chapter 11 molecular composition of gases. avogadro’s law equal volumes of gases at the same...
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Chapter 11Chapter 11
Molecular Molecular Composition of GasesComposition of Gases
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Avogadro’s LawAvogadro’s LawEqual Volumes of Gases at Equal Volumes of Gases at the Same Temperature & the Same Temperature & Pressure contain the Pressure contain the Same Same Number of “Particles.”Number of “Particles.”
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BalloonsBalloons Holding 1.0Holding 1.0 L L
of Gas at 25ºof Gas at 25º C and C and
1 atm.1 atm.
Each balloon containsEach balloon contains
0.0410.041 mole of gas ormole of gas or
2.5 x 102.5 x 102222 molecules molecules..
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Avogadro’s LawAvogadro’s LawFor a gas at constant temperature For a gas at constant temperature and pressure, the volume is directly and pressure, the volume is directly proportional to the number of proportional to the number of moles of gas (at low pressures).moles of gas (at low pressures).
VV = = anan
aa = proportionality constant = proportionality constant
VV = volume of the gas = volume of the gas
nn = number of moles of gas = number of moles of gas
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What are the four quantities needed to describe a gas?
• Pressure• Volume• Temperature• Moles of gas
In Chapter 10, which three of these quantities did we vary?
• Pressure• Volume• Temperature
Now we will consider the moles of gas.
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Moles of GasMoles of Gas
The number of moles of gas will The number of moles of gas will always affect at least one of the always affect at least one of the other three quantities.other three quantities.
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Ideal Gas LawIdeal Gas Law Mathematical relationship Mathematical relationship
among among
pressure,pressure, volume, volume, temperature andtemperature and the number of moles.the number of moles.
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Ideal Gas LawIdeal Gas Law Mathematical equationMathematical equation
coming from the combination coming from the combination ofof
- Boyle’s Law- Boyle’s Law
- Charles’ Law &- Charles’ Law &
- Avogadro’s Law- Avogadro’s Law
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Ideal Gas LawIdeal Gas Law An An equation of state equation of state for a gas.for a gas. ““state” is the condition of the gas at state” is the condition of the gas at
a given time.a given time.
PVPV = = nRTnRT
An Ideal Gas is a hypothetical substance.
Ideal Gas Law is an empirical equation.
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Ideal Gas LawIdeal Gas Law
PPVV = = nnRRTT
RR = proportionality constant = proportionality constant
= ideal gas constant = ideal gas constant
= 0.0821 L atm = 0.0821 L atm mol mol
See Table 11-1 on page 342 for additional values for R.
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Ideal Gas LawIdeal Gas LawPPVV = = nnRRTT
Because of the units of R,Because of the units of R,
PP = pressure in atm = pressure in atm
VV = volume in liters = volume in liters
nn = moles = moles
TT = temperature in Kelvins = temperature in Kelvins
Holds closely at Holds closely at PP < 1 atm < 1 atm
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StandardStandard Temperature Temperature
and Pressureand Pressure““STP”STP”
PP = 1 atmosphere = 1 atmosphere
TT = = CC
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Ideal Gas LawIdeal Gas LawPPVV = = nnRRTT
If gas is at STP,If gas is at STP, RR = 0.0821 L atm = 0.0821 L atm mol mol
PP = 1.00 atm = 1.00 atm TT = 273 Kelvin = 273 Kelvin Then Then VV /n = volume in liters/ mole /n = volume in liters/ mole = RT/ P= RT/ P
= 0.0821 = 0.0821 L atmL atm x 273 K x 273 K mole Kmole K
1.00 atm1.00 atm
= 22.4 L/ mole= 22.4 L/ mole
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Standard Molar Volume of a GasStandard Molar Volume of a Gas
The volume occupied by one mole of The volume occupied by one mole of a gas at STP.a gas at STP.
22.4 L/ mole22.4 L/ mole
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A Mole of Any Gas Occupies a A Mole of Any Gas Occupies a
VolumeVolume of Approximately 22.42 L of Approximately 22.42 L
at STPat STP
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ProblemProblem
What is the volume of 77.0 g of What is the volume of 77.0 g of nitrogen dioxide at STP?nitrogen dioxide at STP?
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SolutionSolution77.0 g NO77.0 g NO22 x x 1 mole NO1 mole NO22 x x 22.4 L22.4 L
46.01 g NO46.01 g NO22 1 mole 1 mole
= 37. 5 L NO= 37. 5 L NO22
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ProblemProblem
What is the mass of 1.33 x 10What is the mass of 1.33 x 1044 mL of oxygen gas at STP?mL of oxygen gas at STP?
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SolutionSolution
1.33 x 101.33 x 1044 mL x mL x 1 L1 L x x 1 mol1 mol
1000 mL 22.4 L1000 mL 22.4 L
x x 32.00 g O32.00 g O22
1 mole O1 mole O22
= 19.0 g O= 19.0 g O22
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ProblemProblemAt STP, 3 L of chlorine is produced during a At STP, 3 L of chlorine is produced during a chemical reaction. chemical reaction.
What is the mass of this gas?What is the mass of this gas?
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SolutionSolution 3 L x 3 L x 1 mole 1 mole x x 70.90 g Cl70.90 g Cl22
22.4 L 1 mole Cl22.4 L 1 mole Cl22
= 9 g Cl= 9 g Cl22
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ProblemProblem
What pressure, in atmospheres, is What pressure, in atmospheres, is exerted by 0.325 mole of hydrogen exerted by 0.325 mole of hydrogen gas in a 4.08 L container at 35gas in a 4.08 L container at 35 C? C?
NotNot at STP!!! at STP!!!
Use Ideal Gas Equation!!!Use Ideal Gas Equation!!!
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Solution: Solution: P V = n R TP V = n R T
P = nRT/VP = nRT/V
T (K) = 35 + 273 = 308 KT (K) = 35 + 273 = 308 K
P = 0.325 mole x 0.0821 P = 0.325 mole x 0.0821 L atm L atm x 308 K x 308 K
mole Kmole K
4.08 L4.08 L
P = 2.01 atmP = 2.01 atm
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ProblemProblemA sample that contains 4.38 mole of a gas A sample that contains 4.38 mole of a gas at 250 K has a pressure of 0.857 atm. at 250 K has a pressure of 0.857 atm. What is the volume? What is the volume?
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Solution: PV = nRTSolution: PV = nRT
V = nRT/PV = nRT/P
V = 4.38 mole x V = 4.38 mole x 0.0821 L atm 0.0821 L atm x 250 K x 250 K
mole Kmole K
0.857 atm0.857 atm
V = 105 LV = 105 L
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Molar Mass of a GasMolar Mass of a Gas
n = # of molesn = # of moles
= = grams of gasgrams of gas = = (m) (m)
molar massmolar mass molar mass molar mass
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P V = n R T
P V = (m/molar mass) R T
Therefore,molar mass = m R T
P V
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ProblemProblemAt 28At 28C and 0.974 atm, 1.00 L of gas has a C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g.mass of 5.16 g.
What is the molar mass of this gas?What is the molar mass of this gas?
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SolutionSolution
molar mass = m R T P V
Molar Mass = 5.16 g x 0.0821 L atm x 301 K mole K
0.974 atm x 1.00 L
Molar Mass = 130.92 g/mole = 131 g/mole
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Also, since density (d) = m
VThen, since
molar mass = m R T P V
molar mass = d R T P
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Density = (molar mass) x P R T
Temperature in K!!
Gas density usually measured in g/L
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ProblemProblem What is the density of a sample of What is the density of a sample of ammonia gas, NHammonia gas, NH33, if the pressure is , if the pressure is 0.928 atm and the temperature is 0.928 atm and the temperature is 63.0 63.0 C?C?
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SolutionSolution
Density = (molar mass) x P R T
Density = 17.04 g/mole x 0.928 atm0.0821 L atm x 336 K
mole K
Density = 0.573 g/ L NH3
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ProblemProblemThe density of dry The density of dry air at sea levelair at sea level
(1 atm) is 1.225 g/L (1 atm) is 1.225 g/L at 15at 15C.C.
What is the What is the average molar average molar mass of the air?mass of the air?
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SolutionSolution
Density = (molar mass) x P R T
Therefore,
Molar Mass = Density x R x T P
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SolutionSolution
Molar Mass = Density x R x T P
Molar Mass = 1.225 Molar Mass = 1.225 g g x 0.0821 x 0.0821 L atmL atm x 288 K x 288 K
L mole KL mole K
______
1 atm1 atm
Molar Mass = 28.96 g/mole = 29.0 g/moleMolar Mass = 28.96 g/mole = 29.0 g/mole
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HomeworkHomeworkRead pages 338-339.Read pages 338-339.
Complete Worksheet.Complete Worksheet.