think about this… gas atmosphere this is a u-tube manometer. the red stuff is a liquid that moves...

Think About This…Think About This…Think About This…Think About This…

Gas Atmosphere This is a U-Tube Manometer. The red stuff is a liquid that moves based on the pressures on each end of the tube. Based on the position of the red liquid, which has a higher pressure, the Gas or the Atmosphere? How do you know?


Gas Atmosphere Which has a higher pressure in this U-Tube, the Gas or the Atmosphere? How do you know?


Gas Atmosphere What about this U-Tube? Which has the higher pressure, the Gas or the Atmosphere? How do you know?

Follow along in your textChapter 12Section 1

Pages 416 - 422

Follow along in your textChapter 12Section 1

Pages 416 - 422

The Physical The Physical Properties of Properties of

GasesGases

Kinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular Theory

Particles in an ideal gas…

• have no volume.

• have elastic collisions.

• are in constant, random, straight-line motion.

• don’t attract or repel each other.

• have an average kinetic energy directly related to Kelvin temperature.

Real GasesReal GasesReal GasesReal Gases

Particles in a REAL gas…• have their own volume• attract each other

Gas behavior is most ideal…• at high temperatures• at low pressures• in nonpolar, basically covalent,

atoms/molecules

Characteristics of GasesCharacteristics of GasesCharacteristics of GasesCharacteristics of GasesGases expand to fill any container.

• random motion, no attraction

Gases are fluids (like liquids).• no attraction

Gases have very low densities.• no volume = lots of empty space

Characteristics of GasesCharacteristics of GasesCharacteristics of GasesCharacteristics of Gases Gases can be compressed.

• no volume = lots of empty space

Gases undergo effusion & diffusion.• random motion

TemperatureTemperatureTemperatureTemperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

K = ºC + 273

Always use absolute temperature (Kelvin) when working with gases.

What is What is PressurePressure??What is What is PressurePressure??

area

forcepressure

Which shoes create the most pressure?

Pressure EquipmentPressure EquipmentPressure EquipmentPressure Equipment

Barometer• measures atmospheric pressure

Mercury Barometer

Aneroid Barometer

Pressure EquipmentPressure EquipmentPressure EquipmentPressure Equipment

Manometer• measures contained gas pressure

U-tube Manometer Bourdon-tube gauge

KEY UNITS AT SEA LEVEL

101,325 Pa

101.325 kPa

1 atm

760 mm Hg

760 torr

14.7 psi

1.01 bar

*These are all equal to each other!*

PressurePressurePressurePressure

STP is the Norm!STP is the Norm!STP is the Norm!STP is the Norm!

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.325 kPa

-OR-

STP

Example: How many torr equals 5.00 psi?

ClosureClosureClosureClosure

1. How many Pascals equals 7.3 atm?

2. How many mmHg equals 19.3 barr?

3. How many atms equals 96.3 kPa?

5.00 psi 760 torr

14.7 psi = 259 torr

Warm-Up :Warm-Up :Complete the following Complete the following pressure conversions .pressure conversions .

Warm-Up :Warm-Up :Complete the following Complete the following pressure conversions .pressure conversions .

1. 15,650 Pa to mm Hg

2. 23 atm to psi

3. 893,000 torr to kPa

4. 6.5 psi to barr

Follow along in your text

Chapter 12 Section 2

Pages 423 - 432


Chapter 12 Section 2

Pages 423 - 432

The Gas LawsThe Gas Laws

Meet the VariablesMeet the VariablesMeet the VariablesMeet the Variables

P = pressure exerted by the gas

T = temperature in kelvins of the gas

V = total volume occupied by the gas

n = number of moles of the gas

k = symbolizes anything constant

Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law

V

P

PV = k

Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law

The pressure and volume of a gas are inversely related at constant mass & temp

P1V1 = P2V2

When the temp & number of particles remains the same, this is the equation

GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

P1V1T2 = P2V2T1

Boyle’s Law ProblemBoyle’s Law ProblemBoyle’s Law ProblemBoyle’s Law Problem

A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.

Remember BOYLE’S LAW!

P V

(150.kPa)(100.mL)=(200.kPa)V2

V2 = 75.0 mL

kT

VV

T

Charles’ LawCharles’ LawCharles’ LawCharles’ Law

Charles’ LawCharles’ LawCharles’ LawCharles’ Law

The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

V1 V2

T1 T2

=

If all conditions are kept constant, then the equation looks like this

GIVEN:

V1 = 473 mL

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

P1V1T2 = P2V2T1

Charles’ Law ProblemCharles’ Law ProblemCharles’ Law ProblemCharles’ Law Problem

A gas occupies 473 mL at 36°C. Find its volume at 94°C.

Remember CHARLES’ LAW!

T V

(473 mL)(367 K)=V2(309 K)

V2 = 562 mL

kT

PP

T

Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law

Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law

The pressure and absolute temperature (K) of a gas are directly related

At constant mass & volume, the equation looks like this

P1 P2

T1 T2

=

GIVEN:

P1 = 765 torr

T1 = 23°C = 296K

P2 = 560. torr

T2 = ?

WORK:

P1V1T2 = P2V2T1

Gay-Lussac’s ProblemGay-Lussac’s ProblemGay-Lussac’s ProblemGay-Lussac’s Problem

A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?

Remember GAY-LUSSAC’S LAW!

P T

(765 torr)T2 = (560. torr)(296K)

T2 = 217 K = -38°C

Warm Up:Warm Up:Warm Up:Warm Up: DO NOT SOLVE! Just determine which Gas Law you should use.

1. What is the new volume if a 5.0L container of gas at 23 °C is heated to 70 °C?

2. If a gas at STP is heated to 900 K, what is the resulting pressure?

3. 15 mL of a gas at 1 atm is compressed to 5 mL. How much pressure was applied?

4. A 2.0 L container at 4.0 atm and 0.0 °C is heated to 50. °C at constant volume. What is the resulting pressure?

5. A 35 L container at 23 °C is expanded to 50 L to obtain a pressure of 4.0 atm. What was the original pressure if temperature was constant?

Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law

This is a combination of the 3 main gas laws: Boyle’s Law, Charles’ Law & Gay-Lussac’s Law

Pressure is opposite temperature & volume

PP VVTT

= kPVT

Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1

GIVEN:

V1 = 7.84 mL

P1 = 71.8 kPa

T1 = 25°C = 298 K

V2 = ?

P2 = 101.325 kPa

T2 = 273 K

WORK:

P1V1T2 = P2V2T1

(71.8 kPa)(7.84 mL)(273 K)

=(101.325 kPa) V2 (298 K)

V2 = 5.09 mL

Combined Gas Law Combined Gas Law ProblemProblemCombined Gas Law Combined Gas Law ProblemProblem

A gas occupies 7.84 mL at 71.8 kPa & 25°C. Find its volume at STP.

P T VCOMBINED GAS LAW

Warm Up:Warm Up:Warm Up:Warm Up:

Stoichiometry Quiz #5Stoichiometry Quiz #5

How many grams of CO2 are produced from 75 L of CO at

STP?

___CO + ___O2 → ___CO2


Chapter 12 Section 3Pages 433 - 442


Chapter 12 Section 3Pages 433 - 442

The Ideal Gas The Ideal Gas LawLaw

PV

TVn

PVnT

Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law

= k

UNIVERSAL GAS CONSTANTR=0.0821 Latm/molKR=8.314 LkPa/molK

R=62.4 LmmHg/molK

= R

Merge the Combined Gas Law with Avogadro’s Principle:


UNIVERSAL GAS CONSTANTR=0.0821 Latm/molKR=8.314 LkPa/molK

R=62.4 LmmHg/molK

PV=nRT


PVM=mRT

If you are given the mass of a gas, you can use this equation instead of converting mass to moles first.

massmolar mass


d=PM/RT

If you are given the molar mass of a gas, you can use this equation to

find the density

molar massdensity

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0821Latm/molK

WORK:

PV = nRT

Ideal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law Problems Calculate the pressure in atmospheres of

0.412 mol of He at 16°C & occupying 3.25 L.

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

(3.25 L)

P = 3.01 atm

GIVEN:

V = ?

m = 85 g

MO2 = 32 g/mol

T = 25°C = 298 K

P = 104.5 kPaR = 8.314 LkPa/molK

Ideal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law Problems

Find the volume of 85 g of O2 at 25°C and 104.5 kPa.

WORK:

PVM = mRT

V= (85 g)(8.314)(298 K)

(104.5 kPa)(32 g/mol)

V = 64 L

Gas StoichiometryGas StoichiometryGas StoichiometryGas Stoichiometry Moles Moles Liters of a Gas: Liters of a Gas:

• STP - use 22.4 L/mol • Non-STP - use ideal gas law

Non-Non-STPSTP• Given liters of gas?

start with ideal gas law• Looking for liters of gas?

start with stoichiometry conversion

1 molCaCO3

100.09g CaCO3

Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem

What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?

5.25 gCaCO3 = 1.26 mol CO2

CaCO3 CaO + CO2

1 molCO2

1 molCaCO3

5.25 g ? Lnon-STPLooking for liters: Start with stoich

and calculate moles of CO2.

Plug this into the Ideal Gas Law to find liters.

WORK:

PV = nRT

(103 kPa)V=(1mol)(8.314 LkPa/molK)(298K)

V = 1.26 dm3 CO2


What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?

GIVEN:

P = 103 kPaV = ?

n = 1.26 molT = 25°C = 298 KR = 8.314 LkPa/molK

WORK:

PV = nRT

(97.3 kPa) (15.0 L)= n (8.314 LkPa/molK) (294K)

n = 0.597 mol O2


How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?

GIVEN:

P = 97.3 kPaV = 15.0 L

n = ?T = 21°C = 294 KR = 8.314 LkPa/molK

4 Al + 3 O2 2 Al2O3 15.0 L

non-STP ? gGiven liters: Start with

Ideal Gas Law and calculate moles of O2.

NEXT

2 mol Al2O3

3 mol O2


How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?

0.597mol O2 = 40.6 g Al2O3

4 Al + 3 O2 2 Al2O3

101.96 g Al2O3

1 molAl2O3

15.0Lnon-STP

? gUse stoich to convert moles of O2 to grams Al2O3.

Closure:Closure:Closure:Closure:

How many grams of CO2 are produced from 75L of CO at

35 ºC and 96.2 kPa?

___CO + ___O2 → ___CO2

Warm Up:Warm Up:Warm Up:Warm Up:

Solve using the Ideal Gas Law & StoichometrySolve using the Ideal Gas Law & Stoichometry

What mass of NHWhat mass of NH33 is created is created

when 20.0 L of Nwhen 20.0 L of N22 is combined is combined

with excess Hwith excess H22 at 42 °C and at 42 °C and

2.50 atm? 2.50 atm?

More Gas Laws!More Gas Laws!

Follow along inyour text

Chapter 12 Sections 2 & 3

Pages 432 - 439

Follow along inyour text

Chapter 12 Sections 2 & 3

Pages 432 - 439

kn V V

n

Avogadro’s LawAvogadro’s LawAvogadro’s LawAvogadro’s Law Equal volumes of gases contain equal

numbers of moles at constant temp & pressure• 22.4 L/mole• 6.02 x 1023 particles/mole

Dalton’s Law of Partial Dalton’s Law of Partial PressurePressureDalton’s Law of Partial Dalton’s Law of Partial PressurePressure

The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + ...

P1 = (Ptotal) (% of Gas1)

GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

Dalton’s Law of Partial Dalton’s Law of Partial PressurePressureDalton’s Law of Partial Dalton’s Law of Partial PressurePressure

Hydrogen gas is collected over water. The water vapor has a pressure of 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Sig Figs: Round to least number of decimal places.

The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.

Graham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of Diffusion

DiffusionDiffusion• Spreading of gas molecules from

high density to low density until even all over.

EffusionEffusion

• Passing of gas molecules under pressure through a tiny opening


Speed of diffusion/effusionSpeed of diffusion/effusion

• Kinetic energy is determined by the temperature of the gas.

• At the same temp & KE, heavier molecules move more slowly. “The smaller the mass, the

faster the gas!”


Graham’s LawGraham’s Law• Rate of diffusion of a gas is inversely related

to the square root of its molar mass.• The equation shows the ratio of Gas A’s

speed to Gas B’s speed.

A

B

B

A

m

m

v

vv

√m

Determine the relative rate of diffusion for krypton and bromine.

1.381

Kr diffuses 1.381 times faster than Br2.

Kr

Br

Br

Kr

m

m

v

v2

2

A

B

B

A

m

m

v

v

g/mol83.80

g/mol159.80


The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.

A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?

A

B

B

A

m

m

v

v

2

2

2

2

H

O

O

H

m

m

v

v

g/mol 2.02

g/mol32.00

m/s 12.3

vH 2


3.980m/s 12.3

vH 2

m/s49.0 vH 2

Put the gas with the unknown

speed as “Gas A”.

think about this… gas atmosphere this is a u-tube manometer. the red stuff is a liquid that moves...

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