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Unit 10Unit 10Gas LawsGas Laws
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I. Kinetic TheoryI. Kinetic Theory Particles in an ideal gas…
1.gases are hard, small, spherical particles
2.don’t attract or repel each other. 3.are in constant, random,
straight-line motion.4.indefinite shape and volume.5.have “perfectly” elastic
collisions.
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A. Graham’s LawA. Graham’s Law• DiffusionDiffusion
– The tendency of molecules to move toward areas of lower concentration.•Ex: air leaving tire when valve is opened
• EffusionEffusion– Passing of gas molecules through a tiny opening in a container
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A. Graham’s LawA. Graham’s Law
Which one is Diffusion and which one is Effusion?
DiffusionEffusion
Tiny opening
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II. Factors Affecting Gas II. Factors Affecting Gas PressurePressure
A. Amount of GasAdd gas - ↑ pressureRemove gas - ↓ pressureEx: pumping up a tire adding air to a balloon aerosol cans
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B. VolumeReduce volume - ↑ pressureIncrease volume - ↓ pressure Ex: piston in a car
II. Factors Affecting Gas II. Factors Affecting Gas Pressure Pressure
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C. TemperatureIncrease Temp. - ↑ pressureDecrease Temp. - ↓
pressureEx: Helium balloon on
cold/hot day, bag of chips
II. Factors Affecting Gas II. Factors Affecting Gas Pressure Pressure
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Gas Gas PressurePressure-- collision of gas collision of gas molecules with the walls of the molecules with the walls of the containercontainer
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Atmospheric Pressure-Atmospheric Pressure- collision collision of air molecules with objectsof air molecules with objects
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Atmospheric pressure is measured with a barometer.
Increase altitude – decrease pressureEx. Mt. Everest – atmospheric pressure is 253 mm Hg
Vacuum- empty space with no particles and no pressure Ex: space
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Gas Pressure (Cont.) -- 3 ways to measure pressure:
»atm (atmosphere)»mm Hg»kPa (kilopascals)
U-tube Manometer
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III. Variables that describe a III. Variables that describe a gasgas
VariableVariabless UnitsUnits
Pressure (P) – kPa, mm Hg, atm
Volume (V) – L , mL , cm3
Temp (T) – °C , K (convert to Kelvin)
K = °C + 273
Mole (n) - mol
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Draw on the Left Side of Draw on the Left Side of Your SpiralYour Spiral
Pressure
Volume
Temperature
kPa
Mole
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How pressure units are How pressure units are related:related:
1 atm = 760 mm Hg = 101.3 kPa
How can we make these into conversion factors?
1 atm 101.3 kPa760 mm Hg 1 atm
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Guided Problem:Guided Problem:1. Convert 385 mm Hg to kPa
385 mm Hg
2. Convert 33.7 kPa to atm 33.7 kPa
x 101.3 kPa= 51.3 kPa
760 mm Hg
x = .33 atm101.3 kPa
1 atm
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Standard Temperature and Pressure
Standard pressure – 1 atm, 760 mmHg, or 101.3 kPa
Standard temp. – 0° C or 273K
STP
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Gases (cont.)Gases (cont.) Kelvin Temperature scale is directly
proportional to the average kinetic energy
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A. Boyle’s Law
IV. Gas LawsIV. Gas Laws
P
V
• The pressure and volume of a gas are inversely related
-at constant mass & temp
•P1 × V1 = P2 × V2
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10. The pressure on 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temp remains constant?
P1 = 105 kPa P2 = 40.5 kPa V1 = 2.5 L V2 = ?
P1 × V1 = P2 × V2
(105) (2.5) = (40.5)(V2) 262.5 = 40.5 (V2) 6.48 L = V2
Example Problems Example Problems pg 335 # 10 &11pg 335 # 10 &11
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11. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temp remains constant? P1 = 205 kPa P2 = ?
V1 = 4.0 L V2 = 12.0 L
P1 × V1 = P2 × V2
(205) (4.0) = (P2)(12)
820 = (P2) 12
68.3 L = P2
Example Problems Example Problems pg 335 # 10 &11pg 335 # 10 &11
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B. Charles’ LawB. Charles’ Law
• The volume and temperature (in Kelvin) of a gas are directly related – at constant mass & pressure
V
T
• V1 = V2
***Temp must be in KelvinK = °C + 273
T1T2
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Example Problems Example Problems pg. 337 # 12 & 13pg. 337 # 12 & 13
12. If a sample of gas occupies 6.80 L at 325°C, what will be its volume at 25°C if the pressure does not change?
V1= 6.8L V2 = ?
T1 = 325°C = 598 K T2 = 25°C = 298 K
6.8 = V2
598 298598 × V2 = 2026.4
598 598 V2 = 3.39 L
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13. Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant?
V1= 5.0L V2 = ?
T1 = -50°C = 223 K T2 = 100°C = 373 K
5 = V2
223 373(223) V2 = 1865 223 223
V2 = 8.36 L
Example Problems Example Problems pg. 337 # 12 & 13pg. 337 # 12 & 13
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P
T
C. Gay-Lussac’s LawC. Gay-Lussac’s Law• The pressure and absolute
temperature (K) of a gas are directly related – at constant mass & volume
P1 = P2
T1 T2
***Temp must be in KelvinK = °C + 273
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Example ProblemsExample Problems1. The gas left in a used aerosol can is at
a pressure of 103 kPa at 25°C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928°C?
P1= 103 kPa P2 = ?
T1 = 25°C = 298 K T2 = 928°C = 1201 K
103 = P2
298 1201298 × P2 = 123,703
P2 = 415 kPa
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Example Problem Example Problem pg. 338 # 14pg. 338 # 14
14. A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change?
P1= 6.58 kPa P2 = ?
T1 = 539 K T2 = 211 K
6.58 = P2
539 211539 × P2 = 1388
539 539 P2 = 2.58 kPa
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D. Combined Gas LawD. Combined Gas Law
P1V1
T1
=P2V2
T2
Combines the 3 gas laws as follows:
•The other laws can be obtained from this law by holding one quantity (P,V or T) constant.
•Use this law also when none of the variables are constant.
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How to remember each Law!How to remember each Law!
PP VV
TT
Gay-Lussac
Boyles
Charles
Cartesian Divers
Balloon and flask Demo
Fizz Keepers
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E. Ideal Gas LawE. Ideal Gas Law• The 4th variable that considers the
amount of gas in the system
P1V1
T1 n=
P2V2
T2 n
• Equal volumes of gases contain equal numbers of moles (varies directly).
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E. Ideal Gas LawE. Ideal Gas Law
You don’t need to memorize this value!
•You can calculate the # of n of gas at standard values for P, V, and T
PVTn
= R (1 atm)(22.4L)(273K)(1 mol) = R
UNIVERSAL GAS CONSTANT
R= 0.0821 atm∙L/mol∙K
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E. Ideal Gas LawE. Ideal Gas Law
P= pressure in atmV = volume in litersn = number of molesR= 0.0821 atm∙L/mol∙KT = temperature in Kelvin
PV=nRT
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E. Example ProblemsE. Example Problems1. At what temperature will 5.00g of Cl2
exert a pressure of 900 mm Hg at a volume of 750 mL?
2. Find the number of grams of CO2 that exert a pressure of 785 mm Hg at a volume of 32.5 L and a temperature of 32 degrees Celsius.
3. What volume will 454 g of H2 occupy at 1.05 atm and 25°C.
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F. Dalton’s Partial Pressure F. Dalton’s Partial Pressure LawLaw
• The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
Ptotal = P1 + P2 + P3 + ...
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F. Dalton’s LawF. Dalton’s Law• Example problem:
1. Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (PO2
) if the total pressure is
101.3 kPa. And the partial pressures of nitrogen, carbon dioxide, and other gases are 79.10 kPa, 0.040 kPa, and 0.94 kPa.
PO2 = Ptotal – (PN2
+ PCO2 + Pothers)
= 101.3 kPa – (79.10 kPa + 0.040 kPa + 0.94 kPa)
= 21.22 kPa
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F. Dalton’s Law F. Dalton’s Law 2. A container holds three gases : oxygen ,
carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm respectively. What is the total pressure of the container?
3. A gas mixture contains oxygen, nitrogen and carbon dioxide. The total pressure is 50.0 kPa. If the carbon dioxide has a partial pressure of 21 kPa and the nitrogen has a partial pressure of 15 kPa, what is the partial pressure of the oxygen?
4. A container contains two gases – helium and argon, at a total pressure of 4.00 atm. Calculate the partial pressure of helium if the partial pressure of the argon is 1.5 atm.