i. physical properties. real vs. ideal gases: b ideal gas = an imaginary gas that conforms perfectly...
TRANSCRIPT
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I. Physical PropertiesI. Physical Properties
Gases
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Real vs. Ideal Gases:Real vs. Ideal Gases:
Ideal gas = an imaginary gas that conforms perfectly to all the assumptions of the kinetic-molecular theory• We will assume that the gases
used for the gas law problems are ideal gases.
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Real Gases vs. Ideal Gases:Real Gases vs. Ideal Gases:
Real gas = a gas that does not behave completely according to the assumptions of the kinetic-molecular theory.
All real gases deviate to some degree from ideal gas behavior. However, most real gases behave nearly ideally when their particles are sufficiently far apart and have sufficiently high kinetic energy.
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Causes of non-ideal behavior:Causes of non-ideal behavior:
The kinetic-molecular theory is more likely to hold true for gases whose particles have NO attraction for each other.
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Causes of non-ideal behavior:Causes of non-ideal behavior:
1. High pressure (low volume): • Space taken up by gas particles
becomes significant• Intermolecular forces are more
significant between gas particles that are closer together
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Causes of non-ideal behavior:Causes of non-ideal behavior:
2. Low temperature: •Gas particles move slower so intermolecular forces become more important
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Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences.
theory developed in the late 19th century to account for the behavior of the atoms and molecules that make up matter
KMT (Kinetic Molecular Theory)KMT (Kinetic Molecular Theory)
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based on the idea that particles in all forms of matter are always in motion and that this motion has consequences
KMT (Kinetic Molecular Theory)KMT (Kinetic Molecular Theory)
SOLID LIQUID GAS
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can be used to explain the properties of solids, liquids, and gases in terms of the energy of particles and the forces that act between them
KMT (Kinetic Molecular Theory)KMT (Kinetic Molecular Theory)
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B. Kinetic Molecular Theory – Ideal gasB. Kinetic Molecular Theory – Ideal gas
KMT describing particles in an IDEAL gas:• have no volume.• have elastic collisions. • are in constant, random motion.• don’t attract or repel each other.• have an avg. KE directly related
to Kelvin temperature.
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C. Kinetic Molecular Theory - Real GasesC. Kinetic Molecular Theory - Real Gases
KMT describing 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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D. Characteristics of Gases- using KMTD. Characteristics of Gases- using KMT
Gases expand to fill any container.KMT - gas particles move rapidly
in all directions without significant attraction or repulsion between particles
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D. Characteristics of Gases- using KMTD. Characteristics of Gases- using KMT
Gases are fluids (like liquids). KMT - No significant attraction or
repulsion between gas particles; glide past each other
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D. Characteristics of Gases- using KMTD. Characteristics of Gases- using KMT
Gases have very low densities.
• KMT - particles are so much farther apart in the gas state
sodium in the solid state:
sodium in the liquid state:
sodium in the gas state:
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Gases can be compressed.• KMT - gas particles are far apart
from one anther with room to be “squished” together
D. Characteristics of Gases- using KMTD. Characteristics of Gases- using KMT
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Gases undergo diffusion & effusion.• KMT – gas particles move in
continuous, rapid, random motion
D. Characteristics of Gases- using KMTD. Characteristics of Gases- using KMT
Effusion
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E. TemperatureE. Temperature
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
32FC 95 K = ºC + 273
Always use Kelvin temperature when working with gases.
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F. PressureF. Pressure
area
forcepressure
Which shoes create the most pressure?
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F. PressureF. Pressure
Why do gases exert pressure? • Gas particles exert a
pressure on any surface with which they collide!More collisions =
increase in pressure!
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F. PressureF. Pressure
Barometer = measures atmospheric pressure
The height of the Hg in the tube depends on the pressure• The pressure of the
atmosphere is proportional to the height of the Hg column, so the height of the Hg can be used to measure atmospheric pressure! Mercury Barometer
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F. PressureF. Pressure
Pressure UNITS
101.3 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
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G. STPG. STP
Standard Temperature & Pressure
0°C (exact) 1 atm (exact)
273 K 101.3 kPa
760 mm Hg (exact)
760 torr (exact)
STP
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II. The Gas Laws
II. The Gas Laws
Gases
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GAS LAW PROBLEMS- MUST USE KELVIN
K = ºC + 273
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A. Boyle’s LawA. Boyle’s Law
P
V
PV = k
Volume (mL)
Pressure (torr)
P·V (mL·torr)
10.0 760.0 7.60 x 103
20.0 379.6 7.59 x 103
30.0 253.2 7.60 x 103
40.0 191.0 7.64 x 103
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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
P1V1 = P2V2
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A. Boyle’s LawA. Boyle’s Law
•Real life application: When you breathe, your diaphragm moves downward, increasing the volume of the lungs. This causes the pressure inside the lungs to be less than the outside pressure so air rushes in.
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A. Boyle’s LawA. Boyle’s Law
•Ex: Halving the volume leads to twice the rate of collisions and a doubling of the pressure.
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A. Boyle’s LawA. Boyle’s Law
• As the volume increases, the pressure decreases!
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kT
VV
T
B. Charles’ LawB. Charles’ Law
Volume (mL)
Temperature (K)
V/T (mL/K)
40.0 273.2 0.146
44.0 298.2 0.148
47.7 323.2 0.148
51.3 348.2 0.147
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V1_ = V2__
T1 T2
V
T
B. Charles’ LawB. Charles’ Law
The volume and temperature (K) of a gas are DIRECTLY related • at constant mass &
pressure
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B. Charles’ LawB. Charles’ Law
Real life application: Bread dough rises because yeast produces carbon dioxide. When placed in the oven, the heat causes the gas to expand, and the bread rises even further.
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B. Charles’ LawB. Charles’ Law
As temperature decreases, the volume of the gas decreases
Liquid nitrogen’s temp. is about 63K or -210 ºC or -346 ºF!
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B. Charles’ LawB. Charles’ Law
As temperature decreases, the volume of the gas decreases
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B. Charles’ LawB. Charles’ Law
NUMBER OF PARTICLES & PRESSURE ARE CONSTANT!!!
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kT
PP
T
C. Guy-Lussac’s LawC. Guy-Lussac’s Law
Temperature (K)
Pressure (torr)
P/T (torr/K)
248 691.6 2.79
273 760.0 2.78
298 828.4 2.78
373 1,041.2 2.79
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_P1_ = P2__
T1 T2
P
T
C. Guy-Lussac’s LawC. Guy-Lussac’s Law
The pressure and temperature (K) of a gas are DIRECTLY RELATED • at constant mass &
volume
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C. Guy-Lussac’s LawC. Guy-Lussac’s Law
When the temp. of a gas increases (KE increases) and gas particles move faster and hit container walls more frequently and collisions are more forceful
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C. Guy-Lussac’s LawC. Guy-Lussac’s Law
Real life application: The air pressure inside a tire increases on a hot summer day.
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= kBoyles PV P
Guy Lussac’sT VCharles’ T
PVT
D. Combined Gas LawD. Combined Gas Law
P1V1
T1
=P2V2
T2
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E. EXAMPLE ProblemsE. EXAMPLE Problems
You need your calculators!
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C. Johannesson
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
Gas Law ProblemsGas Law Problems
1.) A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW V1_ = V2__
T1 T2
T V
V2= (473 cm3)(367 K)
(309 K)
V2 = 562 cm3
V2 = V1T2
T1
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C. Johannesson
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
V2 = P1V1
P2
Gas Law ProblemsGas Law Problems
2.) A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW P1V1 = P2V2
P V
V2 = (150.kPa)(100.mL)
200.kPa
V2 = 75.0 mL
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C. Johannesson
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.3 kPa
T2 = 273 K
WORK:
V2 = P1V1T2
P2T1
V2 = (71.8 kPa)(7.84 cm3)(273 K)
(101.3 kPa) (298 K)
V2 = 5.09 cm3
Gas Law ProblemsGas Law Problems3.) A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its
volume at STP.
P T V
COMBINED GAS LAW P1V1 = __P2V2
T1 T2
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C. Johannesson
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
T2 = P2T1
P1
Gas Law ProblemsGas Law Problems
4.) A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?
P T
T2 = (560. torr)(296K)
765 torr
T2 = 217 K
GUY LUSSAC’S LAW P1_ = P2__
T1 T2
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III. Standard Molar Volume,
Gas Densities & Molar Mass
III. Standard Molar Volume,
Gas Densities & Molar Mass
Gases
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A. Standard Molar VolumeA. Standard Molar Volume
The volume occupied by one mole of a gas at STP • 22.41410 L/mol or about 22.4
L/mol In other words, one mole of any
ideal gas at STP will occupy 22.4 L
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B. Example Problems-standard molar volumeB. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce 0.0680 mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?
0.068 mol O2 22.4 L O2
1 mol O2
=
1.52 L O2
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B. Example Problems-standard molar volumeB. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce 0.0680 mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?
0.068 mol O2 22.4 L O2
1 mol O2
=
1.52 L O2
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B. Example Problems-standard molar volumeB. Example Problems-standard molar volume
2.) A chemical reaction produced 98.0 mL of SO2 at STP. What mass (in grams) of the gas was produced?
98.0 mL SO2 1 L SO2 1 mol O2
=
0.280 g SO2
22.4 L SO21000 mL SO2 1 mol SO2
64.064 g SO2
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C. DensityC. Density
The ratio of an object’s mass to its volume.
D = M
V The volume (and density) of a gas
will change when pressure and temperature change. • Densities are usually given in g/L at
STP
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C. DensityC. Density
For an ideal gas:
Density (at STP) = molar mass
standard molar volume
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D. Example Problems- densityD. Example Problems- density
1.) What is the density of CO2 in g/L at STP?
44.009 g CO2 1 mol CO2
1 mol CO2 22.4 L CO2
=
1.96 g/L CO2
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D. Example Problems- densityD. Example Problems- density
2.) What is the molar mass of a gas whose density at STP is 2.08 g/L.
2.08 g 22.4 L
1 L 1 mol =
46.6 g/mol
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IV. More Gas Laws: Ideal Gas
Law & Avogadro's Law
IV. More Gas Laws: Ideal Gas
Law & Avogadro's Law
Gases
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kn
VV
n
A. Avogadro’s LawA. Avogadro’s Law
Equal volumes of gases contain equal numbers of moles• at constant temp &
pressure
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PV
TVn
PVnT
B. Ideal Gas LawB. Ideal Gas Law
= kUNIVERSAL GAS
CONSTANTR=0.0821
Latm/molKR=8.315
dm3kPa/molK
= R
You don’t need to memorize these values!
• Boyle’s, Charles, and Avogadro’s laws are contained within the ideal gas law!
• Merge the Combined Gas Law with Avogadro’s Law:
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B. Ideal Gas LawB. Ideal Gas Law
GAS CONSTANTR=0.0821
Latm/molKR=8.315
dm3kPa/molK
PV=nRT
You don’t need to memorize these values!
P= pressureV = volume n = molesT = temperature (in K)R = the ideal gas constant
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C. Johannesson
GIVEN:P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 L
R = 0.0821Latm/molK
WORK:
P = nRT
V
P =(0.412 mol)(0.0821Latm/molK) (289K)
3.25L
P = 3.01 atm
C. Ideal Gas Law ProblemsC. Ideal Gas Law Problems
1.) Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.
PV = nRT
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C. Johannesson
GIVEN:
V = ?
n = 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
2.) Find the volume of 85 g of O2 at 25°C
and 104.5 kPa. PV = nRT
= 2.7 mol
WORK:
85 g O2 1 mol O2 = 2.7 mol 31.998 g O2 O2
V = nRT PV =(2.7 mol)(8.315 dm3kPa/molK) (298K)
104.5 kPa
V = 64 dm3
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C. Johannesson
GIVEN:
V = 1.00 L
n = ?
T = 28°C = 301 K
P = 98.7 kPa
R = 8.314 LkPa/molK
C. Ideal Gas Law ProblemsC. Ideal Gas Law Problems3.) At 28C and 98.7 kPa, 1.00 L of an unidentified gas has
a mass of 5.16 g. Calculate the number of moles of gas present and the molar mass of the gas. PV = nRT
WORK:
n = PV RT
(98.7 kPa) (1.00L)
(8.314 LkPa/molK) (301 K)
n = 0.0394 mol
m = 5.16g
0.0394 mol
=
= 131 g/mol
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Gases
IV. Two More Laws: Dalton’s Law & Graham’s Law
IV. Two More Laws: Dalton’s Law & Graham’s Law
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A. 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 + P3 ...
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A. Dalton’s LawA. Dalton’s Law
Ptotal = P1 + P2 + P3 ...
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A. Dalton’s LawA. Dalton’s Law
When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 AND water vapor.
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C. Johannesson
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 Law
1.) 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 gas formula sheet 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.
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C. Johannesson
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
2.) 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?
Look up water-vapor pressure on gas formula sheet for
35.0°C.
Sig Figs: Round to least number of decimal places.
A. 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.
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B. Graham’s LawB. Graham’s Law
Diffusion =• Spreading of gas molecules
throughout a container until evenly distributed.
Effusion =• Passing of gas molecules
through a tiny opening in a container
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B. Graham’s LawB. Graham’s Law
KE = ½mv2
Speed of diffusion/effusion
• Kinetic energy is determined by the temperature of the gas.
• At the same temp & KE, heavier molecules move more slowly.Larger m smaller v
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B. Graham’s LawB. Graham’s Law
KE = kinetic energym= molar massv = velocity
KE = ½mv2
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C. Johannesson
B. Graham’s LawB. Graham’s Law
Graham’s Law = Rate of effusion of a gas is inversely related to the square root of its molar mass.• The equation shows the ratio of
Gas A’s speed to Gas B’s speed.
A
B
B
A
m
m
v
v
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1.) Determine the relative rate of effusion for krypton and bromine.
1.381
Kr effuses 1.381 times faster than Br2.
Kr
Br
Br
Kr
m
m
v
v2
2
A
B
B
A
m
m
v
v
g/mol83.80
g/mol159.80
C. Graham’s Law ProblemsC. Graham’s Law Problems
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.
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C. Johannesson
2.) A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?
A
B
B
A
m
m
v
v
2
2
2
2
H
O
O
H
m
m
v
v
g/mol 2.02
g/mol32.00
m/s 12.3
vH 2
C. Graham’s Law ProblemsC. Graham’s Law Problems
3.980m/s 12.3
vH 2
m/s49.0 vH 2
Put the gas with the unknown
speed as “Gas A”.
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C. Johannesson
3.) An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.
Am
g/mol32.00 16
A
B
B
A
m
m
v
v
A
O
O
A
m
m
v
v2
2
Am
g/mol32.00 4.0
16
g/mol32.00 mA
2
Am
g/mol32.00 4.0
g/mol2.0
C. Graham’s Law ProblemsC. Graham’s Law Problems
The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.
Square both sides to get rid of the square
root sign.
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Gases
VI. Gas Stoichiometry at Non-STP ConditionsVI. Gas Stoichiometry at Non-STP Conditions
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A. Gas StoichiometryA. Gas Stoichiometry
Moles Liters of a Gas:• STP - use 22.4 L/mol • Non-STP - use ideal gas law
Non-STP• Given liters of gas?
start with ideal gas law• Looking for liters of gas?
start with stoichiometry conv.
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B. Volume Mass Stoich.B. Volume Mass Stoich.
Thinking:Gas volume A moles A moles B mass B
If gas A is at STP:
? L “A” 1 mol “A” ? mol “B” molar mass(g) “B”
22.4 L “A” ? mol “A” 1 mol “B”
Mole Ratio
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B. Volume Mass Stoich.B. Volume Mass Stoich.
If gas “A” is NOT at STP: Use PV = nRT to find moles of “A”
? mol “A” ? mol “B” molar mass (g) “B”
? mol “A” 1 mol “B”
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C. Johannesson
WORK:n = PV RT
n= (97.3 kPa) (15.0 L) (8.315dm3kPa/molK) (294K)
n = 0.597 mol O2
B. Volume MassB. Volume Mass
1.) 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
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2 mol Al2O3
3 mol O2
B. Volume MassB. Volume Mass
1.) 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.
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1 mol SO2
22.4L SO2
B. Volume MassB. Volume Mass2.) What mass of sulfur is required to
produce 12.6 L of sulfur dioxide at STP according to the equation:
S8 (s) + 8 O2 (g) 8 SO2 (g)
12.6 L SO2 = 18.0 g
S8
256.582 g S8
1 molS8
12.6L? g
1 mol S8
8 molSO2
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C. Mass VolumeC. Mass Volume
Thinking:Mass A moles A moles B gas volume B
If gas “B” is at STP:
? g “A” 1 mol “A” ? mol “B” 22.4 L “B”
molar mass (g) “A” ? mol “A” 1 mol “B”
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C. Mass Volume C. Mass Volume
If gas “B” is NOT at STP:
? g “A” 1 mol “A” ? mol “B”
molar mass (g) “A” ? mol “A”
• Then use PV = nRT to find volume of “B”
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C. Mass VolumeC. Mass Volume
1.) What volume of chlorine gas at STP is needed to react completely with 10.4 g of sodium to form NaCl?
2 Na + Cl2 2 NaCl 10.4 g ? L
10.4 g Na
22.990 g Na
22.4 L Cl21 mol Cl2
2 mol Na 1 mol Cl2
1 mol Na
= 5. 07 L Cl2
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1 molCaCO3
100.09g CaCO3
C. Mass VolumeC. Mass Volume
2.) What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
5.25 gCaCO3 = 1.26 mol CO2
CaCO3 CaO + CO2
1 molCO2
1 molCaCO3
5.25 g ? Lnon-STPLooking for liters: Start with stoich
and calculate moles of CO2.
Plug this into the Ideal Gas Law to find liters.
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C. Johannesson
WORK:V = nRT
PV = (1.26mol)(8.315dm3kPa/molK) (298K)
(103 kPa)
V = 30.3 dm3 CO2
C. Mass VolumeC. Mass Volume
2.) What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPaV = ?
n = 1.26 molT = 25°C = 298 KR = 8.315 dm3kPa/molK