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GasesGasesChapter 13Chapter 13
Solids, liquids, and GasesSolids, liquids, and Gases
Chapter 13 2
Compare the position and motion of the Compare the position and motion of the three states of matter.three states of matter.
EnergyEnergy• Potential Energy
– Stored energy – due to position
– Particles are attracted to one another. More energy is required to keep particles farther apart.
– Which of the 3 states has the highest potential energy?
Chapter 13 3
• Kinetic Energy– Motion energy – related to temperature
– The faster the particles are moving, the higher the kinetic energy, the higher the temperature (average kinetic energy)
– Which of the 3 states has the highest kinetic energy?
Chapter 13 4
Kinetic-Molecular TheoryKinetic-Molecular Theory- Theory developed to explain gas behavior- To describe the behavior of a gas, we must first
describe what a gas is:– Gases consist of very small particles each of which have a
mass.
– The distance between gas particles are relatively large. Volume of individual molecules is negligible compared to volume of container.
– Gas particles in rapid, constant, random motion.
Chapter 13 5
– Collisions between gas particles are perfectly elastic.
Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.
Kinetic-Molecular Theory (ContKinetic-Molecular Theory (Cont’’d)d)
No energy is lost during collisions.
Chapter 13 6
Kinetic-Molecular Theory (ContKinetic-Molecular Theory (Cont’’d)d)The average kinetic energy of gas particles depends only on the temperature of the gas.
Gas particles exert no forces on each other. Intermolecular forces (forces between gas molecules) are negligible.
What happens as temperature increases?
Which statements in KMT are assumptions?
Chapter 13 7
LetLet’’s generate some gass generate some gas
- Expand to fill a volume (expandability)- Compressible- Takes shape of container- Diffuses and flows
Properties of a gasProperties of a gas
Chapter 13 8
Variables that can be measured for gasesVariables that can be measured for gases
– Temperature
– Volume
– Amount
– Pressure
Temperature (T)Temperature (T)
• Measured in Fahrenheit, Celsius, or Kelvin.• For this chapter, we have to use Kelvin.
◦C = K – 273 K = ◦C + 273
Chapter 13 9
Volume (V)Volume (V)
• Measured in Liters, cubic meters, gallons, etc…
Amount (n)Amount (n)
• Measured in moles
Chapter 13 10
Pressure (P) – Pressure (P) – what causes pressure?what causes pressure?
1 atm = 760 mmHg = 760 torr = 101.3 kPa
The pressure of a gas is measured using a manometer.
*Atomspheric
pressure is
measured
using a
Barometer.
Unit ConversionsUnit Conversions
22◦C = ______ K 37 K = ______ ◦C
18 mL = ______ L 4.3 L = ______ cm3
500 cm3= ______ mL 2.2 dm3 = _______ L
Chapter 13 11
Unit ConversionsUnit Conversions
2.8 g N2 = ______ mol 612 mol SO3 = _______ g
22 kPa = ______ atm 289 mmHg = _______ kPa
4.3 atm = ______ torr 518 kPa = _______ mmHg
Chapter 13 12
Chapter 13 13
The Gas LawsThe Gas Laws- There are four variables required to describe a gas:
- Amount of substance: moles (n)- Volume of substance: volume (V)- Pressures of substance: pressure (P)- Temperature of substance: temperature (T)
- The gas laws will hold two of the variables constant and see how the other two vary (n,V,P,T)
Chapter 13 14
TodayToday’’s Lab – Boyles Lab – Boyle’’s Laws Law
• We will maintain a constant temperature and number of moles of gas.
• So we will vary the Pressure and the Volume and see how they relate.
PV = k or P/V = kinverse direct
BolyleBolyle’’s Law Lab – 16 pts totals Law Lab – 16 pts total• Heading – 1 pt• Purpose – 1 pt• Procedure – 1 pt• Data – 4 pts (make sure they have units for the 3rd and 4th
columns.• Graph – 3 pts (should have a title and labeled axis)• Questions – 6 pts (in complete sentences)
1) 1/2 2) doubled 3) inverse 4) PV=k
5) The pressure and volume of a gas at a constant temperature are inversely proportional to each other.
6) Source of error.
Chapter 13 15
Chapter 13 16
Variables for Gases Discussed BeforeVariables for Gases Discussed Before
• T – Temperature• V – Volume• P – Pressure• n – Amount of a substance (moles)
Chapter 13 17
BoyleBoyle’’s Law – s Law – peeps in bell jar demo peeps in bell jar demo
The Pressure-Volume Relationship Boyle’s Law - The volume of a fixed quantity of gas is
inversely proportional to its pressure at a constant temperature.
2211
kPVor k
) and (constant 1
VPVPV
P
TnV
P
Chapter 13 18
A gas occupies 22 L at 2.43 atm. What is the new volume if the pressure changed to 5.11 atm?
Chapter 13 19
Gay-Lussac’s Law – can crush demo
Gay-Lussac’s Law – As the temperature of an enclosed gas increases, the pressure increases if the volume is constant.
The Pressure-Temperature Relationship:
)constantVandnfor(2
2
1
1
T
P
T
P
kT
PorkTPorTP
Chapter 13 20
A gas has a pressure of 1.47 atm at 303 K. What is the new temperature if the pressure changed to 680 mmHg?
Chapter 13 21
Charles Law – Charles Law – ivory soap demoivory soap demo
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
The Temperature-Volume Relationship:
)constantPandnfor(2
2
1
1
T
V
T
V
kT
VorkTVorTV
Chapter 13 22
A gas occupies 14 L at 275 K. What is the new volume if the temperature changed to 297 K?
Chapter 13 23
AvogadroAvogadro’’s Laws Law
Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
) and (constant TPnV
The Quantity-Volume Relationship:
2
2
1
1
n
V
n
V
kn
V
Chapter 13 24
A balloon contains 1.98 mol of a gas and has a volume of 4.2 L. Some of the gas was let out to give a volume of 3.1 L. What is the amount of gas left in the balloon?
Chapter 13 25
The Gas Laws SummaryThe Gas Laws Summary
Boyle’s Law
2
2
1
1V
) and (constant
n
V
n
TPnV
2
2
1
1
)and(constant
T
V
T
V
PnTV
2211
) and (constant 1
VPVP
TnV
P
Charles’ Law
Avogadro’s Law
2
2
1
1P
) and (constant
T
P
T
VnTP
Guy-Lussac’s Law
Chapter 13 26
The Ideal Gas EquationThe Ideal Gas Equation
Ideal gas equation:
PV = nRT P = pressure (atm or mmHg or kPa)
V = volume (L) n = amount (mol)
R = gas constant T = temperature (K)
Combine the gas laws (Boyle, Charles, Guy-Lussac, Avogadro) yields a new law or equation.
Chapter 13 27
Finding R with Dry Ice Lab
nT
PVR PV = nRT
What units are used for R?
Calculating R with Dry Ice Lab – 20 pts totalCalculating R with Dry Ice Lab – 20 pts total• Heading – 1 pt
• Purpose – 1 pt
• Procedure – 1 pt
• Data – 10 pts (make sure they have units)
• Calculations – 2pts
• Conclusion – 5 pts– Their value for R
– Literature value for R
– % error
– Sources of error
– How they affected their results (higher or lower and must be consistent with their result for R)
Chapter 13 28
Chapter 13 29
Calculating RCalculating RWe define STP (standard temperature and pressure) as
0C = 273 K, and 1 atm = 760 mmHg = 101.3 kPa
Volume of 1 mol of gas at STP is 22.4 L.
nT
PVR PV = nRT
Chapter 13 30
The Ideal Gas Equation-Finding RThe Ideal Gas Equation-Finding R
Kmol
LkPa
Kmol
LkPaR
31.8
2731
4.223.101
Kmol
Latm
Kmol
LatmR
0821.02731
4.221
Kmol
LmmHg
Kmol
LmmHgR
4.62
2731
4.22760
Chapter 13 31
Combined Gas LawCombined Gas Law• So in cases with changing conditions
11
11
Tn
VPR
22
22
Tn
VPR and
22
22
11
11
Tn
VP
Tn
VP
Chapter 13 32
Ideal vs. Combined Gas Law-When UseIdeal vs. Combined Gas Law-When Use• Ideal Gas Law: Conditions not changing
• Combined Gas Law: Conditions Changing
22
22
11
11
Tn
VP
Tn
VP
nRTPV
Chapter 13 33
Example 1Example 1
• How many moles of a gas at 100 degrees C does it take to fill a 1.0 L flask to a pressure of 1.50 atm?
Not a changing situation so use Ideal Gas Law
PV=nRT
At 60 Celsius a 0.10 L sample of a gas has a pressure of 75.6 kPa. What would its volume be at STP?
Chapter 13 34
Changing situation so use combined gas law.
22
22
11
11
Tn
VP
Tn
VP
Example 2Example 2
Chapter 13 35
• What pressure would 3.55 grams of argon gas be under in a 2.40 L cylinder at -35 Celsius?
Not a changing situation so use Ideal Gas Law
PV=nRT
Example 3Example 3
Chapter 13 36
• What is the density of bromine gas the gas fills a 52.5 L cylinder at 145 K and 583 mmHg?
Not a changing situation so use Ideal Gas Law
PV=nRT
Example 4Example 4
Chapter 13 37
Example 5Example 5A gas occupies 4430 mL at 30 Celsius. It was
transferred in a 3.5 L cylinder. What is the new temperature.
Changing situation so use combined gas law.
22
22
11
11
Tn
VP
Tn
VP
Chapter 13 38
DemosDemos
Egg in E-flask
Fountain
Tank CarTank Car
• This tank car was cleaned with steam then all the valves were shut and tank car was sealed. The workers went home and when they came back the next morning this is what they saw.
Chapter 13 39
Chapter 13 40
Chapter 13 41
Chapter 13 42
Chapter 13 43
Ideal vs. Real gasesIdeal vs. Real gasesIdeal gases behave “ideally” according to the kinetic molecular theory and follow the ideal gas law PV = nRT.
But kinetic molecular theory has several assumptions that work most of the time but not always.
Chapter 13 44
Kinetic Molecular Theory Kinetic Molecular Theory AssumptionsAssumptions
1) The volume of gas particles are so small and the spaces between particles are so large.
Therefore KMT assumes that gas particles have no volume.
2) The distance between gas particles are very large.
Therefore KMT assumes that there is no attraction (IMF) between gas particles.
Chapter 13 45
But Real gases have volumeBut Real gases have volumeUnder most conditions the volume of gas particles is negligible.
To compensate, we plug in the following equation for V in PV = nRT
nbVV
But at small volumes or if the gas particles are large, the volume of the gas particles become significant.
Chapter 13 46
And real gases have attractive And real gases have attractive forces (IMF)forces (IMF)
Under most conditions, the intermolecular forces between gas particles are negligible.
To compensate for the pressure difference caused by IMF we plug in the following equation instead of P in PV = nRT
)(2
V
naPP
But at high pressure and low temperatures, the attractive forces become significant.
Chapter 13 47
Ideal vs. Real gasesIdeal vs. Real gasesWhich gas behaves more ideally?
Ne or HCl?
Neon because it’s particles have a smaller volume and weaker intermolecular forces.
Chapter 13 48
Ideal vs. Real gasesIdeal vs. Real gasesIn PV = nRT , plugging in
We get the Van der Waals Equation
)(2
V
naPP
nbVV
ExampleExample• Use the Van der Waals equation to calculate the
temperature of 3.6 moles of nitrogen gas in a 4.6 L cylinder at 2.5 atm if a and b for nitrogen are 1.390 and 0.03910 respectively.
Chapter 13 49
Chapter 13 50
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Dalton’s Law - In a gas mixture the total pressure is given by the sum of partial pressures of each component:
Pt = P1 + P2 + P3 + …
- The pressure due to an individual gas is called a partial pressure.
Dalton’s Law
Chapter 13 51
Partial Pressure of Dry AirPartial Pressure of Dry Air
At top of Mt. Everest, total atmospheric pressure is 33.73 kPa (about 1/3 that of sea level). So partial pressure of O2 is 7.06 kPa. To support respiration the partial pressure must be 10.76 kPa or higher so oxygen tanks are needed.
ExampleExample
Chapter 13 52
What is the total pressure of a mixture of gases containing oxygen, hydrogen, and water vapor if their partial pressures are 120 kPa, 94 kPa, and 137 kPa respectively?
Chapter 13 53
Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Diffusion or Effusion
Effusion – The escape of gas through a small opening.
Diffusion – The movement of gas particles through another gas.
Chapter 13 54
Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion or Diffusion
Chapter 13 55
Kinetic-Molecular TheoryKinetic-Molecular Theory
• As kinetic energy increases, the velocity of the gas molecules increases.
• Average kinetic energy, KE, is related to root mean square speed:
KE= ½mv2
Molecular Speed
Chapter 13 56
GrahamGraham’’s Law of Effusion or Diffusions Law of Effusion or Diffusion
• Gases at same temperature – tell me about them– They have the same kinetic energy (KE)
• Which would move faster? CO2 or O2
– O2 would move faster because KE=1/2 mv2 and O2 has the lower mass
• Thomas Graham noticed how rate of diffusion (or effusion) related to molar mass
Chapter 13 57
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
Graham’s Law of Effusion or Diffusion
1
2
2
1
22
21
1
2
222
211
222
211
21
2
1
2
1
KEKE : tempsameat gases 2for
m
m
v
v
v
v
m
m
vmvm
vmvm
m1 and m2 are molar mass and v1 and v2 are effusion/diffusion rates
Chapter 13 58
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion/Diffusion – The rate of effusion/diffusion of a gas is inversely proportional to the square root of its molecular mass.
Graham’s Law of Effusion/Diffusion
1
2
2
1MM
rr
ExampleExample
Chapter 13 59
Find the relative rate of diffusion for the gases krypton and bromine.
Chapter 13 60
Gas Laws & StoichiometryGas Laws & Stoichiometry
PV = nRT calculations in reactions
with molar ratios
Link?Moles!!
Chapter 13 61
Example 1Example 1- What mass of NaCl could be produced form excess
sodium and 10.0 L of chlorine gas at 23oC and 1.02 atm?
Chapter 13 62
Example 2Example 2- Calcium carbonate decomposes to form carbon
dioxide and calcium oxide. What volume of carbon dioxide at 25oC and 99.8 kPa is produced if 25.0 g of calcium carbonate is decomposed?