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THE GAS LAWS
CHAPTER 10.3
Objectives:1. Use the kinetic-molecular theory to explain the
relationships between gas volume, temperature, and pressure.
2. Use Boyle’s law to calculate volume-pressure changes at constant temperature.
3. Use Charles’s law to calculate volume-temperature changes at constant pressure.
4. Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.
5. Use the combined gas law to calculate volume-temperature-pressure changes.
6. Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.
The Gas Laws Simple mathematical relationships
between Volume Temperature Pressure Amount of gasGas Law Program:
http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
Shows the relationship of all four of the above on gases
Constant :Volume and amount of gasShows:Change in pressure and temperature
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
Boyle’s Law Pressure caused by
Moving molecules hitting container walls Speed of particles (force) and number of
collisions Both increase pressure
Mathematically:
V1
k=P
PVor =k
k is constant
P is pressure
V is volume
Interactive graph
Volume-Pressure Data for Gas SampleVolume Pressure
P x V (mL) (atm) 1200 0.5 600 600 1.0 600 300 2.0 600 200 3.0 600 150 4.0 600 120 5.0 600 100 6.0 600
Boyle’s Law
P1V1=k
Boyle’s Law Equation:
P2V2=k
So:P1V1= P2V2
Sample Problem 1A 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
V1
P2
V2
=
===
0.947 atm150. mL0.987 atm?
P1V1= P2V2
P2P2
P1V1 = V2P2
V2 =(0.947 atm)
(150. mL)0.987
atm=144
mL
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.
Fahrenheit
Celsius Kelvin
Temperature Scales
212
32
100
0
373
273
K = oC + 273
So : How many Kelvin is 10 oC?K = oC + 273
= 10oC + 273 = 283 K
Charles’s LawVolume-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
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 k=V
orkTT
=
k is constant
T is temperature V is volume
Charles’s Law Equation:
V1
T1
=kV2
T2
=k
So: V2
T2
V1
T1
=
Charles’s LawSample Problem 2A 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?V1
T1
V2
T2
=
===
752 mL
25oC ?50oC
**Always convert to Kelvin!!
+ 273 =+ 273 =
298 K 323 K
V2
T2
V1
T1
= x T2
xT2
V2 =V1T
2T1
=(752 mL)
(323 K)298 K
=815 mL
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
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 k= orkTT
=
k is constant
T is temperature P is pressure
P1
T1
=kP2
T2
=k
So: P2P1 =
P
T1 T2
Gay-Lussac’s LawSample Problem 3The 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
T1
P2
T2
=
===
3.00 atm25oC ?52oC
**Always convert to Kelvin!!
+ 273 =+ 273 =
298 K 325 K
T2
T1
= x T2
xT2
P2 =P1T
2T1
=(3.00 atm)
(325 K)298 K
=3.27 atm
P1 P2
The Combined Gas LawSample Problem 4A 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?
P1
T1
P2
T2
==
=
=
1.08 atm25oC
0.855 atm
10.oC
+ 273 =
+ 273 =
298 K
283 K
T2
T1
= x T2
xT2
P2V2 =P1V1T
2T1
(1.08 atm)
(283 K)(298
K)= 60.0 L
P1V
1
P2V
2
V1 = 50.0 L
V2 = ?
P2 P2
V2 =P1V1T
2T1
P2
= (50.0 L)(0.855 atm)
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 Pressurep1 + p2 + p3 = partial pressures
Dalton’s Law of Partial Pressures
PT = p1 + p2 + p3 + ……
So: Patm = pgas + pH2O
Gas collected by water displacement. Must include the pressure exerted by
water vapor
Dalton’s Law of Partial PressuresSample 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 torrPO2
= ?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
