I. Physical Properties 9 (A) describe and calculate the relations between volume, pressure, number...
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Transcript of I. Physical Properties 9 (A) describe and calculate the relations between volume, pressure, number...
I. Physical Properties9 (A) describe and calculate the
relations between volume, pressure, number of moles, and temperature for an ideal gas as described by Boyle's
law, Charles' law, Avogadro's law, Dalton's law of partial pressure, and
the ideal gas law;9 (B) perform stoichiometric
calculations, including determination of mass and volume relationships
between reactants and products for reactions involving gases; and
9 (C) describe the postulates of kinetic molecular theory.
I. Physical Properties9 (A) describe and calculate the
relations between volume, pressure, number of moles, and temperature for an ideal gas as described by Boyle's
law, Charles' law, Avogadro's law, Dalton's law of partial pressure, and
the ideal gas law;9 (B) perform stoichiometric
calculations, including determination of mass and volume relationships
between reactants and products for reactions involving gases; and
9 (C) describe the postulates of kinetic molecular theory.
GasesGases
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.
B. Real GasesB. Real GasesB. 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
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
C. Characteristics of C. Characteristics of GasesGasesC. Characteristics of C. Characteristics of GasesGases Gases can be compressed.
• no volume = lots of empty space
Gases undergo diffusion & effusion.• random motion
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.
E. PressureE. PressureE. PressureE. Pressure
area
forcepressure
Which shoes create the most pressure?
E. PressureE. PressureE. PressureE. Pressure
Barometer• measures atmospheric pressure
Mercury Barometer
Aneroid Barometer
E. PressureE. PressureE. PressureE. Pressure
Manometer• measures contained gas pressure
U-tube Manometer Bourdon-tube gauge
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
F. STPF. STPF. STPF. STP
Standard Temperature & PressureStandard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa-OR-
STP
II. The Gas LawsBOYLES
CHARLESGAY-
LUSSAC
II. The Gas LawsBOYLES
CHARLESGAY-
LUSSAC
Ch. 12 - GasesCh. 12 - Gases
A. Boyle’s LawA. Boyle’s LawA. Boyle’s LawA. Boyle’s Law
P
V
PV = k
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
A. Boyle’s LawA. Boyle’s LawA. Boyle’s LawA. Boyle’s Law
kT
VV
T
B. Charles’ LawB. Charles’ LawB. Charles’ LawB. Charles’ Law
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
B. Charles’ LawB. Charles’ LawB. Charles’ LawB. Charles’ Law
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
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
III. Ideal Gas LawIII. Ideal Gas Law
GasesGases
kn
VV
n
A. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s Principle
kn
VV
n
A. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s PrincipleA. Avogadro’s Principle
Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas
PV
T
Vn
PVnT
B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law
= k
UNIVERSAL GAS CONSTANT
R=0.0821 Latm/molKR=8.315
dm3kPa/molK
= R
B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law
UNIVERSAL GAS CONSTANT
R=0.0821 Latm/molKR=8.315
dm3kPa/molK
PV=nRT
GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 LR = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K
P = 3.01 atm
B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25 L.
IDEAL GAS LAW
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = 104.5 kPaR = 8.315 dm3kPa/molK
B. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas LawB. Ideal Gas Law
Find the volume of 85 g of O2 at 25°C and 104.5 kPa.
= 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
IDEAL GAS LAW
WORK:
PV = nRT
(97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K)
n = 0.597 mol O2
B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
GIVEN:
P = 97.3 kPaV = 15.0 L
n = ?T = 21°C = 294 KR = 8.315 dm3kPa/molK
4 Al + 3 O2 2 Al2O3 15.0 L
non-STP ? gGiven liters: Start with
Ideal Gas Law and calculate moles of O2.
NEXT
2 mol Al2O3
3 mol O2
B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
0.597mol O2 = 40.6 g Al2O3
4 Al + 3 O2 2 Al2O3
101.96 g Al2O3
1 molAl2O3
15.0Lnon-STP
? gUse stoich to convert moles of O2 to grams Al2O3.
A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
Ptotal = P1 + P2 + ...
Patm = PH2 + PH2O
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
Look up water-vapor pressure on p.899 for 22.5°C.
Sig Figs: Round to least number of decimal places.
The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas?
DALTON’S LAW
Look up water-vapor pressure on p.899 for 35.0°C.
Sig Figs: Round to least number of decimal places.
A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.