gases, liquids, and solids · collecting gas by water displacement the gas bubbles through the...

Gases, Liquids, and

Solids

Kinetic Molecular Theory

“Particles of matter are always in motion

and this motion has consequences.”

Liquids and SolidsHow are liquids and solids similar to and different from gases? (in terms of the KMT)

Property Gases Liquids Solids

Spacing

Movement

Avg. KE

Attraction between

particles

Disorder/Order

Volume

Shape

Fluidity

Density

Compressibility

Diffusing Ability

Liquids and Solids


Spacing Far apart much closer tog.

than gases

most closely

packed

Movement Very fast slower, slip by

each other

vibrate in

position

Avg. KE high lower lowest

Attraction between

particles

Very low more effective, e.g.

H-bonding

most effective,

e.g. ionic bond

Disorder Highly

disordered

less disordered

(due to IM forces

less mobility

most ordered

Volume Indefinite definite (like solids) definite

Liquids and Solids


Shape Indefinite Indefinite definite

Fluidity Yes Yes (like

gases)

No (but some

amorphous solids

flow at a very

slow rate, e.g.

glasses)

Density Very low Relatively high

(closer to

solids) – 10%

less than in

their solid state

Highest – more

closely packed

(osmium is the

densest)

Compressibility Very

compressible

(1000 X)

Relatively

incompressible

(~ 4%), like

solids)

less compressible

than liquids

Liquids and Solids


Diffusing Ability Very yes, but much

slower than

gases (closer

tog., attractive

forces get in

the way;

increased T

increased

diffusion)

millions of

times slower in

solids than in

liquids

Gases

Chapters 13.1 and 14


(for gases)

1. Size

Gases consist of large numbers of tiny particles,

which have mass.

The distance between particles is great.

Gas particles are neither attracted to nor repelled

by each other.


(for gases)

Motion

a) Gas particles are in constant, rapid, straight-line, random motion. They possess KE.

b) Gas particles have elastic collisions

(with each other and container walls)

(no net loss of KE)

e.g. elastic : pool balls (do not lose KE)

inelastic: car crash (lose lots of KE)


Energy

3. The average KE of the gas particles is directly

proportional to the Kelvin temperature of the gas.

(Reminder: KE = ½ mv2)

Properties of Gases

very low density

(1/1000 that of solids or liquids)

indefinite volume, expand and contract

fluid

diffuse through each other

have mass

exert pressure

1 mol at STP = 22.4 L

(STP = 273 K, 1 atm)

Effusion vs. Diffusion

Effusion: escape rate of gas

through a small opening

Diffusion: one material moving

through another – gases diffuse

through each other

Graham’s Law

va

vb

mb

ma

v = rate of diffusion (or effusion)

m = molar mass (g/mol)

The kinetic energies of any two gases at the same temperature are equal: recall KE = ½ mv2

Graham’s Law

Rate of gas effusion

is related to MM of

gas

KE= ½ mv2

(m= molar mass,

v=velocity (m/s) )

½ mava2= ½ mbvb

2

Practice Problems

1. The molar mass of gas "b" is 16.04 g/mol and gas "a" is 44.04 g/mol.

If gas "b" is travelling at 5.25 x 109

m/s, how fast is gas "a" travelling?

Both gases have the same KE.

(3.17 x 109 m/s)

Practice Problems

2.An unknown gas effuses through an opening at a rate 3.53 times slower than nitrogen gas. What is the molecular mass of the unknown gas?

(349 g/mol)

Bonding forces: Ionic, Metallic and Covalent

Ionic and metallic bonding forces hold atoms of compounds together:

Intramolecular forces (covalent bonds) hold atoms of individual molecules

together:

Intermolecular forces exist between molecules of covalently-bonded

compounds

- Relatively weak compared to intramolecular forces

3 types:

• Dispersion (London) forces

• Dipole-dipole forces

• Hydrogen bonds

Forces of Attraction

“London” Forces (aka van der Waals or Dispersion forces)

Weakest intermolecular force occurs between all molecules

larger # of electrons larger temporary dipole stronger

attractions between molecules higher m.p. and b.p.

F2 and Cl2 are gases at room T

Br2 is liquid at room T (more electrons than F2 and Cl2)

I2 is solid at room T (largest number of electrons)

London forces are the only intermolecular forces in the noble gases

Dipole Forces

Attractions between polar molecules, stronger than London dispersion forces:

(-) end of one polar molecule attracts the (+) end of another polar molecule

more polar stronger dipole force

closer together stronger dipole force

Hydrogen Bonding

Always involves H attached to an O, F or N (small, high electronegativity)

Strongest intermolecular force: 5% of the strength of a covalent bond!

Increased b.p. and viscosity

Accounts for high b.p. of H2O

Gas Pressure

Pressure = Force/Area

Units

Force in N

Pressure in:

Pascals (N/m2)

Torr

mm Hg

atmospheres

Standard Pressure

101.3 kPa

760 torr

760 mm Hg

1 atm

Pressure = force

area

Units:

Pascals: 1 Pa = 1 N/m2, 1 kPa = 1000 Pa

mm Hg (or Torr)

psi = lbs/in2

atm = atmospheres

Standard Pressure @ sea level

1atm = 101.325 kPa = 760. mm Hg = 760. Torr = 14.7 psi

Dalton’s Law of Partial Pressures

Ptotal= Pa + Pb + …

Dalton’s Law Practice

Problem

A 1 L sample contains 78% N2, 21% O2 and 1.0% Ar.

The sample is at a pressure of 1 atm (760. mm Hg).

a) What is the partial pressure of each gas in mm Hg?

b) What is the partial volume of each gas in mL?

Application of Dalton’s LawCollecting gas by displacement of water

Ptotal = total pressure (given)PH2O varies at different temperatures (see table…)

Ptotal = Pgas + PH2O

Collecting Gas

by Water Displacement

The gas bubbles through the water in the jar and collects at the top due to its lower density.

The gas has water vapor mixed with it.


Pdry gas = Ptotal – PH2O

Ptotal is what is measured (= atmospheric P)

PH2O can be found in standard tables of vapor pressure of water at different temperatures

Vapor pressure of H2O at various temperatures

Practice Problem: Hydrogen gas is collected over water at a total

pressure of 95.0 kPa and temperature of 25oC. What is the partial

pressure of the dry hydrogen gas? (A: 91.8 kPa)


Mole Fraction

mole fraction of gas A = moles of gas A

total moles gas =

Pgas A

Ptotal

1. The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.

2. The partial pressure of nitrogen was observed to be 590 mm Hg in air with a total atmospheric pressure of 760. mm Hg. Calculate the mole fraction of N2 present.

PO2

Ptotal =

156 torr

743 torr = 0.210

PN2

Ptotal =

590. mm Hg

760. mm Hg = 0.78

Partial Pressure problems1. Determine the partial pressure of oxygen (O2) collected over water if

the temperature is 20.0oC and the total (atmospheric) gas pressure is 98.0 kPa.

(95.7 kPa)

2. The barometer at an indoor pool reads 105.00 kPa. If the temperature in the room is 30.0oC, what is the partial pressure of the “dry” air?

(100.76 kPa)

3. What is the mole fraction of hydrogen (H2) in a gas mixture that has a PH2 of 5.26 kPa? The other gases in the mixture are oxygen (O2), with a PO2 of 35.2 kPa and carbon dioxide with a PCO2 of 16.1 kPa.

(0.0930)

Barometer

instrument used

to measure atmospheric

P, using a column of Hg

invented by Evangelista

Torricelli in 1643

Manometer

measures P of an enclosed gas relative to atmospheric P (open end)

Gas P = atmospheric P ± P of liquid in U-tube

Ask: Is the gas P higher or lower than atmospheric P?

- If higher, add the pressure of the liquid to atm P.

- If lower, subtract the pressure of the liquid from atm P.

Manometer practice

problems

Manometer practice problems

Forces of Attraction

Liquids and Solids

Phase Changes

13.2-13.4

Phase Changes

process Phases Involved Endo/Exothermic?

melting solid liquid endothermic

Phase Changes

process Phases Involved Endo/Exothermic?

Melting Solid to liquid Endothermic

Freezing Liquid to solid Exothermic

Vaporization Liquid to gas Endothermic

Condensation Gas to liquid Exothermic

Sublimation Solid to gas Endothermic

Deposition Gas to solid Exothermic

Phase Change Terminology

Melting point (m.p.): T at which a solid becomes a liquid. Note: Amorphous solids act somewhat like liquids even when solid

Vapor Pressure (VP): P exerted by a vapor over a liquid

Boiling point (b.p.): T at which the VP of a liquid = atmospheric P

Evaporation: Vaporization only at surface of liquid, below b.p.

Freezing point: T at which liquid is converted into crystalline solid

Phase Diagrams

Graph showing the relationships between solid, liquid and gaseous phases over a range of conditions, e.g. P vs. T

Triple point = T and P conditions at which the solid, liquid and vapor of a substance can coexist at equilibrium

Critical T = the highest T at which a gas can be liquified by P alone

Critical P = P exerted by a substance at the critical T

Phase Diagrams

Phase Diagrams

1. What variables are plotted on a phase diagram?

2. How many phases of water are represented in its phase diagram?

What are they?

3. What phases of water coexist at each point along the red curve?

Along the yellow curve?

4. Look at the phase diagram for carbon dioxide. Above which

pressure and temperature is carbon dioxide unable to exist as a

liquid?

5. At which pressure and temperature do the solid, liquid, and

gaseous phases of carbon dioxide coexist?

Phase Diagrams

1. What variables are plotted on a phase diagram? Pressure and temperature

2. How many phases of water are represented in its phase diagram?

What are they? Three: solid, liquid, and vapor (gas)

3. What phases of water coexist at each point

along the red curve? Solid and liquid

along the yellow curve? Solid and gas

4. Look at the phase diagram for carbon dioxide. Above which

pressure and temperature is carbon dioxide unable to exist as a

liquid? 73 atm, 31oC

5. At which pressure and temperature do the solid, liquid, and

gaseous phases of carbon dioxide coexist? 5.1 atm, -57oC

Phase Diagrams

Temperature (oC) Pressure (atm) Phase

200 1

-2 1

150 100

-2 0.001

30 0.8

1 Liquid

100.00 vapor

Phase Diagrams

Temperature (oC) Pressure (atm) Phase

200 1 vapor

-2 1 solid

150 100 Liquid

-2 0.001 vapor

30 0.8 liquid

0.00 < T < 100.00 1 Liquid

100.00 < 1.00 atm vapor

Gas vs. Vapor

Gas

State of particles at room temperature

Vapor

Gas formed from a substance that normally exists as a solid or liquid at

room T and P

Vaporization

Conversion of a liquid to gas or vapor

Boiling point

T at which vapor pressure of liquid = atmospheric pressure

Boiling Point

Boiling point at 1 atm = “normal boiling point”

H2O enters vapor state within liquid. Vapor is less dense, rises to surface

Needs constant energy (heat) to keep it boiling – cooling process

Liquid never rises above its boiling point! (at constant P)

If atm P , less E is required for particles to escape atm P

Mountains vs. pressure cooker

Evaporation

Evaporation occurs when particles have enough KE to overcome their I.M. forces

In a contained vessel (closed): “dynamic equilibrium” occurs when rate of vaporization = rate of condensation

VP (which = partial pressure of a vapor above a liquid) depends on:

1. # gas particles: # particles vapor pressure

2. Temperature: as T v.p. (particles have more energy to escape)

3. Intermolecular forces: Stronger I.M. forces v.p. (fewer particles have enough energy to break the I.M. “bonds” and escape)

In an uncontained vessel (open): evaporation is a cooling process: the most

energetic particles leave so average KE of remaining particles is lower

Rate of evaporation with in:

T (more particles have energy to escape the liquid)

air currents

surface area

Evaporation

Evaporation vs. Boiling

Evaporation

takes place at liquid surface

below boiling T

Boiling

occurs throughout liquid

at boiling T

Liquids

Viscosity: a measure of resistance to flow. In liquids, viscosity is determined

by

Intermolecular forces – more attractions greater viscosity

Molecular shape – longer chains greater viscosity

Temperature – colder temp greater viscosity

Surface tension: E required to increase the SA of a liquid by a given amount

• Stronger intermolecular forces greater surface tension

• Surfactants: compounds that lower the surface tension of water (e.g. detergent or soap)

Liquids

Capillary action – the result of cohesion and adhesion

Cohesion = force of attraction between identical molecules

Adhesion = force of attraction between different types of molecules

e.g. Water in capillary tube – adhesion between water molecules and glass >

cohesion between water molecules

SolidsDensity of solids is higher than density of most liquids, and may be

Crystalline or Amorphous

Crystalline solids

Made of crystals: particles arranged in orderly, geometric, repeating

patterns

Have definite geometric shape

Crystal lattice = total 3-D array of points that describe the arrangements

of particles, smallest unit is the unit cell

Crystal has same symmetry as its unit cell

Abrupt melting point

- all bonds break at once

Crystalline Solids7 shapes, based on arrangement of atoms in unit cell, cell lengths and cell

angles:

Categories of Crystalline Solids

1. Atomic (e.g. noble gases)

2. Molecular (e.g. table sugar, proteins)

3. Covalent network (e.g. diamond, quartz)

4. Ionic

4. Metallic

Binding Forces in Crystals

(see Bonding notes)Atomic and Covalent molecular crystals

Weak intermolecular forces, low m.p., easily vaporized, relatively soft, good insulators

Covalent network (e.g. diamond, graphite)

3-D covalent bonds (giant covalent molecules) very hard, brittle, high m.p., nonconductors or semiconductors; some are planar, e.g. graphite – in sheets

Ionic (e.g. NaCl)

strong positive and negative ions, electrostatically attracted to one another hard, brittle, high m.p., good insulators

Metallic (metals)

positive ions surrounded by a cloud of electrons (electrons can move freely through lattice) high electrical conductivity, malleability, ductility

Amorphous solids

Glasses and plastics – particles arranged

randomly nearly any shape, depending on

molding

No definite melting point, gradually soften to

thick, sticky liquids

Cool too fast for crystals to form

Describing Gases

To describe a gas, you need:

Volume

Pressure

Temperature (K)

# particles (moles)

“Gas Laws”

Constant Temperature

What happens to P when V decreases?

Constant Pressure

What happens to V when T increases?

Constant Volume

What happens to P when T increases?

Constant Volume and Temperature

What happens to P when the # of particles is increased?

Constant Temperature and Pressure

What happens to V when the # of particles is increased?

The Combined Gas Law

What happens to a gas when various conditions are changed?

The combined gas law includes Boyle’s, Charles’s, and Gay-Lussac’s Laws….

P1V1

T1

=P2V2

T2

Boyle’s Law (constant T)

Demonstrates an inverse relationship between

pressure and volume:

P1V1

T1

=P2V2

T2

P1V1=P2V2

Charles’s Law (constant P)

Demonstrates a direct relationship between

temperature and volume:

P1V1

T1

=P2V2

T2

V1

T1

=V2

T2

Demonstrates a direct relationship between

temperature and pressure

Gay Lussac’s Law (constant V)

P1V1

T1

=P2V2

T2

P1

T1

=P2

T2

Boyle's Law

P1V1 = P2V2

The pressure on 2.50 L of anaesthetic gas is changed

from 760. mm Hg to 304 mm Hg. What will be the new

volume if the temperature remains constant?

(6.25 L)

Charles's Law

V1 = V2

T1 T2

If a sample of gas occupies 6.8 L at 327oC, what will be

its volume at 27oC if the pressure does not change?

(3.4 L)

Gay-Lussac's Law

P1 = P2

T1 T2

A gas has a pressure of 50.0 mm Hg at 540. K. What will

be the temperature, in oC, if the pressure is 70.0 mm

Hg and the volume does not change?

(483oC)

Combined Gas Law

P1V1 = P2V2

T1 T2

1. If a gas has a pressure of 2.35 atm at 25oC, and fills a

container of 543 mL, what is the new pressure if the

container is increased to 750. mL at 50.1oC?

(1.84 atm)

Combined Gas Law

2. A sample of methane that initially occupies 250. mL

at 500. Pa and 500. K is expanded to a volume of 700.

mL. To what temperature will the gas need to be

heated to lower the pressure of the gas to 200. Pa?

(560. K)

Ideal Gases

Based on kinetic molecular theory

Follows gas laws at all T and P

Assumes particles:

- have no V impossible

- have no attraction to each other

if true, would be

impossible to liquefy

gases

(e.g. CO2 is liquid

at 5.1 atm, < 56.6oC

Real Gases

Because particles of real gases occupy space:

Follow gas laws at most T and P

At high P, individual volumescount

At low T, attractions count

The more polar the molecule, the more attraction counts P decreases

The Ideal Gas Law: PV = nRT

To describe a gas completely you need to identify: V – VolumeP – Pressure T – Temperature in Kn - # of moles

The Ideal Gas Law:

1. Can be used to derive the combined gas law

2. Is usually used to determine a missing piece of information about a gas (requires the ideal gas constant R)

n mass

molar mass m

M

g

g

mol

moles

Ideal Gas Law

Most gases act like ideal gases most of the time.

PV = nRT

P, V inversely related BOYLE

PT directly related GAY-LUSSAC

V, T directly related CHARLES

V, n directly related AVOGADRO

R = universal gas constant

R: The Ideal Gas Constant

PV = nRT R = PV

nT=

1 atm 22.4 L

1 mol 273 K

0.0821 L atm

K mol

Derived from ideal gas law using STP conditions:• standard temperature: 273K• standard pressure: 1 atm• volume of 1 mole of gas: 22.4 L

The value of R depends on units of pressure used:

8.314 L kPa

K mol

62.4 L mm Hg

K mol

Ideal Gas Law

Solve for R at STP:

T = 0oC + 273 = 273 K; P = 1 atm

R = PV = (1 atm)(22.4 L) = 0.0821 Latm

nT (1 mol)(273 K) molK

Note that the value of R depends on the units of

pressure

Ideal Gas Law

Calculate the pressure, in atmospheres, of 1.65 g of helium gas at 16.0oC and

occupying a volume of 3.25 L.

P = ?

PV = nRT

V = 3.25 L

n = 1.65 g (1 mol He) = 0.412 mol He

4.00 g He

R = 0.0821 Latm/molK

T = 16.0oC + 273 = 289 K

P = nRT = (0.412 mol)(0.0821 Latm)(289 K)

V 3.25 L molK

= 3.01 atm

Combined Gas Law Warmup

4. The volume of a gas-filled balloon is 30.0 L at 40oC and 153 kPa. What volume will the balloon have at STP?

(A: 39.5 L)

5. A 3.50-L gas sample at 20oC and a pressure of 86.7 kPa expands to a volume of 8.00 L. The final pressure of the gas is 56.7 kPa. What is the final temperature of the gas, in oC?

(A: 165oC, 438 K)

Ideal Gas Law Practice

1. A sample of carbon dioxide with a mass of 0.250 g

was placed in a 350. mL container at 127oC. What is

the pressure, in kPa, exerted by the gas?

(53.9 kPa)


2. A 500. g block of dry ice (solid CO2)

vaporizes to a gas at room temperature. Calculate

the volume of gas produced at 25oC and 975 kPa.

(29.0 L CO2)


3. At what temperature will 7.0 mol of helium gas exert

a pressure of 1.2 atm in a 25.0 kL tank?

(5.2 x 104 K)


4. What mass of chlorine (Cl2) is contained in a 10.0 L

tank at 27oC and 3.50 atm?

Hint: begin by solving for n.

(101 g)

Density of Gases

Measured in g/L (liquids and solids: g/mL)

1 mole of any gas = 22.4 L at STP (273 K and 1 atmosphere)

You can use 22.4 L = 1 mol as a conversion factor at STP

For non-standard conditions, use the ideal gas law.

Finding Molar Mass of a Gas

Using Its Density, at STP1. What is the molar mass of a gas that has a density of 1.28 g/L at STP?

(28.7 g/mol)

2. A 0.519 g gas sample is found to have a volume of 200. mL at STP. What is the molar mass of this gas?

(58.1 g/mol)

3. A chemical reaction produced 98.0 mL of sulfur dioxide gas (SO2) at STP. What was the mass (in grams) of the gas produced?

(0.280 g SO2)

Finding Molar Mass of a Gas Using

the Ideal Gas Law under non-

standard conditions4. A 1.25 g sample of the gaseous product of a chemical reaction was found to have a volume of 350. mL at 20.0oC and 750. mm Hg. What is the molar mass of this gas?

2 steps –

1. Find the number of moles (n), using the Ideal Gas Law

2. Divide the mass of the gas given in the problem by n g/mol

(86.8 g/mol)

(More practice – Gases WS #5)

Density of Gases (Honors)

Practice: 1. Derive density from the ideal gas law

(RQ 14.3):

PV = nRT n = m/MM

2. Rearrange to isolate M (molar mass)

Apply these variations to HW on p. 438, #46-50

Warmup1. a) What is the density of a gas that has a mass of 0.0256 g

and a volume of 178 mL?

(A: 1.44 x 10-1 g/L or 1.44 x 10-4 g/mL)

b) If this is the density under STP, what is the molar mass

of this gas?

(A: 3.22 or 3.23 g/mol)

2. What is the molar mass of a gas that has a mass of 1.23 g and a volume of 580. mL at STP?

(A: 47.5 g/mol)

3. What is the molar mass of a gas that has a mass of 1.53 g and volume of 825 mL at 55oC and 95 kPa?

(A: 53 g/mol)

Stoichiometry and Gases at

STP(review)Calcium carbonate reacts with phosphoric acid to produce calcium

phosphate, carbon dioxide, and water.

3 CaCO3(s) + 2 H3PO4(aq) Ca3(PO4)2(aq) + 3 CO2(g) + 3 H2O(l)

1. How many grams of phosphoric acid, H3PO4, react with excess calcium carbonate, CaCO3, to produce 3.74 g of Ca3(PO4)2?

(2.36 g H3PO4)

2. Assuming STP, how many liters of carbon dioxide are produced when 5.74 g of H3PO4 reacts with an excess of CaCO3?

(1.97 L)

Stoichiometry of Gases

non-STP conditions

Context is stoichiometry

To get to moles of your known asap, you either perform

a mass conversion, or use PV=nRT – solve for # moles

The last step of stoich is to convert moles of your

unknown into required units. That means either moles

mass or volume, for which you use PV=nRT – solves

for L

Stoichiometry and Gases

under non-STP Conditions

Either stoichiometry first (known = solid or liquid) to find moles of unknown, followed by the Ideal Gas Law to find the volume of a gaseous product (unknown),

OR

use the Ideal Gas Law to find the # moles of a gaseous reactant (known), followed by stoichiometry to find the amount of a solid or liquid product (unknown).

Stoichiometry and Gases

under non-STP Conditions

If water is added to magnesium nitride, ammonia gas is produced when the mixture is heated.

Mg3N2(s) + 3 H2O(l) 3 MgO(s) + 2 NH3(g)

1. If 10.3 g of magnesium nitride is treated with water, what volume of ammonia gas would be collected at 24oC and 752 mm Hg? (A: 5.03 L)

2. When you produce 16.2 L of ammonia gas at 100.oC and 802 mm Hg, how many grams of magnesium oxide are also produced? (A: 33.7 g MgO)

Warm up – Stoichiometry

and Gases at STPThe formation of aluminum oxide from its constituent elements is represented by this equation.

4 Al(s) + 3 O2(g) 2 Al2O3(s)

1. How many grams of aluminum are required to react with excess oxygen, to produce 3.74 g of Al2O3?

(1.98 g Al)

2. Assuming STP, how many liters of oxygen are produced when 5.74 g of Al2O3 reacts with an excess of aluminum?

(1.89 L O2)

Warmup - Stoichiometry

and Gases at STP

If 650. mL of hydrogen gas is produced through a

replacement reaction involving solid iron and sulfuric

acid (H2SO4) at STP, how many grams of iron (II)

sulfate are also produced?

Stoichiometry and Gases non-STP conditions

How many liters of oxygen at 27oC and 188 mm Hg are needed to burn 65.5 g of carbon according to the equation

2 C(s) + O2(g) 2 CO(g)

(A: 272 L)

Stoichiometry and Gas Lawsnon-STP conditions

23. From Gases WS #6:

WO3 (s) + 3 H2 (g) W (s) + 3 H2O (l)

How many liters of hydrogen at 35oC and 745 mm Hg are needed to react completely with 875 g of tungsten oxide?

Begin with stoich or PV = nRT?

(A:292 L)

Stoichiometry and Gas Lawsnon-standard conditions

24. Magnesium will “burn” in carbon

dioxide to produce elemental carbon and magnesium

oxide. What mass of magnesium will react with a 250

mL container of CO2 at 77oC and 65 kPa?

Begin with stoich or PV=nRT?

(A: 0.27 g Mg)

Summary of Gas Laws (check your reference

sheet – 1st page of notes)Grahams Law

Dalton’s Law

Ideal Gas Law

Combined Gas Law

Boyle’s Law (on top)

Charles’s Law (likes T.V.)

Gay Lussac’s Law (what’s left…)

P1V1 P2V2

V1

T1

V2

T2

P1

T1

P2

T2

P1V1

T1

P2V2

T2

PV nRT

Ptotal = Pa + Pb + Pc .... Pn

va

vb

mb

ma

Vapor Pressure

What is the VP of ethanol at 60oC?

of water at the same T (60oC)?

Which compound boils at a lower T? How can you tell?

Which exhibits stronger IMFs? ethanol or water?

Explain your answer.

Vapor Pressure

What is the VP of ethanol at 60oC? about 402 torr

of water at the same T (60oC)? about 180 torr

Which compound boils at a lower T? How can you tell?

Which exhibits stronger IMFs? ethanol or water?

Warmup – Stoichiometry and Gases 1. The formation of aluminum oxide from its constituent elements is represented by this

equation.

4 Al(s) + 3 O2(g) 2 Al2O3(s)

Assuming STP, how many liters of oxygen are produced from 5.74 g of Al2O3?

(1.89 L O2)

2. Consider the following chemical equation:

2 Cu2S(s) + 3 O2(g) 2 Cu2O(s) + 2 SO2(g)

What volume of oxygen gas, measured at 27oC and 0.998 atm, is required to react completely with 25.0 g of copper(I) sulfide?

(A: 5.82 L O2)

3. What mass of NaCl can be produced by the reaction of Na(s) with 3.65 L Cl2(g) at

25oC and 105 kPa? (Hint: Write the chemical equation first.)

(A: 18.1 g NaCl)

Gas Stoichiometry

Volume – volume: Since all gases take up the same volume at the same temperature, the mole ratio can be used as a volume ratio of two gases in the equation.

Volume – mass: Requires conditions under which reaction takes place - use ideal gas law.

gases, liquids, and solids · collecting gas by water displacement the gas bubbles through the...

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