectly elastic

Kinetic Theory of Gases

3.There are no attractive forces between gas particles.

4. The KE of gas molecules is proportional to the Kelvin temperature.

Real Gas Behavior

• At high pressures and low temperatures, gases can not behave ideally.

–Particles will attract

• Nonpolar gases, and gases will little mass, almost always behave ideally

Temperature

• The higher the temperature the greater the particle speed.

• Temperature helps measure kinetic Energy

• SI base unit is Kelvin (K)

Converting from Celsius

• 0 degree Celsius is equal to 273 K

• K = 0C + 273

• 0C = K – 273

• Convert 191 K to Celsius

• 0C = 191 – 273 = -82 0C

Ch 11.2 Gas Pressure

– The result of simultaneous collisions of billions of gas particles with an object.

– The more collisions, the greater the pressure

Vacuum

•A controlled condition where no gas particles are present.

•So no gas pressure can exist.

Atmospheric Pressure

•Results from collisions of air molecules with objects.

• Decreases as you climb a mountain because the air thins out at higher elevations

•Measured by a barometer

Measuring Pressure

•STP (Standard Temp. Pressure)

•Standard Temperature at sea level is 00C or 273 K

•Standard Pressure is 101.3 kPa, 760 torr, 760 mm Hg, or 1 atm

Pressure Conversions

• How many kPa’s are in 1.50 atm?

1 atm = 101.3 kPa

1.50 atm x 101.3kPa =

1 atm

• How many kPa’s are in 690 mm Hg?

690 mm Hg x 101.3 kPa =

760 mm Hg

152 kPa

92 kPa

Ch 11.3 Gas Relationships

• Relationships between pressure, volume, temperature and number of moles (amount of particles)

• While examining relationships, two measurements will always be constant (unchanged)

Pressure vs Volume

Real Gas

Pressure vs Volume

• Boyle’s Law P1V1=P2V2

• For a given mass of gas at a constant temperature, the volume of the gas varies inversely with pressure.

• Pressure increases, volume decreases

• Reason:

Practice Problem

• You had a gas that exerted 202 kPa of pressure and took up a space of 3000.0 mL. If you decided to expand the tank to 7.00 L, what would be the new pressure? (Assume constant temperature)

• P1V1=P2V2

• Check units

• 202 kPa x 3.0000 liters = P2 x 7.00 liters

• 606 kPa L = P2 x 7.00 liters

• P2 = 86.6 kPa

Peeps and Boyle’s Law

BOYLES LAW: DP and DV

Ideal gasCh 11.4

Temperature vs Volume

• Charles Law- V1/T1=V2/T2 or V1T2=V2T1

• For a given mass of gas at a constant pressure, the volume of the gas varies directly with its Kelvin temperature.

• Temperature increases, volume increases

• Reason:

Practice Problem• If you took a balloon outside that was at

20.00C at 2.0 liters and heated up to 29.00C, what volume would the balloon occupy now? (Assume constant pressure)

• V1/T1=V2/T2

• Check units(Remember KELVIN)

• 2.0 L / 293 K = V2 / 302 K

• 604 L K= V2 293 K

• V2 = 2.1 L

FUN WITH ANIMAL BALLOONS

Ch.11.5Temperature vs Pressure

• Gay-Lussac’s Law - P1/T1=P2/T2 or P1T2=P2T1

• For a given mass of gas at a constant volume, the pressure of a gas varies directly with its Kelvin temperature.

• Temperature increases, pressure increases

• Reason:

Ostrich Egg in Microwave

Ch. 11.6 Combined Gas Law

• Combines all three gas laws into one expression. (only moles is constant)

Practice Problem

• You have a 2.0 liter balloon that was at 20.00C and 1.5 atm. If you take this balloon and place it in a room at STP conditions, what volume would the balloon occupy?

• P1V1/T1=P2V2/T2(Remember KELVIN)• 1.5 atm x 2.0 L / 293 K = 1atm x V2 / 273 K

• 819 atm L K = V2 293 K x 1 atm

• V2 = 2.8 L

Ch. 11.7 Avogadro’s Law

• 1 mole of gas at STP= 22.4 liters of any gas

• Amount (moles) is directly proportional to the space occupied.

• Density of a gas is measured at STP.

– Molar mass / molar volume

Practice Problem

• How many liters of Hydrogen gas are in 6.2 grams of hydrogen gas at STP?

• Molar of mass of is H2 2 gram/mole

• 6.2 grams H2 x 22.4 L H2/ 2 gram of H2

• 69 L of H2

Moles (amount) vs Temp

• moles increases, temp. decreases

• Inverse relationship

• Reason:

• Ex. Compress tanks become colder as you fill them

Ideal Gas

Moles (amount) vs pressure

• moles increases, pressure increases

• Reason:

• Ex. Think of a super soaker or simply filling your tire

Ch. 11.8 Ideal Gas Law

•PV = nRT

•R = constant 0.0821 (L*atm)/(mol*K)

8.31 (L*kPa)/(mol*K)

Practice Problem

• A propane tank holds 3000. g of C3H8. How much larger a container would be needed to hold the same amount of propane if the gas is at 250C and a pressure of 3.0 atm?

Solution

• PV=nRT.

• Check for any units to convert

• First solve for n

• 3000. g of C3H8. x 1 mole =

44 grams C3H8

• V*3.0 atm= 68.18 moles x 0.0821x298K

• V = 560 L of propane

68.18 moles

Practice Problem

•2.0 grams of Nitrogen gas is kept under a pressure of 722 torr, and a temperature of 30.00C. What is the density of the gas under these conditions?

Solution• PV=nRT.• Convert units first• 2.0 g of N2 x 1 mole =

28 grams N2

• 720 torr x 1 atm/760 torr = 0.95 atm

• V*0.95 atm= 0.071 moles x 0.0821x303K• = 1.9 L of nitrogen

• D=m/v• D =2.0 g / 1.9 L = 1.1g/L of nitrogen

0.071 moles

Ch. 11.9 Gas Laws and Reactions

• Gas Stoichiometry

• Most check to see if reaction is occurring at STP or not.

• If not, you most use the Ideal Gas Law

• Examples on page 378-379

Ch. 11.10

Dalton’s Law

• Law deals with mixtures of gases.

–SCUBA tanks

LAST LAW!I PROMISE

(ch18)

Graham’s Law of Diffusion

•Diffusion is the random scattering of gas molecules.

–The longer they diffuse the more evenly distributed they will become in the container.

•The heavier the gas the slower the rate of diffusion.

Gas Law Review