A. Kinetic Molecular TheoryA. Kinetic Molecular Theory
Particles in an ideal gas…
•have no volume.
•have elastic collisions.
•are in constant, random, straight-
line motion.
•don’t attract or repel each other.
•have an avg. KE directly related to
Kelvin temperature.
C. Characteristics of GasesC. Characteristics of Gases
Gases expand to fill any container.
•random motion, no attraction
Gases are fluids (like liquids).
•no attraction
Gases have very low densities.
•no volume = lots of empty space
C. Characteristics of GasesC. Characteristics of Gases
Gases can be compressed.
•no volume = lots of empty space
Gases undergo diffusion & effusion.
•random motion
D. TemperatureD. Temperature
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
32FC
9
5
K = ºC + 273
Always use absolute temperature
(Kelvin) when working with gases.
E. PressureE. Pressure
area
force
pressure
Which shoes create the most pressure?
E. PressureE. Pressure
Manometer
•measures contained gas pressure
U-tube Manometer Bourdon-tube gauge
E. PressureE. Pressure
2
m
N
kPa
KEY UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi
F. STPF. STP
Standard Temperature & PressureStandard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa
-OR-
STP
II. The Gas
Laws
BOYLES
CHARLES
GAY-
LUSSAC
Gas LawsGas Laws
A. Boyle’s LawA. Boyle’s Law
P
V
PV = k
A. Boyle’s LawA. Boyle’s Law
The pressure and volume of
a gas are inversely related
•at constant mass & temp
P
V
PV = k
k
T
V
V
T
B. Charles’ LawB. Charles’ Law
k
T
V
V
T
B. Charles’ LawB. Charles’ Law
The volume and absolute
temperature (K) of a gas are directly
related
•at constant mass & pressure
k
T
P
P
T
C. Gay-Lussac’s LawC. Gay-Lussac’s Law
k
T
P
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
= k
PV
T
D. Combined Gas LawD. Combined Gas Law
P
1V
1
T
1
=
P
2V
2
T
2
P
1V
1T
2 =
P
2V
2T
1
GIVEN:
V
1 = 473 cm
3
T
1 = 36°C = 309K
V
2 = ?
T
2 = 94°C = 367K
WORK:
P
1V
1T
2 = P
2V
2T
1
E. Gas Law ProblemsE. Gas Law Problems
A gas occupies 473 cm
3
at 36°C.
Find its volume at 94°C.
CHARLES’ LAW
TV
(473 cm
3
)(367 K)=V
2(309 K)
V
2
= 562 cm
3
GIVEN:
V
1 = 100. mL
P
1 = 150. kPa
V
2 = ?
P
2 = 200. kPa
WORK:
P
1V
1T
2 = P
2V
2T
1
E. Gas Law ProblemsE. Gas Law Problems
A gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
PV
(150.kPa)(100.mL)=(200.kPa)V
2
V
2 = 75.0 mL
GIVEN:
V
1
= 7.84 cm
3
P
1
= 71.8 kPa
T
1
= 25°C = 298 K
V
2
= ?
P
2
= 101.325 kPa
T
2
= 273 K
WORK:
P
1
V
1
T
2
= P
2
V
2
T
1
(71.8 kPa)(7.84 cm
3
)(273 K)
=(101.325 kPa)
V
2 (298 K)
V
2 = 5.09 cm
3
E. Gas Law ProblemsE. Gas Law Problems
A gas occupies 7.84 cm
3
at 71.8 kPa &
25°C. Find its volume at STP.
P T V
COMBINED GAS LAW