i. physical properties ch. 11 - gases. a. kinetic molecular theory b particles in an ideal gas…...
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I. Physical PropertiesI. Physical Properties
Ch. 11 - Gases
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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.
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B. Real GasesB. Real Gases
Particles in a REAL gas…• have their own volume• attract each other
Gas behavior is most ideal…• at low pressures• at high temperatures• in nonpolar atoms/molecules
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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
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C. Characteristics of GasesC. Characteristics of Gases
Gases can be compressed.• no volume = lots of empty space
Gases undergo diffusion & effusion.• random motion
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D. 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. Pressure
area
forcepressure
Which shoes create the most pressure?
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E. PressureE. Pressure
Barometer• measures atmospheric pressure
Mercury Barometer
Aneroid Barometer
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E. PressureE. Pressure
Manometer• measures contained gas pressure
U-tube Manometer Bourdon-tube gauge
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E. 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. STP
Standard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa-OR-
STP
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II. The Gas Laws
II. The Gas Laws
Ch. 11 - Gases
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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
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A. Boyle’s LawA. Boyle’s Law
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kT
VV
T
B. Charles’ LawB. Charles’ Law
The volume and absolute temperature (K) of a gas are directly related • at constant mass &
pressure
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B. Charles’ LawB. Charles’ Law
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kT
PP
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
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= kPVPTVT
PVT
D. 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 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 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 Problems
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
P T VCOMBINED GAS LAW
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GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1V1T2 = P2V2T1
E. Gas Law ProblemsE. Gas Law Problems
A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?
GAY-LUSSAC’S LAW
P T
(765 torr)T2 = (560. torr)(296K)
T2 = 217 K
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END OF QUIZ I MATERIALEND OF QUIZ I MATERIAL
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III. Ideal Gas LawIII. Ideal Gas Law
Ch. 11 - Gases
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kn
VV
n
A. Avogadro’s PrincipleA. Avogadro’s Principle
Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas
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PV
TVn
PVnT
A. Ideal Gas Law p. 384A. Ideal Gas Law p. 384
= kUNIVERSALAS
CONSTANTR=0.0821 Latm/molK
R = 62.4 Lmm Hg/molK (torr)
R=8.315 dm3kPa/molK
= R
You don’t need to memorize these values!
Merge the Combined Gas Law with Avogadro’s Principle:
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A. Ideal Gas LawA. Ideal Gas Law
UNIVERSAL GAS CONSTANT
R=0.0821 Latm/molKR = 62.4 Lmm Hg/molK (torr)
R=8.315 dm3kPa/molK
PV=nRT
You don’t need to memorize these values!
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GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 L
R = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K
P = 3.01 atm
C. Ideal Gas Law ProblemsC. Ideal Gas Law Problems
Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.
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GIVEN:
V = ?
m = 85 g
T = 25°C = 298 K
P = 104.5 kPa
R = 8.315 dm3kPa/molK
C. Ideal Gas Law ProblemsC. Ideal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and 104.5 kPa.
n= 2.7 mol
WORK:
85 g 1 mol = 2.7 mol
32.00 g
PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K
V = 64 dm3
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Let’s Practice!! Handout p.24Let’s Practice!! Handout p.24
Things to remember:OMIT 5 & 6Convert temperatures to Kelvin
(add 273)Convert all volumes to Liters (mL to
L divide by 1000)Use R value that matches pressure
unit