Using PV = nRT (Honors)

The Ideal Gas LawThe Ideal Gas Law

PV = nRT PV = nRT Adds in the factor of number of moles of gas “n”.Adds in the factor of number of moles of gas “n”.

P = Pressure (atm)P = Pressure (atm)

V = Volume (Liters)V = Volume (Liters)

T = Temperature (Kelvin)T = Temperature (Kelvin)

n = number of molesn = number of moles Remember when dealing with moles:

# moles (n) = grams of substance Gram Formula Mass

What is “R”?What is “R”?R is a constant, called the “Universal Gas Constant”R is a constant, called the “Universal Gas Constant”

It is derived by plugging in the values for 1 mole ofIt is derived by plugging in the values for 1 mole of

gas at STP. (22.4 Liters, 273K, 1 atm)gas at STP. (22.4 Liters, 273K, 1 atm)

R = 0.0821R = 0.0821

Instead of learning a different value for R for all the Instead of learning a different value for R for all the possible unit combinations, we can just memorize possible unit combinations, we can just memorize one value and convert the units to match R.one value and convert the units to match R.

L • atm Mol • K

GIVEN:GIVEN:

P = ? atmP = ? atm

n = 0.412 moln = 0.412 mol

T = 16°C = 289 KT = 16°C = 289 K

V = 3.25 LV = 3.25 L

R = R = 0.08210.0821LLatm/molatm/molKK

WORK:WORK:

PV = nRTPV = nRT

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

P = 3.01 atmP = 3.01 atm

Ideal Gas Law Ideal Gas Law (Honors)(Honors) Calculate the pressure in atmospheres of Calculate the pressure in atmospheres of

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

IDEAL GAS LAWIDEAL GAS LAW

ExampleExample# 1 Honors Packet# 1 Honors Packet

A 9.81 L cylinder contains 23.5 moles of A 9.81 L cylinder contains 23.5 moles of nitrogen at 23° C. What pressure is nitrogen at 23° C. What pressure is exerted by the gas?exerted by the gas?


A pressure of 850 mmHg is exerted by A pressure of 850 mmHg is exerted by 28.6 grams of sulfur dioxide at a temp of 28.6 grams of sulfur dioxide at a temp of 40 °C. Calculate the volume of the vessel 40 °C. Calculate the volume of the vessel holding the gas.holding the gas.

Density is Hidden in this Density is Hidden in this FormulaFormula

Density = Density = Mass (g)Mass (g)

VolumeVolume

# moles = # moles = Mass (g)Mass (g)

Gram formula massGram formula mass

Can you rework PV = nRT to solve for Can you rework PV = nRT to solve for Density?Density?

ExampleExample# 7 in Honors Packet# 7 in Honors Packet

What is the density of neon at 40 °C and What is the density of neon at 40 °C and 1.23 atmospheres?1.23 atmospheres?

Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

PPtotal total == PP11+P+P22+….+….

Total pressure of a mixture of gases in a Total pressure of a mixture of gases in a container is the container is the sum sum of the individual of the individual pressures (pressures (partial pressurespartial pressures) of each gas, ) of each gas, as if each took up the total space alone.as if each took up the total space alone.

This is often useful when gases are This is often useful when gases are collected “over water”collected “over water”

Collecting Gas over WaterCollecting Gas over Water

http://www.kentchemistry.com/links/GasLaws/dalton.htmCrash Course: Partial Pressure and Vapor PressureCrash Course: Partial Pressure and Vapor Pressurehttp://www.youtube.com/watch?v=JbqtqCunYzA&safe=active

Pressureatm = Pressure (O2) + Pressure (H2O vapor)

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 LawDalton’s Law HH22 gas is collected over gas is collected over

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

Look up water-vapor pressure on for 22.5°C.

GIVEN:

Pgas = ?

Ptotal = 742.0 torr

PH2O = 42.2 torr

WORK:

Ptotal = Pgas + PH2O

742.0 torr = PH2 + 42.2 torr

Pgas = 699.8 torr

A gas is collected over water at a temp of 35.0°C A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? What is the partial pressure of the dry gas?

Dalton’s LawDalton’s Law

DALTON’S LAW

Look up water-vapor pressure for 35.0°C.

Examples Examples of Vapor of Vapor Pressure Pressure TablesTables

You can even use Table H to You can even use Table H to find VP of Waterfind VP of Water

Using Mole Using Mole Fraction Fraction (Honors)(Honors)

Moles gas (X) x P (total) = P (X)

Total Moles Gas


A mixture of 2.00 moles HA mixture of 2.00 moles H22, 3.00 moles , 3.00 moles

NHNH33, 4.00 moles CO, 4.00 moles CO2,2, and 5.00 moles N and 5.00 moles N22

exert a total pressure of 800 mmHg. What exert a total pressure of 800 mmHg. What is the partial pressure of each gas?is the partial pressure of each gas?

Graham’s LawGraham’s LawDiffusionDiffusion

Spreading of gas molecules Spreading of gas molecules throughout a container until throughout a container until evenly distributed.evenly distributed.

EffusionEffusionPassing of gas molecules Passing of gas molecules

through a tiny opening in a through a tiny opening in a containercontainer

Smaller and lighter gas particles Smaller and lighter gas particles do this faster!do this faster!

https://www.youtube.com/watch?feature=player_embedded&v=H7QsDs8ZRMIwww.youtube.com/watch?feature=player_embedded&v=H7QsDs8ZRMIhttps://www.youtube.com/watch?feature=player_embedded&v=L41KhBPBymAhttps://www.youtube.com/watch?feature=player_embedded&v=L41KhBPBymA

Crash Course: Grahams LawCrash Course: Grahams Law http://www.youtube.com/watch?v=TLRZAFU_9Kg&safe=activehttp://www.youtube.com/watch?v=TLRZAFU_9Kg&safe=active

Graham’s LawGraham’s Law

Speed of diffusion/effusionSpeed of diffusion/effusionAt the same temp & KE, At the same temp & KE, heavier molecules heavier molecules

move more slowlymove more slowly..

Ex: Which of the following gases will diffuse Ex: Which of the following gases will diffuse most rapidly?most rapidly?

A.) NA.) N22 B.) COB.) CO22 C.) CHC.) CH44

Graham’s Law FormulaGraham’s Law Formula

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

to the square root of its molar mass.to the square root of its molar mass.

A

B

B

A

m

m

v

v

Ratio of gas Ratio of gas A’s speed to A’s speed to gas B’s speedgas B’s speed

Determine the relative rate of diffusion for Determine the relative rate of diffusion for krypton and bromine.krypton and bromine.

1.381

KrKr diffuses 1.381 times faster than Brdiffuses 1.381 times faster than Br22..

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”. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vRelative rate mean find the ratio “vAA/v/vBB”.”.

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

speed as speed as “Gas A”.“Gas A”.

An unknown gas diffuses 4.0 times faster than OAn unknown gas diffuses 4.0 times faster than O22. Find . Find

its molar mass.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 first gas is “Gas A” and the second gas is “Gas B”. The ratio “vThe ratio “vAA/v/vBB” is 4.0.” is 4.0.

Square both Square both sides to get rid sides to get rid of the square of the square

root sign.root sign.


Under the same conditions of temp and Under the same conditions of temp and pressure, how many times faster will pressure, how many times faster will hydrogen effuse compared to carbon dioxide?hydrogen effuse compared to carbon dioxide?

ExampleExample#7 Honors Packet#7 Honors Packet

If COIf CO22 diffuses from a flask in 30 seconds diffuses from a flask in 30 seconds

and an unknown gas diffuses from the and an unknown gas diffuses from the same flask in 5 minutes, calculate the same flask in 5 minutes, calculate the molecular weight of this unknown gas.molecular weight of this unknown gas.

V

n

Avogadro’s Principle

Volume of a gas is directly proportional to the number of moles of gas particles present.

Equal volumes of gases contain equal numbers of moles of particlesat constant temp & pressuretrue for any gas

Avogadros Law: https://www.youtube.com/watch?feature=player_embedded&v=fexEvn0ZOpoMolar Volume of a Gas: https://www.youtube.com/watch?feature=player_embedded&v=4b852VIEkHQ

Practice QuestionsPractice QuestionsAt the same temperature and pressure, 1.0

liter of CO(g) and 1.0 liter of CO2(g) have

A. equal masses and the same number of molecules

B. different masses and a different number of molecules

C. equal volumes and the same number of molecules

D. different volumes and a different number of molecules

Which rigid cylinder contains the same number of gas molecules at STP as a 2.0-liter rigid cylinder containing H2(g) at STP?

(1) 1.0-L cylinder of O2(g)

(2) 2.0-L cylinder of CH4(g)

(3) 1.5-L cylinder of NH3(g)

(4) 4.0-L cylinder of He(g

Which two samples of gas at STP contain the same total number of molecules?

(1) 1 L of CO(g) and 0.5 L of N2(g)

(2) 2 L of CO(g) and 0.5 L of NH3(g)

(3) 1 L of H2(g) and 2 L of Cl2(g)

(4) 2 L of H2(g) and 2 L of Cl2(g)

Gas StoichiometryGas Stoichiometry

Moles Moles Liters of a Gas Liters of a Gas STP - use 22.4 L/mol STP - use 22.4 L/mol Non-STP - use ideal gas lawNon-STP - use ideal gas law

Non-Non-STP ProblemsSTP Problems Given liters of gas? Given liters of gas?

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

start with stoichiometry conv.start with stoichiometry conv.

1 mol1 molCaCOCaCO33

100.09g 100.09g CaCOCaCO33

Gas Stoichiometry ProblemGas Stoichiometry Problem What volume of COWhat volume of CO22 forms from 5.25 g of forms from 5.25 g of

CaCOCaCO33 at at 1.017atm 1.017atm & 25ºC?& 25ºC?

5.25 g5.25 gCaCOCaCO33 = = 1.26 mol CO1.26 mol CO22

CaCOCaCO33 CaO + CO CaO + CO22

1 mol1 molCOCO22

1 mol1 molCaCOCaCO33

5.25 g5.25 g ? L? Lnon-STPnon-STPLooking for liters: Start with stoich Looking for liters: Start with stoich

and calculate moles of COand calculate moles of CO22..

Plug this into the Ideal Plug this into the Ideal Gas Law to find liters.Gas Law to find liters.

WORK:WORK:

PV = nRTPV = nRT

(1.017atm)V(1.017atm)V=(1mol=(1mol)(.)(.08210821 L Latm/molatm/molKK))(298K)(298K)

V = 1.26 L COV = 1.26 L CO22

Gas Stoichiometry ProblemGas Stoichiometry ProblemWhat volume of COWhat volume of CO22 forms from 5.25 g of forms from 5.25 g of

CaCOCaCO33 at 1.017atm & 25ºC? at 1.017atm & 25ºC?

GIVEN:GIVEN:

PP == 1.017atm1.017atm

nn == 1.26 mol1.26 molTT == 25°C = 298 K25°C = 298 KRR == .0821.0821 L Latm/molatm/molKK

WORK:WORK:

PV = nRTPV = nRT

((0.96 atm ) (15.0 L)) (15.0 L)= n (= n (.0821.0821 L Latm/molatm/molKK) (294K)) (294K)

n = 0.597 mol On = 0.597 mol O22

Gas Stoichiometry ProblemGas Stoichiometry Problem How many grams of AlHow many grams of Al22OO33 are formed from 15.0 L of are formed from 15.0 L of

OO22 at 0.96 atm & 21 at 0.96 atm & 21°C?°C?

GIVEN:GIVEN:

PP == 0.96 atm

V = 15.0 LV = 15.0 L

nn == ??TT == 21°C = 294 K21°C = 294 K

RR = .0821= .0821 L Latm/molatm/molKK

4 Al + 3 O4 Al + 3 O22 2 Al 2 Al22OO33 15.0 L 15.0 L

non-STPnon-STP ? g? gGiven liters: Start with Given liters: Start with

Ideal Gas Law and Ideal Gas Law and calculate moles of Ocalculate moles of O22..

NEXT NEXT

2 mol 2 mol AlAl22OO33

3 mol O3 mol O22

Gas Stoichiometry ProblemGas Stoichiometry Problem How many grams of AlHow many grams of Al22OO33 are formed from are formed from

15.0 L of O15.0 L of O22 at at 0.96 atm 0.96 atm & 21& 21°C?°C?

0.5970.597mol Omol O22 = = 40.6 g Al40.6 g Al22OO33

4 Al + 3 O4 Al + 3 O22 2 Al 2 Al22OO33

101.96 g 101.96 g AlAl22OO33

1 mol1 molAlAl22OO33

15.0L15.0Lnon-STPnon-STP

? g? gUse stoich to convert moles Use stoich to convert moles of Oof O22 to grams Al to grams Al22OO33..

ExampleExample # 10 Honors Multiple Choice# 10 Honors Multiple Choice

Ni(CO)Ni(CO)44 (l) →(l) → Ni Ni (s) (s) + 4CO + 4CO (g)(g)

What volume of CO is formed from the What volume of CO is formed from the complete decomposition of 444g of Ni(CO)complete decomposition of 444g of Ni(CO)44 at at

752 torr and 22.0 °C752 torr and 22.0 °C

Do stoich for 444g to find moles of CO.Do stoich for 444g to find moles of CO.Plug the moles into Ideal Gas Law to get V at non Plug the moles into Ideal Gas Law to get V at non

standard conditions.standard conditions.

Some Cool VideosSome Cool Videos Crash Course: Ideal Gas LawsCrash Course: Ideal Gas Laws http://www.youtube.com/watch?v=BxUS1K7xu30&safe=activehttp://www.youtube.com/watch?v=BxUS1K7xu30&safe=active

Crash Course: Ideal Gas Law ProblemsCrash Course: Ideal Gas Law Problems http://www.youtube.com/watch?v=8SRAkXMu3d0http://www.youtube.com/watch?v=8SRAkXMu3d0

Crash Course: Real GasesCrash Course: Real Gases http://www.youtube.com/watch?v=GIPrsWuSkQc&safe=activehttp://www.youtube.com/watch?v=GIPrsWuSkQc&safe=active

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Using PV = nRT (Honors)