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# Applied Chapter 12.3 : Molecular Composition of Gases

## on Mar 31, 2010

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## Applied Chapter 12.3 : Molecular Composition of GasesPresentation Transcript

• Chapter 12.3
Molecular Composition of Gases
• Solve problems using ideal gas law.
Describe the relationships between gas behavior and chemical formulas, such as those expressed by Graham’s law of diffusion, law of combining gas volumes, and Dalton’s law of partial pressures.
Objectives:
• The Ideal Gas Law
• The mathematical relationship among pressure, volume, temperature, and the number or moles of a gas.
Here’s how it works!!!
View slide
• Deriving the Ideal gas law
• Boyle’s law PV = k View slide
• Charles’s law V/T = k
• Avogadro’s law V/n = k
• V = 1/P x T x n
• V = R x 1/P x T x n
• V = nRT or PV = nRT
• P
The Ideal Gas Constant
• Based on: 1 mol of gas at STP
• R = PV = (1 atm)(22.4 L) = 0.0821 L*atm
• nT(1 mol)(273.15 K) mol*K
• Problems
• What is the pressure in atmospheres exerted by 0.500 mol sample of nitrogen gas in 10.0 L container at 298 K?
P =
?
PV = nRT
V =
10.0 L
V
V
n =
0.500 mol
nRT
(0.500 mol)(0.0821 )(298 K)
=
P =
L*atm
L*atm
R =
o.0821
V
10.0 L
mol*K
mol*K
T =
298 K
1.22 atm
=
***Make sure all units match Gas Constant
***Solve for P
• Problem 2
• What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0 ⁰C and 0.974 atm pressure?
PV = nRT
P =
0.974 atm
V =
?
P
P
nRT
(0.250 mol)(0.0821 )(293 K)
n =
0.250 mol
=
V =
L*atm
L*atm
o.0821
P
R =
0.974 atm
mol*K
mol*K
T =
20.0oC
6.17 L of O2
=
+ 273 =
293 K
***Temperature must be in Kelvin!!!
***Solve for V
• Problem 3
• How many moles of chlorine gas, Cl2, in grams, is contained in a 10.0 L tank at 27 ⁰C and 3.50 atm of pressure?
PV = nRT
P =
3.50 atm
RT
RT
V =
10.0 L
PV
(3.50 atm)(10.0 L)
n =
?
n
=
=
R =
RT
(0.0821 )(300.k)
L*atm
L*atm
o.0821
mol*K
mol*K
T =
27 oC
+ 273 =
300. K
=
1.42 mol Cl2
***Temperature must be in Kelvin!!!
***Solve for n
• Effusion
Process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.
Effusion
• Diffusion
• The gradual mixing of two gases due to their sp0ontaneous, random motion
• Graham’s Law of Effusion
Rate of effusion:
Depends on:
Velocity of gas molecules
Mass of molecules
1
1
MAvA2
=
MBvB2
Rate of effusion of A
MB
2
2
=
MA
Rate of effusion of B
SO:
MAvA2
=
MBvB2
vA2
MB
• Defined as:
• The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.
=
vB2
MA
vA
MB
=
vB
MA
• Application of Graham’s Law
Lighter gases (lower Molar mass or densities) diffuse faster than heavier gases.
Also provides a method for determining molar masses.
Rates of effusion of known and unknown gases can be compared to one another
Rates of effusion of different gases
• Problem 1
Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.
32.00 g/mol
Rate of effusion of H2
MO2
=
3.98
=
=
Rate of effusion of O2
2.02 g/mol
MH2
***Remember that the molar masses are inversely related
***Find the molar masses of each
Hydrogen effuses 3.98 times faster than oxygen
***Expressed like this
• Problem 2
A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Rate of effusion of H2
Munknown
=
Rate of effusion of unknown
MH2
2
Munknown
2
=
9
2.02 g/mol
Munknown
81
=
x 2.o2 g/mol
2.o2 g/mol x
2.02 g/mol
Munknown
=
160 g/mol
• Dalton’s Law of Partial Pressures
States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
Partial pressure: pressure of each gas in a mixture
PT = p1 + p2 + p3 + ……
PT= Total Pressure
p1 + p2 + p3 = partial pressures
• Dalton’s Law of Partial Pressures
Gas collected by water displacement.
Must include the pressure exerted by water vapor
PT = p1 + p2 + p3 + ……
So: Patm = pgas + pH2O
• Dalton’s Law of Partial Pressures
Sample Problem 5
Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0oC, respectively. What was the partial pressure of the oxygen collected?
Patm = pO2 + pH2O
Patm = 731.0 torr
PO2 = ?
PH2O = 17.5 torr (from appendix in table A-8, pg. 899)
pO2 = Patm - pH2O
pO2 = 731.0 torr – 17.5 torr
= 713.5 torr