1 chapter 5 gases. 2 characteristics of gases there are three phases for all substances: solid,...
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CHAPTER 5CHAPTER 5
GASES GASES
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Characteristics of GasesCharacteristics of Gases
There are three phases for all substances: There are three phases for all substances: solid, liquid and gases.solid, liquid and gases.
Gases are highly compressible and occupy Gases are highly compressible and occupy the full volume of their containers.the full volume of their containers.
When a gas is subjected to pressure, its When a gas is subjected to pressure, its volume decreases.volume decreases.
Gases always form homogeneous mixtures Gases always form homogeneous mixtures with other gases.with other gases.
Gases only occupy about 0.1 % of the volume Gases only occupy about 0.1 % of the volume of their containers.of their containers.
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PressurePressure
Pressure is the force acting on an Pressure is the force acting on an object per unit area:object per unit area:
Units: psiUnits: psi pounds per square inchpounds per square inchAF
P
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PressurePressure
SI Units: 1 N = 1 kg.m/sSI Units: 1 N = 1 kg.m/s22; 1 Pa = 1 N/m; 1 Pa = 1 N/m22.. Atmospheric pressure is measured Atmospheric pressure is measured
with a barometer.with a barometer. If a tube is inserted into a container of If a tube is inserted into a container of
mercury open to the atmosphere, the mercury open to the atmosphere, the mercury will rise 760 mm up the tube.mercury will rise 760 mm up the tube.
Standard atmospheric pressure is the Standard atmospheric pressure is the pressure required to support 760 mm pressure required to support 760 mm of Hg in a column.of Hg in a column.
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PressurePressure
force per unit areaforce per unit area standard pressurestandard pressure
– 76 cm Hg76 cm Hg– 760 mm Hg760 mm Hg– 760 torr760 torr– 1 atmosphere1 atmosphere– 101.3 kPa101.3 kPa
Units: 1 atm = 760 Units: 1 atm = 760 mmHg = 760 torr = mmHg = 760 torr = 1.01325 1.01325 10 1055 Pa = Pa = 101.325 kPa.101.325 kPa.
density = 13.6 g/mL
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PressurePressure
The pressures of gases not open to The pressures of gases not open to the atmosphere are measured in the atmosphere are measured in manometers.manometers.
A manometer consists of a bulb of A manometer consists of a bulb of gas attached to a U-tube containing gas attached to a U-tube containing Hg.Hg.
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PressurePressure
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PressurePressure
If the U-tube is closed, then the If the U-tube is closed, then the pressure of the gas is the difference pressure of the gas is the difference in height of the liquid (usually Hg).in height of the liquid (usually Hg).
If the U-tube is open to the If the U-tube is open to the atmosphere, a correction term atmosphere, a correction term needs to be added:needs to be added:– If If PPgasgas < < PPatmatm then then PPgasgas + + PPhh22 = = PPatmatm..
– If If PPgasgas > > PPatmatm then then PPgasgas = = PPatmatm + + PPhh22..
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The Gas LawsThe Gas Laws
Weather balloons are used as a practical Weather balloons are used as a practical consequence to the relationship between consequence to the relationship between pressure and volume of a gas.pressure and volume of a gas.
As the weather balloon ascends, the volume As the weather balloon ascends, the volume decreases.decreases.
As the weather balloon gets further from the As the weather balloon gets further from the earth’s surface, the atmospheric pressure earth’s surface, the atmospheric pressure decreases.decreases.
Boyle’s Law: the volume of a fixed quantity of Boyle’s Law: the volume of a fixed quantity of gas is inversely proportional to its pressure.gas is inversely proportional to its pressure.
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Boyle’s LawsBoyle’s Laws
Mathematically:Mathematically:
A plot of V versus P is a hyperbola.A plot of V versus P is a hyperbola. Similarly, a plot of V versus 1/P must be Similarly, a plot of V versus 1/P must be
a straight line passing through the a straight line passing through the origin.origin.
PV
1constant constant PV
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Boyle’s LawsBoyle’s Laws
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Boyle’s LawBoyle’s Law
V1/P V= k (1/P) or PV = k P1V1 = k1 P2V2 = k2
kk1 1 = k= k2 2 (for same sample of gas @ (for same sample of gas @ same T)same T)
PP11VV11 = P = P22VV22 Boyle’s Law (math form) Boyle’s Law (math form) think in terms of balloons!!!think in terms of balloons!!!
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Boyle’s LawBoyle’s Law Example At 25Example At 2500C a sample of He has a C a sample of He has a
volume of 400 mL under a pressure of volume of 400 mL under a pressure of 760 torr. What volume would it occupy 760 torr. What volume would it occupy under a pressure of 2.00 atm at the under a pressure of 2.00 atm at the same T?same T?
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Boyle’s LawBoyle’s Law Example At 25Example At 2500C a sample of He has a C a sample of He has a
volume of 400 mL under a pressure of volume of 400 mL under a pressure of 760 torr. What volume would it occupy 760 torr. What volume would it occupy under a pressure of 2.00 atm at the under a pressure of 2.00 atm at the same T?same T?
2
112
2211
P
V PV
or V PV P
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Boyle’s LawBoyle’s Law Example At 25Example At 2500C a sample of He has a C a sample of He has a
volume of 400 mL under a pressure of volume of 400 mL under a pressure of 760 torr. What volume would it occupy 760 torr. What volume would it occupy under a pressure of 2.00 atm at the under a pressure of 2.00 atm at the same T?same T?
P V P V or
VP V
P
760 torr mL
1520 torr
mL
1 1 2 2
21 1
2
400
200
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The Gas LawsThe Gas Laws
We know that hot air balloons expand We know that hot air balloons expand when they are heated.when they are heated.
Charles’ Law: the volume of a fixed Charles’ Law: the volume of a fixed quantity of gas at constant pressure quantity of gas at constant pressure increases as the temperature increases.increases as the temperature increases.
Mathematically:Mathematically:
TV constant constant TV
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Charles’ LawCharles’ Law
A plot of V versus T is a straight line.A plot of V versus T is a straight line. When T is measured in When T is measured in C, the intercept C, the intercept
on the temperature axis is -273.15on the temperature axis is -273.15C. C. We define absolute zero, 0 K = -We define absolute zero, 0 K = -
273.15273.15C.C. Note the value of the constant reflects Note the value of the constant reflects
the assumptions: amount of gas and the assumptions: amount of gas and pressure.pressure.
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Charles’ LawCharles’ Law
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Charles’ LawCharles’ Law
volume of a gas is directly proportional volume of a gas is directly proportional to the absolute temperature at constant to the absolute temperature at constant pressurepressureK = C + 273o
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Charles’ LawCharles’ Law
volume of a gas is directly proportional volume of a gas is directly proportional to the absolute temperature at constant to the absolute temperature at constant pressurepressureK = C + 273
mathematical forms of Charles's Law
V T or V = kT or V
Tk
o
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Charles’ LawCharles’ Law
volume of a gas is directly proportional volume of a gas is directly proportional to the absolute temperature at constant to the absolute temperature at constant pressurepressureK = C + 273
mathematical forms of Charles's Law
V T or V = kT or V
Tk
V
Tk and
V
Tk however the k's are equal so
V
T
V
T in the most useful form
o
1
1
2
2
1
1
2
2
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Charles’ LawCharles’ Law
Example A sample of hydrogen, HExample A sample of hydrogen, H22, , occupies 100 mL at 25occupies 100 mL at 2500C and 1.00 atm. C and 1.00 atm. What volume would it occupy at 50What volume would it occupy at 5000C C under the same pressure?under the same pressure?
TT1 1 = 25 + 273 = 298 = 25 + 273 = 298
TT2 2 = 50 + 273 = 323= 50 + 273 = 323
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Charles’ LawCharles’ Law
Example: A sample of hydrogen, HExample: A sample of hydrogen, H22, , occupies 100 mL at 25occupies 100 mL at 2500C and 1.00 atm. C and 1.00 atm. What volume would it occupy at 50What volume would it occupy at 5000C C under the same pressure?under the same pressure?V
T
V
T so V =
V T
T
V =100 mL 323 K
298 K mL
1
1
2
22
1 2
1
2
108
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Standard Temperature & Standard Temperature & PressurePressure
STPSTP– P = 1.00000 atm or 101.3 kPaP = 1.00000 atm or 101.3 kPa– T = 273 K or 0T = 273 K or 000CC
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The Gas LawsThe Gas Laws
Avogadro’s Hypothesis: equal volumes Avogadro’s Hypothesis: equal volumes of gas at the same temperature and of gas at the same temperature and pressure will contain the same number pressure will contain the same number of molecules.of molecules.
Avogadro’s Law: the volume of gas at a Avogadro’s Law: the volume of gas at a given temperature and pressure is given temperature and pressure is directly proportional to the number of directly proportional to the number of moles of gas.moles of gas.
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Avogadro’s LawAvogadro’s Law
Mathematically:Mathematically:
V = constant V = constant nn.. We can show that 22.4 L of any gas at We can show that 22.4 L of any gas at
00C contain 6.02 C contain 6.02 10 102323 gas molecules. gas molecules.
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The Ideal Gas EquationThe Ideal Gas Equation Summarizing the Gas LawsSummarizing the Gas Laws
Boyle: (constant n, T)Boyle: (constant n, T)
Charles: Charles: V V T T (constant n, P) (constant n, P)
Avogadro:Avogadro: V V n n (constant P, T). (constant P, T). Combined:Combined:
Ideal gas equation:Ideal gas equation:
PnT
V
PnT
RV
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The Ideal Gas EquationThe Ideal Gas Equation
Ideal gas equation: Ideal gas equation:
PV = nRTPV = nRT.. R = gas constant = 0.08206 L•atm/mol-K.R = gas constant = 0.08206 L•atm/mol-K. We define STP (standard temperature and We define STP (standard temperature and
pressure) = 0pressure) = 0C, 273.15 K, 1 atm.C, 273.15 K, 1 atm. Volume of 1 mol of gas at STP is 22.4 L.Volume of 1 mol of gas at STP is 22.4 L.
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Ideal Gas LawIdeal Gas Law
R has several values depending upon the R has several values depending upon the choice of unitschoice of unitsR = 8.314 J/mol KR = 8.314 kg m2/s2 K molR = 8.314 dm3 kPa/K molR = 1.987 cal/K mol
R = PV
nT
1.00 atm L
1.00 mol K
L atm
mol K
22 4
273
0 0821
.
.
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Ideal Gas LawIdeal Gas Law Example: What volume would 50.0 g of Example: What volume would 50.0 g of
ethane, Cethane, C22HH66, occupy at 140, occupy at 140ooC under a C under a pressure of 1820 torr?pressure of 1820 torr?
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Example: What volume would 50.0 g of Example: What volume would 50.0 g of ethane, Cethane, C22HH66, occupy at 140, occupy at 140ooC under a C under a pressure of 1820 torr?pressure of 1820 torr?
T = 140 + 273 = 413 KP = 1820 torr (1 atm/760 torr) = 2.39 atm50 g (1 mol/30 g) = 1.67 mol
Ideal Gas LawIdeal Gas Law
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V = n R T
P
Ideal Gas LawIdeal Gas Law
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V = n R T
P
mol L atmmol K
K
torr1 atm
760 torr
L
167 0 0821 413
1820
236
. .
.
Ideal Gas LawIdeal Gas Law
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Example: Calculate the number of moles Example: Calculate the number of moles in, and the mass of, an 8.96 L sample of in, and the mass of, an 8.96 L sample of methane, CHmethane, CH44, measured at standard , measured at standard conditions.conditions.
Ideal Gas LawIdeal Gas Law
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n =
PV
RT
1.00 atm L
0.0821L atm
mol K K
0.400 mol CH
g CH mol16.0 g
mol 6.40 g
4
4
8 96
273
0 400
.
? .
Ideal Gas LawIdeal Gas Law
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Example: Calculate the pressure exerted Example: Calculate the pressure exerted by 50.0 g of ethane, Cby 50.0 g of ethane, C22HH66, in a 25.0 L , in a 25.0 L container at 25container at 25ooC.C.
Ideal Gas LawIdeal Gas Law
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atm 63.1
L 25.0
K 298K mol
atm L0821.0mol 67.1
V
T Rn = P
K 298 = T & mol 1.67 =n :Example From
Ideal Gas LawIdeal Gas Law
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If If PVPV = = nRTnRT and and nn and and T T are constant, are constant, then then PV PV = constant and we have = constant and we have Boyle’s law.Boyle’s law.
Other laws can be generated similarly.Other laws can be generated similarly. In general, if we have a gas under two In general, if we have a gas under two
sets of conditions, thensets of conditions, then
The Ideal Gas LawThe Ideal Gas Law
22
22
11
11TnVP
TnVP
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Ideal Gas LawIdeal Gas Law Example: A sample of nitrogen gas, NExample: A sample of nitrogen gas, N22, ,
occupies 750 mL at 75occupies 750 mL at 7500C under a C under a pressure of 810 torr. What volume pressure of 810 torr. What volume would it occupy at standard conditions?would it occupy at standard conditions?V = 750 mL V = ?
T = 348 K T = 273 K
P = 810 torr P = 760 torr
V =P V T
P T
1 2
1 2
1 2
21 1 2
2 1
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Example: A sample of nitrogen gas, NExample: A sample of nitrogen gas, N22, , occupies 750 mL at 75occupies 750 mL at 7500C under a C under a pressure of 810 torr. What volume pressure of 810 torr. What volume would it occupy at standard conditions?would it occupy at standard conditions?
V = 750 mL V = ?
T = 348 K T = 273 K
P = 810 torr P = 760 torr
V =P V T
P T
torr mL K
760 torr K
mL
1 2
1 2
1 2
21 1 2
2 1
810 750 273
348
627
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Ideal Gas LawsIdeal Gas Laws
A sample of methane, CHA sample of methane, CH44, occupies 260 , occupies 260 mL at 32mL at 32ooC under a pressure of 0.500 C under a pressure of 0.500 atm. At what temperature would it atm. At what temperature would it occupy 500 mL under a pressure of 1200 occupy 500 mL under a pressure of 1200 torr?torr?
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Ideal Gas LawsIdeal Gas Laws
V = 260 mL V = 500 mL
P = 0.500 atm = 380 torr P = 1200 torr
T = 305 K T = ?
T =T P V
P V
K 1200 torr mL
380 torr mL
= 1852 K or 1579 C
1 2
1 2
1 2
21 2 2
1 1
o
305 500
260
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Density has units of mass over volume. Density has units of mass over volume. Rearranging the ideal-gas equation with Rearranging the ideal-gas equation with
MM as molar mass we get as molar mass we get
Molar Mass and DensityMolar Mass and Density
RTP
dV
nRTP
Vn
MM
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The molar mass of a gas can be The molar mass of a gas can be determined as follows:determined as follows:
Molar Mass and DensityMolar Mass and Density
PdRTM
Volumes of Gases in Chemical ReactionsVolumes of Gases in Chemical Reactions The ideal-gas equation relates The ideal-gas equation relates PP, , VV, and , and TT to to
number of moles of gas.number of moles of gas. The The nn can then be used in stoichiometric can then be used in stoichiometric
calculations.calculations.
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Molar Mass and DensityMolar Mass and Density Example: One mole of a gas occupies Example: One mole of a gas occupies
36.5 L and its density is 1.36 g/L at a 36.5 L and its density is 1.36 g/L at a given T & P.given T & P.(a)(a) What is its molar mass?What is its molar mass?
(b)(b) What is its density at STP?What is its density at STP?
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Molar Mass and DensityMolar Mass and Density Example: One mole of a gas occupies Example: One mole of a gas occupies
36.5 L and its density is 1.36 g/L at a 36.5 L and its density is 1.36 g/L at a given T & P.given T & P.(a)(a) What is its molar mass?What is its molar mass?
(b)(b) What is its density at STP?What is its density at STP?? . ..
gmol
Lmol
gL
g / mol 36 5 13649 6
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Molar Mass and DensityMolar Mass and Density Example: One mole of a gas occupies Example: One mole of a gas occupies
36.5 L and its density is 1.36 g/L at a 36.5 L and its density is 1.36 g/L at a given T & P.given T & P.(a)(a) What is its molar mass?What is its molar mass?
(b)(b) What is its density at STP?What is its density at STP?? . ..
.
..
gmol
Lmol
gL
g / mol
? g
L
g
mol
mol
L g / L
STP
36 5 13649 6
49 6 1
22 42 21
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Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
Example: A compound that contains only Example: A compound that contains only carbon & hydrogen is 80.0% C and 20.0% carbon & hydrogen is 80.0% C and 20.0% H by mass. At STP 546 mL of the gas has H by mass. At STP 546 mL of the gas has a mass of 0.732 g . What is the a mass of 0.732 g . What is the molecular (true) formula for the molecular (true) formula for the compound?compound?
100 g of compound contains:100 g of compound contains:
80 g of C & 20 g of H80 g of C & 20 g of H
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H mol 8.19H g 1.01
H mol 1H g 20.0 = atoms H mol ?
C mol 67.6C g 12.0
C mol 1C g 80.0 = atoms C mol ?
Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
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mol 0244.0K 273
K molatm L
0821.0
L 546.0atm 00.1
RT
PV=n
15 = mass with CH 36.67
19.8is formulasimplest
H mol 8.19H g 1.01
H mol 1H g 20.0 = atoms H mol ?
C mol 67.6C g 12.0
C mol 1C g 80.0 = atoms C mol ?
3
Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
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molar mass = 0.732 g
0.0244 mol
g
mol
CH C H3 2 6
30 0
30 0
1502 2
.
.
.
Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
52
Example: A 1.74 g sample of a compound Example: A 1.74 g sample of a compound that contains only carbon & hydrogen that contains only carbon & hydrogen contains 1.44 g of C and 0.300 g of H. At contains 1.44 g of C and 0.300 g of H. At STP 101 mL of the gas has a mass of STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular (true) 0.262 gram. What is its molecular (true) formula?formula?
Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
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mol 00451.0K 273
K molatm L
0.0821
L 0.101atm 00.1
RT
PV=n
29 = mass with HC5.20.120
0.297
H mol 297.0H g 1.01
H mol 1H g 0.300 = atoms H mol ?
C mol 120.0C g 0.12
C mol 1C g 1.44 = atoms C mol ?
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Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
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? g
mol
g
0.00451 mol g / mol
58.1
29C H C H2 5 4 10
0 262581
2 2
..
Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
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Example: A 250 mL flask contains 0.423 Example: A 250 mL flask contains 0.423 g of vapor at 100g of vapor at 10000C and 746 torr. What C and 746 torr. What is molar mass of compound?is molar mass of compound?
n =
PVRT
atm L
L atmmol K
K mol
746760
0 250
0 0821 3730 00801
.
..
Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
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Example: A 250 mL flask contains 0.423 Example: A 250 mL flask contains 0.423 g of vapor at 100g of vapor at 10000C and 746 torr. What C and 746 torr. What is molar mass of compound?is molar mass of compound?
n =
PV
RT
atm L
L atm
mol K K
mol
? g
mol
g
0.00801 mol g / mol
746
7600 250
0 0821 3730 00801
0 42352 8
.
..
..
Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds
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Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Since gas molecules are so far apart, Since gas molecules are so far apart, we can assume they behave we can assume they behave independently.independently.
Dalton’s Law: in a gas mixture the total Dalton’s Law: in a gas mixture the total pressure is given by the sum of partial pressure is given by the sum of partial pressures of each component:pressures of each component:
PPtt = = PP11 + + PP22 + + PP33 + … + …
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Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Each gas obeys the ideal gas equation:Each gas obeys the ideal gas equation:
Combining equations:Combining equations:
VRT
nP ii
VRT
nnnPt 321
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Mole FractionsMole Fractions
Let Let nnii be the number of moles of gas be the number of moles of gas ii exerting a partial pressure exerting a partial pressure PPii, then, then
Pi = iPt
where where ii is the mole fraction ( is the mole fraction (nnii//nntt).).
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It is common to synthesize gases and It is common to synthesize gases and collect them by displacing a volume of collect them by displacing a volume of water.water.
To calculate the amount of gas To calculate the amount of gas produced, we need to correct for the produced, we need to correct for the partial pressure of the water:partial pressure of the water:
Ptotal = Pgas + Pwater
Collecting Gases over waterCollecting Gases over water
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Collecting Gases over waterCollecting Gases over water
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Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures
Example: If 100 mL of hydrogen, Example: If 100 mL of hydrogen, measured at 25measured at 2500C & 3.00 atm pressure, C & 3.00 atm pressure, and 100 mL of oxygen, measured at 25and 100 mL of oxygen, measured at 2500C C & 2.00 atm pressure, were forced into & 2.00 atm pressure, were forced into one of the containers at 25one of the containers at 2500C, what would C, what would be the pressure of the mixture of gases?be the pressure of the mixture of gases?
63
Example: If 100 mL of hydrogen, Example: If 100 mL of hydrogen, measured at 25measured at 2500C & 3.00 atm pressure, C & 3.00 atm pressure, and 100 mL of oxygen, measured at 25and 100 mL of oxygen, measured at 2500C C & 2.00 atm pressure, were forced into & 2.00 atm pressure, were forced into one of the containers at 25one of the containers at 2500C, what would C, what would be the pressure of the mixture of gases?be the pressure of the mixture of gases?
P P P
3.00 atm + 2.00 atm
= 5.00 atm
Total H O2 2
Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures
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Vapor PressureVapor Pressure
Pressure exerted by gas over the liquid Pressure exerted by gas over the liquid at equilibrium.at equilibrium.
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Example: A sample of hydrogen was Example: A sample of hydrogen was collected by displacement of water at collected by displacement of water at 252500C. The atmospheric pressure was 748 C. The atmospheric pressure was 748 torr. What pressure would the dry torr. What pressure would the dry hydrogen exert in the same container?hydrogen exert in the same container?
Vapor PressureVapor Pressure
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Example: A sample of hydrogen was Example: A sample of hydrogen was collected by displacement of water at collected by displacement of water at 252500C. The atmospheric pressure was 748 C. The atmospheric pressure was 748 torr. What pressure would the dry torr. What pressure would the dry hydrogen exert in the same container?hydrogen exert in the same container?
P P P P P P
P torr = 724 torr
P from table
Total H H O H Total H O
H
H O
2 2 2 2
2
2
748 24
Vapor PressureVapor Pressure
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Example: A sample of oxygen was Example: A sample of oxygen was collected by displacement of water. The collected by displacement of water. The oxygen occupied 742 mL at 27oxygen occupied 742 mL at 27ooC. The C. The barometric pressure was 753 torr. What barometric pressure was 753 torr. What volume would the dry oxygen occupy at volume would the dry oxygen occupy at STP?STP?
Vapor PressureVapor Pressure
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V 742 mL V ?
T K T K
P = 726 torr P torr
V mL273 K300 K
726 torr torr
645 mL @ STP
1 2
1 2
1 2
2
300 273
753 27 760
742760
Vapor PressureVapor Pressure
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Kinetic-Molecular TheoryKinetic-Molecular Theory
Theory developed to explain gas behavior.Theory developed to explain gas behavior. Theory of moving molecules.Theory of moving molecules. Assumptions:Assumptions:
– Gases consist of a large number of molecules in constant random Gases consist of a large number of molecules in constant random motion.motion.
– Volume of individual molecules negligible compared to volume of Volume of individual molecules negligible compared to volume of container.container.
– Intermolecular forces (forces between gas molecules) negligible.Intermolecular forces (forces between gas molecules) negligible.– Energy can be transferred between molecules, but total kinetic Energy can be transferred between molecules, but total kinetic
energy is constant at constant temperature.energy is constant at constant temperature.– Average kinetic energy of molecules is proportional to Average kinetic energy of molecules is proportional to
temperature.temperature.
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Kinetic Molecular TheoryKinetic Molecular Theory
Postulate 1Postulate 1– gases are discrete molecules that are far gases are discrete molecules that are far
apartapart– have few intermolecular attractionshave few intermolecular attractions– molecular volume small compared to gas’s molecular volume small compared to gas’s
volumevolume Proof - Gases are easily compressible.Proof - Gases are easily compressible.
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Postulate 2Postulate 2– molecules are in constant straight line molecules are in constant straight line
motionmotion– varying velocitiesvarying velocities– molecules have elastic collisionsmolecules have elastic collisions
Proof - A sealed, confined gas exhibits no Proof - A sealed, confined gas exhibits no pressure drop over time.pressure drop over time.
Kinetic Molecular TheoryKinetic Molecular Theory
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Postulate 3Postulate 3– kinetic energy is proportional to absolute Tkinetic energy is proportional to absolute T– average kinetic energies of molecules of average kinetic energies of molecules of
different gases are equal at a given Tdifferent gases are equal at a given T Proof - Motion increases as temperature Proof - Motion increases as temperature
increases.increases.
Kinetic Molecular TheoryKinetic Molecular Theory
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Boyle’s & Dalton’s LawBoyle’s & Dalton’s Law– PP 1/V as V increases collisions decrease so 1/V as V increases collisions decrease so
P dropsP drops
– PPtotaltotal = P = PAA + P + PBB + P + PCC + .....because few + .....because few intermolecular attractionsintermolecular attractions
Charles’ LawCharles’ Law– V V T increased T raises molecular velocities, T increased T raises molecular velocities,
V increases to keep P constantV increases to keep P constant
Kinetic Molecular TheoryKinetic Molecular Theory
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Kinetic-Molecular TheoryKinetic-Molecular Theory
Kinetic molecular theory Kinetic molecular theory gives us an understanding gives us an understanding of pressure and of pressure and temperature on the temperature on the molecular level.molecular level.
Pressure of a gas results Pressure of a gas results from the number of from the number of collisions per unit time on collisions per unit time on the walls of container.the walls of container.
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Kinetic-Molecular TheoryKinetic-Molecular Theory
Magnitude of pressure given by how often Magnitude of pressure given by how often and how hard the molecules strike. and how hard the molecules strike.
Gas molecules have an average kinetic Gas molecules have an average kinetic energy.energy.
Each molecule has a different energy.Each molecule has a different energy. There is a spread of individual energies of There is a spread of individual energies of
gas molecules in any sample of gas.gas molecules in any sample of gas. As the temperature increases, the average As the temperature increases, the average
kinetic energy of the gas molecules kinetic energy of the gas molecules increases.increases.
76
Kinetic-Molecular TheoryKinetic-Molecular Theory
77
Kinetic-Molecular TheoryKinetic-Molecular Theory
As kinetic energy increases, the velocity of As kinetic energy increases, the velocity of the gas molecules increases.the gas molecules increases.
Root mean square speed, Root mean square speed, uu, is the speed of a , is the speed of a gas molecule having average kinetic energy.gas molecule having average kinetic energy.
Average kinetic energy, Average kinetic energy, , is related to root , is related to root mean square speed:mean square speed:
= ½m= ½muu22
78
Kinetic-Molecular TheoryKinetic-Molecular Theory
As volume increases at constant temperature, As volume increases at constant temperature, the average kinetic of the gas remains the average kinetic of the gas remains constant. Therefore, constant. Therefore, uu is constant. However, is constant. However, volume increases so the gas molecules have volume increases so the gas molecules have to travel further to hit the walls of the to travel further to hit the walls of the container. Therefore, pressure decreases.container. Therefore, pressure decreases.
If temperature increases at constant volume, If temperature increases at constant volume, the average kinetic energy of the gas the average kinetic energy of the gas molecules increases. Therefore, there are molecules increases. Therefore, there are more collisions with the container walls and more collisions with the container walls and the pressure increases.the pressure increases.
79
Effusion & DiffusionEffusion & Diffusion
diffusion - intermingling of gasesdiffusion - intermingling of gases effusion - gases pass through porous effusion - gases pass through porous
containerscontainers rates are inversely proportional to square rates are inversely proportional to square
roots of molecular weights or densitiesroots of molecular weights or densities
80
effusion - gases pass through porous containerseffusion - gases pass through porous containers– Effusion is the escape of a gas through a tiny hole (a Effusion is the escape of a gas through a tiny hole (a
balloon will deflate over time due to effusion).balloon will deflate over time due to effusion). rates are inversely proportional to square roots rates are inversely proportional to square roots
of molecular weights or densitiesof molecular weights or densities
R
R
M
M or
R
R
D
D
1
2
2
1
1
2
2
1
Effusion & DiffusionEffusion & Diffusion
81
Diffusion of a gas is the spread of the gas Diffusion of a gas is the spread of the gas through space.through space.
Diffusion is faster for light gas molecules.Diffusion is faster for light gas molecules. Diffusion is significantly slower than rms Diffusion is significantly slower than rms
speed (consider someone opening a perfume speed (consider someone opening a perfume bottle: it takes while to detect the odor but bottle: it takes while to detect the odor but rms speed at 25rms speed at 25C is about 1150 mi/hr).C is about 1150 mi/hr).
Diffusion is slowed by gas molecules colliding Diffusion is slowed by gas molecules colliding with each other.with each other.
Average distance of a gas molecule between Average distance of a gas molecule between collisions is called mean free path.collisions is called mean free path.
Effusion & DiffusionEffusion & Diffusion
82
At sea level, mean free path is about 6 At sea level, mean free path is about 6 10 10-6-6 cm. cm.
Effusion & DiffusionEffusion & Diffusion
83
Calculate the ratio of the rate of effusion of He to Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SOthat of sulfur dioxide, SO22, at the same T & P., at the same T & P.
Effusion & DiffusionEffusion & Diffusion
84
Calculate the ratio of the rate of effusion of He to Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SOthat of sulfur dioxide, SO22, at the same T & P., at the same T & P.
R
R
M
M
g / mol
4.0 g / mol
R R
He
SO
SO
He
He SO
2
2
2
641
16 4 4
.
Effusion & DiffusionEffusion & Diffusion
85
Example: A sample of hydrogen, HExample: A sample of hydrogen, H22, was , was found to effuse through a pinhole 5.2 found to effuse through a pinhole 5.2 times as rapidly as the same volume of times as rapidly as the same volume of unkown gas ( at the same T & P). What unkown gas ( at the same T & P). What is the molecular weight of the unknown is the molecular weight of the unknown gas?gas?
Effusion & DiffusionEffusion & Diffusion
86
g/mol 54 = g/mol) 0.2(27M
g/mol 0.2
M27
g/mol 0.2
M2.5
M
M
R
R
unk
unk
unk
H
unk
unk
H
2
2
Effusion & DiffusionEffusion & Diffusion
87
As kinetic energy increases, the velocity of As kinetic energy increases, the velocity of the gas molecules increases.the gas molecules increases.
Average kinetic energy of a gas is related to Average kinetic energy of a gas is related to its mass:its mass:
= ½mu2
Average SpeedAverage Speed
88
Consider two gases at the same temperature: Consider two gases at the same temperature: the lighter gas has a higher rms than the the lighter gas has a higher rms than the heavier gas.heavier gas.
Mathematically:Mathematically:
The lower the molar mass, The lower the molar mass, MM, the higher the , the higher the rms for that gas at a constant temperature.rms for that gas at a constant temperature.
MRT
u3
Average SpeedAverage Speed
89
Average SpeedAverage Speed
90
R = 8.314 kg mR = 8.314 kg m22/s/s22 K mol K mol M must be in kg/molM must be in kg/mol
uRT
Mrmsm
3Average SpeedAverage Speed
91
What is velocity of NWhat is velocity of N22 molecules at room molecules at room T, 25T, 2500C?C?
u
3 8.314kg m
sec K mol K
kg / mol
m / s = 1159 mi / hr
rms
2
2
298
0 028
515
.
Average SpeedAverage Speed
92
What is the velocity of He at room T, What is the velocity of He at room T, 252500C?C?
Average SpeedAverage Speed
93
What is the velocity of He at room T, What is the velocity of He at room T, 252500C?C?
u
3 8.314kg m
sec K mol K
kg / mol
m / s = 3067 mi / hr
rms
2
2
298
0 004
1363
.
Average SpeedAverage Speed
94
Real GasesReal Gases
real gases behave ideally at ordinary T’s real gases behave ideally at ordinary T’s & P’s& P’s
at low T and high P - large deviationsat low T and high P - large deviations– molecules are on top of one anothermolecules are on top of one another– molecular interactions become importantmolecular interactions become important
95
As the pressure on a gas increases, the As the pressure on a gas increases, the molecules are forced closer together.molecules are forced closer together.
As the molecules get closer together, the As the molecules get closer together, the volume of the container gets smaller.volume of the container gets smaller.
The smaller the container, the more space The smaller the container, the more space the gas molecules begin to occupy.the gas molecules begin to occupy.
Therefore, the higher the pressure, the less Therefore, the higher the pressure, the less the gas resembles an ideal gas.the gas resembles an ideal gas.
As the gas molecules get closer together, the As the gas molecules get closer together, the smaller the intermolecular distance.smaller the intermolecular distance.
Real GasesReal Gases
96
Real GasesReal Gases
97
Real GasesReal Gases
98
The smaller the distance between gas The smaller the distance between gas molecules, the more likely attractive forces molecules, the more likely attractive forces will develop between the molecules.will develop between the molecules.
Therefore, the less the gas resembles and Therefore, the less the gas resembles and ideal gas.ideal gas.
As temperature increases, the gas molecules As temperature increases, the gas molecules move faster and further apart.move faster and further apart.
Also, higher temperatures mean more energy Also, higher temperatures mean more energy available to break intermolecular forces.available to break intermolecular forces.
Therefore, the higher the temperature, the Therefore, the higher the temperature, the more ideal the gas.more ideal the gas.
Real GasesReal Gases
99
Real GasesReal Gases
100
van der Waals’s equationvan der Waals’s equation
a & b are van der Waal’s constantsa & b are van der Waal’s constants
different values for each gasdifferent values for each gas
P + n a
VV nb nRT
2
2
Real GasesReal Gases
101
van der Waals’s equationvan der Waals’s equation– a accounts for intermolecular attractionsa accounts for intermolecular attractions– b accounts for volume of gas moleculesb accounts for volume of gas molecules– large V’s a & b reduce to small quantitieslarge V’s a & b reduce to small quantities
van der Waal’s equation reduces to van der Waal’s equation reduces to ideal gas law at high temperatures and ideal gas law at high temperatures and low pressureslow pressures
Real GasesReal Gases
102
attractive forces in nonpolar gases attractive forces in nonpolar gases – London ForcesLondon Forces
• noble gases- instantaneous induced dipole noble gases- instantaneous induced dipole momentsmoments
• Chapter 9Chapter 9
attractive forces in polar gases attractive forces in polar gases – dipole-dipole or H-bondsdipole-dipole or H-bonds
• water molecules can H-bondwater molecules can H-bond• why ice floats - Chapter 9why ice floats - Chapter 9
Real GasesReal Gases
103
Real GasesReal Gases
104
Example: Calculate the pressure exerted Example: Calculate the pressure exerted by 84.0 g of ammonia, NHby 84.0 g of ammonia, NH33, in a 5.00 L at , in a 5.00 L at 20020000C using the ideal gas law.C using the ideal gas law.
Real GasesReal Gases
105
n = 84.0 g NH mol
17.0 g mol
P = nRTV
molL atmmol K
K
5.00 L atm
3
14 94
4 94 0 0821 473
38 4
.
. .
.
Real GasesReal Gases
106
Example: Solve previous example using Example: Solve previous example using the van der Waal’s equation.the van der Waal’s equation.
Real GasesReal Gases
107
Example: Solve previous example using Example: Solve previous example using the van der Waal’s equation.the van der Waal’s equation.
n = 4.94 mol a = 4.17 L atm
mol b = 0.0371L
mol
P + n a
VV - nb nRT
P =nRT
V - nbn a
V
2
2
2
2
2
2
Real GasesReal Gases
108
P
P
P atm
a 7.6% difference
4 94 0 0821 473
5 00 4 94 0 0371
4 94 417
5 00
19184 817
4 07 39 8 41
35 7
2
2
. .
. ( . )( . )
. .
.
..
. ( . . )
.
Real GasesReal Gases
109
We can’t forget about moles!We can’t forget about moles!
Example: What volume of oxygen Example: What volume of oxygen measured at STP, can be produced by measured at STP, can be produced by the thermal decomposition of 120 g of the thermal decomposition of 120 g of KClOKClO33??
110
Mass-Volume Relationships in Mass-Volume Relationships in Reactions Involving GasesReactions Involving Gases
2 mol 2 mol 2 mol 3 mol2 mol 3 mol
2(122.6g) 2(122.6g) 2(74.6g) 3(32g)2(74.6g) 3(32g) 3 moles of oxygen 3 moles of oxygen
equal to 3(22.4L) equal to 3(22.4L) oror
67.2 L at STP67.2 L at STP
2 2 KClO (s) KCl(s) + 3 O g)3MnO
22 & (
111
? L O 120 g KClO mol KClO
122.6 g KClO
mol O
2 mol KClO
L O
mol O L O
STP 2 33
3
2
3
STP 2
2STP 2
1 3
22 4
132 9
..
Mass-Volume Relationships in Mass-Volume Relationships in Reactions Involving GasesReactions Involving Gases
112
Synthesis QuestionSynthesis Question
The lethal dose for hydrogen sulfide is 6 ppm. The lethal dose for hydrogen sulfide is 6 ppm. In other words, if in 1 million molecules of air In other words, if in 1 million molecules of air there are six hydrogen sulfide molecules then there are six hydrogen sulfide molecules then that air would be deadly to breathe. How many that air would be deadly to breathe. How many hydrogen sulfide molecules would be required hydrogen sulfide molecules would be required to reach the lethal dose in a room that is 77 to reach the lethal dose in a room that is 77 feet long, 62 feet wide and 50 feet tall at 1.0 feet long, 62 feet wide and 50 feet tall at 1.0 atm and 25.0atm and 25.0oo C? C?
113
363
39
39
cm 106.76cm 1000
L1cm 106.76
cm 106.76cm 1524cm 1890cm 2347V
cm 1524in
cm 54.2
ft
in 12ft 05
cm 1890in
cm 54.2
ft
in 12ft 26
cm 2347in
cm 54.2
ft
in 12ft 77
Synthesis QuestionSynthesis Question
114
S Hof molecules 1066.1air of molecules 10
S Hof molecule 1air of molecules 1066.1 dose Lethal
air of molecules 1066.1mol
molecules 106.022mol 276,000
air of mol 276,000 K2980.0821
L106.76atm 1
RT
PV n
K298 25 273 T
L106.76cm 1000
L1cm 106.76
223
6229
2923
Kmolatm L
6
63
39
Synthesis QuestionSynthesis Question
115
Group QuestionGroup Question
Tires on a car are typically filled to a Tires on a car are typically filled to a pressure of 35 psi at 300K. A tire is 16 pressure of 35 psi at 300K. A tire is 16 inches in radius and 8.0 inches in inches in radius and 8.0 inches in thickness. The wheel that the tire is thickness. The wheel that the tire is mounted on is 6.0 inches in radius. What mounted on is 6.0 inches in radius. What is the mass of the air in a tire?is the mass of the air in a tire?