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AP Chemistry
Ms. Grobsky
We have already considered 4 laws that describe the behavior of
gases
β’ Boyleβs law
π =π
π
(at constant T and n)
β’ Charlesβ law
π = ππ (at constant P and n)
β’ Guy-Lussacβs law
π = ππ
(at constant V and n)
β’ Avogadroβs law
π = ππ (at constant T and P)
You may think a small gas molecule would take up less space than a large gas molecule,
but it doesnβt at the same temperature and pressure!
The previous 4 relationships, which show how the volume of a gas depends on pressure, temperature, and number of moles of gas present, can be combined as follows:
V = RTn
P
β’ R is the combined proportionality constant called the universal gas constant
Always use the value 0.08206 (0.0821) πΏβππ‘π
πΎβπππ for R
Equation can be rearranged to yield the more familiar ideal gas
law:
PV = nRT
The ideal gas law is an equation of state for a gas
β’ State of a gas is its condition at a given time
A particular state of a gas is described by its pressure, volume,
temperature, and number of moles
A gas that obeys this equation is said to behave ideally
β’ Expresses behavior that real gases approach at low pressures
and high temperatures
Thus, an ideal gas is a hypothetical substance
However, most gases obey the ideal gas equation closely enough at
pressure below 1 atm so assume ideal behavior unless stated otherwise
A sample of hydrogen gas (H2) has a
volume of 8.56 L at a temperature of 0Β°C
and a pressure of 1.5 atm. β’ Calculate the moles of H2 molecules present in
the sample.
A sample of diborane gas (B2H6), a
substance that bursts into flame when
exposed to air, has a pressure of 345 torr
at a temperature of -15Β°C and a volume of
3.48 L. β’ If conditions are changed so that the
temperature is 36Β°C and the pressure is 468 torr,
what will be the volume of the sample?
One very important use of the ideal gas law is the calculation of the molar mass of a gas from its measured density
π = πππππ ππ πππ
πππππ πππ π =
πππ π
πππππ πππ π =
π
πππππ πππ π
Substituting the above into the ideal gas equation
gives:
π = ππ π
π=
ππππππ πππ π
π π
π=
π(π π)
π(πππππ πππ π )
However, m/V is the gas density (d) in units of g/L
Substituting and rearranging for molar
mass:
πππππ πππ π = ππ π
π
βMolar Mass Kitty Catβ
All good cats put dirt
[dRT] over their pee [P]
The density of a gas was measured at
1.50 atm and 27Β°C and found to be 1.95
g/L. β’ Calculate the molar mass of the gas and give its
identity.
Use PV = NRT to solve for the volume of one mole of gas at STP:
Look familiar? This is the molar volume of a gas at STP
β’ Use stoichiometry to solve gas problems only if gas is at STP conditions
β’ Use the ideal gas law to convert quantities that are NOT at STP
Quicklime (CaO) is produced by the
thermal decomposition of calcium
carbonate (CaCO3). β’ Calculate the volume of CO2 at STP produced
from the decomposition of 152 g CaCO3 by the
reaction:
πΆππΆπ3 π β πΆππ π + πΆπ2(π)
A sample of methane gas (CH4) having a
volume of 2.80 L at 25Β°C and 1.65 atm
was mixed with a sample of oxygen gas
having a volume of 35.0 L at 31Β°C and
1.25 atm. The mixture was then ignited to
form carbon dioxide and water. β’ Calculate the volume of CO2 formed at a
pressure of 2.50 atm and a temperature of 125Β°C
The pressure of a mixture of gases is the sum of the pressures of the different components of the mixture:
ππ‘ππ‘ππ = π1 + π2 + β― ππ
Daltonβs Law of Partial Pressures
β’ Uses the concept of mole fractions! Recall:
Οπ΄ =moles A
moles A + moles B + moles C + β―
β’ So now:
ππ΄ = Οπ΄ππ‘ππ‘ππ
β’ What does this mean? The partial pressure of each gas in a mixture of gases
in a container depends on the number of moles of that
gas!
Therefore, the total pressure is the SUM of the partial
pressures and depends on the total moles of gas
present β no matter what their identity is!
Mixtures of helium and oxygen are used
in scuba diving tanks to help prevent βthe
bends.β For a particular dive, 46 L He at
25Β°C and 1.0 atm and 12 L O2 at 25Β°C and
1.0 atm were pumped into a tank with a
volume of 5.0 L. β’ Calculate the partial pressure of each gas and
the total pressure in the tank at 25Β°C.
The mole fraction of nitrogen in the air is
0.7904. The mole fraction of oxygen in
the air is .2093. The mole fraction of
carbon dioxide in the air is .0003.
Calculate the partial pressures of all
three major components in air when the
atmospheric pressure is 760. torr.
It is common to collect a gas by water
displacement which means some of the
pressure is due to water vapor collected as
the gas was passing through! β’ You must correct for this by looking up the partial
pressure of water at that particular temperature!
THIS IS A VERY POPULAR TYPE
OF PROBLEM ON THE AP EXAM!
A sample of solid potassium chlorate was heated in a test tube and decomposed by the following reaction:
2 πΎπΆππ3 π β 2 πΎπΆπ π + 3 π2 π
The oxygen produced was collected by
displacement of water at 22Β°C at a total
pressure of 754 torr. The volume of the
gas collected was 0.650 L, and the vapor
pressure of water at 22Β°C is 21 torr. β’ Calculate the partial pressure (in atm) of O2 in
the gas collected and the mass of KClO3 in the
sample that was decomposed.
All particles are in constant, random
motion
All collisions between particles are
perfectly elastic
The volume of the particles in a gas is
negligible
The average kinetic energy of the
molecules is in its Kelvin temperature
The KMT model neglects any
intermolecular forces as well β’ Gases expand to fill their containers
Solids/liquids do not
β’ Gases are compressible
Solids/liquids are not appreciably compressible
Assumptions of the KMT successfully
account for the observed behavior of an
ideal gas β’ Recall that there are five gas laws describing the
behavior of gases that were derived from
experimental observations
If the volume is decreased that means that the
gas particles will hit the wall more often Pressure is increased!
When a gas is heated,
the speed of its
particles increase
and thus, hit the walls
more often and with
more force Only way to keep pressure
constant is to INCREASE
the VOLUME of the
container!
When the temperature of a gas increases, the speeds of its particles increase
The particles are hitting the wall with greater force and greater frequency
Since the volume remains the same, this would result in INCREASED gas pressure
An increase in the number of particles at
the same temperature would cause the
pressure to increase if the volume were
held constant The only way to keep constant pressure is to vary the
volume!
The pressure exerted by a mixture of
gases is the SUM of the partial pressures β’ This is because gas particles are acting
independent of each other and the volumes of
the individual particles DO NOT matter
Root Mean Square Velocity, Effusion, and Diffusion
From the KMT, Kelvin temperature indicates the average kinetic energy of the gas particles
ππ
π= π π =
2
3(πΎπΈ)ππ£π
β’ 2/3 KE comes from the application of velocity, momentum, force, and pressure
when deriving an expression for pressure See Appendix 2 in your book for a complete mathematical explanation!
Thus,
(πΎπΈ)ππ£π = 3
2π π
ALL gases have the same average kinetic
energy at the same temperature!
This mathematical relationship is very
important because it shows that with
higher temperature comes greater
motion of the gas particles β’ HEAT βEM UP β SPEED βEM UP!
Look at the graph at right β’ How do the number
of gaseous molecules with a given velocity change with increasing temperature? By drawing a vertical
line from the peak of each bell curve to the x-axis, the AVERAGE velocity of the sample is derived
Average velocity of a specific gas molecule at a specific temperature is also root mean square velocity (ΞΌrms)
β’ Can be calculated using Maxwellβs equation:
π2 = ππππ =3π π
ππ
β’ Where: R is βenergy Rβ = 8.314 J/Kβmol
T = temperature in Kelvin
MM = molar mass of a single gas particle in KILOGRAMS per mole!
β’ ππππ has units of m/s!
This equation is important because it shows that molar mass is inversely proportional to velocity
β’ Massive particles move slowly
β’ Light particles move quickly
But remember - ALL gases have the same average kinetic energy at the same temperature!
Calculate the root mean square velocity
for the atoms in a sample of helium gas at
25Β°C.
If we could monitor the path of a single
molecule, it would be very erratic β’ The average distance a particle travels between
collisions is called the mean free path
Itβs on the order of a tenth of a micrometer β
waaaaayyyyy small!
Thomas Graham experimentally showed that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles β’ Stated in another way, the relative rates of effusion of two
gases at the same temperature and pressure are given by the inverse ratio of the square roots of the masses of the gas particles
π ππ‘π ππ ππππ’π πππ πππ πππ 1
π ππ‘π ππ ππππ’π πππ πππ πππ 2=
ππππ πππ πππ 1
ππππ πππ πππ 2=
3π ππ1
3π ππ2
=π2
π1
M1 and M2 = molar masses of the gases in g/mol
We have seen that the postulates of the KMT, when combined with appropriate physical principles, produce an equation that successfully fits the experimentally observed behavior of gases
There are 2 further tests of this model: β’ Effusion Describes the passage of a gas through a tiny orifice into an
evacuated chamber Rate of effusion measures the speed at which the gas is transferred into
the chamber
β’ Diffusion Describes the mixing of gases Rate of diffusion is the rate of the mixing!
Calculate the ratio of the effusion rates of
hydrogen gas (H2) and uranium
hexafluoride (UF6), a gas used in the
enrichment process to produce fuel for
nuclear reactors
Quite complicated to describe
theoretically because so many collisions
occur when gases mix
corrected pressure corrected volume
Pideal Videal
At high pressure (smaller volume) and low
temperature (attractive forces become
important) you must adjust for non-ideal gas
behavior using van der Waalβs equation.
2
( )obs
nP a x V nb nRT
V