chem 106, prof. t. l. heise 1 che 106: general chemistry chapter ten copyright © tyna l. heise...
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Chem 106, Prof. T. L. Heise 11
CHE 106: General Chemistry
CHAPTER TEN
Copyright © Tyna L. Heise 2001
All Rights Reserved
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Chem 106, Prof. T. L. Heise 22
How is matter encountered?How is matter encountered?
–NOT on the atomic or molecular scale
–As a large collection of atoms or molecules
–Recognized as solids liquids and gases
GasesGases
Chapt. 10.1
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Chem 106, Prof. T. L. Heise 33
Characteristics of GasesCharacteristics of Gases
–Expands to fill container completely
–Highly compressible
–Form homogeneous mixtures regardless of identities or proportions
GasesGases
Chapt. 10.1
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Chem 106, Prof. T. L. Heise 44
Why exhibit these characteristics?Why exhibit these characteristics?
–Individual particles are far apart
–Act as if they are only molecule present
GasesGases
Chapt. 10.1
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Chem 106, Prof. T. L. Heise 55
Properties of Gases
Chapt. 10.2
When measuring gases, easiest properties are
1) Temperature - thermochemistry
2) Volume - solution chemistry
3) Pressure
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Chem 106, Prof. T. L. Heise 66
Pressure
Chapt. 10.2
- conveys idea of force
- P = F P = pressure
A F = force
A = area
- pressure is exerted on any surface a gas contacts
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Chem 106, Prof. T. L. Heise 77
Atmospheric pressureAtmospheric pressure
Gases of our atmosphere exert a force on surface of the earth due to gravity.
- force exerted is equal to mass times acceleration due to gravity
F = ma
since P = F then P = ma a = 9.8 m
A A s2
Chapt. 10.2
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Chem 106, Prof. T. L. Heise 88
Atmospheric pressureAtmospheric pressure
Chapt. 10.2
Atmospheric pressure is measured using a mercury barometer
•Tube is filled with mercury
•Small portion falls back into dish when inverted, vacuum exists above liquid in column
•Column moves up or down depending on atmospheric force on surface of mercury in dish
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Chem 106, Prof. T. L. Heise 99
Atmospheric pressureAtmospheric pressure
Chapt. 10.2
Units of Measurement:
1 atm
760 mmHg
760 torr
1.01325 x 105 Pa
101.325 kPa
Convert 13.33 kPa into atm
13.33 kPa 1 atm = 0.1316 atm
101.325 kPa
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Chem 106, Prof. T. L. Heise 1010
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Manometers
•Similar in operation to barometer
•Two types, closed tube and open tube
•Closed tube - measures pressures below atmospheric
•Difference in tube heights = pressure
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Chem 106, Prof. T. L. Heise 1111
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
•Open tube - measures pressures near atmospheric
•Difference in tube heights relates pressure of gas to atmospheric pressure
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Chem 106, Prof. T. L. Heise 1212
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
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Chem 106, Prof. T. L. Heise 1313
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
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Chem 106, Prof. T. L. Heise 1414
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
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Chem 106, Prof. T. L. Heise 1515
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
0.835 atm 760 torr =
1 atm
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Chem 106, Prof. T. L. Heise 1616
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
0.835 atm 760 torr = 634 torr
1 atm * which is less than atmospheric pressure
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Chem 106, Prof. T. L. Heise 1717
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
atmospheric pressure makes arm open to air lower and arm attached to gas container higher
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Chem 106, Prof. T. L. Heise 1818
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
b) Pgas = Patm + difference in heights
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Chem 106, Prof. T. L. Heise 1919
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
b) Pgas = Patm + difference in heights
Pgas = 634 torr
Patm = 755 torr
634 = 755 + difference
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Chem 106, Prof. T. L. Heise 2020
Pressures of Enclosed gasesPressures of Enclosed gases
Chapt. 10.2
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
b) Pgas + difference in heights = Patm
Pgas = 634 torr
Patm = 755 torr
634 + X = 755 X = 121 torr
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Chem 106, Prof. T. L. Heise 2121
The Gas LawsThe Gas Laws
Chapt. 10.3
Four variables needed to adequately describe a gas
•Temperature (T)
•Pressure (P)
•Volume (V)
•Number of moles (n)
•Equations that express relationships between these variables are the gas laws.
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Chem 106, Prof. T. L. Heise 2222
Boyle’s LawBoyle’s Law
Chapt. 10.3
Boyle investigated the relationship between volume
and pressure.
As pressure increased, the volume decreased; proves an inverse relationship
PV = constant
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Chem 106, Prof. T. L. Heise 2323
Charles’s LawCharles’s Law
Chapt. 10.3
Charles investigated the relationship between volume
and temperature.
As temp decreased, the volume decreased; proves a direct relationship
V = constant T
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Chem 106, Prof. T. L. Heise 2424
Avogadro’s LawAvogadro’s Law
Chapt. 10.3
Avogadro investigated the relationship between volume
and amount of substance.
As number of molecules doubles, the volume as doubles - proves a direct relationship
Avogadro’s Hypothesis - equal volumes of gases at same temperature and pressure contain equal numbers
of molecules 22.4 L
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Chem 106, Prof. T. L. Heise 2525
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Combining the previous three laws, allows for a better
mathematical look at gases.
The term R in the gas equation is called the gas constant.
The conditions of 0°C and 1 atm are referred to as standard temperature and pressure!
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Chem 106, Prof. T. L. Heise 2626
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
The ideal gas law allows us to isolate all variables that are to be held constant and set up proportionalities.
P1 P2
T1 =
T2
V1 V2
T1 =
T2
P1V1 = P2V2
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Chem 106, Prof. T. L. Heise 2727
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24°C?
1) PV = nRT
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Chem 106, Prof. T. L. Heise 2828
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24°C?
2) PV = nRT
P = x R = 0.08206 L-atm/mol-K
V = 144 cm3 T = 24°C
n = 0.33 g N2
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Chem 106, Prof. T. L. Heise 2929
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24°C?
3) PV = nRT
P = x R = 0.08206 L-atm/mol-K
V = 144 cm3 = 144 mL = 0.144 L T = 24°C = 297 K
n = 0.33 g N2 1 mol N2 = 0.012 mol N2
28 g N2
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Chem 106, Prof. T. L. Heise 3030
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24°C?
4) P = nRT = 0.012 mol N2 (0.08206 L-atm/mol-K) (297 K)
V 0.144 L
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Chem 106, Prof. T. L. Heise 3131
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24°C?
2) P = nRT = 0.012 mol N2 (0.08206 L-atm/mol-K) (297 K)
V 0.144 L
P = 2.0 atm
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Chem 106, Prof. T. L. Heise 3232
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?
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Chem 106, Prof. T. L. Heise 3333
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?
1) V1 = V2
T1 T2
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Chem 106, Prof. T. L. Heise 3434
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?
1) V1 = V2 V1 = 28,500 ft3
T1 T2 T1 = -15°C
V2 = x
T2 = 31°C
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Chem 106, Prof. T. L. Heise 3535
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?
2) V1 = V2
T1 T2
V1 = 28,500 ft3 (12)3 in3 16.4 cm3 1 L = 8.08 x 105 L
1 ft3 1 in3 103 cm3
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Chem 106, Prof. T. L. Heise 3636
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?
3) V1 = V2 V1 = 8.08 x 105 L
T1 T2 T1 = -15°C = 258 K
V2 = x
T2 = 31°C = 304 K
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Chem 106, Prof. T. L. Heise 3737
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?
4) V1 = V2 8.08 x 105 L = X
T1 T2 258 K 304 K
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Chem 106, Prof. T. L. Heise 3838
The Ideal Gas EquationThe Ideal Gas Equation
Chapt. 10.4
Sample Exercise:
A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?
4) V1 = V2 8.08 x 105 L = X
T1 T2 258 K 304 K
X = 9.52 x 105 L
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Chem 106, Prof. T. L. Heise 3939
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Ideal gas equation can be used to determine
- density of gas
- molar mass of gas
- volumes of gases formed or consumed
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Chem 106, Prof. T. L. Heise 4040
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Gas Densities and Molar Mass
n = moles
V L
* multiply both sides by molar mass, M
n M = P M
V R T
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Chem 106, Prof. T. L. Heise 4141
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Gas Densities and Molar Mass
nM = moles g
V L moles
* multiplying both sides by molar mass, M,
will give us g , so that
LDensity = PM
RT
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Chem 106, Prof. T. L. Heise 4242
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Gas Densities and Molar Mass
this equation can algebraically be rearranged to solve for molar mass
Density = PM
RT
Molar Mass = d R T
P
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Chem 106, Prof. T. L. Heise 4343
Further Applications of LawFurther Applications of Law
Chapt. 10.4
Sample Exercise:
Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr
1) M = d R T
P
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Chem 106, Prof. T. L. Heise 4444
Further Applications of LawFurther Applications of Law
Chapt. 10.4
Sample Exercise:
Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr
2) M = d R T d = 1.17 g/L
P R = 62.36 L-torr/mol-K
T = 21°C = 294 K
P = 740.0 torr
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Chem 106, Prof. T. L. Heise 4545
Further Applications of LawFurther Applications of Law
Chapt. 10.4
Sample Exercise:
Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr
3) M = d R T d = 1.17 g/L
P R = 62.36 L-torr/mol-K
T = 21°C = 294 K
P = 740.0 torr
= 1.17 g/L ( 62.36 L-torr/mol-K) (294 K)
740.0 torr
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Chem 106, Prof. T. L. Heise 4646
Further Applications of LawFurther Applications of Law
Chapt. 10.4
Sample Exercise:
Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr
4) M = d R T d = 1.17 g/L
P R = 62.36 L-torr/mol-K
T = 21°C = 294 K
P = 740.0 torr
= 1.17 g/L ( 62.36 L-torr/mol-K) (294 K)
740.0 torr
= 29.0 g/L
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Chem 106, Prof. T. L. Heise 4747
Further Applications of LawFurther Applications of Law
Chapt. 10.4
Sample Exercise:
The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere.
1) d = P M
R T
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Chem 106, Prof. T. L. Heise 4848
Further Applications of LawFurther Applications of Law
Chapt. 10.4
Sample Exercise:
The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere.
2) d = P M Pressure = 1.6 atm
R T M = 28.6 g/mol
R = 0.08206 L-atm/mol-K
T = 95 K
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Chem 106, Prof. T. L. Heise 4949
Further Applications of LawFurther Applications of Law
Chapt. 10.4
Sample Exercise:
The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere.
3) d = P M = 1.6 atm(28.6 g/mol)
R T 0.08206 L-atm/mol-K ( 95 K)
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Chem 106, Prof. T. L. Heise 5050
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Sample Exercise:
The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere.
4) d = P M = 1.6 atm(28.6 g/mol)
R T 0.08206 L-atm/mol-K ( 95 K)
= 5.9 g/L
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Chem 106, Prof. T. L. Heise 5151
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Volumes of Gases in Chemical Reactions
A knowledge of gases is often important because gases are often reactants or products in chemical reactions
- coefficients in balanced equations is again going to be very important
n = PV RT
Gas Data
Moles of Gas A
Moles of Gas B
g of Gas B
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Chem 106, Prof. T. L. Heise 5252
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Sample Exercise:
In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to react with 1.00 mol of O2 in this reaction?
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Chem 106, Prof. T. L. Heise 5353
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Sample Exercise:
In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to react with 1.00 mol of O2 in this reaction?
1) No need to convert into moles of O2, we were given that information
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Chem 106, Prof. T. L. Heise 5454
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Sample Exercise:
In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to react with 1.00 mol of O2 in this reaction?
2) Convert from mol O2 into mol NH3
1.00 mol O2 4 mol NH3 = 0.800 mol NH3
5 mol O2
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Chem 106, Prof. T. L. Heise 5555
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Sample Exercise:
In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to react with 1.00 mol of O2 in this reaction?
3) Use ideal gas equation and given info to solve for final variable
PV = nRT
P = 5.00 atm n = 0.800 mol NH3 T = 850°C = 1123 K
V = X R = 0.08206 L-atm/mol-K
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Chem 106, Prof. T. L. Heise 5656
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Sample Exercise:
In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to react with 1.00 mol of O2 in this reaction?
4) Use ideal gas equation and given info to solve for final variable
PV = nRT V = nRT = 0.800 mol ( 0.08206) (1123 K)
P 5.00 atm
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Chem 106, Prof. T. L. Heise 5757
Further Applications of LawFurther Applications of Law
Chapt. 10.5
Sample Exercise:
In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to react with 1.00 mol of O2 in this reaction?
4) Use ideal gas equation and given info to solve for final variable
PV = nRT V = nRT = 0.800 mol ( 0.08206) (1123 K)
P 5.00 atm
V = 14.7 L
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Chem 106, Prof. T. L. Heise 5858
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
John Dalton’s work with air allowed him to make the following observation:
total pressure of a mixture equals the sum of the pressures that each gas would exert if it were present alone
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Chem 106, Prof. T. L. Heise 5959
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
This equation implies that each gas behaves independently, so each gas has a unique mole quantity, and total moles equals sums of each individual amount...
nt = n1 + n2 + n3 + . . .
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Chem 106, Prof. T. L. Heise 6060
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
1) Solve for H2
PV = nRT
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Chem 106, Prof. T. L. Heise 6161
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
2) Solve for H2 PV = nRT
P = X
V = 10.0 L
n = 2.00 g H2 1 mol H2 = + 8.00 g N2 1 mol N2 = 2 g H2 28 g N2
R = 0.08206 L-atm/mol-K
T = 273 K
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Chem 106, Prof. T. L. Heise 6262
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
3) Solve for H2 PV = nRT
P = X P = nRT
V = 10.0 L V
n = 1.28 moles total = 1.28 mol (0.08206) ( 273 K) 10.0 L
R = 0.08206 L-atm/mol-K = 2.88 atm
T = 273 K
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Chem 106, Prof. T. L. Heise 6363
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
4) Solve for N2
PV = nRT
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Chem 106, Prof. T. L. Heise 6464
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
5) Solve for N2 PV = nRT
P = X
V = 10.0 L
n = 8.00 g N2 1 mol N2
28 g N2
R = 0.08206 L-atm/mol-K
T = 273 K
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Chem 106, Prof. T. L. Heise 6565
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
6) Solve for N2 PV = nRT
P = X P = nRT
V = 10.0 L V
n = 8.00 g N2 1 mol N2 = 0.286 mol (0.08206) ( 273 K) 28 g N2 10.0 L
R = 0.08206 L-atm/mol-K = 0.641 atm
T = 273 K
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Chem 106, Prof. T. L. Heise 6666
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
7) Sum individual pressures to identify total pressurePt = P(H2) + P(N2)
= 2.24 atm + 0.641 atm
= 2.88 atm
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Chem 106, Prof. T. L. Heise 6767
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Partial pressures and Mole Fractions
- each gas behaves independently and so it is easy to relate the amount of a given gas to its partial pressure
- since P = nRT then P1 = n1RT and P1 = n1
V V Pt nt
Pt = ntRT
V
- the ratio n1/nt is denoted the mole fraction of gas 1
- mole fractions are unitless values expressing ratio
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Chem 106, Prof. T. L. Heise 6868
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol % CH4. Calculate the partial pressure of each gas.
1) Solve for N2
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Chem 106, Prof. T. L. Heise 6969
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol % CH4. Calculate the partial pressure of each gas.
2) Solve for N2
Total pressure = 1220 torr
Mole fraction = 82/100 = 0.82
PN = 0.82 (1220 torr) = 1.0 x 103 torr
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Chem 106, Prof. T. L. Heise 7070
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol % CH4. Calculate the partial pressure of each gas.
3) Solve for Ar
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Chem 106, Prof. T. L. Heise 7171
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol % CH4. Calculate the partial pressure of each gas.
4) Solve for Ar
Total pressure = 1220 torr
Mole fraction = 12/100 = 0.12
PAr = 0.12 (1220 torr) = 150 torr
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Chem 106, Prof. T. L. Heise 7272
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol % CH4. Calculate the partial pressure of each gas.
5) Solve for CH4
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Chem 106, Prof. T. L. Heise 7373
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol % CH4. Calculate the partial pressure of each gas.
2) Solve for CH4
Total pressure = 1220 torr
Mole fraction = 6.0/100 = 0.060
PCH4 = 0.060(1220 torr) = 73 torr
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Chem 106, Prof. T. L. Heise 7474
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Collecting gases over water
- when determining moles of produced gases, the best way to collect a gas sample is over water
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Chem 106, Prof. T. L. Heise 7575
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Collecting gases over water
- when determining moles of produced gases, the best way to collect a gas sample is over water
- the volume of gas is measured by raising and lowering collecting container until water levels are equal inside and out…when this occurs, atmospheric pressures are equal
- total pressure inside is equal to sum of pressure of gas collected and pressure of water vapor
Ptotal = Pgas + PH2O
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Chem 106, Prof. T. L. Heise 7676
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
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Chem 106, Prof. T. L. Heise 7777
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
Pt = PN + Pwater
Pt = 745 torr
Pwater = 25.21 torr
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Chem 106, Prof. T. L. Heise 7878
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
Pt = PN + Pwater
Pt = 745 torr
Pwater = 25.21 torr
745 torr = X + 25.21 torr
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Chem 106, Prof. T. L. Heise 7979
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
Pt = PN + Pwater
Pt = 745 torr
Pwater = 25.21 torr
745 torr = X + 25.21 torr
PN = 720. torr
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Chem 106, Prof. T. L. Heise 8080
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
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Chem 106, Prof. T. L. Heise 8181
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT
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Chem 106, Prof. T. L. Heise 8282
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT P = 720. Torr
V = 0.511 L
n = X
R = 62.36 L-torr/mol-K
T = 26°C = 299 K
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Chem 106, Prof. T. L. Heise 8383
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT n = PV P = 720. torr RT
V = 0.511 L = 720. torr(0.511 L)
n = X 62.36 (299 K)
R = 62.36 L-torr/mol-K
T = 26˚C = 299 K
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Chem 106, Prof. T. L. Heise 8484
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT n = PV P = 720. Torr RT
V = 0.511 L = 720. torr(0.511 L)
n = X 62.36 (299 K)
R = 62.36 L-torr/mol-K = 0.0197 mol N2
T = 26˚C = 299 K
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Chem 106, Prof. T. L. Heise 8585
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
3) Convert from moles of N2 to g NH4NO2
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Chem 106, Prof. T. L. Heise 8686
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
3) Convert from moles of N2 to g NH4NO2
0.0197 mol N2 1 mol NH4NO2 64.0 g NH4NO2
1 mol N2 1 mol NH4NO2
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Chem 106, Prof. T. L. Heise 8787
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Chapt. 10.6
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?
3) Convert from moles of N2 to g NH4NO2
0.0197 mol N2 1 mol NH4NO2 64.0 g NH4NO2
1 mol N2 1 mol NH4NO2
= 1.26 g NH4NO2
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Chem 106, Prof. T. L. Heise 8888
Kinetic Molecular TheoryKinetic Molecular Theory
Chapt. 10.7
Ideal gas equation describes how gases behave, but not why they do . . .
Kinetic molecular theory explains why
- gases consist of large numbers of molecules that are in continuous, random motion
- volume of the gas molecules themselves is negligible when compared to the volume of the gas as a whole
- attractive and repulsive forces between gases molecules is negligible
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Chem 106, Prof. T. L. Heise 8989
Kinetic Molecular TheoryKinetic Molecular Theory
Chapt. 10.7
Ideal gas equation describes how gases behave, but not why they do . . .
Kinetic molecular theory explains why
- energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature is remaining constant -
i.e. collisions are elastic
- average kinetic energy is proportional to the absolute temperature
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Chem 106, Prof. T. L. Heise 9090
Kinetic Molecular TheoryKinetic Molecular Theory
Chapt. 10.7
This theory is helpful in explaining the pressure and temperature at a molecular level:
pressure of gas is caused by the collisions of themolecules against the wallof a container
magnitude of pressureis determined by bothhow often and howforcefully the moleculesstrike the wall
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Chem 106, Prof. T. L. Heise 9191
Kinetic Molecular TheoryKinetic Molecular Theory
Chapt. 10.7
This theory is helpful in explaining the pressure and temperature at a molecular level:
absolute temperature ofa gas is the measure ofthe average kinetic energyof its molecules
molecular motionincreases withincreasing temperature
this is on average, individual moleculeshave individual speeds
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Chem 106, Prof. T. L. Heise 9292
Kinetic Molecular TheoryKinetic Molecular Theory
Chapt. 10.7
Applications of the Gas Laws
1. Effect of a volume increase at constant temp
- molecules must move a longer distance between collisions
- fewer collisions per unit time with wall of container
- pressure decreases
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Chem 106, Prof. T. L. Heise 9393
Kinetic Molecular TheoryKinetic Molecular Theory
Chapt. 10.7
Applications of the Gas Laws
2. Effect of a temperature increase at constant volume
- increase in speed and u
- more collisions per unit time with wall of container
- pressure increases
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Chem 106, Prof. T. L. Heise 9494
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Chapt. 10.8
According to the kinetic-molecular theory, the average kinetic energy of any collection of gases,
u = 3RT R = gas constant
M T = temp
M = molar mass
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Chem 106, Prof. T. L. Heise 9595
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Chapt. 10.8
has a specific value of u at any given temperature
-two gases at same temp have same avg. kinetic energy
-if masses are different, than the speed of particles will be different because of the inclusion of M in formula
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Chem 106, Prof. T. L. Heise 9696
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Chapt. 10.8
The dependence of speed on mass, has several implications:
Effusion - the escape of a molecule through a tiny hole
Effusion rate is inversely proportional to the square root of molar mass
Normally rates are compared as a ratio
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Chem 106, Prof. T. L. Heise 9797
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Chapt. 10.8
The dependence of speed on mass, has several implications:
Diffusion - spreading of one substance throughout a space or through another substance
Diffusion is also related to size of particle, however, molecule collisions make diffusion much more difficult
- average distance traveled by a molecule as it diffuses is called the mean free path, which varies with pressure
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Chem 106, Prof. T. L. Heise 9898
Deviations from an Ideal GasDeviations from an Ideal Gas
Chapt. 10.9
Although the ideal gas equation is useful, all real gases fail to obey the relationships to some degree
- deviation from an ideal gas occurs most at high pressure and low temperature
- to ensure as much compliance as possible to the ideal gas equation, a real gas should be considered when it is at high temperatures and low pressure
WHY?
Real gases DO - have molecular attractions
- lose energy when they collide
- have volume
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Chem 106, Prof. T. L. Heise 9999
Deviations from an Ideal GasDeviations from an Ideal Gas
Chapt. 10.9
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Chem 106, Prof. T. L. Heise 100100
Deviations from an Ideal GasDeviations from an Ideal Gas
Chapt. 10.9
Van der Waals’ Equation: takes into account volumeand molecular attraction
* constants (a) and (b) are different for each gas and must be identified using a table
P = nRT - n2a V - nb V2
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Chem 106, Prof. T. L. Heise 101101
Characteristics of gasesCharacteristics of gases PressurePressure Boyle’s, Charles’s, & Avogadro’s LawsBoyle’s, Charles’s, & Avogadro’s Laws Ideal gas equationIdeal gas equation Gas densities and Molar massGas densities and Molar mass Dalton’s Law of partial pressuresDalton’s Law of partial pressures Mole fractionsMole fractions Kinetic Molecular theoryKinetic Molecular theory Effusion, Diffusion,Effusion, Diffusion, u u and mean free path and mean free path Deviations from ideal gas behaviorDeviations from ideal gas behavior Van der Waals equationVan der Waals equation
Chapter Ten; ReviewChapter Ten; Review