physical properties gases. kinetic molecular theory b particles in an ideal gas… have no volume...

Physical PropertiesPhysical

Properties

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

• average kinetic energy is directly proportional to the absolute temperature

To change pressure you can:• Change temperature• Change volume• Change amount

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

• random motion, no attraction

Gases are fluid (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 diffusion & effusion• random motion

State Changes

Describing GasesDescribing GasesDescribing GasesDescribing Gases

Gases can be described by their:

• Temperature

• Pressure

• Volume

• Number of molecules/moles

• K

• atm

• L

• #

TemperatureTemperatureTemperatureTemperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

Temperature: Every time temperature is used in a gas law equation it must be stated in Kelvin (an absolute temperature)

PressurePressurePressurePressure

area

forcepressure

Which shoes create the most pressure?

Pressure of a gas is the force of its particles exerted over a unit area (number of collisions against a container’s walls)

PressurePressurePressurePressure Pressure may have different units: torr,

millimeters of mercury (mm Hg), atmospheres (atm), Pascals (Pa), or kilopascals (kPa).

KEY EQUIVALENT UNITS • 760 torr • 760 mm Hg• 1 atm

• 101,325 Pa

• 101.325 kPa (kilopascal)

• 14.7 psi

STPSTPSTPSTP

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.325 kPa

The volume of 1 mole of gas at STP is 22.4 L

-OR-

STP

Pressure Problem 1Pressure Problem 1Pressure Problem 1Pressure Problem 1

The average pressure in Denver, Colorado, is 0.830 atm. Express this in kPa.

0.830 atm

1 atm

101.325 kPa= 84.1

kPa

Pressure Problem 2Pressure Problem 2Pressure Problem 2Pressure Problem 2

Convert a pressure of 1.75 atm to psi.

1.75 atm

1 atm

14.7 psi= 25.7 psi

The Gas Laws

The Gas Laws

The Behavior of GasesThe Behavior of Gases

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

P

V

Investigated the relationship between pressure and volume on a gas.

He found that pressure and volume of a gas are inversely related

• at constant mass & temp


As pressure increases

Volume decreases

Boyles LAW:

P

V

P1V1 = k

GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

P1V1 = P2V2

Gas Law ProblemsGas Law ProblemsGas Law ProblemsGas Law Problems

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

BOYLE’S LAW

P V

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

V2 = 75.0 mL

k1

1

T

VV

T

Charles’ LawCharles’ LawCharles’ LawCharles’ Law

Jacques Charles investigated the relationship of temperature & volume in a gas. He found that the volume of a gas and the Kelvin temperature of a gas are directly related.

V

T


As volume increases, temperature increases

Charles Law:

kT

V

1

1

GIVEN:

V1 = 473 cm3

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

V1T2 = V2T1


A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.

CHARLES’ LAW

T V

(473 cm3)(367 K)=V2(309 K)

V2 = 562 cm3

GIVEN:

V1 = 43.5 L

T1 = 0°C = 273K

V2 = ?

T2= 35.4°C= 308.4K

WORK:

V1T2 = V2T1

Gas Law ProblemsGas Law ProblemsGas Law ProblemsGas Law Problems Example: The volume of a sample of gas is

43.5 L at STP. What will the volume of the gas be if the temperature is increased to 35.4°C?

CHARLES’ LAW

T V

(43.5 L)(308.4 K)=V2(273 K)

V2 = 49.1 L

k1

1

T

PP

T

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

Since temperature and volume are related, it would make sense that temperature and pressure are related. The equation for this will resemble Charle’s law:

P

T

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

k1

1

T

P

GIVEN:

P1 = 765 torr

T1 = 23°C = 296K

P2 = 560. torr

T2 = ?

WORK:

P1T2 = P2T1


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

GAY-LUSSAC’S LAW

P T

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

T2 = 217 K = -56°C

= kPVPTVT

PVT

Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1

GIVEN:

V1 = 7.84 cm3

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 cm3)(273 K)

=(101.325 kPa) V2 (298 K)

V2 = 5.09 cm3


A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

P T VCOMBINED GAS LAW

The First 4 Gas LawsThe First 4 Gas LawsThe First 4 Gas LawsThe First 4 Gas Laws

The Gas Laws Table

GasesGases

II. Ideal Gas Law

II. Ideal Gas Law

Ideal Gas Law and Gas Ideal Gas Law and Gas StoichiometryStoichiometry

Part 1Part 1Ideal Gas LawIdeal Gas Law

Avogadro’s LawAvogadro’s LawAvogadro’s LawAvogadro’s Law

The volume of a gas increases or reduces, as its number of moles is being increased or decreased.

The volume of an enclosed gas is directly proportional to its number of moles.

k n

V

1

1V

n

Avogadro’s PrincipleAvogadro’s Principle Avogadro’s PrincipleAvogadro’s Principle

Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas• n = number of moles

PV

TVn

PVnT

Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law

= kUNIVERSAL GAS

CONSTANTR=0.08206

Latm/molKR=8.315

dm3kPa/molK

= R

Merge the Combined Gas Law with Avogadro’s Principle:

Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law

UNIVERSAL GAS CONSTANTR=0.08206

Latm/molKR=8.315

dm3kPa/molK

PV=nRT

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.08206Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

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

Calculate the pressure in atmospheres of 0.412 mol of Heat 16°C & occupying 3.25 L.

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/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.

= 2.7 mol

WORK:

85 g 1 mol O2 = 2.7 mol

32.00 g O2

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

Ideal Gas Law and Molar Ideal Gas Law and Molar Mass Mass Ideal Gas Law and Molar Ideal Gas Law and Molar Mass Mass

The Ideal Gas Law can be used to calculate the mass or the molar mass of a gas sample if the mass of the sample is known.

number of molesgas = mass gas or ngas = mgas_

Molar Massgas MMgas

substituting this for ngas in PV = nRT yields

P V = mgas__ R T

MMgas

GIVEN:P = 1.00 atm

T = 22°C = 295. K

V = .600 L

R = 0.08206 Latm/molKMM = 28.01 g

mN2 = ?

WORK:

MM P V = mgas R T

(28.01)(1.0)(.60)=m (.0821) (295) g/mol atm L Latm/molK K

mN2 = .69 g of N2

Applications of Ideal Gas Applications of Ideal Gas LawLaw Applications of Ideal Gas Applications of Ideal Gas LawLaw

Calculate the grams of N2 present in a 0.60 L sample kept at 1.00 atm pressure and a temperature of 22.0oC.

Applications of Ideal Gas LawApplications of Ideal Gas LawApplications of Ideal Gas LawApplications of Ideal Gas Law

Can be used to calculate the molar mass of a gas and the density

Substitute this into ideal gas law

And m/V = d in g/L, so

MMmassmolar

mass

massmolar

gas a of grams gas a of moles

mn

)(

)(

VMM

RTm

V

nRTP

P

dRTMM

MM

dRTP or

GIVEN:

P = 1.50 atm

T = 27°C = 300. K

d = 1.95 g/LR = 0.08206 Latm/molK

MM = ?

WORK:

MM = dRT/P

MM=(1.95)(0.08206)(300.)/1.50 g/L Latm/molK K atm

MM = 32.0 g/mol


The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas.

GIVEN:

d = ? g/L CO2

T = 25°C = 298 KP = 750. torr

R = 0.08206 Latm/molK

MM = 44.01 g/mol

MM = dRT/P →d = MM P/RT

d=(44.01 g/mol)(.987 atm)

(0.08206 Latm/molK )(298K)

d = 1.78 g/L CO2


Calculate the density of carbon dioxide gas at 25°C and 750. torr.

WORK:

750 torr 1 atm = .987 atm 760 torr

= .987 atm

Part 2Part 2

Gas Gas StoichiometryStoichiometry

* Stoichiometry Steps * Stoichiometry Steps Review *Review ** Stoichiometry Steps * Stoichiometry Steps Review *Review *

1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

• Mole ratio - moles moles• Molar mass - moles grams• Molarity - moles liters soln• Molar volume - moles liters gas

Core step in all stoichiometry problems!!

• Mole ratio - moles moles

4. Check answer.

1 mol of a gas=22.4 Lat STP

Molar Volume at STPMolar Volume at STP Molar Volume at STPMolar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

Molar Volume at STPMolar Volume at STP Molar Volume at STPMolar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

Gas StoichiometryGas Stoichiometry Gas StoichiometryGas Stoichiometry

Liters of one Gas Liters of one Gas Liters of another Gas: Liters of another Gas:• Avogadro’s Principle • Coefficients give mole ratios and volume

ratios Moles (or grams) of A Moles (or grams) of A Liters of B: Liters of B:

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

Non-Non-STPSTP• Given liters of gas?

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

start with stoichiometry conversion

How many grams of KClO3 are req’d to

produce 9.00 L of O2 at STP?

9.00 LO2

1 molO2

22.4 L O2

= 32.8 g KClO3

2 molKClO3

3 molO2

122.55g KClO3

1 molKClO3

? g 9.00 L

Gas Stoichiometry Problem – Gas Stoichiometry Problem – STPSTPGas Stoichiometry Problem – Gas Stoichiometry Problem – STPSTP

2KClO3 2KCl + 3O2

1 molCaCO3

100.09g CaCO3

Gas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STP

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

5.25 gCaCO3 = 0.0525 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 for n to find liters

NEXT P = 103 kPa

V = ?

n = ?

R = 8.315 dm3kPa/molK

T = 298K

WORK:

PV = nRT

(103 kPa)V=(0.0525mol)(8.315dm3kPa/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 = 0.0525 molT = 25°C = 298 KR = 8.315 dm3kPa/molK

Gas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STP

WORK:

PV = nRT

(97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K)

n = 0.597 mol O2

Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem

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 = 8.315 dm3kPa/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

Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem

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

Ch. 14 - Gases

III. Two III. Two More LawsMore LawsIII. Two III. Two More LawsMore Laws

Dalton’s Law & Graham’s LawDalton’s Law & Graham’s Law

A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law

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

Ptotal = P1 + P2 + ...Pn

GIVEN:

PO2 = ?

Ptotal = 1.00 atm

PCO2 = 0.12 atm

PN2 = 0.70 atm

WORK:Ptotal = PCO2 + PN2 + PO2

1.00 atm = 0.12 atm + .70 atm + PO2

PO2 = 0.18 atm

A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law A sample of air has a total pressure of

1.00 atm. The mixture contains only CO2, N2, and O2. If PCO2 = 0.12 atm and PN2 = 0.70, what is the partial pressure of O2?


Mole Fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture, represented by chi ( )

...321

1

TOTAL

1

nnn

n

n

n

TOTAL

A

TOTAL

AA P

P

n

nχ

Because n is directly proportional to P according to the ideal gas law, we can derive

GIVEN:

χN2 = 0.7808

Ptotal = 760. torr

PN2 = ?

WORK:

χN2 = PN2 /PTOTAL

0.7808 = (PN2)/(760. torr)

PN2 = 593 torr


The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760. torr.


•When H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor

•Water exerts a pressure known as water-vapor pressure•So, to determine total pressure of gas and water vapor inside a container, make total pressure inside bottle = atmosphere and use this formula:

Ptotal = Pgas + PH2O

GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 20.4 mm Hg

= 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa


Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Look up water-vapor pressure for 22.5°C, convert to kPa

Sig Figs: Round to lowest 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.

B. Graham’s LawB. Graham’s LawB. Graham’s LawB. Graham’s Law

DiffusionDiffusion• Spreading of gas molecules

throughout a container until evenly distributed

EffusionEffusion

• Passing of gas molecules through a tiny opening in a container


KE = ½mv2

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.Larger m smaller v


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

MM

MM

rate

rate

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 means 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”.

An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.

Am

g/mol32.00 16

A

B

B

A

m

m

v

v

A

O

O

A

m

m

v

v2

2

Am

g/mol32.00 4.0

16

g/mol32.00 mA

2

Am

g/mol32.00 4.0

g/mol2.0


The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.

Square both sides to get rid of the square

root sign.

physical properties gases. kinetic molecular theory b particles in an ideal gas… have no volume...

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