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 4 Kinetic Kinetic- Molecular Theory of Matter Molecular Theory of Matter Particles of Matter are always in Particles of Matter are always in motion motion Ideal Gases Ideal Gases- an imaginary gas that an imaginary gas that fits all the assumptions of the theory fits all the assumptions of the theory Kinetic Energy (KE) formula Kinetic Energy (KE) formula Physical properties of gases Physical properties of gases Real gases Real gases- gases in our daily lives gases in our daily lives 5 Ideal Gas Ideal 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 apart Are 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 temperatures Have energy increases at higher temperatures 6 Kinetic Energy of Gases Kinetic 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 energy energy – Lighter particles move faster than Lighter particles move faster than heavier particles heavier particles – KE = KE = mv mv 2 KE =kinetic energy KE =kinetic energy 2 v = velocity (speed) v = velocity (speed) m = mass m = mass
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1
GasesGases
Properties of GasesProperties of Gases
Gas PressureGas Pressure
2
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.
3
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
4
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
5
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
2
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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
�� 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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Learning Check G2Learning Check G2
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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Solution G2Solution G2
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
Learning Check G3Learning Check G3
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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Solution G3Solution G3
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
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
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:
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:
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:
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:
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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Learning Check GL2Learning Check GL2
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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Solution GL2Solution GL2
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.
38
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?
40
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
42
1)1)1)1) determine which variables you determine which variables you determine which variables you determine which variables you have:have:have:have:
1)1)1)1) determine which variables you determine which variables you determine which variables you determine which variables you have:have:have:have:
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
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:
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:
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:
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
44
Learning Check GL3Learning Check GL3
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.
45
Solution GL3Solution GL3
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.
46
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?
47
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
Learning Check GL4Learning Check GL4
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) 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
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!
52
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?
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
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?
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
1)1)1)1) determine which variables you determine which variables you determine which variables you determine which variables you have:have:have:have:
1)1)1)1) determine which variables you determine which variables you determine which variables you determine which variables you have:have:have:have:
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:
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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:
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
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).).
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Solution GL5Solution GL5
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)
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
?
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Learning Check GL6Learning Check GL6
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
Solution GL6Solution GL6
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)
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
Learning Check C2Learning Check C2
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?
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Solution G9Solution G9
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
Learning Check C3Learning Check C3
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
Solution C3Solution C3
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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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
Learning Check C4Learning Check C4
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
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Solution C4Solution C4
PP1 1 = = 1.0 1.0 atmatm VV11 = 15 L = 15 L TT11 = = 273 K273 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
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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
Learning Check G15Learning Check G15
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
Solution G15Solution G15
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
Learning Check G16Learning Check G16
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?
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Solution G16Solution G16
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
Learning Check G17Learning Check G17
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
Solution G17Solution G17
Solve ideal gas equation for n (moles)Solve ideal gas equation for n (moles)n n = = PVPV
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
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Learning Check C5Learning Check C5
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
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Solution C5Solution C5
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)
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
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Solution C6Solution C6
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
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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.
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Learning Check C7Learning Check C7
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
Solution C7Solution C7
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
�� 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
�� 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?