Unit 10Unit 10Gas LawsGas Laws
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.
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
A. Graham’s LawA. Graham’s Law
Which one is Diffusion and which one is Effusion?
DiffusionEffusion
Tiny opening
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
B. VolumeReduce volume - ↑ pressureIncrease volume - ↓ pressure Ex: piston in a car
II. Factors Affecting Gas II. Factors Affecting Gas Pressure Pressure
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
Gas Gas PressurePressure-- collision of gas collision of gas molecules with the walls of the molecules with the walls of the containercontainer
Atmospheric Pressure-Atmospheric Pressure- collision collision of air molecules with objectsof air molecules with objects
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
Gas Pressure (Cont.) -- 3 ways to measure pressure:
»atm (atmosphere)»mm Hg»kPa (kilopascals)
U-tube Manometer
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
Draw on the Left Side of Draw on the Left Side of Your SpiralYour Spiral
Pressure
Volume
Temperature
kPa
Mole
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
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
Standard Temperature and Pressure
Standard pressure – 1 atm, 760 mmHg, or 101.3 kPa
Standard temp. – 0° C or 273K
STP
Gases (cont.)Gases (cont.) Kelvin Temperature scale is directly
proportional to the average kinetic energy
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
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
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
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
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
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
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
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
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
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.
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
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).
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
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
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.
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 + ...
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
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.