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Lecture #14
The Ideal Gas Law
and
Stoichiometry
Chemistry 142 B
Autumn Quarter, 2004
J. B. Callis, Instructor
Ideal Gas Law
• An ideal gas is defined as one for which both the volume
of molecules and forces between the molecules are so
small that they have no effect on the behavior of the gas.
• The ideal gas equation is:
PV=nRT
R = Ideal gas constant = 8.314 J / mol K = 8.314 J mol-1 K-1
• R = 0.08206 L atm mol-1 K-1
The Ideal Gas Law Subsumes
the Other Gas Laws
• During chemical and physical processes, any of the four
variables in the ideal gas equation may be fixed.
• Thus, PV=nRT can be rearranged for the fixed variables:
– for a fixed amount at constant temperature
• PV = nRT = constant
Boyle’s Law
– for a fixed amount at constant volume
• P/T = nR/V = constant
Amonton’s Law
– for a fixed amount at constant pressure
• V/T = nR/P = constant
Charles’s Law
– for a fixed volume and temperature
• P/n = RT/V = constant
Avogadro’s Law
Many gas law problems involve a change of
conditions, with no change in the amount of gas.
PxV
T
= constant
P1 x V1
T1
Therefore, for a change
of conditions :
=
P2 x V2
T2
Problem 14-1: Change of Three Variables - I
A gas sample in the laboratory has a volume of 45.9
L at 25 oC and a pressure of 743 mm Hg. If the
temperature is increased to 155 oC by compressing
the gas to a new volume of 31.0 L what is the
pressure?
P1=
P2 =
V1 =
T1 =
T2 =
V2 =
Problem 14-1: Change of Three Variables - II
P1 x V1
T1
=
P2 x V2
T2
=
P2 =
Problem 14-2: Gas Law
Problem: Calculate the pressure in a container whose Volume is 87.5 L
and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC.
Plan: Convert all information into the units required, and substitute into
the Ideal Gas equation ( PV=nRT ).
Solution:
nXe =
T=
P=
Problem 14-3: Ideal Gas Calculation - Nitrogen
Calculate the pressure in a container holding 375 g of
Nitrogen gas. The volume of the container is 0.150 m3
and the temperature is 36.0 oC.
Problem 14-4: Sodium Azide Decomposition - I
Sodium Azide (NaN3) is used in some air bags in
automobiles. Calculate the volume of Nitrogen gas
generated at 21 oC and 823 mm Hg by the decomposition
of 60.0 g of NaN3 .
2 NaN3 (s)
2 Na (s) + 3 N2 (g)
Problem 14-4: Sodium Azide Decomposition - II
Problem 14-5: Ammonia Density
Calculate the Density of ammonia gas (NH3) in grams
per liter at 752 mm Hg and 55 oC.
Density = mass per unit volume = g / L
P=
T=
n = mass / Molar mass = g / M
d=
Calculation of Molar Mass
n=
n=
Mass
Molar Mass
PxV =
Mass
RxT
Molar Mass
Molar Mass = MM =
Mass x R x T
PxV
Problem 14-6: Dumas Method of Molar Mass
Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml
and allowed to boil until all of the liquid is gone, and only vapor fills the
flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass
of the flask before and after the experiment was 148.375g and 149.457 g,
what is the molar mass of the liquid?
Plan: Use the ideal gas law to calculate the molar mass of the liquid.
Solution:
Problem 14-7: Calculation of Molecular Weight of a
Natural Gas - Methane
Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml
flask. If the sample had a mass of 0.118 g at a pressure of 550.0 torr,
what is the molecular weight of the gas?
Plan: Use the ideal gas law to calculate n, then calculate the molar mass.
Solution:
Gas Mixtures
• Gas behavior depends on the number, not
the identity, of gas molecules.
• The ideal gas equation applies to each gas
individually and to the mixture as a whole.
• All molecules in a sample of an ideal gas
behave exactly the same way.
Dalton’s Law of Partial Pressures - I
• Definition: In a mixture of gases, each gas
contributes to the total pressure: the pressure it
would exert if the gas were present in the
container by itself.
• To obtain a total pressure, add all of the partial
pressures: Ptotal = P1+P2+P3+…PN
Dalton’s Law of Partial Pressure - II
• Pressure exerted by an ideal gas mixture is
determined by the total number of moles:
P=(ntotal RT)/V
• ntotal = sum of the amounts of each gas pressure
• the partial pressure is the pressure of gas if it was
present by itself.
• P = (n1 RT)/V + (n2 RT)/V + (n3RT)/V + ...
• the total pressure is the sum of the partial
pressures.
Problem 14-8: Dalton’s Law of Partial Pressures
A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of
Helium at a temperature of 17.0 oC.
What are the Partial Pressures of each gas, and the total
Pressure?
Problem 14-8: Dalton’s Law of Partial Pressures
cont.
Problem 14-9: Dalton’s Law using mole fractions
A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar
and 2.15 mol Xe. What are the partial pressures of
the gases if the total pressure is 2.00 atm ?
Total # moles =
XNe =
PNe = XNe PTotal
XAr =
PAr =
XXe =
Relative Humidity
Pressure of Water in Air
Rel Hum =
x 100%
Maximum Vapor Pressure of Water
Example : the partial pressure of water at 15oC is 6.54 mm
Hg, what is the relative humidity?
Relative Humidity
Pressure of Water in Air
Rel Hum =
x 100%
Maximum Vapor Pressure of Water
Example : the partial pressure of water at 15oC is 6.54 mm
Hg, what is the relative humidity?
Rel Hum =(6.54 mm Hg/ 12.788 mm Hg )x100%
= 51.1 %
Problem 14-10: Collection of Hydrogen
gas over Water - Vapor pressure - I
2 HCl (aq) + Zn(s)
ZnCl2 (aq) + H2 (g)
Calculate the mass of Hydrogen gas collected over water if
156 ml of gas is collected at 20oC and 769 mm Hg.
Problem 14-10: Collection of Hydrogen
gas over Water - Vapor pressure - II
PV = nRT
n=
n=
mass =
n = PV / RT
Chemical Equation Calc - III
Mass
Atoms (Molecules)
Avogadro’s
Number
6.02 x 1023
Reactants
Molarity
moles / liter
Solutions
Molecules
Moles
Molecular
g/mol
Weight
Products
PV = nRT
Gases
Problem 14-11: Gas Law Stoichiometry
Problem: A slide separating two containers is removed, and the gases
are allowed to mix and react. The first container with a volume of 2.79 L
contains Ammonia gas at a pressure of 0.776 atm and a temperature of
18.7 oC. The second with a volume of 1.16 L contains HCl gas at a
pressure of 0.932 atm and a temperature of 18.7 oC. What mass of solid
ammonium chloride will be formed, and what will be remaining in the
container, and what is the pressure?
Plan: This is a limiting reactant problem, so we must calculate the moles
of each reactant using the gas law to determine the limiting reagent. Then
we can calculate the mass of product, and determine what is left in the
combined volume of the container, and the conditions.
Solution:
Equation:
NH3 (g) + HCl (g)
TNH3 = 18.7 oC + 273.15 = 291.9 K
NH4Cl (s)
Problem 14-11: Gas Law Stoichiometry
n = PV
RT
RRNH3 =
RRHCl =
Therefore the product will be
Answers to Problems in Lecture #14
1.
2.08 atm
2.
10.5 atm
3.
2.26 atm
4.
30.8 liters
5.
0.626 g / L
6.
58.03 g/mol
7.
15.9 g/mol
8.
PCO2 = 0.812 atm, PHe = 0.30 atm, PTotal = 1.11 atm
9.
1.21 atm for Ne, 0.20 atm for Ar, 0.586 atm for Xe
10. 0.0129 g hydrogen
11. 2.28 g NH4Cl made; remaining NH3 at a pressure of 0.274 atm