1 1 Gases Gases Properties of Gases Properties of Gases Gas Pressure Gas Pressure 2 Gases Gases What gases are important for each of the What gases are important for each of the following: O following: O 2 , CO , CO 2 and/or He? and/or He? A. B. C. D. A. B. C. D. 3 Gases Gases What gases are important for each of the What gases are important for each of the following: O following: O 2 , CO , CO 2 and/or He? and/or He? A. CO A. CO 2 B. O B. O 2 /CO /CO 2 C. O C. O 2 D. He D. He
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Gases Properties of Gases Gas Pressure - Parkway Schools · heavier particles – KE = mv 2 KE ... Gases exert a uniform pressure on all inner surfaces of their containers ... C.
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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
�� Gases have low densitiesGases have low densities
�� Gases fill a container completely and uniformlyGases fill a container completely and uniformly��Gases exert a uniform pressure on all inner Gases exert a uniform pressure on all inner
surfaces of their containerssurfaces of their containers�� Fluidity: gas particles can slide past one another Fluidity: gas particles can slide past one another
(gases and liquids are (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
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==14.7psi14.7psiWhat 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==14.7psi14.7psiWhat is 2.00 What is 2.00 atmatm expressed in expressed in torrtorr
�� 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??
32.0psi1
101.3kPa14.7psi = 221kPa
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Gas LawsGas Laws
�� BoyleBoyle ’’s Laws Law�� CharlesCharles ’’ LawLaw�� GayGay--LussacLussac ’’s Laws Law�� Combined Gas LawCombined Gas Law
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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
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
4. Get the unknown variable alone4. Get the unknown variable alone
�� (2.0atm)(3.0mL) (2.0atm)(3.0mL) = V= V22
4.0atm4.0atm
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5. Plug in your calculator and solve5. Plug in your calculator and solve
�� VV22 = 1.5L= 1.5L
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Learning Check Learning Check
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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SolutionSolution
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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Learning CheckLearning Check
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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SolutionSolution
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’’ 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
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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
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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Learning CheckLearning Check
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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Solution GL3Solution GL3
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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Learning Check Learning Check
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?
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) 170°C
T2 =(291K)(640mL)=443 K
420 mL
T2 =443 K - 273 K = 170°C
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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
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...
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?
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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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)
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)
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Solution Solution
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)
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Combined Gas LawCombined Gas Law
PP11VV11 = P= P22VV22
TT11 TT22
OROR
PP11VV11TT22 = P= P22VV22TT11
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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??
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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 = ??= ????
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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
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Learning Check Learning Check
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.
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Solution Solution
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.
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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)