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Chapter 12.3
Molecular Composition of Gases
1. Solve problems using ideal gas law.2. 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 LawThe mathematical relationship among pressure, volume,
temperature, and the number or moles of a gas.
Here’s how it works!!!
Deriving the Ideal gas lawo Boyle’s law PV = ko Charles’s law V/T = ko Avogadro’s law V/n = k
o V = 1/P x T x no V = R x 1/P x T x no V = nRT or PV = nRT P
The Ideal Gas ConstantBased 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
ProblemsWhat is the pressure in atmospheres exerted by 0.500
mol sample of nitrogen gas in 10.0 L container at 298 K?
P =V =n =
R =T =
?
10.0 L0.500 molo.0821 mol*K
L*atm
298 K
PV = nRT
***Make sure all units match Gas Constant***Solve for P
V V
P =
nRT
V=
(0.500 mol)(0.0821 )(298 K)
mol*K
L*atm
10.0 L
1.22 atm=
Problem 2What is the volume, in liters, of 0.250 mol of
oxygen gas at 20.0 ⁰C and 0.974 atm pressure?P =V =n =
R =T =
0.974 atm?
0.250 molo.0821 mol*K
L*atm
20.0o
C
PV = nRT
P P
V =
nRT
P=
(0.250 mol)(0.0821 )(293 K)
mol*K
L*atm
0.974 atm
6.17 L of O2
=
***Temperature must be in Kelvin!!!
***Solve for V
+ 273 =
293 K
Problem 3How 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?
P =V =n =
R =T =
3.50 atm
10.0 L?
o.0821 mol*K
L*atm
27 oC
PV = nRTRT
***Temperature must be in Kelvin!!!
***Solve for n
+ 273 =
300. K
RT
n =PVRT
=(3.50 atm)(10.0 L)
mol*K
L*atm(0.0821 )(300.k)
=1.42 mol Cl2
EffusionProcess whereby the molecules of a gas
confined in a container randomly pass through a tiny opening in the container.
Diffusion The gradual mixing of two gases due to
their sp0ontaneous, random motion
Effusion
Graham’s Law of EffusionRate of effusion:
Depends on:Velocity of gas moleculesMass of molecules
12
MAvA2
=1
2MBvB2
MAvA2 = MBvB
2
vA2
vB2
=MA
MB
vA
vB
=MA
MB
SO:
Rate of effusion of ARate of effusion of B
=MA
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.
Application of Graham’s Law1. Lighter gases (lower Molar mass or
densities) diffuse faster than heavier gases.
2. Also provides a method for determining molar masses.
a) Rates of effusion of known and unknown gases can be compared to one another
Rates of effusion of different gases
Problem 1Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.
Rate of effusion of H2Rate of effusion of O2
=MH2
MO2 =32.00 g/mol
2.02 g/mol= 3.9
8
***Remember that the molar masses are inversely related ***Find the molar masses of each
***Expressed like this
Hydrogen effuses 3.98 times faster than oxygen
Problem 2A 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 H2Rate of effusion of
unknown
=MH2
Munknown
2.02 g/mol
Munknown=9
22
81
=2.02 g/mol
Munknownx 2.o2 g/mol2.o2 g/mol
x
Munknown =160 g/mol
Dalton’s Law of Partial PressuresStates 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 Pressurep1 + p2 + p3 = partial pressures
Dalton’s Law of Partial Pressures
PT = p1 + p2 + p3 + ……
So: Patm = pgas + pH2O
Gas collected by water displacement. Must include the pressure exerted by water
vapor
Dalton’s Law of Partial PressuresSample Problem 5Oxygen 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