Post on 23-Dec-2015
Think About This…Think About This…Think About This…Think About This…
Gas Atmosphere This is a U-Tube Manometer. The red stuff is a liquid that moves based on the pressures on each end of the tube. Based on the position of the red liquid, which has a higher pressure, the Gas or the Atmosphere? How do you know?
Think About This…Think About This…Think About This…Think About This…
Gas Atmosphere Which has a higher pressure in this U-Tube, the Gas or the Atmosphere? How do you know?
Think About This…Think About This…Think About This…Think About This…
Gas Atmosphere What about this U-Tube? Which has the higher pressure, the Gas or the Atmosphere? How do you know?
Follow along in your textChapter 12Section 1
Pages 416 - 422
Follow along in your textChapter 12Section 1
Pages 416 - 422
The Physical The Physical Properties of Properties of
GasesGases
Kinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular Theory
Particles in an ideal gas…
• have no volume.
• have elastic collisions.
• are in constant, random, straight-line motion.
• don’t attract or repel each other.
• have an average kinetic energy directly related to Kelvin temperature.
Real GasesReal GasesReal GasesReal Gases
Particles in a REAL gas…• have their own volume• attract each other
Gas behavior is most ideal…• at high temperatures• at low pressures• in nonpolar, basically covalent,
atoms/molecules
Characteristics of GasesCharacteristics of GasesCharacteristics of GasesCharacteristics of GasesGases expand to fill any container.
• random motion, no attraction
Gases are fluids (like liquids).• no attraction
Gases have very low densities.• no volume = lots of empty space
Characteristics of GasesCharacteristics of GasesCharacteristics of GasesCharacteristics of Gases Gases can be compressed.
• no volume = lots of empty space
Gases undergo effusion & diffusion.• random motion
TemperatureTemperatureTemperatureTemperature
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
K = ºC + 273
Always use absolute temperature (Kelvin) when working with gases.
What is What is PressurePressure??What is What is PressurePressure??
area
forcepressure
Which shoes create the most pressure?
Pressure EquipmentPressure EquipmentPressure EquipmentPressure Equipment
Barometer• measures atmospheric pressure
Mercury Barometer
Aneroid Barometer
Pressure EquipmentPressure EquipmentPressure EquipmentPressure Equipment
Manometer• measures contained gas pressure
U-tube Manometer Bourdon-tube gauge
KEY UNITS AT SEA LEVEL
101,325 Pa
101.325 kPa
1 atm
760 mm Hg
760 torr
14.7 psi
1.01 bar
*These are all equal to each other!*
PressurePressurePressurePressure
STP is the Norm!STP is the Norm!STP is the Norm!STP is the Norm!
Standard Temperature & PressureStandard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa
-OR-
STP
Example: How many torr equals 5.00 psi?
ClosureClosureClosureClosure
1. How many Pascals equals 7.3 atm?
2. How many mmHg equals 19.3 barr?
3. How many atms equals 96.3 kPa?
5.00 psi 760 torr
14.7 psi = 259 torr
Warm-Up :Warm-Up :Complete the following Complete the following pressure conversions .pressure conversions .
Warm-Up :Warm-Up :Complete the following Complete the following pressure conversions .pressure conversions .
1. 15,650 Pa to mm Hg
2. 23 atm to psi
3. 893,000 torr to kPa
4. 6.5 psi to barr
Follow along in your text
Chapter 12 Section 2
Pages 423 - 432
Follow along in your text
Chapter 12 Section 2
Pages 423 - 432
The Gas LawsThe Gas Laws
Meet the VariablesMeet the VariablesMeet the VariablesMeet the Variables
P = pressure exerted by the gas
T = temperature in kelvins of the gas
V = total volume occupied by the gas
n = number of moles of the gas
k = symbolizes anything constant
Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
V
P
PV = k
Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
The pressure and volume of a gas are inversely related at constant mass & temp
P1V1 = P2V2
When the temp & number of particles remains the same, this is the equation
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
Boyle’s Law ProblemBoyle’s Law ProblemBoyle’s Law ProblemBoyle’s Law Problem
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
Remember BOYLE’S LAW!
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
kT
VV
T
Charles’ LawCharles’ LawCharles’ LawCharles’ Law
Charles’ LawCharles’ LawCharles’ LawCharles’ Law
The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure
V1 V2
T1 T2
=
If all conditions are kept constant, then the equation looks like this
GIVEN:
V1 = 473 mL
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
Charles’ Law ProblemCharles’ Law ProblemCharles’ Law ProblemCharles’ Law Problem
A gas occupies 473 mL at 36°C. Find its volume at 94°C.
Remember CHARLES’ LAW!
T V
(473 mL)(367 K)=V2(309 K)
V2 = 562 mL
kT
PP
T
Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law
Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law
The pressure and absolute temperature (K) of a gas are directly related
At constant mass & volume, the equation looks like this
P1 P2
T1 T2
=
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1V1T2 = P2V2T1
Gay-Lussac’s ProblemGay-Lussac’s ProblemGay-Lussac’s ProblemGay-Lussac’s Problem
A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?
Remember GAY-LUSSAC’S LAW!
P T
(765 torr)T2 = (560. torr)(296K)
T2 = 217 K = -38°C
Warm Up:Warm Up:Warm Up:Warm Up: DO NOT SOLVE! Just determine which Gas Law you should use.
1. What is the new volume if a 5.0L container of gas at 23 °C is heated to 70 °C?
2. If a gas at STP is heated to 900 K, what is the resulting pressure?
3. 15 mL of a gas at 1 atm is compressed to 5 mL. How much pressure was applied?
4. A 2.0 L container at 4.0 atm and 0.0 °C is heated to 50. °C at constant volume. What is the resulting pressure?
5. A 35 L container at 23 °C is expanded to 50 L to obtain a pressure of 4.0 atm. What was the original pressure if temperature was constant?
Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law
This is a combination of the 3 main gas laws: Boyle’s Law, Charles’ Law & Gay-Lussac’s Law
Pressure is opposite temperature & volume
PP VVTT
= kPVT
Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law
P1V1
T1
=P2V2
T2
P1V1T2 = P2V2T1
GIVEN:
V1 = 7.84 mL
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 mL)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 mL
Combined Gas Law Combined Gas Law ProblemProblemCombined Gas Law Combined Gas Law ProblemProblem
A gas occupies 7.84 mL at 71.8 kPa & 25°C. Find its volume at STP.
P T VCOMBINED GAS LAW
Warm Up:Warm Up:Warm Up:Warm Up:
Stoichiometry Quiz #5Stoichiometry Quiz #5
How many grams of CO2 are produced from 75 L of CO at
STP?
___CO + ___O2 → ___CO2
Follow along in your text
Chapter 12 Section 3Pages 433 - 442
Follow along in your text
Chapter 12 Section 3Pages 433 - 442
The Ideal Gas The Ideal Gas LawLaw
PV
TVn
PVnT
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
= k
UNIVERSAL GAS CONSTANTR=0.0821 Latm/molKR=8.314 LkPa/molK
R=62.4 LmmHg/molK
= R
Merge the Combined Gas Law with Avogadro’s Principle:
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
UNIVERSAL GAS CONSTANTR=0.0821 Latm/molKR=8.314 LkPa/molK
R=62.4 LmmHg/molK
PV=nRT
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
PVM=mRT
If you are given the mass of a gas, you can use this equation instead of converting mass to moles first.
massmolar mass
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
d=PM/RT
If you are given the molar mass of a gas, you can use this equation to
find the density
molar massdensity
GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 LR = 0.0821Latm/molK
WORK:
PV = nRT
Ideal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law Problems Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25 L.
P=(0.412 mol)(0.0821)(289 K)
(3.25 L)
P = 3.01 atm
GIVEN:
V = ?
m = 85 g
MO2 = 32 g/mol
T = 25°C = 298 K
P = 104.5 kPaR = 8.314 LkPa/molK
Ideal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and 104.5 kPa.
WORK:
PVM = mRT
V= (85 g)(8.314)(298 K)
(104.5 kPa)(32 g/mol)
V = 64 L
Gas StoichiometryGas StoichiometryGas StoichiometryGas Stoichiometry Moles Moles Liters of a Gas: Liters of a Gas:
• STP - use 22.4 L/mol • Non-STP - use ideal gas law
Non-Non-STPSTP• Given liters of gas?
start with ideal gas law• Looking for liters of gas?
start with stoichiometry conversion
1 molCaCO3
100.09g CaCO3
Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem
What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
5.25 gCaCO3 = 1.26 mol CO2
CaCO3 CaO + CO2
1 molCO2
1 molCaCO3
5.25 g ? Lnon-STPLooking for liters: Start with stoich
and calculate moles of CO2.
Plug this into the Ideal Gas Law to find liters.
WORK:
PV = nRT
(103 kPa)V=(1mol)(8.314 LkPa/molK)(298K)
V = 1.26 dm3 CO2
Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem
What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPaV = ?
n = 1.26 molT = 25°C = 298 KR = 8.314 LkPa/molK
WORK:
PV = nRT
(97.3 kPa) (15.0 L)= n (8.314 LkPa/molK) (294K)
n = 0.597 mol O2
Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
GIVEN:
P = 97.3 kPaV = 15.0 L
n = ?T = 21°C = 294 KR = 8.314 LkPa/molK
4 Al + 3 O2 2 Al2O3 15.0 L
non-STP ? gGiven liters: Start with
Ideal Gas Law and calculate moles of O2.
NEXT
2 mol Al2O3
3 mol O2
Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
0.597mol O2 = 40.6 g Al2O3
4 Al + 3 O2 2 Al2O3
101.96 g Al2O3
1 molAl2O3
15.0Lnon-STP
? gUse stoich to convert moles of O2 to grams Al2O3.
Closure:Closure:Closure:Closure:
How many grams of CO2 are produced from 75L of CO at
35 ºC and 96.2 kPa?
___CO + ___O2 → ___CO2
Warm Up:Warm Up:Warm Up:Warm Up:
Solve using the Ideal Gas Law & StoichometrySolve using the Ideal Gas Law & Stoichometry
What mass of NHWhat mass of NH33 is created is created
when 20.0 L of Nwhen 20.0 L of N22 is combined is combined
with excess Hwith excess H22 at 42 °C and at 42 °C and
2.50 atm? 2.50 atm?
More Gas Laws!More Gas Laws!
Follow along inyour text
Chapter 12 Sections 2 & 3
Pages 432 - 439
Follow along inyour text
Chapter 12 Sections 2 & 3
Pages 432 - 439
kn V V
n
Avogadro’s LawAvogadro’s LawAvogadro’s LawAvogadro’s Law Equal volumes of gases contain equal
numbers of moles at constant temp & pressure• 22.4 L/mole• 6.02 x 1023 particles/mole
Dalton’s Law of Partial Dalton’s Law of Partial PressurePressureDalton’s Law of Partial Dalton’s Law of Partial PressurePressure
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
Ptotal = P1 + P2 + ...
P1 = (Ptotal) (% of Gas1)
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Dalton’s Law of Partial Dalton’s Law of Partial PressurePressureDalton’s Law of Partial Dalton’s Law of Partial PressurePressure
Hydrogen gas is collected over water. The water vapor has a pressure of 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
Sig Figs: Round to least number of decimal places.
The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.
Graham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of Diffusion
DiffusionDiffusion• Spreading of gas molecules from
high density to low density until even all over.
EffusionEffusion
• Passing of gas molecules under pressure through a tiny opening
Graham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of Diffusion
Speed of diffusion/effusionSpeed of diffusion/effusion
• Kinetic energy is determined by the temperature of the gas.
• At the same temp & KE, heavier molecules move more slowly. “The smaller the mass, the
faster the gas!”
Graham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of Diffusion
Graham’s LawGraham’s Law• Rate of diffusion of a gas is inversely related
to the square root of its molar mass.• The equation shows the ratio of Gas A’s
speed to Gas B’s speed.
A
B
B
A
m
m
v
vv
√m
Determine the relative rate of diffusion for krypton and bromine.
1.381
Kr diffuses 1.381 times faster than Br2.
Kr
Br
Br
Kr
m
m
v
v2
2
A
B
B
A
m
m
v
v
g/mol83.80
g/mol159.80
Graham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of Diffusion
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.
A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?
A
B
B
A
m
m
v
v
2
2
2
2
H
O
O
H
m
m
v
v
g/mol 2.02
g/mol32.00
m/s 12.3
vH 2
Graham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of DiffusionGraham’s Law of Diffusion
3.980m/s 12.3
vH 2
m/s49.0 vH 2
Put the gas with the unknown
speed as “Gas A”.