gases kinetic -molecular theory of matter - parkway · pdf filekinetic -molecular theory of...

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GasesGases

Properties of GasesProperties of Gases

Gas PressureGas Pressure

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GasesGases

What gases are important for each of the What gases are important for each of the following: Ofollowing: O22, CO, CO22 and/or He?and/or He?

A. B. C. D.A. B. C. D.

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GasesGases

What gases are important for each of the What gases are important for each of the following: Ofollowing: O22, CO, CO22 and/or He?and/or He?

A. COA. CO22 B. OB. O22/CO/CO22 C. OC. O22 D. HeD. He

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KineticKinetic--Molecular Theory of MatterMolecular Theory of Matter

�� Particles of Matter are always in Particles of Matter are always in motionmotion

�� Ideal GasesIdeal Gases --an imaginary gas that an imaginary gas that fits all the assumptions of the theoryfits all the assumptions of the theory

�� Kinetic Energy (KE) formulaKinetic Energy (KE) formula�� Physical properties of gasesPhysical properties of gases�� Real gasesReal gases -- gases in our daily livesgases in our daily lives

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Ideal GasIdeal Gas--imaginary gas that fits all imaginary gas that fits all the following assumptions.the following assumptions.

Particles in gases: Particles in gases:

��Are very far apartAre very far apart

��Have collisions that are elastic (no KE loss)Have collisions that are elastic (no KE loss)

��Move rapidly Move rapidly

��Have no attraction (or repulsion) Have no attraction (or repulsion)

��Have energy increases at higher temperaturesHave energy increases at higher temperatures

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Kinetic Energy of GasesKinetic Energy of Gases

�� At the same temperature, all gas At the same temperature, all gas particles have the same amount of particles have the same amount of energyenergy–– Lighter particles move faster than Lighter particles move faster than

heavier particlesheavier particles–– KE = KE = mvmv 22 KE =kinetic energy KE =kinetic energy

22 v = velocity (speed)v = velocity (speed)m = massm = mass

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Physical Properties of GasesPhysical Properties of Gases

�� Gases are compressibleGases are compressible

Why can you put more air in a tire, but canWhy can you put more air in a tire, but can’’t t

add more water to a glass full of water? add more water to a glass full of water?

�� Gases have low densitiesGases have low densities

DsolidDsolid or liquid = 2 g/mL or liquid = 2 g/mL

DgasDgas 2 g/L2 g/L

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Physical Properties of GasesPhysical Properties of Gases

1. Why does a round balloon 1. Why does a round balloon

become spherical when filled become spherical when filled

with air?with air?

2. Suppose we filled this room halfway with 2. Suppose we filled this room halfway with water. Where would pressure be exerted?water. Where would pressure be exerted?

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Physical Properties of GasesPhysical Properties of Gases�� Gases fill a container completely and Gases fill a container completely and

uniformlyuniformly

��Gases exert a uniform pressure on all Gases exert a uniform pressure on all inner surfaces of their containersinner surfaces of their containers

�� Fluidity: gas particles can slide past one Fluidity: gas particles can slide past one another (gases and liquids are another (gases and liquids are ““fluidsfluids””))

� Diffusion: gases move from high concentration to low concentration

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Real GasesReal Gases

�� Close to ideal gas at standard Close to ideal gas at standard conditionsconditions

�� Have volume Have volume �� Attraction between particlesAttraction between particles�� Deviation from ideal gas is greater Deviation from ideal gas is greater

whenwhen–– Particles are close togetherParticles are close together

»» Low temperaturesLow temperatures»» High pressuresHigh pressures

–– Gas is a compound rather than an Gas is a compound rather than an elementelement

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Some Gases in Our LivesSome Gases in Our Lives

Air:Air:

oxygen Ooxygen O22 nitrogen Nnitrogen N22 ozone Oozone O33

argon argon ArAr carbon dioxide COcarbon dioxide CO2 2 water Hwater H22OONoble gasesNoble gases: :

helium He neon Ne krypton Kr xenon helium He neon Ne krypton Kr xenon XeXeOther gases:Other gases:

fluorine Ffluorine F22 chlorine Clchlorine Cl22 ammonia NHammonia NH3 3

methane CHmethane CH44 carbon monoxide COcarbon monoxide CO

nitrogen dioxide NOnitrogen dioxide NO22 sulfur dioxide SOsulfur dioxide SO22 12

PressurePressure�� Force per unit areaForce per unit area�� P = P = forceforce

areaarea�� Gas pressure is a result of collisions Gas pressure is a result of collisions

of gas particles. Depends on:of gas particles. Depends on:–– Number of gas particlesNumber of gas particles–– Temperature Temperature –– VolumeVolume

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Instruments used to measure pressureInstruments used to measure pressure

�� Barometer Barometer –– measures atmospheric measures atmospheric pressurepressure

�� Manometer Manometer –– measures gas pressure measures gas pressure of a containerof a container

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BarometersBarometers

760 mmHg760 mmHg

atmatm

pressurepressure

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ManometerManometer

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Unit of Pressure Unit of Pressure

One atmosphere (1 One atmosphere (1 atmatm) )

��Is the average pressure of the atmosphere at Is the average pressure of the atmosphere at

sea level sea level

��Is the standard of pressureIs the standard of pressure

�� P = P = ForceForce

AreaArea

1.00atm=760mmHg=760torr=101.3kPa=14.7psi1.00atm=760mmHg=760torr=101.3kPa=14.7psi

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Types of Pressure UnitsTypes of Pressure Units

PressurePressure Used inUsed in

760 mm Hg or 760 760 mm Hg or 760 torrtorr ChemistryChemistry

14.7 lb/in.14.7 lb/in.22 U.S. pressure gaugesU.S. pressure gauges

29.9 in. Hg29.9 in. Hg U.S. weather reportsU.S. weather reports

101.3 101.3 kPakPa (kilopascals)(kilopascals) Weather in all Weather in all countries except U.S.countries except U.S.

1.013 bars1.013 bars Physics and Physics and astronomyastronomy

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ConversionsConversions�� 760.mmHg760.mmHg ==760.torr760.torr ==1.00atm1.00atm==101.3kPa101.3kPa==14.7psi14.7psi

What is 2.00 What is 2.00 atmatm expressed in expressed in torrtorr??

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ConversionsConversions�� 760.mmHg760.mmHg ==760.torr760.torr ==1.00atm1.00atm==101.3kPa101.3kPa==14.7psi14.7psi

�� The pressure of a tire is measured The pressure of a tire is measured as 32.0 psi.as 32.0 psi.

What is this pressure in What is this pressure in kPakPa??

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Learning Check G1Learning Check G1

A.The downward pressure of the Hg in a A.The downward pressure of the Hg in a barometer is _____ than (as) the weight of the barometer is _____ than (as) the weight of the atmosphere.atmosphere.

1) greater1) greater 2) less2) less 3) the same3) the same

B.A water barometer has to be 13.6 times taller B.A water barometer has to be 13.6 times taller than Hg barometer (than Hg barometer (DDHgHg = 13.6 g/mL)= 13.6 g/mL)becausebecause

1) H1) H22O is less dense O is less dense 2) Hg is heavier2) Hg is heavier

3) air is more dense than H3) air is more dense than H22OO

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Solution G1Solution G1

A.The downward pressure of the Hg in a A.The downward pressure of the Hg in a barometer is barometer is 3) the same3) the same (as) the weight of the (as) the weight of the atmosphere.atmosphere.

B.A water barometer has to be 13.6 times taller B.A water barometer has to be 13.6 times taller than Hg barometer (than Hg barometer (DDHgHg = 13.6 g/mL)= 13.6 g/mL)

becausebecause

1) H1) H22O is less dense O is less dense

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When you drink through a straw you reduce the When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up pressure in the straw. Why does the liquid go up the straw? the straw?

1) the weight of the atmosphere pushes it1) the weight of the atmosphere pushes it

2) the liquid is at a lower level2) the liquid is at a lower level

3) there is empty space in the straw3) there is empty space in the straw

Could you drink a soda this way on the moon?Could you drink a soda this way on the moon?

1) yes 2) no 1) yes 2) no 3) maybe3) maybe

Why or why not?Why or why not?

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When you drink through a straw you reduce the When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up pressure in the straw. Why does the liquid go up the straw? the straw?

1) the weight of the atmosphere pushes it1) the weight of the atmosphere pushes it

3) there is empty space in the straw3) there is empty space in the straw

Could you drink a soda this way on the moon?Could you drink a soda this way on the moon?

2) no 2) no

Why or why not? Why or why not? Low atmospheric pressureLow atmospheric pressure24


A. What is 475 mm Hg expressed in A. What is 475 mm Hg expressed in atmatm??

1) 475 1) 475 atmatm 2) 0.625 2) 0.625 atmatm 3) 3.61 x 103) 3.61 x 1055 atmatm

B. The pressure of a tire is measured as 29.4 psi.B. The pressure of a tire is measured as 29.4 psi.

What is this pressure in mm Hg?What is this pressure in mm Hg?

1) 1) 2.00 mm Hg2.00 mm Hg

2) 2) 1520 mm Hg1520 mm Hg

3)3) 22,300 mm Hg22,300 mm Hg

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A. What is 475 mm Hg expressed in A. What is 475 mm Hg expressed in atmatm??

485 mm Hg x 485 mm Hg x 1 1 atmatm = 0.625 = 0.625 atmatm (B)(B)

760. mm Hg760. mm Hg

B. The pressure of a tire is measured as 29.4 psi.B. The pressure of a tire is measured as 29.4 psi.

What is this pressure in mm Hg?What is this pressure in mm Hg?

29.4 psi x 29.4 psi x 760. mmHg760. mmHg = 1.52 x 10= 1.52 x 1033 mmHgmmHg

14.7 psi 14.7 psi (B)(B)

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Gas LawsGas Laws

�� BoyleBoyle ’’s Laws Law�� CharlesCharles ’’ LawLaw�� GayGay--LussacLussac ’’s Laws Law�� Combined Gas LawCombined Gas Law�� Ideal Gas LawIdeal Gas Law�� DaltonDalton ’’s Partial Pressures Partial Pressure�� GrahamGraham ’’s Law of Effusions Law of Effusion

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BoyleBoyle’’ s Laws Law

�� Reducing the volume by oneReducing the volume by one --half half doubles the pressuredoubles the pressure

�� The volume of a fixed mass of gas The volume of a fixed mass of gas varies inversely with the pressure at varies inversely with the pressure at constant temperatureconstant temperature

�� PP11VV11 –– PP22VV22

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As the pressure As the pressure As the pressure As the pressure increasesincreasesincreasesincreases

As the pressure As the pressure As the pressure As the pressure increasesincreasesincreasesincreases

VolumeVolumeVolumeVolumedecreasesdecreasesdecreasesdecreases

VolumeVolumeVolumeVolumedecreasesdecreasesdecreasesdecreases

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Pressure and VolumePressure and Volume

ExperimentExperimentPressurePressure Volume P x VVolume P x V

((atmatm) (L) () (L) (atmatm x L)x L)

11 8.0 8.0 2.0 2.0 16 16

22 4.04.0 4.04.0 __________

33 2.02.0 8.08.0 __________

44 1.01.0 1616 __________

Boyle's Law P x V = k (constant) when Boyle's Law P x V = k (constant) when

T remains constantT remains constant 30

How does Pressure and Volume of gases relate graphically?

How does Pressure and Volume of gases relate graphically?

Volume

Volume

PressurePressure

PV = kPV = k

Temperature, Temperature, Temperature, Temperature, # of particles# of particles# of particles# of particlesremain constantremain constantremain constantremain constant

Temperature, Temperature, Temperature, Temperature, # of particles# of particles# of particles# of particlesremain constantremain constantremain constantremain constant

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BoyleBoyleBoyleBoyle’’’’s Mathematical Law:s Mathematical Law:s Mathematical Law:s Mathematical Law:BoyleBoyleBoyleBoyle’’’’s Mathematical Law:s Mathematical Law:s Mathematical Law:s Mathematical Law:

since PV = ksince PV = ksince PV = ksince PV = ksince PV = ksince PV = ksince PV = ksince PV = k

P1V1 = P2V2P1V1 = P2V2

EgEgEgEg: A gas has a volume of 3.0 L at : A gas has a volume of 3.0 L at : A gas has a volume of 3.0 L at : A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 2 atm. What is its volume at 4 2 atm. What is its volume at 4 2 atm. What is its volume at 4 atmatmatmatm????

EgEgEgEg: A gas has a volume of 3.0 L at : A gas has a volume of 3.0 L at : A gas has a volume of 3.0 L at : A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 2 atm. What is its volume at 4 2 atm. What is its volume at 4 2 atm. What is its volume at 4 atmatmatmatm????

What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?

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1)1)1)1) determine which variables you determine which variables you determine which variables you determine which variables you have:have:have:have:


P and V = BoyleP and V = BoyleP and V = BoyleP and V = Boyle’’’’s Laws Laws Laws LawP and V = BoyleP and V = BoyleP and V = BoyleP and V = Boyle’’’’s Laws Laws Laws Law

2)2)2)2)determine which law is being determine which law is being determine which law is being determine which law is being represented:represented:represented:represented:


� PPPP1111 = 2 = 2 = 2 = 2 atmatmatmatm� VVVV1111 = 3.0 L= 3.0 L= 3.0 L= 3.0 L� PPPP2222 = 4 = 4 = 4 = 4 atmatmatmatm� VVVV2222 = ?= ?= ?= ?

� PPPP1111 = 2 = 2 = 2 = 2 atmatmatmatm� VVVV1111 = 3.0 L= 3.0 L= 3.0 L= 3.0 L� PPPP2222 = 4 = 4 = 4 = 4 atmatmatmatm� VVVV2222 = ?= ?= ?= ?

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3) Rearrange the equation for 3) Rearrange the equation for 3) Rearrange the equation for 3) Rearrange the equation for the variable you donthe variable you donthe variable you donthe variable you don’’’’t knowt knowt knowt know

3) Rearrange the equation for 3) Rearrange the equation for 3) Rearrange the equation for 3) Rearrange the equation for the variable you donthe variable you donthe variable you donthe variable you don’’’’t knowt knowt knowt know

4) Plug in the variables and 4) Plug in the variables and 4) Plug in the variables and 4) Plug in the variables and chug it on a calculator:chug it on a calculator:chug it on a calculator:chug it on a calculator:

4) Plug in the variables and 4) Plug in the variables and 4) Plug in the variables and 4) Plug in the variables and chug it on a calculator:chug it on a calculator:chug it on a calculator:chug it on a calculator:

PPPP1111VVVV1111 = V= V= V= V2222PPPP1111VVVV1111 = V= V= V= V2222

PPPP2222PPPP2222

(2.0 atm)(3.0L) = V(2.0 atm)(3.0L) = V(2.0 atm)(3.0L) = V(2.0 atm)(3.0L) = V2222(2.0 atm)(3.0L) = V(2.0 atm)(3.0L) = V(2.0 atm)(3.0L) = V(2.0 atm)(3.0L) = V2222

(4atm)(4atm)(4atm)(4atm)(4atm)(4atm)(4atm)(4atm)VVVV2222 = 1.5L= 1.5L= 1.5L= 1.5LVVVV2222 = 1.5L= 1.5L= 1.5L= 1.5L

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Learning Check GL1Learning Check GL1

A sample of nitrogen gas is 6.4 L at a pressure A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 the pressure is changed to 1.40 atmatm? (T ? (T constant) Explain.constant) Explain.

1) 3.2 L1) 3.2 L

2) 6.4 L2) 6.4 L

3) 12.8 L3) 12.8 L

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Solution GL1Solution GL1

A sample of nitrogen gas is 6.4 L at a pressure A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 the pressure is changed to 1.40 atmatm? (T ? (T constant)constant)

6.4 L x 6.4 L x 0.70 0.70 atmatm = = 3.2 L (1)3.2 L (1)

1.40 1.40 atmatm

Volume must decrease to cause an increase Volume must decrease to cause an increase in the pressurein the pressure

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A sample of helium gas has a volume of A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? is needed to change the volume to 36.0 L? (T constant) Explain.(T constant) Explain.

1) 200. mmHg 1) 200. mmHg

2) 400. mmHg 2) 400. mmHg

3) 1200 mmHg3) 1200 mmHg

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A sample of helium gas has a volume of A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? is needed to change the volume to 36.0 L? (T constant) Explain.(T constant) Explain.

600. mm Hg x 600. mm Hg x 12.0 L12.0 L = 200. mmHg (1) = 200. mmHg (1)

36.0 L36.0 L

Pressure decrease when volume increases.Pressure decrease when volume increases.

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Volume of balloon Volume of balloon Volume of balloon Volume of balloon at room at room at room at room

temperaturetemperaturetemperaturetemperature

Volume of balloon Volume of balloon Volume of balloon Volume of balloon at room at room at room at room

temperaturetemperaturetemperaturetemperature

Volume of balloon Volume of balloon Volume of balloon Volume of balloon at 5at 5at 5at 5°°°°CCCC

Volume of balloon Volume of balloon Volume of balloon Volume of balloon at 5at 5at 5at 5°°°°CCCC

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CharlesCharles’’ LawLaw

V V = 125 mL V = 125 mL V = 250 mL= 250 mL

T T = 273 K T = 273 K T = 546 K = 546 K

Observe the V and T of the balloons. How does Observe the V and T of the balloons. How does volume change with temperature?volume change with temperature?

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CharlesCharles’’ Law: V and TLaw: V and T

At constant pressure, the volume of a gas isAt constant pressure, the volume of a gas is

directly related to its absolute (K) temperaturedirectly related to its absolute (K) temperature

VV11 = = VV22 oror VV11TT22 = V= V22TT11

TT11 TT22

K = K = °°C + 273C + 273

41 Temp

How does Temperature and How does Temperature and How does Temperature and How does Temperature and Volume of gases relate graphically?Volume of gases relate graphically?Volume of gases relate graphically?Volume of gases relate graphically?

How does Temperature and How does Temperature and How does Temperature and How does Temperature and Volume of gases relate graphically?Volume of gases relate graphically?Volume of gases relate graphically?Volume of gases relate graphically?

Volume

V/T = k

Pressure, Pressure, Pressure, Pressure, # of particles# of particles# of particles# of particlesremain constantremain constantremain constantremain constant

Pressure, Pressure, Pressure, Pressure, # of particles# of particles# of particles# of particlesremain constantremain constantremain constantremain constant

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T and V = CharlesT and V = CharlesT and V = CharlesT and V = Charles’’’’s Laws Laws Laws LawT and V = CharlesT and V = CharlesT and V = CharlesT and V = Charles’’’’s Laws Laws Laws Law



� TTTT1111 = 127= 127= 127= 127°°°°C + 273 = 400KC + 273 = 400KC + 273 = 400KC + 273 = 400K� VVVV1111 = 3.0 L= 3.0 L= 3.0 L= 3.0 L� TTTT2222 = 227= 227= 227= 227°°°°C + 273 = 5ooKC + 273 = 5ooKC + 273 = 5ooKC + 273 = 5ooK� VVVV2222 = ?= ?= ?= ?

� TTTT1111 = 127= 127= 127= 127°°°°C + 273 = 400KC + 273 = 400KC + 273 = 400KC + 273 = 400K� VVVV1111 = 3.0 L= 3.0 L= 3.0 L= 3.0 L� TTTT2222 = 227= 227= 227= 227°°°°C + 273 = 5ooKC + 273 = 5ooKC + 273 = 5ooKC + 273 = 5ooK� VVVV2222 = ?= ?= ?= ?

8

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4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:

(500K)(3.0L) = V(500K)(3.0L) = V(500K)(3.0L) = V(500K)(3.0L) = V2 2 2 2 (400K)(400K)(400K)(400K)(500K)(3.0L) = V(500K)(3.0L) = V(500K)(3.0L) = V(500K)(3.0L) = V2 2 2 2 (400K)(400K)(400K)(400K)

VVVV2222 = 3.8L= 3.8L= 3.8L= 3.8LVVVV2222 = 3.8L= 3.8L= 3.8L= 3.8L

3.0L V3.0L V3.0L V3.0L V22223.0L V3.0L V3.0L V3.0L V2222

400K 500K400K 500K400K 500K400K 500K400K 500K400K 500K400K 500K400K 500K========

5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug

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Use CharlesUse Charles’’ Law to complete the statements Law to complete the statements

below:below:

1. If final T is higher than initial T, final V1. If final T is higher than initial T, final V

is (is (greater, or lessgreater, or less) than the initial V.) than the initial V.

2. If final V is less than initial V, final T is2. If final V is less than initial V, final T is

((higher, or lowerhigher, or lower) than the initial T.) than the initial T.

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VV11 = = VV22 OR OR VV11TT22 == VV22TT11

TT11 TT22

1. If final T is higher than initial T, final V1. If final T is higher than initial T, final V

is (is (greatergreater) than the initial V.) than the initial V.

2. If final V is less than initial V, final T is (2. If final V is less than initial V, final T is (lowerlower) ) than the initial T.than the initial T.

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V and T ProblemV and T Problem

A balloon has a volume of 785 A balloon has a volume of 785 mLmL on a Fall day when the on a Fall day when the temperature is 21temperature is 21°°C. In the C. In the winter, the gas cools to 0winter, the gas cools to 0°°C. C. What is the new volume of the What is the new volume of the balloon? balloon?

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VT CalculationVT Calculation

Complete the following setup:Complete the following setup:

Initial conditionsInitial conditions Final conditionsFinal conditionsVV11 = 785 mL= 785 mL VV22 = ?= ?

TT11 = 21= 21°°C = 294 KC = 294 K TT22 = 0= 0°°C = 273 KC = 273 K

VV22 = _______ mL x = _______ mL x __ K __ K = _______ mL= _______ mL

VV11 KK

Check your answer: If temperature decreases, Check your answer: If temperature decreases,

V should decrease.V should decrease. 48


A sample of oxygen gas has a volume of A sample of oxygen gas has a volume of 420 mL at a temperature of 18420 mL at a temperature of 18°°C. What C. What temperature (in temperature (in °°C) is needed to change C) is needed to change the volume to 640 mL?the volume to 640 mL?

1) 4431) 443°°CC 2) 1702) 170°°C 3) C 3) -- 8282°°CC

9

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A sample of oxygen gas has a volume of A sample of oxygen gas has a volume of 420 mL at a temperature of 18420 mL at a temperature of 18°°C. What C. What temperature (in temperature (in °°C) is needed to change C) is needed to change the volume to 640 mL?the volume to 640 mL?

2) 1702) 170°°CC

TT22 = 291 K x = 291 K x 640 mL640 mL = 443 K= 443 K

420 mL420 mL

= 443 K = 443 K -- 273 K 273 K = 170= 170°°CC

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GayGay--LussacLussac’’ s Law: P and Ts Law: P and T

Doubling the Kelvin temperature doubles the pressure

The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature

P1 = P2 or P1T2 = P2T1T1 T2

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Pressure Pressure Pressure Pressure GaugeGaugeGaugeGauge


Car before a tripCar before a tripCar before a tripCar before a tripCar before a tripCar before a tripCar before a tripCar before a trip

Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...

LetLetLetLet’’’’s get ons get ons get ons get onthe road the road the road the road Dude!Dude!Dude!Dude!

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Car after a long tripCar after a long tripCar after a long tripCar after a long tripCar after a long tripCar after a long tripCar after a long tripCar after a long trip

Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...Think of a tire...

WHEW!WHEW!WHEW!WHEW!



53 Temp

Pressure

How does Pressure and How does Pressure and How does Pressure and How does Pressure and Temperature of gases relate Temperature of gases relate Temperature of gases relate Temperature of gases relate

graphically?graphically?graphically?graphically?

How does Pressure and How does Pressure and How does Pressure and How does Pressure and Temperature of gases relate Temperature of gases relate Temperature of gases relate Temperature of gases relate

graphically?graphically?graphically?graphically?

P/T = k

Volume, Volume, Volume, Volume, # of particles# of particles# of particles# of particlesremain constantremain constantremain constantremain constant

Volume, Volume, Volume, Volume, # of particles# of particles# of particles# of particlesremain constantremain constantremain constantremain constant

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LussacLussacLussacLussac’’’’ssss Mathematical Law:Mathematical Law:Mathematical Law:Mathematical Law:LussacLussacLussacLussac’’’’ssss Mathematical Law:Mathematical Law:Mathematical Law:Mathematical Law:

What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?What if we had a change in conditions?

since P/T = ksince P/T = ksince P/T = ksince P/T = ksince P/T = ksince P/T = ksince P/T = ksince P/T = k

P1 P2

T1 T2=

EgEgEgEg: A gas has a pressure of 3.0 : A gas has a pressure of 3.0 : A gas has a pressure of 3.0 : A gas has a pressure of 3.0 atmatmatmatm at at at at 127127127127ºººº C. What is its pressure at 227C. What is its pressure at 227C. What is its pressure at 227C. What is its pressure at 227ºººº C?C?C?C?

EgEgEgEg: A gas has a pressure of 3.0 : A gas has a pressure of 3.0 : A gas has a pressure of 3.0 : A gas has a pressure of 3.0 atmatmatmatm at at at at 127127127127ºººº C. What is its pressure at 227C. What is its pressure at 227C. What is its pressure at 227C. What is its pressure at 227ºººº C?C?C?C?

10

55T and P = GayT and P = GayT and P = GayT and P = Gay----LussacLussacLussacLussac’’’’s Laws Laws Laws LawT and P = GayT and P = GayT and P = GayT and P = Gay----LussacLussacLussacLussac’’’’s Laws Laws Laws Law

� TTTT1111 = 127= 127= 127= 127°°°°C + 273 = 400KC + 273 = 400KC + 273 = 400KC + 273 = 400K

� PPPP1111 = 3.0 = 3.0 = 3.0 = 3.0 atmatmatmatm

� TTTT2222 = 227= 227= 227= 227°°°°C + 273 = 500KC + 273 = 500KC + 273 = 500KC + 273 = 500K

� PPPP2222 = ?= ?= ?= ?

� TTTT1111 = 127= 127= 127= 127°°°°C + 273 = 400KC + 273 = 400KC + 273 = 400KC + 273 = 400K

� PPPP1111 = 3.0 = 3.0 = 3.0 = 3.0 atmatmatmatm

� TTTT2222 = 227= 227= 227= 227°°°°C + 273 = 500KC + 273 = 500KC + 273 = 500KC + 273 = 500K

� PPPP2222 = ?= ?= ?= ?



2)2)2)2)determine which law is determine which law is determine which law is determine which law is being represented:being represented:being represented:being represented:

2)2)2)2)determine which law is determine which law is determine which law is determine which law is being represented:being represented:being represented:being represented:

56

4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:4) Plug in the variables:

(500K)(3.0atm) = P(500K)(3.0atm) = P(500K)(3.0atm) = P(500K)(3.0atm) = P2 2 2 2 (400K)(400K)(400K)(400K)(500K)(3.0atm) = P(500K)(3.0atm) = P(500K)(3.0atm) = P(500K)(3.0atm) = P2 2 2 2 (400K)(400K)(400K)(400K)

PPPP2222 = 3.8atm= 3.8atm= 3.8atm= 3.8atmPPPP2222 = 3.8atm= 3.8atm= 3.8atm= 3.8atm

3.0atm P3.0atm P3.0atm P3.0atm P22223.0atm P3.0atm P3.0atm P3.0atm P2222

400K 500K400K 500K400K 500K400K 500K400K 500K400K 500K400K 500K400K 500K========

5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug5) Cross multiply and chug

57V, nV, nPP11/T/T11 = P= P22/T/T22PP↑↑↑↑↑↑↑↑ TT↑↑↑↑↑↑↑↑GayGay--LussacLussac ’’ss

P, nP, nVV11/T/T11 = V= V22/T/T22VV↑↑↑↑↑↑↑↑ TT↑↑↑↑↑↑↑↑CharlesCharles ’’

T, nT, nPP11VV1 1 = P= P22VV22PP↑↑↑↑↑↑↑↑ VV↓↓↓↓↓↓↓↓BoyleBoyle ’’ss

CONCON--STANTSTANTLAWLAWRELATRELAT--IONSHIPIONSHIPLAWLAW

58


Use GayUse Gay--LussacLussac’’s law to complete the statements s law to complete the statements below: below:

1. When temperature decreases, the1. When temperature decreases, the

pressure of a gas (pressure of a gas (decreases or increasesdecreases or increases).).

2. When temperature increases, the pressure 2. When temperature increases, the pressure

of a gas of a gas (decreases or increases(decreases or increases).).

59


1. When temperature decreases, the1. When temperature decreases, the

pressure of a gas (pressure of a gas (decreasesdecreases).).

2. When temperature increases, the2. When temperature increases, the

pressure of a gas (pressure of a gas (increasesincreases).).

60

PT ProblemPT Problem

A gas has a pressure at 2.0 A gas has a pressure at 2.0 atmatm at 18at 18°°C. C. What will be the new pressure if the What will be the new pressure if the temperature rises to 62temperature rises to 62°°C? (V constant)C? (V constant)

T = 18T = 18°°C T = 62C T = 62°°CC

11

61

PT CalculationPT Calculation

PP11 = 2.0 = 2.0 atmatm TT11 = 18= 18°°C + 273 = 291 KC + 273 = 291 K

PP22 = ?= ? TT22 = 62= 62°°C + 273 = 335 KC + 273 = 335 K

What happens to P when T increases?What happens to P when T increases?

P increases (directly related to T)P increases (directly related to T)

PP22 = P= P11 x x TT22

TT11

PP22 = 2.0 = 2.0 atmatm x x KK = = atmatm

KK

?

62


Complete with Complete with 1) Increases 2) Decreases 1) Increases 2) Decreases

3) Does not change3) Does not change

A. Pressure _____, when V decreasesA. Pressure _____, when V decreases

B. When T decreases, V _____.B. When T decreases, V _____.

C. Pressure _____ when V changes from 12.0 L to C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T)24.0 L (constant n and T)

D. Volume _____when T changes from 15.0 D. Volume _____when T changes from 15.0 °°C to C to 45.045.0°°C (constant P and n)C (constant P and n)

63


A. Pressure A. Pressure 1) Increases1) Increases, when V decreases, when V decreases

B. When T decreases, V B. When T decreases, V 2) Decreases2) Decreases

C. Pressure C. Pressure 2) Decreases 2) Decreases when V changes when V changes

from 12.0 L to 24.0 L (constant n and T)from 12.0 L to 24.0 L (constant n and T)

D. Volume D. Volume 1) Increases 1) Increases when T changes from 15.0 when T changes from 15.0 °°C to 45.0C to 45.0°°C (constant P and n)C (constant P and n)

64

The Combined Gas LawThe Combined Gas Law

Volume and Moles Volume and Moles

(Avogadro(Avogadro’’s Law)s Law)

Partial PressuresPartial Pressures

65

Combined Gas LawCombined Gas Law

PP11VV11 = P= P22VV22 oror PP11VV11TT22 = P= P22VV22TT11

TT11 TT22

Rearrange the combined gas law to solve for VRearrange the combined gas law to solve for V22

PP11VV11TT22 = P= P22VV22TT11

VV22 = P= P11VV11TT22

PP22TT11

66

Combined Gas LawCombined Gas Law

PP11VV11 = P= P22VV22 oror PP11VV11TT22 = P= P22VV22TT11

TT11 TT22

Isolate VIsolate V22


VV22 = P= P11VV11TT22

PP22TT11

12

67

Learning Check C1Learning Check C1

Solve the combined gas laws for TSolve the combined gas laws for T22..

68

Solution C1Solution C1

Solve the combined gas law for TSolve the combined gas law for T22. .

(Hint: cross(Hint: cross--multiply first.)multiply first.)

PP11VV11 = P= P22VV22

TT11 TT22


TT22 = P= P22VV22TT11

PP11VV11

69

Combined Gas Law ProblemCombined Gas Law Problem

A sample of helium gas has a volume of 0.180 A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 L, a pressure of 0.800 atmatm and a temperature of and a temperature of 2929°°C. What is the new temperature(C. What is the new temperature(°°C) of the C) of the gas at a volume of 90.0 mL and a pressure of gas at a volume of 90.0 mL and a pressure of 3.20 3.20 atmatm??

70

Data TableData Table

Set up Data TableSet up Data Table

PP1 1 = 0.800 = 0.800 atmatm VV11 = 0.180 L= 0.180 L TT11 = 302 K= 302 K

PP22 = 3.20 = 3.20 atmatm VV22= 90.0 mL T= 90.0 mL T2 2 = ??= ????

71

SolutionSolution

Solve for TSolve for T22Enter dataEnter data

TT22 = 302 K x = 302 K x atmatm x x mL mL = K= K

atmatm mLmL

TT22 = K = K -- 273 = 273 = °°CC

72

CalculationCalculation

Solve for TSolve for T22

TT22 = 302 K x = 302 K x 3.20 3.20 atmatm x x 90.0 mL 90.0 mL = 604 K= 604 K

0.800 0.800 atmatm 180.0 mL180.0 mL

TT22 = 604 K = 604 K -- 273 = 331 273 = 331 °°CC

13

73


A gas has a volume of 675 mL at 35A gas has a volume of 675 mL at 35°°C and C and 0.850 0.850 atmatm pressure. What is the temperature pressure. What is the temperature in in °°C when the gas has a volume of 0.315 L C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?and a pressure of 802 mm Hg?

74


TT11 = 308 K= 308 K TT22 = ?= ?

VV11 = 675 mL= 675 mL VV22 = 0.315 L = 315 mL= 0.315 L = 315 mL

PP11 = 0.850 = 0.850 atmatm PP22 = 802 mm Hg = 802 mm Hg = 646 mm Hg = 646 mm Hg

TT22 = 308 K x = 308 K x 802 mm Hg 802 mm Hg x x 315 mL315 mL

646 mm Hg 675 mL646 mm Hg 675 mL

P inc, T inc V P inc, T inc V decdec, T , T decdec

= 178 K = 178 K -- 273 = 273 = -- 9595°°C C

75

Volume and MolesVolume and Moles

How does adding more molecules of a gas How does adding more molecules of a gas change the volume of the air in a tire?change the volume of the air in a tire?

If a tire has a leak, how does the loss of air (gas) If a tire has a leak, how does the loss of air (gas) molecules change the volume? molecules change the volume?

76


True (1) or False(2)True (1) or False(2)

1.___The P exerted by a gas at constant V is not 1.___The P exerted by a gas at constant V is not affected by the T of the gas.affected by the T of the gas.

2.___ At constant P, the V of a gas is directly 2.___ At constant P, the V of a gas is directly proportional to the absolute Tproportional to the absolute T

3.___ At constant T, doubling the P will cause the V 3.___ At constant T, doubling the P will cause the V

of the gas sample to decrease to oneof the gas sample to decrease to one--half its half its

original V.original V.

77


True (1) or False(2)True (1) or False(2)

1. 1. (2)(2)The P exerted by a gas at constant V is not The P exerted by a gas at constant V is not affected by the T of the gas.affected by the T of the gas.

2. 2. (1)(1) At constant P, the V of a gas is directly At constant P, the V of a gas is directly proportional to the absolute Tproportional to the absolute T

3. 3. (1)(1) At constant T, doubling the P will cause the V At constant T, doubling the P will cause the V

of the gas sample to decrease to oneof the gas sample to decrease to one--half its half its

original V.original V.

78

AvogadroAvogadro’’ s Laws Law

When a gas is at constant T and P, the V is When a gas is at constant T and P, the V is directly proportional to the number of moles (n) directly proportional to the number of moles (n) of gasof gas

VV11 = = VV22

nn11 nn22

initial initial final final

14

79

STPSTP

The volumes of gases can be compared when The volumes of gases can be compared when they have the same temperature and pressure they have the same temperature and pressure (STP).(STP).

Standard temperature 0Standard temperature 0°°C or 273 KC or 273 K

Standard pressureStandard pressure 1 1 atmatm (760 mm Hg)(760 mm Hg)

80


A sample of neon gas used in a neon sign has a A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 the neon gas at 2.0 atmatm and and ––2525°°C?C?

PP1 1 = = VV11 = = TT11 = = KK

PP22 = = VV22 = ?? = ?? TT22 = = KK

VV22 = 15 L x = 15 L x atmatm x x KK = = 6.8 L6.8 L

atmatm KK

81


PP1 1 = = 1.0 1.0 atmatm VV11 = 15 L = 15 L TT11 = = 273 K273 K

PP22 = = 2.0 2.0 atmatm VV22 = ?? = ?? TT22 = = 248 K248 K

VV22 = 15 L x = 15 L x 1.0 1.0 atmatm x x 248 K248 K = = 6.8 L6.8 L

2.0 2.0 atmatm 273 K273 K

82

Ideal Gas LawIdeal Gas Law

The equality for the four variables involved in The equality for the four variables involved in BoyleBoyle’’s Law, Charless Law, Charles’’ Law, GayLaw, Gay--LussacLussac’’s Law s Law and Avogadroand Avogadro’’s law can be writtens law can be written

PV = PV = nnRRTT

R = ideal gas constant R = ideal gas constant

83

PV = PV = nRTnRT

R is known as the universal gas constantR is known as the universal gas constant

Using STP conditionsUsing STP conditionsP V P V

R = R = PV PV = = (1.00 atm)(22.4 L) (1.00 atm)(22.4 L) nTnT (1mol) (273K)(1mol) (273K)

n Tn T

= = 0.0821 0.0821 LL--atmatm

molmol--KK 84


What is the value of R when the STP value for What is the value of R when the STP value for P is 760 mmHg? P is 760 mmHg?

15

85


What is the value of R when the STP value for What is the value of R when the STP value for P is 760 mmHg? P is 760 mmHg?

R = R = PV PV = = (760(760 mm Hg) (22.4 L) mm Hg) (22.4 L)

nTnT (1mol) (273K)(1mol) (273K)

= = 62.4 L62.4 L--mm Hgmm Hgmolmol--KK

86


DinitrogenDinitrogen monoxide (Nmonoxide (N22O), laughing gas, is O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23of gas occupies a 20.0 L tank at 23°°C, what is C, what is the pressure (mmHg) in the tank in the dentist the pressure (mmHg) in the tank in the dentist office?office?

87


Set up data for 3 of the 4 gas variables Set up data for 3 of the 4 gas variables

Adjust to match the units of R Adjust to match the units of R

V = 20.0 LV = 20.0 L 20.0 L20.0 L

T = 23T = 23°°C + 273 C + 273 296 K296 K

n = 2.86 moln = 2.86 mol 2.86 mol2.86 mol

P = ? P = ? ??

88

Rearrange ideal gas law for unknown PRearrange ideal gas law for unknown P

P = P = nRTnRT

VV

Substitute values of n, R, T and V and solve Substitute values of n, R, T and V and solve for Pfor P

P P = (= (2.86 mol)(62.42.86 mol)(62.4LL--mmHgmmHg)(296)(296 KK))

(20.0(20.0 LL) ) (K(K--molmol))

= 2.64 x 10= 2.64 x 1033 mm Hgmm Hg

89


A 5.0 L cylinder contains oxygen gas at A 5.0 L cylinder contains oxygen gas at 20.020.0°°C and 735 mm Hg. How many C and 735 mm Hg. How many grams of oxygen are in the cylinder?grams of oxygen are in the cylinder?

90


Solve ideal gas equation for n (moles)Solve ideal gas equation for n (moles)n n = = PVPV

RTRT

= = (735 mmHg)(5.0 L(735 mmHg)(5.0 L)(mol K)(mol K))

(62.4 (62.4 mmHg LmmHg L)(293 K))(293 K)

= 0. 20 mol O= 0. 20 mol O2 2 x x 32.0 g O32.0 g O2 2 = = 6.4 g O6.4 g O22

1 mol O1 mol O22

16

91

Molar VolumeMolar Volume

At STPAt STP

4.0 g He4.0 g He 16.0 g CH16.0 g CH44 44.0 g CO44.0 g CO22

1 mole 1 mole 1 mole1 mole 1mole1mole(STP)(STP) (STP)(STP) (STP)(STP)

V = 22.4 L V = 22.4 L V = 22.4 LV = 22.4 L V = 22.4 L V = 22.4 L

92

Molar Volume FactorMolar Volume Factor

1 mole of a gas at STP = 22.4 L1 mole of a gas at STP = 22.4 L

22.4 L 22.4 L and and 1 mole 1 mole

1 mole 22.4 L1 mole 22.4 L

93


A.What is the volume at STP of 4.00 g of A.What is the volume at STP of 4.00 g of

CHCH44??

1) 5.60 L1) 5.60 L 2) 11.2 L2) 11.2 L 3) 44.8 L3) 44.8 L

B. How many grams of He are present in 8.0 B. How many grams of He are present in 8.0

L of gas at STP? L of gas at STP?

1) 25.6 g1) 25.6 g 2) 0.357 g2) 0.357 g3) 1.43 g3) 1.43 g

94


A.What is the volume at STP of 4.00 g of CHA.What is the volume at STP of 4.00 g of CH44??

4.00 g CH4.00 g CH44 x x 1 mole CH1 mole CH44 xx 22.4 L (STP)22.4 L (STP) = 5.60 L= 5.60 L

16.0 g CH16.0 g CH44 1 mole CH1 mole CH44

B. How many grams of He are present in 8.0 L of gas at B. How many grams of He are present in 8.0 L of gas at

STP? STP?

8.00 L x 8.00 L x 1 mole He 1 mole He x x 4.00 g He4.00 g He = 1.43 g He= 1.43 g He

22.4 He 1 mole He 22.4 He 1 mole He

95

DaltonsDaltons’’ Law of Partial PressuresLaw of Partial Pressures

Partial Pressure Partial Pressure

Pressure each gas in a mixture would exert if it Pressure each gas in a mixture would exert if it were the only gas in the containerwere the only gas in the container

Dalton's Law of Partial PressuresDalton's Law of Partial Pressures

The total pressure exerted by a gas mixture is The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in the sum of the partial pressures of the gases in that mixture.that mixture.

PPTT = P= P11 + P+ P22 + P+ P33 + .....+ .....96

Gases in the AirGases in the Air

The % of gases in air Partial pressure (STP) The % of gases in air Partial pressure (STP)

78.08% N78.08% N22 593.4 mmHg593.4 mmHg

20.95% O20.95% O22 159.2 mmHg159.2 mmHg

0.94% 0.94% ArAr 7.1 mmHg7.1 mmHg

0.03% CO0.03% CO22 0.2 mmHg0.2 mmHg

PPAIRAIR = P= PN N + P+ POO + + PPArAr + P+ PCO CO = 760 mmHg= 760 mmHg2 2 2 2 22

17

97


A.A.If the atmospheric pressure today is 745 mm If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of OHg, what is the partial pressure (mm Hg) of O22

in the air?in the air?

1) 35.61) 35.6 2) 156 2) 156 3) 7603) 760

B. At an atmospheric pressure of 714, what is the B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) Npartial pressure (mm Hg) N22 in the air?in the air?

1) 557 1) 557 2) 9.142) 9.14 3) 0.1093) 0.109

98


A.A.If the atmospheric pressure today is 745 mm If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of OHg, what is the partial pressure (mm Hg) of O22

in the air?in the air?

2) 156 2) 156

B. At an atmospheric pressure of 714, what is the B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) Npartial pressure (mm Hg) N22 in the air?in the air?

1) 5571) 557

99

Partial PressuresPartial Pressures

The total pressure of a gas mixture dependsThe total pressure of a gas mixture depends

on the total number of gas particles, on the total number of gas particles, notnot onon

the types of particles.the types of particles.

P = 1.00 P = 1.00 atmatm P = 1.00 P = 1.00 atmatm

0.5 mole O 2+ 0.3 mole He+ 0.2 mole Ar

1 mole H 2

100

Health NoteHealth Note

When a scuba diver is several hundred feet When a scuba diver is several hundred feet

under water, the high pressures cause Nunder water, the high pressures cause N2 2 from the from the tank air to dissolve in the blood. If the diver rises tank air to dissolve in the blood. If the diver rises too fast, the dissolved Ntoo fast, the dissolved N22 will form bubbles in the will form bubbles in the blood, a dangerous and painful condition called blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with and does not dissolve in the blood, is mixed with OO22 in scuba tanks used for deep descents. in scuba tanks used for deep descents.

101


A A 5.00 L scuba tank contains 1.05 mole of O5.00 L scuba tank contains 1.05 mole of O2 2

and 0.418 mole He at 25and 0.418 mole He at 25°°C. What is the C. What is the partial pressure of each gas, and what is the partial pressure of each gas, and what is the total pressure in the tank?total pressure in the tank?

102


PP = = nRTnRT PPTT = P= POO + + PPHeHe

V V 22

PPTT = = 1.47 mol x 0.0821 L1.47 mol x 0.0821 L--atmatm x 298 Kx 298 K

5.00 L5.00 L (K mol)(K mol)

== 7.19 7.19 atmatm

18

103

Gases Collected by Displacement of Gases Collected by Displacement of WaterWater

�� Atmospheric pressure equals the Atmospheric pressure equals the pressure of the gas plus the pressure pressure of the gas plus the pressure of the water vaporof the water vapor

�� PPatmosphereatmosphere = P= Pgasgas + P+ PH2OH2O

�� Water vapor pressure tableWater vapor pressure table

104

Table of Vapor Pressures for Water Table of Vapor Pressures for Water

105

PracticePractice�� Find the pressure of oxygen gas Find the pressure of oxygen gas

collected over water at 25.0collected over water at 25.0 °°C if the C if the barometer reading is 752mmHg.barometer reading is 752mmHg.

106

Gases and the MoleGases and the Mole

107

GasesGases�� Many of the chemicals we deal with Many of the chemicals we deal with

are gases.are gases.�� They are difficult to weigh, so weThey are difficult to weigh, so we ’’ ll ll

measure volumemeasure volume�� Need to know how many moles of gas Need to know how many moles of gas

we have.we have.�� Two things affect the volume of a gasTwo things affect the volume of a gas�� Temperature and pressureTemperature and pressure�� Compare at the same temp. and Compare at the same temp. and

pressure.pressure.108

Standard Temperature and Standard Temperature and PressurePressure

�� Avogadro's HypothesisAvogadro's Hypothesis -- at the same at the same temperature and pressure equal temperature and pressure equal volumes of gas have the same number volumes of gas have the same number of particles.of particles.

�� 00ººC and 1 atmosphere pressure C and 1 atmosphere pressure �� Abbreviated Abbreviated atmatm�� 273 K and 101.3 273 K and 101.3 kPakPa�� kPakPa is is kiloPascalkiloPascal

19

109

AtAt SStandardtandardTTemperatureemperatureandandPPressureressure

�� abbreviated STPabbreviated STP�� At STP 1 mole of gas occupies 22.4 LAt STP 1 mole of gas occupies 22.4 L�� Called the Called the molar volumemolar volume�� Used for conversion factorsUsed for conversion factors�� Moles to Liter and L to molMoles to Liter and L to mol

110

Conversion factorsConversion factors

�� Used to change units.Used to change units.�� Three questionsThree questions

–– What can you change the given What can you change the given into?into?

–– What unit do you want to get rid of?What unit do you want to get rid of?–– Where does it go to cancel out?Where does it go to cancel out?

111

ExamplesExamples

��What is the volume of 4.59 What is the volume of 4.59 mole of COmole of CO 22 gas at STP?gas at STP?

112

Density of a gasDensity of a gas�� D = m /VD = m /V�� for a gas the units will be g / Lfor a gas the units will be g / L�� We can determine the density of any We can determine the density of any

gas at STP if we know its formula.gas at STP if we know its formula.�� To find the density we need the mass To find the density we need the mass

and the volume.and the volume.�� If you assume you have 1 mole than If you assume you have 1 mole than

the mass is the molar mass (PT)the mass is the molar mass (PT)�� At STP the volume is 22.4 L.At STP the volume is 22.4 L.

113

ExamplesExamples�� Find the density of COFind the density of CO 22 at STP.at STP.

114

Quizdom�� Find the density of CHFind the density of CH 44 at STP.at STP.

20

115

The other wayThe other way�� Given the density, we can find the Given the density, we can find the

molar mass of the gas.molar mass of the gas.�� Again, pretend you have a mole at Again, pretend you have a mole at

STP, so V = 22.4 L.STP, so V = 22.4 L.�� m = D x Vm = D x V�� m is the mass of 1 mole, since you m is the mass of 1 mole, since you

have 22.4 L of the stuff.have 22.4 L of the stuff.

116

What is the molar mass of a gas What is the molar mass of a gas with a density of 1.964 g/L?with a density of 1.964 g/L?

117

Gases and StoichiometryGases and Stoichiometry2 H2 H22OO2 2 (l) (l) ------> 2 H> 2 H22O (g) + OO (g) + O2 2 (g)(g)Decompose 1.1 g of HDecompose 1.1 g of H 22OO22 in a flask with a in a flask with a

volume of 2.50 L. What is the volume of Ovolume of 2.50 L. What is the volume of O 22 at at STP? STP?

SolutionSolution1.1 g 1.1 g HH22OO22 1 mol H1 mol H 22OO22 1 mol O1 mol O 22 22.4 L O22.4 L O22

34 g H34 g H22OO22 2 mol H2 mol H 22OO22 1 mol O1 mol O 22

= 0.36 L O2 at STP118

Gas Stoichiometry: Practice!Gas Stoichiometry: Practice!

A. What is the volume at STP of 4.00 g of A. What is the volume at STP of 4.00 g of

CHCH44??

B. How many grams of He are present in B. How many grams of He are present in

8.0 L of gas at STP? 8.0 L of gas at STP?

119

What if itWhat if it’’s NOT at STP?s NOT at STP?

��1. Do the problem like it was 1. Do the problem like it was at STP. (Vat STP. (V11))

��2. Convert from STP (V2. Convert from STP (V 11, P, P11, , TT11) to the stated conditions ) to the stated conditions (P(P22, T, T22))

120

GAS DIFFUSION & EFFUSIONGAS DIFFUSION & EFFUSION

�� diffusiondiffusion is the is the gradual mixing gradual mixing of molecules of of molecules of different gases.different gases.

�� effusioneffusion is the is the movement of movement of molecules through molecules through a small hole into an a small hole into an empty container.empty container.

21

121

GAS DIFFUSION & EFFUSIONGAS DIFFUSION & EFFUSIONGrahamGraham ’’s law s law

governs effusion governs effusion and diffusion of and diffusion of gas molecules.gas molecules.

Thomas Graham, 1805Thomas Graham, 1805 --1869. 1869. Professor in Glasgow and London.Professor in Glasgow and London.

Rate of effusion is inversely proportional to its molar mass.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

M of AM of B

Rate for B

Rate for A

122

GAS DIFFUSION & EFFUSIONGAS DIFFUSION & EFFUSION

Molecules effuse thru holes Molecules effuse thru holes in a rubber balloon, for in a rubber balloon, for example, at a rate (= example, at a rate (= moles/time) that ismoles/time) that is

�� proportional to Tproportional to T�� inversely proportional to inversely proportional to

M.M.Therefore, He effuses more Therefore, He effuses more

rapidly than Orapidly than O 22 at same T.at same T.

HeHe

123

Gas Diffusionrelation of mass to rate of diffusion

Gas DiffusionGas Diffusionrelation of mass to rate of diffusionrelation of mass to rate of diffusion

� HCl and NH 3 diffuse from opposite ends of tube.

� Gases meet to form NH4Cl

� HCl heavier than NH 3

� Therefore, NH 4Cl forms closer to HClend of tube.

�� HClHCl and NHand NH 33 diffuse diffuse from opposite ends of from opposite ends of tube. tube.

�� Gases meet to form Gases meet to form NHNH44ClCl

�� HClHCl heavier than NHheavier than NH 33

�� Therefore, NHTherefore, NH 44Cl Cl forms closer to forms closer to HClHClend of tube.end of tube.

124

If neon travels at 400. m/s, estimate the average speed of butane (C4H10) at the same temperature. 235 m/s

125

Chlorine has a velocity of 0.0380 m/s. What is the average velocity of sulfur dioxide under the same conditions?

0.0400 m/s

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