chem unit7

States of Matter

Particle vibration fluid motion rapid, random motionRigid positions move past independent of each other

each otherFixed volume fixed volume volume of container

Phases and Transitions

Sublimation

Condensation

EvaporationMelting

Freezing

Intermolecular Forces

• Strongest = IONIC FORCES

• High melting points• Oppositely charged ions

Dipole-Dipole Forces

• Dipole: contains both positively and negatively charged regions:

Hydrogen Bonding

• Special case of dipole-dipole forces

• Causes water to have higher than expected boiling point

Water Properties

• Surface tension

Water Properties

• Capillary action

Boiling PointsMolecule Boiling

Point

H2O 100°C

H2S –60.7 °C

H2Se –41°C

H2Te –2°C

Hydrogen Bonding

• Weak bond between H and electronegative atom:

• Recall

Hydrogen Bonding

Hydrogen Bonding

• DNA

Dipole-Dipole Forces

London (Dispersion) Forces

• Weak attractions between non-polar

• Increases with size of molecule (number of electrons)

• Molecular Shape (less compact > compact)

> >

London (Dispersion) Forces

• Weak attractions between non-polar

• “Temporary” or instantaneous dipoles

Intermolecular Forces

IONIC >> Dipole – Dipole >> London

Intermolecular Forces

• Which intermolecular force is expected?

CH4

Intermolecular Forces

• Which intermolecular force is expected?

CH4

Non-polar covalent molecule

London (dispersion) forces

Intermolecular Forces

• Which intermolecular force is expected?

Butanol

CH3CH2CH2CH2OH

Intermolecular Forces

• Which intermolecular force is expected?

Butanol

CH3CH2CH2CH2OH

Dipole-Dipole

Energy of Phase Changes

• Energy is required for all phase changes

Phase Diagram (H2O)

Phase Diagram (CO2)

Triple Point

Triple Point: where all three phases co-exist (T, p)

Phase Diagrams (carbon)

Properties of Matter

ForcePressure = area

h = 760 mm = 1 atm

Torricelli barometer

Robert Boyle

Boyle’s Law

• For a gas at constant T and nV and p are inversely proportional

pV = constant

Charles’ Law

• At constant pressure, V/T = constant

Charles’ Law

Gay – Lussac’s Law

• In a constant volume:P/T = constant

Gay-Lussac’s Law

Combined Gas Law

• Boyle’s Law pV = constant• Charles’ Law V/T = constant• Gay-Lussac’s Law p/T = constant

Combining all three:p1V1 p2V2

So: T1 = T2

Combined Gas Law

• A sample of gas has a volume of 400 liters when its temperature is 20°C and its pressure is 300 mm Hg. What volume will the gas occupy at STP?

Combined Gas Law

p1V1 p2V2

T1 = T2 T in Kelvin

(300/760 mm Hg)(400 L) = (760 mm Hg) (V2)

(293 K) (273 K)

V2 = 147 Liters

Combined Gas Law

• A sample of He gas has a volume of 250 mL at 456 torr and 25°C. At what temperature does this gas have a volume of 150 mL and 561 torr?

p1V1 p2V2

T1 = T2

Combined Gas Law

• A sample of He gas has a volume of 250 mL at 456 torr and 25°C. At what temperature does this gas have a volume of 150 mL and 561 torr?

p1V1 p2V2 T2 = p2V2 T1

T1 = T2 p1V1

Combined Gas Law

• A sample of He gas has a volume of 250 mL at 456 torr and 25°C. At what temperature does this gas have a volume of 150 mL and 561 torr?

p1V1 p2V2 (561/760)(0.15L)(298K)

T1 = T2 T2 = (456/760)(0.25L)

= 220 K = -53°C

Ideal Gases

• Non-interacting

• Point particles

• Randomly moving with elastic collisions (no energy lost)

Ideal Gases

• Avogadro’s Law:Equal volumes of gas contain the same number of molecules at the same T & p.

n = number of moles

p1V1 = constant

n T1

Ideal Gas Law

p1V1 = constant = 0.082 L atm = R

n T1 K mole

= Universal Gas Constant

One mole of gas at STP:Volume = nRT/p = (1 mole)(0.082 Latm)

(273K)/1 atm

= 22.4 Liters

Ideal Gas Law

• pV = nRT

• How many moles of Helium are present in a balloon that has a volume of 65 L at 20° C and 705 torr?Given Needed

V, T, p, R n n = pV/RT

Ideal Gas Law

• pV = nRT

• How many moles of Helium are present in a balloon that has a volume of 65 L at 20° C and 705 torr?

n = pV/RT

= (705/760 atm)(65 L)

(0.082 Latm/K)(293)

= 2.5 moles He

Ideal Gas

6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How many moles of this gas are present?

Ideal Gas

6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How many moles of this gas are present?

n = pV/RT

= (3 atm)(6.2 L)(0.082 L atm/mole K)(310 K)

= 0.73 moles

Ideal Gases and Density

Density

• Gas density increases with molecular mass.

Density

• What is the density of NO2 gas at 0.97 atm and 35°C?

MW = 46 g/mole Molar mass p (46 g/mole)(0.97 atm)

• Density = RT (0.082 L atm/mole K) (308 K)

= 1.767 g/L

Gas Diffusion

• Movement of particles from region of Higher density to lower density

Gas Diffusion

• Movement of particles from region of Higher density to lower density

Depends on density (molar mass)

Graham’s Law proportionality

Rate of effusion inversely to square root of molar mass

Smaller molecules escape FASTER than larger molecules

Dalton’s Law

PT = P1 + P2 + P3 + . . . .

Total pressure of a gas sample is the sun of the partial pressures.

Dalton’s Law

• A mixture of O2, CO2 and N2 has a pT of 0.97 atm; if pO2 = 0.7 atm and pN2 = 0.12 atm, what is pCO2?

Dalton’s Law

• A mixture of O2, CO2 and N2 has a pT of 0.97 atm; if pO2 = 0.7 atm and pN2 = 0.12 atm, what is pCO2?

• pT = pO2 + pN2 + pCO2 = 0.97 atm

• pCO2 = 0.97 atm - (0.7 atm + 0.12 atm)

= 0.15 atm

Dalton’s Law

• The partial pressures of CH4 and O2 are 0.175 atm and 0.25 atm.

At 65°C in a volume of 2 L, how many moles of each gas are present?

Dalton’s Law

• The partial pressures of CH4 and O2 are 0.175 atm and 0.25 atm.

At 65°C in a volume of 2 L, how many moles of each gas are present?

nCH4 = pV/RT = 0.175 atm (2L)/0.082 L atm/mole K

(338 K = 0.126 moles

nO2 = pV/RT = 0.25 atm (2L)/ 0.082 L atm/mole K (338 K)

= 0.018 moles

Unit 7 Review

• Phases of Matter and Transitions• Intermolecular Forces• Phase Diagrams• Boyle’s, Charles’, Gay Lussac’s Laws• Ideal Gas Law pV = nRT• Graham’s Law of Diffusion• Partial Pressure