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Chapter 10.3 : The Gas Laws

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• 1. The Gas lawsChapter 10.3
Objectives:
Use the kinetic-molecular theory to explain the relationships between gas volume, temperature, and pressure.
Use Boyle’s law to calculate volume-pressure changes at constant temperature.
Use Charles’s law to calculate volume-temperature changes at constant pressure.
Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.
Use the combined gas law to calculate volume-temperature-pressure changes.
Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.
• 2. The Gas Laws
Simple mathematical relationships between
Volume
Temperature
Pressure
Amount of gas
Gas Law Program:
Shows the relationship of all four of the above on gases
http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
Constant :
Volume and amount of gas
Shows:
Change in pressure and temperature
• 3. Boyle’s Law
States – volume of a fixed mass of gas varies inversely with the pressure at constant temperature.
Constant temperature and amount of gas
If you double volume, pressure is cut in half
If you cut volume in half, pressure doubles
• 4. Boyle’s Law
Pressure caused by
Moving molecules hitting container walls
Speed of particles (force) and number of collisions
Both increase pressure
Mathematically:
Volume-Pressure Data for Gas Sample
1
Volume Pressure P x V
(mL) (atm)
V
k
=
PV
or
=
k
P
k is constant
1200 0.5 600
P is pressure
600 1.0 600
V is volume
300 2.0 600
200 3.0 600
150 4.0 600
120 5.0 600
100 6.0 600
Interactive graph
• 5. Boyle’s Law
Boyle’s Law Equation:
P1V1
=
k
P2V2
=
k
So:
P1V1
=
P2V2
Sample Problem 1
A sample of oxygen gas has a volume of 150. mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant.
P1
=
0.947 atm
P1V1
=
P2V2
V1
=
150. mL
(0.947 atm)
(150. mL)
P2
P2
V2
=
P2
=
0.987 atm
0.987 atm
P1V1
V2
=
?
=
V2
=
144 mL
P2
• 6. Charles’s Law
Volume-temperature Relationship
When using temperature, Must use absolute zero, Kelvin Temperature scale
Temperature -273.15oC is absolute zero
All molecular movement would stop.
K = oC + 273
212
100
373
So : How many Kelvin is 10 oC?
K = oC + 273
= 10oC + 273
32
0
273
= 283 K
Fahrenheit
Celsius
Kelvin
Temperature Scales
• 7. Charles’s Law
Volume-Temperature Data for Gas Sample
Temperature Kelvin Volume V/T
(oC) (K) (mL) (mL/K)
273 546 1092 2
100 373 746 2
10 283 566 2
1 274 548 2
0 273 546 2
-1 272 544 2
-73 200 400 2
-173 100 200 2
-223 50 100 2
• 8. Charles’s Law
States that the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.
Constant Pressure and amount of gas
If you double temperature, the volume will double
If you cut the temperature in half, the volume will be half as much.
V
V
k
=
or
kT
=
T
k is constant
T is temperature
V is volume
Charles’s Law Equation:
V1
V2
=
k
=
k
T1
T2
So:
V2
V1
=
T2
T1
• 9. Charles’s Law
Sample Problem 2
A sample of neon gas occupies a volume of 752 mL at 25oC. What volume will the gas occupy at 50oC if the pressure remains constant?
V2
V1
V1
=
752 mL
T2
=
T2
x
x
T2
T1
T1
=
25oC
+ 273 =
298 K
V2
=
?
V1
T2
(752 mL)
(323 K)
T2
=
50oC
+ 273 =
323 K
V2
=
=
T1
298 K
**Always convert to Kelvin!!
=
815 mL
• 10. Gay-Lussac’s Law
Pressure-temperature Relationship
Increasing temperature, increases the speed of the gas particles
Thus, more collisions with the container walls
Causing an increase in pressure
• 11. Gay-Lussac’s Law
States that the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature.
Constant volume and amount of gas
If you double temperature, pressure doubles
If you cut temperature in half, pressure is also cut in half
P
P
k
=
or
kT
=
T
k is constant
T is temperature
P is pressure
P1
P2
=
k
=
k
T1
T2
So:
P2
P1
=
T1
T2
• 12. Gay-Lussac’s Law
Sample Problem 3
The gas in an aerosol can is at a pressure of 3.00 atm at 25oC. Directions on the can warn the user not to keep the can in a place where the temperature exceed 52oC. What would the gas pressure in the can be at 52oC?
P1
P2
P1
=
3.00 atm
T2
=
T2
x
x
T2
T1
T1
=
25oC
+ 273 =
298 K
P2
=
?
P1
T2
(3.00 atm)
(325 K)
T2
=
52oC
+ 273 =
325 K
P2
=
=
T1
298 K
**Always convert to Kelvin!!
=
3.27 atm
• 13. The Combined Gas Law
Sample Problem 4
A helium-filled balloon has a volume of 50.0 L at 25oC and 1.08 atm. What volume will it have at 0.855 atm and 10.oC?
P1V1
P2V2
P1
1.08 atm
=
T2
=
x
T2
x
T2
T1
T1
=
25oC
+ 273 =
298 K
V1
=
50.0 L
P1V1
T2
P2V2
=
P2
=
0.855 atm
T1
P2
P2
T2
=
10.oC
+ 273 =
283 K
P1V1
T2
V2
=
?
V2
=
T1
P2
(1.08 atm)
(283 K)
(50.0 L)
=
(298 K)
(0.855 atm)
60.0 L
=
• 14. Dalton’s Law of Partial Pressures
States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
Partial pressure: pressure of each gas in a mixture
PT = p1 + p2 + p3 + ……
PT= Total Pressure
p1 + p2 + p3 = partial pressures
• 15. Dalton’s Law of Partial Pressures
Gas collected by water displacement.
Must include the pressure exerted by water vapor
PT = p1 + p2 + p3 + ……
So: Patm = pgas + pH2O
• 16. Dalton’s Law of Partial Pressures
Sample Problem 5
Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0oC, respectively. What was the partial pressure of the oxygen collected?
Patm = pO2 + pH2O
Patm = 731.0 torr
PO2 = ?
PH2O = 17.5 torr (from appendix in table A-8, pg. 899)
pO2 = Patm - pH2O
pO2 = 731.0 torr – 17.5 torr
= 713.5 torr