We can use the ideal gas to link moles of a
substance to P,V,T and n.
Announcements
--Exam 2 Sept 5
6:00-7:30PM
Coverage Chapter 4-6. Please see blog for
skipped material.
Chapter 5 Problems
2, 8, 14, 17, 22, 5.31*, 5.41*, 5.43*, 44*, 48, 5.52*, 5.54,
58, 69*, 70*, 72*, (Principles of Chemistry)
2, 8, 11, 16, 19, 21, 24, 25, 27, 39, 41, 44, 46, 52,
56, 60*, 63, 69, 80,81 (4th Edition)
Quiz 5 Thursday Aug 23 or next Tuesday
P,V,T
of gas A
moles
of gas A
n = PV
RT
mole
of gas B
C6H12O6 (s) + 6O2 (g)
n = PV
RT
stoichiometric
factors from
balanced
equation
What is the volume of CO2 produced at 37˚C and 1.00
atm when 5.60 g of glucose is consumed in the
following reaction?
C6H12O6 (s) + 6O2 (g)
What is the volume of CO2 produced at 37˚C and 1.00
atm when 5.60 g of glucose is consumed in the
following reaction?
P,V,T
of gas B
6CO2 (g) + 6H2O (l)
At constant T and P, the volumes of reacting
gases can be expressed as ratios of simple
whole numbers....just like moles and molecules!
6CO2 (g) + 6H2O (l)
H2(g) + Cl2(g) ==> 2HCl(g)
g C6H12O6
mol C6H12O6
5.60 g C6H12O6 x
6 mol CO2
1 mol C6H12O6
x
= 0.187 mol CO2
180 g C6H12O6
1 mol C6H12O6
V=
nRT
=
P
mol CO2
V CO2
L•atm
x 310.15 K
mol•K
= 4.76 L
1.00 atm
0.187 mol x 0.0821
What is asked for?
2 mol Cl2
2 vol Cl2
2 L Cl2
1 mol H2
1 vol H2
1 L H2
2 mol HCl
2 vol HCl
2 L HCl
Gay-Lussac’s Law of Combining Volumes
What is asked for? gram KCl = ?
Translates nomenclature to a balanced equation
Translates words to balanced equations
2 NaN3(s) ! 2 Na(l) + 3 N2(g)
Determine moles of N2:
nN2 = 70 g N3 H
1 mol NaN3
3 mol N2
= 1.62 mol N2
X
65.01 g N3/mol N3 2 mol NaN3
Determine volume of N2:
nRT
(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)
V=
=
P
1.00 atm
(735 mm Hg)
760 mm Hg
= 41.1 L
2K(s) + Cl2(g)
P = 0.950 atm
V = 5.25 L
2KCl(s)
T = 293K
nCl2 = unknown
Determine moles of Cl2:
nCl = PV
2
RT
=
0.950 atm
0.0821
atm*L
mol*K
2 mol KCl
mol KCl formed = 0.207 mol Cl2
mol KCl formed = 17.0 g X
x 5.25L = 0.207 mol Cl2
x 293K
1 mol Cl2
1 mol K
39.10 g K
= 0.414 mol KCl formed
2 mol KCl
2 mol K
= 0.435 mol KCl formed
Cl2 is the limiting reactant.
g KCl formed = 0.414 mol KCl X
74.55 g KCl
mol KCl
= 30.9 g KCl
•
•
•
•
•
Big To Little Questions About Gases
From Big (Macro) To Little (Micro)
How do molecules make pressure in a container?
•
•
A change in pressure in one direction causes a change in
volume in the other (Boyle’s Law)?
How does Dalton’s Partial law arise microscopically or how
is pressure proportional to a gas mole fraction?
How does a temperature change give a proportional change
in volume of a gas or pressure of a gas? What is
temperature on a molecular scale?
Why does gas volume only depend on number of molecules
(moles) and not the size of them? Shouldn’t one mole of a
•
PV = nRT gives a macroscopic description of
the behavior of gases.
How do we relate P, V, n, T to atoms and
molecules that we know exist on a microscopic
level?
The math and physics that bridges the
macroscopic PV = nRT to the microscopic world
is called: Kinetic Molecular Theory
larger molecule occupy more volume (Avogadro’s Law)?
Suppose a rectangular box of length, l.
Pressure arises from change in
momentum resulting from collisions of
A.N. molecules
Kinetic Molecular Theory of Gases
Postulates:
1. Volume occupied by gas is negligible
compared to the volume of container.
P =
N m v2
nRT
=
3V
V
where N is the number of molecules
and v 2 avg velocity
2. Gas molecules exert neither attractive nor
N = n × NA .
repulsive forces on one another (i.e. there are
no intermolecular forces).
P =
NA . m v 2
RT
=
3V
V
solving f or v 2 in terms of T gives
3. Gas molecules are in constant random motion.
v �2 =
4. Collisions among molecules are perfectly
elastic, no kinetic energy is lost in collisions.
Results of kinetic molecular theory:
1. Pressure of an Ideal gas is due to the number and
frequency of gas collisions in a container.
N m v2
nRT
P =
=
3V
V
2. The average speed of gas molecule is proportional to
3RT proportional to molar
gas temperature and
inversely
v2 =
mN
A
mass.
�
�
2 3RT
2 3RT
v =
=
mNA
M
v =
�
2
3RT
m� NA
�
3RT
=
mNA
2
3RT
M
The average kinetic energy of a collection of any gas
molecules is determined by the absolute temperature of
the gas. Same temp ==> same KE!
KE = Ek =
3
1 mu2
RT =
2
2
R = 8.314 J/K . mol
urms =
"
3RT
M
Gas at the same
temperature have
the same kinetic
energy!
T = Kelvin
Average speed of a gas particle is
inversely proportional to molar
mass and proportional to T!
Lighter gases have faster speeds for a given temperature!
Lighter gases have faster speeds for a given temperature!
Molecular Origins of Gas Law
1. Origin of Pressure--arises from gas molecule moving at
high speeds colliding with the container walls. The more
molecules the more frequent the # of collisions.
2. Boyle’s Law (P∝1/V) Decreasing V increases collisions
between gas particles and walls of container.
3. Charles Law(V∝T) Higher T at constant n and P = faster
particles velocity and higher KE. V must increase to
reduce collisions of particles to keep P constant.
A molecular description of Boyle’s Law.
P ! 1/V
Constant T and n
If volume is decreased, then the frequency of collisions
in the container increases and pressure increases
(constant T).
4. Avogardro’s Law (V∝n) If n increases at constant P and T
the more collisions unless volume increases.
5. Dalton’s Law of Partial Pressure: Total P of a fixed
volume of gas depends only T and number of moles of
particles.
A molecular description of Charles’s Law.
V ! T
A molecular description of Avogadro’s Law.
V ! n
Constant P and n
Higher temperature increases KE of gas molecules which
would increase pressure---but its held constant
experimentally. The volume must increase.
A molecular description of Dalton’s law of partial
pressures.
Each gas acts independently of one another. The more
moles of the more collisions the higher the pressure.
Constant P and T
More added molecules would increase collisions and
therefore pressure---but its held constant
experimentally---so the volume must increase.
}
PV = nRT
Pobserved
correction
for
intermolecular
attraction
forces
}
The ideal gas law fails at moderate to high
pressures and low temperatures. Two reasons
why!
Van der Waals Equation
2
Ideal Gas Law
(V – nb) = nRT
(P + an
V2 )
Vcontainer
correction due to
actual volume of
molecules
1. At high pressure molecule volume becomes significant
portion of total volume. (the assumption that a gas
volume is negligible compared to container volume)
2. At moderate pressures, intermolecular forces between
gas particles alters elastic collisions reducing Pideal
The Van der Waals Equation is used to predict the
behavior of non-ideal gases at low T and high P.
(V – nb) = nRT
(P + an
V2 )
2
•
correction
for
intermolecular
attraction
forces
The van der waals equation accounts for
two types of deviations from ideality--intermolecular forces and molar volume.
n=
PV = 1.0
RT
}
}
Pobserved
A yardstick for how “ideal” a gas is called the
“compressibility factor”. For 1 mole of any gas:
Vcontainer
correction due to
actual volume of
molecules
Molar Volume
Repulsive Forces
Dominant
The van der Waals equation corrects
the ideal gas law for two things:
Attractive
Intermolecular Forces
Dominant
1. a accounts for intermolecular
forces of attraction
2. b accounts for volume of gas
molecules
At VERY HIGH PRESSURES molecular volume effects
dominates the van deer waals
Low Pressure
Ideal Gas Law Obeyed
Volume is of the container
Ideal Volume = Videal
High Pressure
Ideal gas law fails because
volume is actually much smaller
than the non-compressed “ideal
volume”.
Actual Volume = Vacutal < Videal
Ideal Gas Line
At MODERATE PRESSURE (lower than HIGH
PRESSURE) intermolecular forces of attraction
between molecules dominate reducing collision force
and pressure.
Low Pressure
Pideal = higher-no interactions
Moderate Pressure
Pacutal < Pideal because of
interactions, we add a factor to
account for it.