11
GASESGASESChemistry I – Chapter 14Chemistry I – Chapter 14
Chemistry I Honors – Chapter 13Chemistry I Honors – Chapter 13SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print
"Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!
22
GAS: One of Three GAS: One of Three STATES OF MATTERSTATES OF MATTERGAS: One of Three GAS: One of Three
STATES OF MATTERSTATES OF MATTER
33General General Properties of Properties of
GasesGases• There is a lot of “free” space There is a lot of “free” space
in a gas.in a gas.• Gases can be expanded Gases can be expanded
infinitely.infinitely.• Gases fill containers Gases fill containers
uniformly and completely.uniformly and completely.• Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.
44
Importance of GasesImportance of Gases
• Airbags fill with NAirbags fill with N22 gas in gas in an accident. an accident.
• Gas is generated by the Gas is generated by the decomposition of sodium decomposition of sodium azide, NaNazide, NaN33..
• 2 NaN2 NaN33 > 2 Na + 3 N > 2 Na + 3 N22
55Our AtmosphereOur Atmosphere: : Gravity holds the Gravity holds the atmosphere close to the Earth. It is divided atmosphere close to the Earth. It is divided into various layers according to differences in into various layers according to differences in temperature:temperature:
• The Troposphere: The Troposphere: – We live in the Troposphere We live in the Troposphere – It contains more than 75% of the atmosphere’s mass It contains more than 75% of the atmosphere’s mass
• The StratosphereThe Stratosphere: : – Most of the clouds and rain are in this layer Most of the clouds and rain are in this layer – It extends up to 48km above the earth!It extends up to 48km above the earth!
• The ozone layer is between the Stratosphere and The ozone layer is between the Stratosphere and MesosphereMesosphere
• The Mesosphere: The Mesosphere: – This layer extends to 80km above the Earth!This layer extends to 80km above the Earth!
• The ThermosphereThe Thermosphere – Here the air is very thin Here the air is very thin – Over 99.99% of the atmosphere lies below this layer Over 99.99% of the atmosphere lies below this layer – It is comprised of: It is comprised of:
66
77
• http://library.thinkquest.org/C003124/en/atmos.htm
88• The Ionosphere:The Ionosphere:
Reflects Reflects radiowaves back radiowaves back to Earth so to Earth so signals can be signals can be sent around the sent around the world world
• The Exosphere:The Exosphere: This layer begins This layer begins at ~480km above at ~480km above the Earth and the Earth and fades away into fades away into space space
99Did you know...Did you know... • About 99% of the atmosphere is made up of About 99% of the atmosphere is made up of
oxygen (21%) and nitrogen (78%). oxygen (21%) and nitrogen (78%). • The remainder is made up of Ar (argon), COThe remainder is made up of Ar (argon), CO22
and very small amounts of Hand very small amounts of H22, NH, NH33, O, O33 (ozone),(ozone), CH CH44, , CO, He, Ne, Kr and Xe. CO, He, Ne, Kr and Xe. also gases such as SOalso gases such as SO22, NO, NO22 are put into the air from are put into the air from vehicles & industryvehicles & industry
• About 20 % of the Earth's population breathes About 20 % of the Earth's population breathes severely contaminated air, largely COseverely contaminated air, largely CO22 and and SOSO22 resulting from industrial processes. This increases resulting from industrial processes. This increases the number of respiratory conditions, especially the number of respiratory conditions, especially amongst children and elders. 13 % of the British amongst children and elders. 13 % of the British children experience asthma caused by air children experience asthma caused by air contaminationcontamination
1010
Properties of Properties of GasesGases
Gas properties can be modeled Gas properties can be modeled using math. Model depends using math. Model depends on—on—
• V = volume of the gas (L)V = volume of the gas (L)• T = temperature (K)T = temperature (K)
–ALL temperatures in the ALL temperatures in the entire chapter MUST be in entire chapter MUST be in Kelvin!!! No Exceptions!Kelvin!!! No Exceptions!
• n = amount (moles)n = amount (moles)• P = pressureP = pressure
(atmospheres) (atmospheres)
1111
TemperatureTemperature• Average kinetic energy of the Average kinetic energy of the
particles of a substanceparticles of a substance
• Without particles there is no Without particles there is no temperaturetemperature
• No temperature in a vacuumNo temperature in a vacuum
• Must be measured in Kelvin to Must be measured in Kelvin to avoid negative values while avoid negative values while doing calculations Cdoing calculations C°° + 273 = K + 273 = K
1212PRESSURE:
A FORCE per unit areaA FORCE per unit area• Measured in pascals (Pa) = 1 N/mMeasured in pascals (Pa) = 1 N/m22 = 100 g / m= 100 g / m22
• A normal day has a pressure of 100 kPaA normal day has a pressure of 100 kPa= 100 x = 100 x 1000 Pa1000 Pa x 100 g / m x 100 g / m22 1 kPa1 kPa= 10 000 kg / m= 10 000 kg / m22
Atmospheric pressure varies with altitude Atmospheric pressure varies with altitude – the lower the altitude, the longer and the lower the altitude, the longer and
heavier is the column of air above an area heavier is the column of air above an area of the earth.of the earth.
1313
1414
PressurPressuree
Pressure of air is Pressure of air is measured with a measured with a BAROMETERBAROMETER (developed by (developed by Torricelli in 1643)Torricelli in 1643)
Hg rises in tube until force of Hg Hg rises in tube until force of Hg (down) balances the force of (down) balances the force of atmosphere (pushing up). atmosphere (pushing up). (Just like a straw in a soft (Just like a straw in a soft drink)drink)
P of Hg pushing down related to P of Hg pushing down related to
• Hg densityHg density
• column heightcolumn height
1515PressurPressureeColumn height measures Column height measures Pressure of atmospherePressure of atmosphere
• * 1 standard atmosphere * 1 standard atmosphere (atm) (atm)
= 760 mm Hg (or torr) = 760 mm Hg (or torr)
= 29.92 inches Hg = 29.92 inches Hg
= 14.7 pounds/in= 14.7 pounds/in2 2 (psi)(psi)
= * 101.3 kPa (SI unit is = * 101.3 kPa (SI unit is PASCAL) PASCAL)
= about 34 feet of water!= about 34 feet of water!
* Memorize these!* Memorize these!
1616
Pressure Conversions
A.The air pressure in a cave underwater is 2.30 atm, what is the pressure in kPa?
101.3 kPa
B. What is 475 mm Hg expressed in kPa?
101.3 kPa 101.3 kPa 760 mm Hg
= 233 kPa
= 63.31 kPa475 mm Hg x
2.30 atm x1 atm
1717
Pressure Conversions
A.The air pressure in a cave underwater is 2.30 atm, what is the pressure in kPa?
101.3 kPa
B. What is 475 mm Hg expressed in kPa?
101.3 kPa101.3 kPa 760 mm Hg
= 233 kPa
= 63.31 kPa475 mm Hg x
2.30 atm x
1 atm
1818
Pressure Conversions
A. What is 3.71 atm expressed in kPa?
B. The pressure of a tire is measured as 830 torr.
What is this pressure in atm?
1919•Crushed Can DemoCrushed Can Demo:: Why did the can implode? What happened Why did the can implode? What happened
to the pressure inside once the temperature to the pressure inside once the temperature decreased?decreased?
Lower temperature = slower moving particlesLower temperature = slower moving particles = less collisions with the wall= less collisions with the wall
T P T P
Did the # of particles/moles change?Did the # of particles/moles change? Did the air pressure change(# particles colliding)?Did the air pressure change(# particles colliding)? Did the volume change? Did the volume change? Only because the pressure outside is much greater Only because the pressure outside is much greater
then inside – the ouside pressure crushed the can.then inside – the ouside pressure crushed the can.http://phet.colorado.edu/en/simulation/gasproperties
T α P
2020
GayLussac’s LawGayLussac’s Law
If n and V are If n and V are constant, constant, then P then P αα T T
P and T are directly P and T are directly proportional.proportional.
PP11 P P22
==
TT11 T T22
• If temperature goes up, If temperature goes up,
the pressure goes up!the pressure goes up!
Joseph Louis GayJoseph Louis GayLussac (17781850)Lussac (17781850)
P1 = T1
P2 T2 Charles’ Law http://
www.youtube.com/watch?v
=tPH57yp0x1U
2121Gas Pressure, Gas Pressure, Temperature, and Temperature, and Kinetic Molecular Kinetic Molecular
TheoryTheory
Gas Pressure, Gas Pressure, Temperature, and Temperature, and Kinetic Molecular Kinetic Molecular
TheoryTheory
P proportional to TP proportional to T
As the temperature As the temperature increases the increases the
particles move particles move faster faster
increasing # of increasing # of collisions collisions
= ↑ pressure= ↑ pressure
2222A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C?
TT11 = 127°C + 273 = 400K = 127°C + 273 = 400K
PP11 = 3.0 atm = 3.0 atm
TT22 = 327°C + 273 = 600K = 327°C + 273 = 600K
PP22 = ? = ?What law applies to this situation?What law applies to this situation? T and P – GayLussac’s Law: GayLussac’s Law:
3.0 atm3.0 atm == PP22 PP2 2 = = 3.0 atm x 600 K3.0 atm x 600 K P P2 2 = 4.50 atm= 4.50 atm
400K 600K 400 K400K 600K 400 K
P1 = P2
T1 T2
2323
Charles’s Charles’s LawLaw
If n and P are If n and P are constant, constant, then V then V αα T T
V and T are directly V and T are directly proportional.proportional.
VV11 V V22
==
TT11 T T22
• If temperature goes up, If temperature goes up,
the volume goes up!the volume goes up!
Jacques Charles (1746Jacques Charles (17461823). Isolated boron 1823). Isolated boron and studied gases. and studied gases. Balloonist.Balloonist.
V1 = T1
V2 T2
2424
Charles’s original balloonCharles’s original balloon
Modern longdistance balloonModern longdistance balloon
2525
Charles’s LawCharles’s Law
Boyle’s Law – Vacuum Demo  http://www.youtube.com/watch?v=N5xft2fIqQU
0 K273 °C
2626
A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C?
TT11 = 127°C + 273 = 400K = 127°C + 273 = 400K
VV11 = 3.0 L = 3.0 L
TT2 2 = 227°C + 273 = 500K= 227°C + 273 = 500K
VV22 = ? = ?
Which law applies to this situation?Which law applies to this situation?
T and V  Charles’s Law: Charles’s Law:
3.0 L3.0 L = = VV22 VV2 2 = = 3.0 L x 500 K3.0 L x 500 K V V22 = 3.75 L = 3.75 L
400 K 500 K 400 K400 K 500 K 400 K
V1 = V2
T1 T2
2727Boyle’s Boyle’s LawLawP P αα 1/V 1/VThis means Pressure This means Pressure
and Volume are and Volume are INVERSELY INVERSELY PROPORTIONAL if PROPORTIONAL if moles and moles and temperature are temperature are constant (do not constant (do not change). For change). For example, P goes up example, P goes up as V goes down.as V goes down.
PP11VV11 = P = P22 V V22
Robert Boyle Robert Boyle (16271691). (16271691). Son of Earl of Son of Earl of Cork, Ireland.Cork, Ireland.
2828
Volumedecreases
As Pressure Increases
2929
Boyle’s Law and Boyle’s Law and Kinetic Molecular Kinetic Molecular
TheoryTheory
Boyle’s Law and Boyle’s Law and Kinetic Molecular Kinetic Molecular
TheoryTheoryP proportional to 1/VP proportional to 1/V
Smaller volume Smaller volume means the particles means the particles
are forced closer are forced closer together – together –
increasing collisions increasing collisions with each other and with each other and
with the walls of with the walls of the containerthe container..
3030
Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
A bicycle pump is a A bicycle pump is a good example of good example of Boyle’s law. Boyle’s law.
As the volume of the As the volume of the air trapped in the air trapped in the pump is reduced, its pump is reduced, its pressure goes up, pressure goes up, and air is forced into and air is forced into the tire.the tire.
3131
Boyle’s Law:Boyle’s Law:

Volu
me
Volu
me
PressurePressure
PV = kPV = k TemperatureTemperature is constant is constant
3232
3333A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm?
PP11 = 2 atm = 2 atmVV11 = 3.0 L = 3.0 LPP22 = 4 atm = 4 atmVV22 = ? = ?Which law applies to this situation?Which law applies to this situation? P and V = P and V = Boyle’s LawBoyle’s Law
2 2 atmatm x 3.0 L = 4 x 3.0 L = 4 atm atm x Vx V22 VV22 = = 2 2 atmatm x 3.0 L x 3.0 L 4 4 atm atm
VV2 2 == 1.5 L1.5 L
PP1 1 VV11 = P = P22 V V22
3434
SUMMARY of Gas LawsSUMMARY of Gas Laws
LAWLAW RELATRELATIONSHIPIONSHIP LAWLAW CONSTANTCONSTANT
Charles’Charles’ VV T T VV11/T/T11 = V = V22/T/T22 P, nP, n
GayGayLussac’sLussac’s PP T T PP11/T/T11 = P = P22/T/T22 V, nV, n
Boyle’sBoyle’s PP V V PP11VV1 1 = P= P22VV22 T, nT, n
3535
Combined Gas Law
• The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
V1 P1 V2 P2
=
T1 T2
3636
Combined Gas Law
If you only need one of the gas laws, you can cover up the item that is constant and you will get that gas law!
= P1 V1
T1
P2 V2
T2
Boyle’s Law
Charles’ Law
GayLussac’s Law
3737
Combined Gas Law Problem
A sample of helium gas has a volume of A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?gas at a volume of 90.0 mL and a pressure of 3.20 atm?
Given:
P1 = 0.800 atm V1 = 180 mL T1 = 29°C + 273
= 302 K
P2 = 3.20 atm V2= 90 mL T2 = ??
3838
CalculationPP1 1 = 0.800 atm V= 0.800 atm V11 = 180 mL = 180 mL T T11 = 302 K = 302 K
PP22 = 3.20 atm V = 3.20 atm V22= 90 mL T= 90 mL T2 2 = ??= ??
P1 V1 P2 V2
= P1 V1 T2 = P2 V2 T1
T1 T2
T2 = P2 V2 T1
P1 V1
T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
TT22 = 604 K  273 = 331 °C = 604 K  273 = 331 °C
= 604 K
3939
Learning Check
A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?
4040
One More Practice Problem
A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?
4141And now, we pause for this commercial message from STP
STP in chemistry stands for Standard Temperature and
Pressure
Standard Pressure = 1 atm (or an equivalent)
Standard Temperature = 0 deg
C (273 K)
STP allows us to compare amounts of gases between
different pressures and temperatures
STP allows us to compare amounts of gases between
different pressures and temperatures
4242
Try This One
A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?
4343Ideal Gases, Avogadro’s Ideal Gases, Avogadro’s Theory and the Ideal Gas Theory and the Ideal Gas
LawLawWhat are three things that make What are three things that make
the behaviour of a gas ideal?the behaviour of a gas ideal?1)1) How do they collide?How do they collide?2) Does their size matter? 2) Does their size matter? 3) How do they move? 3) How do they move?
4444What are three things that make the What are three things that make the
behaviour of a gas ideal?behaviour of a gas ideal?1)1)How do they collide?How do they collide? ELASTICALLY ELASTICALLY no attraction, no attraction, no energy loss, perfect bounces no energy loss, perfect bounces 2) Does their size matter? 2) Does their size matter? NO NO too small, spaces between too bigtoo small, spaces between too big3) How do they move? 3) How do they move? straight lines and randomlystraight lines and randomly
4545
FYI: What you will see in university
4646
Avogadro’s TheoryAvogadro’s TheoryEqual volumes of gases at the same Equal volumes of gases at the same
T and P have the same number ofT and P have the same number ofmolecules.molecules.
• Volume is proportional to molesVolume is proportional to moles
• V and n are directly relatedV and n are directly related
• V V αα n n
twice the volumetwice the volume
twice as many twice as many moleculesmolecules
4747
Ideal Gas Law:Ideal Gas Law:Generally speaking, an ideal gas obeys all Generally speaking, an ideal gas obeys all gas laws perfectly under all conditions:gas laws perfectly under all conditions:
1) Charles’s Law V 1) Charles’s Law V αα T (V / T = k ) T (V / T = k )
2) GayLussac’s Law P 2) GayLussac’s Law P αα T (P / T = k ) T (P / T = k )
3) Boyle’s Law V 3) Boyle’s Law V αα 1 / p ( V x P = k ) 1 / p ( V x P = k )
4) Avogadro’s Theory V 4) Avogadro’s Theory V αα n n
In other words, V In other words, V αα n n α α T T α α 1/p 1/p
All these variables can be combined to All these variables can be combined to include an equation with moles, n using a include an equation with moles, n using a constant… constant… V = constant x n x T x 1/pV = constant x n x T x 1/p
The constant is the letter RThe constant is the letter R
4848
The Constant R:The Constant R:•Is called the Ideal Is called the Ideal GAS CONSTANTGAS CONSTANT•Relates P, T, V, and molesRelates P, T, V, and moles•Is found by substituting in known values for P, T, V Is found by substituting in known values for P, T, V and n (moles) and n (moles) Ex. You could sub in values Ex. You could sub in values at STP: T = 273 K, P = 1 atm, 1 mole at STP: T = 273 K, P = 1 atm, 1 molegasgas= 22.4 L= 22.4 L
R with different units : These #’s are always provided .R with different units : These #’s are always provided .•8.314 J / K 8.314 J / K • • mol or 8.314 J Kmol or 8.314 J K1 1 molmol1 1
• 0.0821 L0.0821 L••atm/Katm/K••molmol mainly used mainly used•8.31451 Pa m8.31451 Pa m33 K K11 mol mol11
•62.364 L Torr K62.364 L Torr K11 mol mol11
Ideal Gas Law:Ideal Gas Law:
4949
Brings together gas properties.Brings together gas properties.
Can be derived from experiment and theory.Can be derived from experiment and theory.
BE SURE YOU KNOW THIS EQUATION!BE SURE YOU KNOW THIS EQUATION!
P V = n R TP V = n R T
V= R x n x T x 1/pV= R x n x T x 1/pMultiply both sides by p, Multiply both sides by p,
rearrange n…rearrange n…
IDEAL GAS LAWIDEAL GAS LAW
http://jersey.uoregon.edu/vlab/Piston/index.html
5050
Using PV = nRTUsing PV = nRTP = PressureP = Pressure
V = VolumeV = Volume
T = TemperatureT = Temperature
N = number of molesN = number of moles
R is the constant called the R is the constant called the Ideal Gas ConstantIdeal Gas Constant
R = 0.0821R = 0.0821L • atm
Mol • K
5151REMINDER:REMINDER:
1 standard atmosphere (atm) 1 standard atmosphere (atm)
1 atm = 760 mm Hg (or torr) 1 atm = 760 mm Hg (or torr)
= 101.3 kPa (SI unit is PASCAL) = 101.3 kPa (SI unit is PASCAL)
Ex. 202.6 kPa = Ex. 202.6 kPa = 202.6 kPa 202.6 kPa
101.3 kPa / atm101.3 kPa / atm
= 2.000 atm= 2.000 atm
Ex. 1040 torr = Ex. 1040 torr = 1040 torr1040 torr 760 mm Hg / atm 760 mm Hg / atm
= 1.50 atm= 1.50 atm
5252
Using PV = nRTUsing PV = nRTHow much NHow much N22 is required to fill a small room with a is required to fill a small room with a
volume of 27,000 L to 745 mm Hg at 25 volume of 27,000 L to 745 mm Hg at 25 ooC?C? SolutionSolution1. Given: convert to proper units1. Given: convert to proper units V = 27 000 LV = 27 000 L T = 25 T = 25 ooC + 273 = 298 KC + 273 = 298 K P = 745 mm Hg (760 mm Hg/atm) P = 745 mm Hg (760 mm Hg/atm)
= 0.98 atm = 0.98 atmAnd we always know R = 0.0821 LAnd we always know R = 0.0821 L••atm / molatm / mol••KK
5353
Using PV = nRTUsing PV = nRTHow much NHow much N22 is required to fill a small room with a volume is required to fill a small room with a volume
of 27 000 L to P = 745 mm Hg at 25 of 27 000 L to P = 745 mm Hg at 25 ooC?C?
SolutionSolution
1. Divide both sides by RT to solve for n. 1. Divide both sides by RT to solve for n. 2. Now plug in those values and solve for the 2. Now plug in those values and solve for the unknown.unknown. PVPV = = nRTnRT
n = PV / RTn = PV / RT
n = (0.98 atm)(2.7 x 10 4 L)
(0.0821 L • atm/K • mol)(298 K)n =
(0.98 atm)(2.7 x 10 4 L)
(0.0821 L • atm/K • mol)(298 K)
n = 1.1 x 10n = 1.1 x 1033 mol (or about 30 kg of gas) mol (or about 30 kg of gas)
RT RTRT RT
5454
Learning Check
EX. Dinitrogen monoxide (NDinitrogen monoxide (N22O), laughing gas, O), laughing gas, is used by dentists as an anesthetic. is used by dentists as an anesthetic.
If 2.86 mol of gas occupies a 20.0 L tank at If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure in atm in the tank in 23°C, what is the pressure in atm in the tank in the dentist office?the dentist office?
Given: n=2.86 mol, V=20.0L T=23°C + 273 K P=?Given: n=2.86 mol, V=20.0L T=23°C + 273 K P=?
PV = nRTPV = nRT P = nRT / V P = nRT / V P = P = 2.86 mol (0.0821 L atm/mol K)(296 K)2.86 mol (0.0821 L atm/mol K)(296 K) 20.0L 20.0L
P = 3.48 atm P = 3.48 atm
5555
Using PV = nRT When Given Mass
0.78 grams of hydrogen gas is produced at 220.78 grams of hydrogen gas is produced at 22°°C and C and 125 kPa. What volume of hydrogen is expected?125 kPa. What volume of hydrogen is expected?
Given: T = 22Given: T = 22°C + 273 = 295 K, °C + 273 = 295 K, P = 125 kPa = 125 kPa/101.3 = 1.23 atm P = 125 kPa = 125 kPa/101.3 = 1.23 atm
n = find n using molar mass first n = find n using molar mass first
n = n = massmass PV = nRTPV = nRT Mm V = nRT/PMm V = nRT/P n = n = 0.78g 0.78g = = 0.03860.0386molmol(0.0821 (0.0821 L atm/mol KL atm/mol K)295K )295K
2.02 2.02g/mol g/mol 1.23 atm1.23 atm
= 0.0386 mol = 0.76 L= 0.0386 mol = 0.76 L
5656A 5.0 L cylinder contains oxygen gas at A 5.0 L cylinder contains oxygen gas at
20.0°C and 735 mm Hg. How many 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?grams of oxygen are in the cylinder?
Step 1) find moles using PV = nRTStep 1) find moles using PV = nRT n = 0.201 moln = 0.201 molStep 2) find mass using m = n x MmStep 2) find mass using m = n x Mm
m = 40.3 gm = 40.3 g
5757Deviations from Deviations from Ideal Gas LawIdeal Gas Law
• Real molecules have
volume.
The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves.
• There are intermolecular forces.
An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.
– Otherwise a gas could not condense to become a liquid.
5858
Gases in the AirThe % of gases in air Partial pressure (STP)
78.08% N2 593.4 mm Hg
20.95% O2 159.2 mm Hg
0.94% Ar 7.1 mm Hg
0.03% CO2 0.2 mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg 2 2 2
Total Pressure 760 mm Hg
5959Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures
What is the total pressure in the flask?What is the total pressure in the flask?
PPtotaltotal in gas mixture = P in gas mixture = PAA + P + PBB + ... + ...Therefore, Therefore,
PPtotaltotal = P = PHH22OO + P + POO22 = 0.48 atm = 0.48 atm
Dalton’s Law: total P is sum ofDalton’s Law: total P is sum of PARTIALPARTIAL pressures.pressures.
2 H2 H22OO2 2 (l) > 2 H(l) > 2 H22O (g) + OO (g) + O2 2 (g)(g)
0.32 atm 0.32 atm 0.16 0.16 atmatm
6060
Dalton’s Dalton’s LawLaw
John DaltonJohn Dalton1766184417661844
6161Health NoteWhen a scuba diver is several hundred feet under water, the high pressures cause N2 from
the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form
bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in
scuba tanks used for deep descents.
6262Collecting a gas “over water”
• Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.
6363Table of Vapor Pressures for Water
6464
Solve This!
A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H2 gas?
768 torr – 17.5 torr = 750.5 torr
6565
GAS DENSITYGAS DENSITYGAS DENSITYGAS DENSITY
HighHigh densitydensity
Low Low densitydensity
22.4 L of ANY gas AT STP = 1 mole
6666Gases and Gases and StoichiometryStoichiometry
2 H2 H22OO2 2 (l) > 2 H(l) > 2 H22O (g) + OO (g) + O2 2 (g)(g)
Decompose 1.1 g of HDecompose 1.1 g of H22OO22 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 O22 at STP?at STP?
Bombardier beetle Bombardier beetle uses decomposition uses decomposition of hydrogen peroxide of hydrogen peroxide to defend itself.to defend itself.
6767Gases and Gases and StoichiometryStoichiometry
2 H2 H22OO2 2 (l) > 2 H(l) > 2 H22O (g) + OO (g) + O2 2 (g)(g)
Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with a volume of 2.50 L. in a flask with a volume of 2.50 L. What is the volume of OWhat is the volume of O22 at STP? at STP?
SolutionSolution1.1 g1.1 g HH22OO22 1 mol H 1 mol H22OO22 1 mol O 1 mol O22 22.4 L O 22.4 L O22
34 g H34 g H22OO22 2 mol H 2 mol H22OO22 1 mol O 1 mol O22
= 0.36 L O2 at STP
6868
Gas Stoichiometry: Practice!
A. What is the volume at STP of 4.00 g of CH4?
B. How many grams of He are present in 8.0 L of gas at
STP?
6969
What if it’s NOT at STP?
• 1. Do the problem like it was at STP. (V1)
• 2. Convert from STP (V1, P1, T1) to the stated conditions (P2, T2)
7070
Try this one!
How many L of O2 are needed to react 28.0 g NH3
at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
7171
GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
• diffusiondiffusion is the is the gradual mixing of gradual mixing of molecules of molecules of different gases.different gases.
• effusioneffusion is the is the movement of movement of molecules through a molecules through a small hole into an small hole into an empty container.empty container.
HONORS HONORS onlyonly
7272GAS DIFFUSION AND GAS DIFFUSION AND
EFFUSIONEFFUSION
Graham’s law governs Graham’s law governs effusion and diffusion effusion and diffusion of gas molecules.of gas molecules.
Thomas Graham, 18051869. Thomas Graham, 18051869. Professor in Glasgow and London.Professor in Glasgow and London.
Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.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
HONORS HONORS onlyonly
7373
GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION
Molecules effuse thru holes in a Molecules effuse thru holes in a rubber balloon, for example, at a rubber balloon, for example, at a rate (= moles/time) that israte (= moles/time) that is
• proportional to Tproportional to T
• inversely proportional to M.inversely proportional to M.
Therefore, He effuses more rapidly Therefore, He effuses more rapidly than Othan O22 at same T. at same T.
HeHe
HONORS HONORS onlyonly
7474
Gas DiffusionGas Diffusionrelation of mass to rate of relation of mass to rate of
diffusiondiffusion
Gas DiffusionGas Diffusionrelation of mass to rate of relation of mass to rate of
diffusiondiffusion
• HCl and NH3 diffuse from opposite ends of tube.
• Gases meet to form NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms closer to HCl end of tube.
• HCl and NH3 diffuse from opposite ends of tube.
• Gases meet to form NH4Cl
• HCl heavier than NH3
• Therefore, NH4Cl forms closer to HCl end of tube.
HONORS HONORS onlyonly