i. physical properties. a. kinetic molecular theory b particles in an ideal gas… have no volume....
TRANSCRIPT
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I. Physical PropertiesI. Physical Properties
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A. Kinetic Molecular A. Kinetic Molecular TheoryTheoryA. Kinetic Molecular A. Kinetic Molecular TheoryTheory
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.
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C. Characteristics of C. Characteristics of GasesGasesC. Characteristics of C. Characteristics of GasesGasesGases 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
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C. Characteristics of C. Characteristics of GasesGasesC. Characteristics of C. Characteristics of GasesGasesGases can be compressed.
• no volume = lots of empty space
Gases undergo diffusion & effusion.• random motion
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D. TemperatureD. TemperatureD. TemperatureD. Temperature
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
32FC 95 K = ºC + 273
Always use absolute temperature (Kelvin) when working with gases.
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E. PressureE. PressureE. PressureE. Pressure
area
forcepressure
Which shoes create the most pressure?
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E. PressureE. PressureE. PressureE. Pressure
Barometer• measures atmospheric pressure
Mercury Barometer
Aneroid Barometer
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E. PressureE. PressureE. PressureE. Pressure
Manometer• measures contained gas pressure
U-tube Manometer Bourdon-tube gauge
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E. PressureE. PressureE. PressureE. Pressure
2m
NkPa
KEY UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi
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F. STPF. STPF. STPF. STP
Standard Temperature & PressureStandard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa-OR-
STP
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II. The Gas LawsBOYLES
CHARLESGAY-
LUSSAC
II. The Gas LawsBOYLES
CHARLESGAY-
LUSSAC
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A. Boyle’s LawA. Boyle’s LawA. Boyle’s LawA. Boyle’s Law
P
V
PV = k
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A. Boyle’s LawA. Boyle’s LawA. Boyle’s LawA. Boyle’s Law
The pressure and volume of a gas are inversely related
• at constant mass & temp
P
V
PV = k
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kT
VV
T
B. Charles’ LawB. Charles’ LawB. Charles’ LawB. Charles’ Law
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kT
VV
T
B. Charles’ LawB. Charles’ LawB. Charles’ LawB. Charles’ Law
The volume and absolute temperature (K) of a gas are directly related • at constant mass & pressure
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kT
PP
T
C. Gay-Lussac’s LawC. Gay-Lussac’s LawC. Gay-Lussac’s LawC. Gay-Lussac’s Law
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kT
PP
T
C. Gay-Lussac’s LawC. Gay-Lussac’s LawC. Gay-Lussac’s LawC. Gay-Lussac’s Law
The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume
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= kPVPTVT
PVT
D. Combined Gas LawD. Combined Gas LawD. Combined Gas LawD. Combined Gas Law
P1V1
T1
=P2V2
T2
P1V1T2 = P2V2T1
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GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law Problems
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
T V
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
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GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law Problems
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
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GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 cm3
E. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law ProblemsE. Gas Law Problems
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
P T VCOMBINED GAS LAW