NAME ____________________________________ AP NOTES: UNIT 2 (1): GAS LAWS
This unit deals with the gas laws and the work is a focused application of Big Idea 2:
Big Idea 2:
Chemical and physical properties of materials can be explained by the structure & the
arrangement of the species (atoms, ions, or molecules) and the forces between them.
Topics will include work on the:
1) measurement of gases
2) Kinetic Molecular Theory (KMT)
3) simple gas laws
4) molar volume of gases and stoichiometry (Avogadro’s Law)
5) ideal gas law
6) Dalton’s Law of Partial Pressure
7) Grahams Law of Effusion and Diffusion
8) real gases and deviations from the ideal gas law
The AP curriculum focuses upon the student mastery regarding the ideas that the transformation
of matter can be observed in multiple ways that are generally categorized as either chemical or
physical change.
These categories can generally be distinguished through consideration of the electrostatic
(Coulombic) forces that are associated with a given change at the particulate level. The strength of
such forces falls along a continuum, with the strongest forces generally being chemical bonds.
Chemical changes involve the making and breaking of chemical bonds.
For physical changes, the forces being overcome are weaker intermolecular interactions, which
are also Coulombic in nature. The shapes of the particles involved, and the space between them,
are key factors in determining the nature of these physical changes. Using only these general
concepts of varying strengths of chemical bonds and weaker intermolecular interactions, many
properties of a wide range of chemical systems can be understood
There is a relationship between the macroscopic properties of solids, liquids, and gases, and the
structure of the constituent particles of those materials on the molecular and atomic scale. The
properties of solids, liquids, and gases also reflect the relative orderliness of the arrangement of
particles in those states, their relative freedom of motion, and the nature and strength of the
interactions between them.
For gases, volumetric relationships can be used to describe ideal behavior, and a conceptual
understanding of that behavior can be constructed based on the atomic model and a relatively
simple Kinetic Molecular Theory (KMT).
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Learning Objective: The student is able to use KMT and concepts of intermolecular forces to make
predictions about the macroscopic properties of gases, including both ideal and
non-ideal behaviors.
Learning Objective: The student is able to refine multiple representations of a sample of matter in the gas
phase to accurately represent the effect of changes in macroscopic properties on the
sample.
Learning Objective: The student can apply mathematical relationships or estimation to determine
macroscopic variables for ideal gases.
The gaseous state can be effectively modeled with a mathematical equation relating various
macroscopic properties. A gas has neither a definite volume nor a definite shape; because the effects
of attractive forces are minimal, we usually assume that the particles move independently.
1. Ideal gases exhibit specific mathematical relationships among the number of particles present, the
temperature, the pressure, and the volume.
2. In a mixture of ideal gases, the pressure exerted by each component (the partial pressure) is
independent of the other components. Therefore, the total pressure is the sum of the partial pressures.
3. Graphical representations of the relationships between P, V, and T are useful to describe gas behavior.
4. Kinetic molecular theory combined with a qualitative use of the Maxwell- Boltzmann distribution
provides a robust model for qualitative explanations of these mathematical relationships.
5. Some real gases exhibit ideal or near-ideal behavior under typical laboratory conditions. Laboratory
data can be used to generate or investigate the relationships in 2.A.2.a and to estimate absolute zero
on the Celsius scale.
6. All real gases are observed to deviate from ideal behavior, particularly under conditions that are close
to those resulting in condensation. Except at extremely high pressures that are not typically seen in
the laboratory, deviations from ideal behavior are the result of intermolecular attractions among gas
molecules.
7. These forces are strongly distance-dependent, so they are most significant during collisions.
8. Observed deviations from ideal gas behavior can be explained through an understanding of the
structure of atoms and molecules and their intermolecular interactions.
9. The presence of intermolecular forces among gaseous particles, including noble gases, leads to
deviations from ideal behavior, and it can lead to condensation at sufficiently low temperatures and/or
sufficiently high pressures.
10. Graphs of the pressure-volume relationship for real gases can demonstrate the deviation from ideal
behavior; these deviation can be interpreted in terms of the presence and strengths of intermolecular
forces.
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I) Summary of Gas Laws: The gas laws are a varied grouping of mathematical expressions which can help to
Gas Law
Simple Gas Laws
Combined Gas Law
Boyle’s Law
Basic Tenet(s)
inter-relates changes in pressure, Kelvin temperature and volume
assumes constant temperature, and predicts changes in volume as a function of changes in pressure
(an inverse relationship)
assumes constant pressure, and predicts changes in volume as a function of changes in Kelvin
temperature (a direct relationship) The conversion for Kelvin is 273.15
Charles’ Law
assumes a constant volume (e.g. a sealed canister) and predicts changes in pressure as a function of
changes in Kelvin temperature (a direct relationship).
Gay-Lussac’s Law
(Amonton’s Law)
The central gas law … Very important … It accounts for changes in pressure, Kelvin temperature,
volume, and moles of gas species, of an ideal gas.
Ideal Gas Law
draws upon the relationship between the number of gas species and the volume of gas
Avogadro’s Law
draws upon the relationship between 22.4[2] L of volume at STP and 1 mol of gas
Molar Volume
Dalton’s Law of
Partial Pressure
states that the total pressure of a gas mixture is due to the sum of the partial pressures of each
component gas, and is proportional to the number of moles. The partial pressure of each
component gas is independent of the other partial pressures.
Graham’s Law of
Diffusion & Effusion
states the relative relationship between the molecular mass of a gas, and its rate of diffusion and
effusion
Henry’s Law
assumes constant temperature and states that the amount of gas that can dissolve into a fluid is
directly proportional to the partial pressure of the gas in equilibrium with the fluid.
predict &/or explain the behavior of species existing in the gaseous (vapor) state of matter … which is that
state exhibiting the greatest possible entropy, as a fluid, at a given pressure and temperature.
Gases are the simplest state of matter, and so the connections between the properties of individual
molecules and those of bulk mater are relatively easy to identify. (Atkins 133)
A) The theoretical basis of these laws is the Kinetic Molecular Theory (KMT).
1) The behavior of gases at the molecular level is responsible for their macroscopic properties.
a) *The species that make up a gaseous sample of matter are in constant random motion,
moving with an average kinetic energy that is proportional to the temperature of the
gas.
Assignment: Prep-Reading: B & L: p. 400 - 402
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2) The gas laws are seriously important aspects touching upon topics from organismal
respiratory processes of vertebrates and invertebrates, to the composition of the Earth’s
atmosphere, to rocket propulsion, the internal combustion engine, meteorology, ocean health,
& aerosol spray cans.
a) Where kinetics describes the rates of change, this work focuses upon one state of
matter that experiences these changes. In our next unit, we will go back to our
study of processes, (equilibrium), in light of both kinetics and matter. But for now,
onto the gases ….
b) In spite of their different chemical properties, gases behave quite similarly as far as
physical properties are concerned
c) As a point of order, the “gas phase” of a substance can be called a vapor
d) a sample of gas expands spontaneously to fill its container
e) a sample of gas is compressible (the gas is mostly *empty space between the
molecules of a gas)
f) Since most of the volume of a gas sample is indeed “empty space”, each molecule
behaves (essentially) as though the other molecules were not present. (This of course
is not the case, when the intermolecular forces of attraction come into play …but more
on this phenomenon later).
i) However, the point is that due to this idea of “lone” existence, each of the
gas samples of chemically different gases, tend to behave similarly in terms
of physical behavior.
II) Pressure: The force exerted per unit area by gas molecules as the strike the surfaces they encounter (such
as the interior surface of a container, in which the gas is stored)
Essentially, pressure is the sum of the constant collisions between the molecules of a gas and the
surrounding surfaces.
Variations in pressure; allow us to breath, allow us to use drinking straws, cause the winds of the
Earth, lead to weather predictions etc ….
It is as the point of collision (as in: with the side of
the container) that pressure is exerted. That
pressure is dependent upon a number of factors.
When volume and mols are constant, then
changes in pressure are directly related to changes
in the absolute temperature.
Imagine a single gas
molecule in a closed
canister.
As that single molecule moves
it may collide with the side of
the canister and create point of pressure
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Given the causation of gas pressure, it is not a huge leap to
surmise that gas pressure depends on 3 things: The mass of,
the number of and velocity of … the gas molecules.
Yeah, Read on, Macduff
From P.W. Atkins (p141) The pressure of a gas arises
from the impact of its molecules on the walls of the
container. (b) When the volume of the sample is
decreased, there are more molecules in a given
volume and so there are more collisions with the same
area of the wall in a given time interval. Because the
impact on the walls is now greater, so is the pressure.
A) Pressure = Force
Area
where
Force = (mass)(acceleration)
1) The total pressure exerted by a gas depends on several factors including; mass,
temperature (linking to velocity), and the concentration (e.g. moles of gas species per unit
volume).
a) Question: Given the above equation relationships complete the following:
i) The greater the concentration, the greater / lesser the gas pressure
ii) The greater the temperature, the greater / lesser the gas pressure
2) For those with a broader background in physics, Dr. P.W. Atkins provides a really nice
graphic (p. 135) … where g =local gravitational field (free fall acceleration), d = density,
and h = height
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3) Units: The SI unit for pressure = Pascal (Pa)
1 atmosphere = 14.7 psi
1 mmHg = 1 torr
1 atmosphere = 760 mmHg
1 Pascal = 1 N/m2
1 atmosphere = 101.325 Pa or 101.325 kPa
4) Measurement of pressure: The Barometer and the Manometer
a) Extremely small gas molecules (e.g. H2(g) and He(g) can escape the gravitational field
of the Earth (although they cannot escape the Jovian field …which is some 2.64 times
greater than Earth’s … )
While most of the gas molecules on Earth bow to the extant gravitational field, each
experiences some level of gravitational acceleration. The resulting kinetic energies
overwhelm the gravitational forces, keeping the gases, well above the ground, and
we still have a gaseous atmosphere … Thus leading us to a very clear conclusion….
WE LIVE AT THE BOTTOM OF AN OCEAN … OF GAS
b) Evangelista Torricelli is credited with inventing the barometer. He used (liquid)
mercury in a tube, evacuated of air, and sat that tube upside down in a pool of even
more mercury.
Standard Atmospheric Pressure =
101.3 kPa
1 atm
760. mmHg
760 torr
c) The Manometer: Similar to a barometer, a manometer
is a U-shaped tube filled with mercury. One end is
open to the atmosphere, and the other end is attached to
a flask containing the gas, whose pressure is to be
measured.
Brown, Lemay et al. 13th ed. p 402
149
When the gas pressure is exactly equal to atmospheric
pressure, the mercury levels *on both sides of the tube
are the same.
When the atmospheric pressure is superior
to (greater than) the test gas pressure, the
level of mercury on the left side of the Ushaped tube will be * lower than the level
of mercury, on the right hand side of the tube.
In the diagram from Tro, is the pressure of the
test gas, greater than or less than atmospheric
pressure? *greater than
Tro p.
198
Hence the manometer measures a relative pressure, with the mercury column’s height being the
difference between the known atmospheric pressure and that of the test gas. What do we use to
measure the atmospheric pressure at the time of testing? * a barometer
II) Kinetic Molecular Theory for an Ideal Gas
A) Ideal gases do NOT exist. But we do study how closely real gases approach ideality...or rather,
how closely the behavior of real gases obey the Kinetic Molecular Theory. The concept of ideal
gases is a mathematical paradigm, complete with assumptions, designed to predict and/or explain
the physical behavior of an imaginary gas, which obeys KMT. We then compare this ideal
behavior to that of a real gas.
As you will see, some of these assumptions are pretty “bizzaro”… like, assuming no gas molecule
has the ability to cause / exert intermolecular forces of attraction. With this (false) assumption we
can conclude that no ideal gas molecule can affect another, by its presence, or proximity ….and we
will see how chemists deal with such deviations, later in the unit … bizzaro (?) (…go read a comic book….)
150
B) Basic Tenets & Assumptions of the Kinetic Molecular Theory
1) A gas consists of a collection of molecules in continuous, random motion. The average
kinetic energy of the entire gas sample is dependent only upon the absolute (Kelvin)
temperature and NOT upon the identity of the gas.
2) Gas molecules are infinitesimally small points. The individual molecular volume is virtually
nothing compared to the total volume occupied by the gas.
3) Gas molecules move in straight lines, until they collide with each other, or the wall of the
container. Collisions are perfectly elastic and no energy is lost upon collision.
4) The molecules do not influence each other, except during collisions …This means that
it is part of the model to suggest that there are no attractive forces between ideal gas
molecules, and no repulsive forces, between them, except during collisions.
★★Real gases, that are nonpolar & very small, best mimic ideal gas behavior, especially under
low pressure and high temperature. Small, nonpolar molecular gases mimic the behavior, in
large part because they have so few electrons, which in turn create very weak intermolecular
forces of attraction (IMF). Additionally, real gases best approximate ideal gas behavior when the
pressure is low and the temperature high, because the molecules are allowed to expand as far
away from each other as is maximally possible at these condition. This meshes nicely with
tenet 4 of the Kinetic Molecular Theory.
The Trivedi team does a solid job summarizing these issues in section 5.23. Postulates 5 and 6 of the
flash drive introduce Et … Take a look.
Check Out: An interesting website, with lots of information about thrust, the ideal gas law,
the simple gas laws http://www.chemistryland.com/CHM151S/05-Gases/Gases/PuttingGasesToWork151.htm#
151
III) The Simple Gas Laws: Combined Gas Law, Boyle’s Law, Charles’ Law, and Avogadro’s Law
These work quite nicely when you wish to compare the P, V and/or T of a sample of gas to ITSELF,
under new conditions of P,V and/or T.
No variable is held constant
PorigVorig = Pnew Vnew
TK orig
TK new
Compares the P, V, TK of a
single sample of gas to new
conditions, for the same sample
COMBINED GAS LAW
BOYLE’S LAW
(temperature is constant)
scuba, breathing, ears
popping,
the bends
Linked closely with
Henry’s Law
CHARLES’S LAW
(pressure is constant)
hot air ballooning,
GAY-LUSSAC’S LAW
(Amonton’s Law)
(volume is constant)
bic lighters, exploding aerosol
spray cans
A) COMBINED GAS LAW:
1) EQUATION: (Pressure1) (Volume1) = (Pressure2) (Volume2)
TempKelvin1
TempKelvin2
assume a closed system
unless noted otherwise
a) Notice that each variable changes … no variable is held constant.
2) The equation compares the properties of a gaseous system in its "original state" to the
its properties after changes to P,V, and T have been made.
a) Pressure may be in atmospheres (atm), kilopascals (kPa) or even in torr and mmHg
b) Volume may be in mL or in L
c) The temperature for all gas laws * must be in Kelvin
You may need to convert from Celsius.
152
3) e.g1) A 10.0 mL sample of He(g) at 0.750 atm and 27.0 °C is heated to 82.0°C with a
reduction in pressure to 0.500 atm. What is the new volume of the gas?
E:
S:
A: *17.8 mL
e.g2) A 40.0 mL sample of a gas is at 546 K and has a pressure of 607.8 kPa. Calculate
the new volume of the gas if the temperature were changed to 373 K and the pressure
were changed to 302.9 kPa
E:
S:
A: *54.8 mL
e.g.3) A 35 mL sample of a gas originally at 100. K and 0.500 atm., was changed to STP
conditions. Calculate the new volume, in liters.
A: *0.048 L
Assignment: Brown and LeMay: p. 404 -406. Read and notate (on the next page of this note packet) the
important issues regarding Boyle’s Law and Charles’ Law. I will point out a few issues, tomorrow,
regarding the laws and ask you about Gay-Lussac’s Law (or Amonton’s Law) …based upon your grasp
of these two other gas laws. Include the parameters of the law, an equation, a graphical relationship and any
other important factors
153
Ideas about Boyle’s Law:
Ideas about Charles’ Law:
Ideas about Gay-Lussac’s Law (Amonton’s Law): Pressure varies in direct proportion with changes in the Kelvin
temperature, at constant volume.
154
BASIC PRACTICE: THE SIMPLE GAS LAWS
_____1 A 20. mL sample of a gas is at 546 K and has a pressure of 607.8 kPa As the temperature is changed to
273.15.15 K and the pressure to 202.6 kPa, the new volume of the gas will be: (THINK: Anything constant ?)
if no
if yes
use Combined GL
1)
2)
3)
4)
Alter the Combined GL
3.3 mL
13 mL
30. mL
120 mL
_____2 A 35.0 mL sample of a gas is at STP. If the temperature is changed to 300. K and the pressure to
3.00 atm, the new volume of the gas will be:
(THINK: Anything constant ?)
if no
if yes
use Combined GL
1)
2)
3)
4)
Alter the Combined GL
25.0 mL
12.8 mL
32.5 mL
9.05 mL
_____3 A 40.0 L sample of a gas is at 400.K and has a pressure of 106.3.kPa. If the conditions were changed to
STP, the new volume of the gas would be :
(THINK: Anything constant ?)
if no
use Combined GL
1)
2)
3)
4)
if yes
Alter the Combined GL
33.4 L
17.8 L
8.20L
28.6 L
_____4 600. mL of a gas is at STP. What is the gas volume when the temperature is raised to 100.0 ºC at
constant pressure ?
(Anything constant ?) Hey, that temp. is in Celsius !!
if no
use CGL
1)
2)
3)
4)
if yes
Alter CGL
220. mL
0.00589 mL
820. mL
137 mL
____5 At 66.6 kPa the volume of a gas is 20.0 mL. As temperature is held constant, what will be the volume of
the gas at 80.0 kPa? (How do you know which equation to use? Are there any key words?)
1)
2)
3)
4)
16.7 mL
24.0 mL
150. mL
12.9 mL
155
_____6 The volume of a sample of He is 40.0 mL at 15.0 ˚C and 100. kPa. What will be the volume at STP?
(What are the key words? Do you need to do anything with the temp? )
1) 41.8 mL
2) 37.4 mL
3) 8.02 mL
4) 14.9 mL
_____7 At constant pressure, 88 liters of He(g) at 27ºC are compressed to 24 liters. What is the new
temperature? (What do you need to do to the temperature in this problem?)
1)
2)
3)
4)
7.4 ºC
78 K
82 K
5.0 ºC
_____8 Given 2.0 L of a sample of H2 at 0.263 atm. and assuming constant temperature, what volume will the
gas occupy at 0.526 atm? (Any key words that help you?)
1)
2)
3)
4)
1.0 L
2.0 L
3.0 L
4.0 L
____9 A student threw an aerosol spray can into a bon fire. The volume of the can (and therefore the gas in it)
equals 750 mL. The original pressure was 1.70 atm. The original temperature of the can (before being
tosse4) was 18ºC. Calculate the pressure of the gas as it is heated in the fire to a temperature of 500. ºC.
1)
2)
3)
4)
4.52 atm
0.912 atm
2.8 atm
3.77 atm
(By the way, the pressure generated inside the can is equal, approximately, to the experienced external pressure when plunged 150 feet below the
surface of the ocean)
____10 The volume of 50.0 mL of a gas at STP increases to 100. mL. If the pressure remains constant, the new
temperature must be
( Key words ...?)
1) 0 K
2) 100.30 K
3) 273.15 K
4) 546.30 K
_____11 A gas sample is at 10.00 ºC. As pressure remains constant, the volume will increase when the
temperature is changed to:
1) 263.15 K
2) 273.15 K
3) 283.15 K
4) 293.15 K
156
_____12 A 100. mL sample of He gas is enclosed in a cylinder under a pressure of 1.00 atm. What volume
would the gas occupy at a pressure of 2.00 atm, assuming temperature remains constant ?
1) 50.0 mL
2) 100. mL
3) 200. mL
4) 380. mL
_____13 Assume 650. mL of oxygen at 172 ºC and 1.50 atm are cooled to a temperature 0.00 ˚C and a volume
of 325 mL what will be the resulting pressure of gas ?
1) 2.15 atm
2) 0.347 atm
3) 1.84 atm
4) 159 atm
_____14 Use the following chart of data to complete the question. The chart below shows the changes in the
volume of a gas as the Kelvin temperature changes at constant pressure. It may help to "picture" what is
going on, by using the diagram.
constant pressure
Volume (mL)
moveable piston
gas
1.
2.
3.
4.
Temperature (K)
600
1200
300
150
100
200
50
25
Which equation best expresses the relationship between the volume of the gas and the Kelvin temperature ?
1) VT = 6 mL•K
2) V/T = 6 mL/K
3) T/V = 1/6 mL
4) VT = 1/6 mL•K
_____15 Which of the following would be the most approximate a graph of the data in question 14, and the gas law being
propounded, using volume as a function of temperature at constant pressure?
1
2
3
4
157
_____16 At constant temperature, which line best shows the relationship between the volume of an ideal
gas and its pressure assuming constant temperature?
D
C
A
B
_____17) The intermolecular attractive forces between particles are weakest when their kinetic energies are
1) high and the distance between particles is large
2) high and the distance between particles is small
3) low and the distance between particles is large
4) low and the distance between particles is small
_____18 If you wanted to get a real gas to behave like an ideal gas, you would have to get the particles as
far away from each other as possible (That is, you need to weaken the IMF). Under what conditions
would a real gas behave most like an ideal gas ?
1)
2)
3)
4)
at high temperatures
at high temperatures
at low temperatures
at low temperatures
and
and
and
and
low pressures
high pressures
low pressures
high pressures
_____19 Under the same conditions of temperature and pressure, which of the following REAL gases would
behave most like an ideal gas, predicated upon the concepts of the Kinetic-Molecular Theory ?
1) NH3(g)
2) He(g)
3) Cl2(g)
4) Ne(g)
Answers
1. 3 Key word(s) : There are none.... Each variable is being changed and THIS IS IN ITSELF YOUR CLUE that this is a combined gas law
problem….. Be sure you convert any Celsius temperature to kelvin : K = 273.15 + ˚C
P1V1 = P2V2
T1Kelvin T2Kelvin
2. 2 Combined Gas Law : See #1
3. 4 Combined Gas Law : See # 1
158
4. 3 Key word(s) : at constant pressure ..... This means that using the combined gas law, you may omit the pressure variables
becomes V1 =
V2
.
So, P1V1 = P2V2
T1Kelvin T2Kelvin
T1Kelvin
T2Kelvin
5. 1 Key word(s) : temperature is held constant ...This means that using the combined gas law, you may omit the temperature variables
becomes P1V1 = P2V2
.
So, P1V1 = P2V2
T1Kelvin
T2Kelvin
6. 2 Key word(s) : well thar ain't none....nothing is held constant. Use the combined gas law:
P1V1 = P2V2
Remember to change the ˚C to K by adding 273.15, then substitute into the equation.
T1Kelvin T2Kelvin
7. 3 Key word(s) : At constant pressure ..... This means that using the combined gas law, you may
becomes V1 = V2
omit the pressure variables . So, P1V1 = P2V2
T1Kelvin T2Kelvin
T1Kelvin T2Kelvin
8. 1 Key word(s) Assuming constant temperature .This means that using the combined gas law, you may
becomes P1V1 = P2V2
omit the temperature variables . So, P1V1 = P2V2
T1Kelvin
T2Kelvin
9. 1 10. 4
Key word(s) : At constant pressure ..... This means that using the combined gas law, you may
becomes V1 =
V2
omit the pressure variables . So, P1V1 = P2V2
T1Kelvin T2Kelvin
T1Kelvin T2Kelvin
11. 4 Key word(s) : At constant pressure ..... This means that using the combined gas law, you may
omit the pressure variables . So, P1V1 = P2V2
T1Kelvin
T2Kelvin
becomes
V1 =
V2
T1Kelvin T2Kelvin
But, you really don't need to do any math... Any temp. over 10 ˚C ( or 283 K) will cause an increase in volume of the gas.
12. 1 Key word(s) Assuming constant temperature .This means that using the combined gas law, you
becomes P1V1 = P2V2
may omit the temperature variables . So, P1V1 = P2V2
T1Kelvin
T2Kelvin
13. 3
14. 2
15. 3
16. curve B
17. 1
18. 1 this would move the species as far away from each other as possible, and thus limit the effects of intermolecular forces of attraction …
bringing the sample more in line with the “ideality” of the Kinetic Molecular Theory
19. 2 it is the smallest of the molecules … hence, the fewest electrons … You may infer then that the fewness of electrons means that the
intermolecular forces of attraction are incredibly weak… This brings He just about as close to an ideal gas as is possible (except for H2)
Now: Use the B & L text: Most of the answers are on pages A-11 and A-12 … Then take a gander at the
reading and questions on the next page of this note packet: 10.5, 10.19, 10.23, 10.25, 10.26 (answer = a)2.69
L b) 7.80L), 10.28 (answer = 2.4L NH3)
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III) The Simple Gas Laws (Continued …)
B) Avogadro’s Law: (Avogadro’s Principle): This is an extension of Avogadro’s Hypothesis.
1) Avogadro’s Hypothesis states: When two contained gas samples are at the same temperature,
pressure and volume, they also have the same number of
molecules.
2) Avogadro’s Law states: The volume of a gas maintained at constant temperature and
pressure is directly proportional to the number of moles of gas.
….volume depends on # of moles at constant P and T.
V = (constant)(n) where n = mols …OR …
V = constant
n
a) This leads us to an equation that looks a fair bit like Boye’s Law, but with volume
and moles: Vinitial/ninitial = Vfinal/nfinal
ii) for instance, to maintain the ratio, it must be true that doubling the number of
moles of gas, causes the volume to double, assuming a constant temperature
and pressure.
b) Essentially this is from where we get: 1 mol of an ideal gas at STP = 22.4 L
Review: 1.00 mol = 22.4 L = 6.02 x 1023 molecules
or 0.250 mol = 5.60 L = 1.505 x 1023 molecules
or 0.0400 mol = 0.896 L = 2.408 x 1022 molecules
3) Avogadro’s Law is also associated with determining the density of 1 mol of a gas at STP
e.g.) Calculate the density of 1 mol of helium at STP:
Molar Density = Mole Mass
22.4L
ans: 0.1787 g/L
e.g) What is the density of xenon gas at STP?
a) 2.75 g/L
b) 4.10 g/L
c) 5.86 g/L
d) 7.22 g/L
ans: c
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4) It is important to note for those going on (although, not for the AP exam), that the
conversion factor of 1 mol = 22.4[2] L at STP is for gases approaching ideal gas behavior.
Not all real gases obey this exact relationship …But(!) They come pert’darn close!
Gas
Molar Volume (L/mol)
Ideal Gas at STP
22.4[2]
Argon
22.397
Nitrogen (N2)
22.402
Oxygen (O2)
22.397
Hydrogen (H2)
22.433
Helium
22.434
Carbon Dioxide
22.260
Ammonia
22.079
The van der Waals equation (which is a correction factor for the ideal gas law), would
be used to adjust experimental results. The van der Waals equation uses constants
(a & b) which are unique to each gas. The constants can be used to find approximate
radii, and the effects of intermolecular forces of attraction. It is NOT on the AP exam …
but you may see it in college.
The van der Waals equation is often looked at as a “mixed bag of helpful” … It corrects
for the intermolecular forces and volume issues of gases, but the constants a and b are
peculiar to a gas, and thus make the use of the equation a bit “clunky”, as the generality
of the ideal gas law is lost.
P + a n2 (V – nb) = nRT
V2
a is about the effect of the attractions between molecules, and b represents the role of
of repulsions, (e.g. the volume of an individual molecule) TNAtkins in lists some constants (p161).
For more, see your notes on Deviations of Real Gases …
Assignment: Attack Sample Exercise 10.3 on page 407 of the B & L text. Work at envisioning a piston …
and the changes the gas there within undergoes. “Seeing the chemistry” is quite important at
this stage of the game.
Then move onto the next page and the TRY THIS and/or the Optional Assignment
Optional Assignment: Go to section 5.13 of the Trivedi Flash … This is a generalized review of gas laws,
but works its way to Avogadro’s Law. It is important to note that Avogadro Law
problems, can, in all reality be solved using the Ideal Gas Law.
161
TRY THIS … a bit more advanced/conceptual than the basic practice
1) Three identical 5 Liter flasks are filled with F(g), Cl2(g), and Ar(g), respectively, at STP. Which statement
is TRUE?
1)
2)
3)
4)
The densities of all the gases are the same.
The velocities of the gas species are the same in each of the three flasks
The same number of gas species is present in each flask
The average kinetic energy of Cl2 molecules is greater than that of either F2 molecules or Ar atoms
2) Imagine that an ideal gas is heated in a steel container from 25°C to 100°C. Which quantity will remain
UNCHANGED?
1)
2)
3)
4)
average kinetic energy
density
collision frequency
pressure
3) Select the correct statement about ideal gases.
1) At constant temperature, the pressure is directly proportional to the volume for a given amount of gas
2) At constant pressure and temperature, the volume is inversely proportional to the number of moles of
gas.
3) Deviations from ideal gas behavior are observed at low pressures and high temperatures
4) At STP, one mole of any gas occupies a volume of 22.4 L
4) A 6.0 L sample at 25 °C and 2.00 atm of pressure contains 0.5 moles of a gas. If an additional 0.25 moles of
gas at the same pressure and temperature were added, what would be the final total volume of the gas?
Hint 1: * Note that the question is asking
about moles and volume at constant temp
and pressure ….Does this bring any gas
law concept to mind?
Hint 2:* Think Avogadro’s Law …using
Vinitial/ninitial = Vfinal/nfinal
Hint 3: * The final moles is the only
tricky idea …. the initial #mole is 0.50
mol. An additional 0.25 mol were
added, for a total new or final number of
moles of 0.75 moles
Ans: 1) 3 2) 2 3) 4
4) 9.0 Liters
162
5) Consider an ideal gas trapped in a cylinder with a moveable piston. Assume the pressure is 1 atm. As the
pressure is tripled, the volume of the gas is
1) reduced to 1/3 the original volume, and the density triples
2) tripled and the density is reduced to 1/3 of the original density
3) tripled and the density is tripled
4) reduced to 1/3 the original, and the density is reduced to 1/3 the original density
6) Which statement about the kinetic molecular theory of gases is TRUE?
1) It does NOT give a satisfactory explanation of Boyles’ Law
2) It explains why small molecules exert a smaller pressure than do the same number of larger
molecules at the same temperature and volume.
3) It explains why gas molecules consume a finite portion of the container in which they are contained.
4) It assumes that small and large molecules have the same average kinetic energy at the same
temperature.
7) The density of a gas is 0.08987 g/L at STP, what is the molar mass of the gas?
1)
2)
3)
4)
0.08987 grams
2.01 grams
249 grams
89.87 grams
Ans: 5) 1 6) 4
7) 2
Just a Quick Cut To The Chase:
There are so many variations at hand, re the Gas Laws that I could go on writing and writing … but I believe if
we were to keep in mind 4 to 5 “facts” so much of the gas law work becomes, finding a way … thinking and
analyzing, as opposed to memorizing.
1) Macroscopic behavior of gases is affected by pressure, temp and volume.
2) Assuming all other factors are “equal”, pressure is in large part due to # of moles
3) Molecular mass can affect pressure and velocity
4) Average Kinetic Energy is proportional to the Kelvin Temperature
5) Ideal gas behavior ≠ Real Gas behavior
163
IV) The Ideal Gas Law (Universal Gas Law)
A) PV =nRT
(found on your reference charts)
P = the pressure of the gas (atmospheres or atm)
V = the volume of the gas (Liters or L)
n = the number of moles of gas
T = the absolute temperature of the gas (Kelvin or K)
R = the gas constant: 0.082057 L∙atm ∙ mol-1 ∙ K-1
1) Pressure Conversion Factors: 1 atmosphere = 760 torr = 760 mmHg = 101.3 kPa
When pressure is given in, torr, mm Hg, or kPa consider converting to atmospheres.
2) About the Universal Gas Constant: You can think of R as a converter that changes the units
on the right side of the above equation to the units on the left side of the "=" sign.
The values 0.08206 L∙atm &
mol∙K
8.314 J get the most use.
mol∙K
I really like sticking with 0.08206 L∙atm∙mol-1∙K-1 & converting to the necessary units.
a) However, I want to point out that 1 Joule = 1 kg•m2/s2 …. it is related to a force!
This allows us to make some neat substitutions … and you will (I think) need to know
that 8.314 J/mol•K can be re-written as 8.314 kg•m2/s2/mol•K
B) The ideal gas law is essentially drawn from the relationships between the volume of gases as a
function of pressure (Boyle’s Law), absolute temperature (Charles’ Law) and number of moles
(Avogadro’s Law)
V = k/P,
V = bT,
V = an
By combining the constants k, b, a, the universal gas law constant is derived.
V = (k)(b)(a)
Tn
P
thus
V = nRT
P
or
PV = nRT
1) Limitations of the Ideal Gas law
a) It predicts and explains phenomena quite well at low pressures and high temperatures
but not at high pressure and low temperature ….
b) Most gases do not behave ideally at pressures above 1 atm
c) Does not work well at the conditions of condensation for
gases (high pressure and low temperature)
2) In spite of the limitations, the Ideal Gas Law equation can be manipulated to calculate all
sorts of information… In fact, when we apply it to mole theory, we can determine the
number of molecules in a gas sample, the density of a gas, or the molecular mass [hence
molar mass] of a gas. This is precisely what Stanislao Cannizzaro did.
164
In the 1860s, Cannizzaro refined the process of determining relative atomic weights by using
Avogadro’s Hypothesis….that at the same temperature, pressure, and volume, different gas samples
have the same number of molecules….
If gas “X” and gas “Y” were heated to the same temperature, placed in equal volume containers under
the same pressure (gas would be released from one container until the two containers had the same
pressure), Avogadro’s conditions would be present.
As a result, Cannizzaro could conclude that the two containers had exactly the same number of
molecules.
The mass of each gas was then determined with a balance (subtracting the masses of the containers),
and the relationship between the masses would be the same as the mass relationship of one molecule of
“X” to one molecule of “Y”. That is, if the total mass of gas “X” was 10.0 grams and the total mass of
gas “Y” was 40 grams, then Cannizzaro knew that one molecule of Y must have four times as much
mass as one molecule of X .
If an arbitrary value such as 1.0 dalton were assigned as the mass of gas X , then the mass of gas Y on
that same scale would be 4.0 daltons! From: http://www.ck12.org/book/CK-12-Chemistry---Second-Edition/r13/section/15.5/
Dang! Stan’s the Man! That’s pretty cool thinking. Read that again … the guy is friggin’ brilliant! It
really was Stansilao’s recognition of Avogadro’s genius that put Amedeo back on the map of the
chemical world, as it were. So, Cannizzaro’s work begins on the road to the modern day understanding
of a common number of species in what we call a molar mass (M).
Some of this theory is expressed in our work with gases, and on the next few pages, I have tried to
provide you with a smattering of Ideal Gas Law problems, in which you are asked to determine, moles,
number of molecules, temperatures, etc… There is nothing too extravagant here …. However, you will
be required to use the Ideal Gas Law equation, and use unit cancellation to make a number of
conversions (or basic conversion equations), in terms of atmosphere, temperature, mass ….
165
PRACTICE:
1) The dinitrogen gas in an automobile air bag, with a volume of 65.0 Liters, exerts 1.09 atm of
pressure at 25.0 °C. Calculate the number of molecules of N2 in the air bag.
Step 1: Take stock of what you are given and what you need.
Make any necessary conversions.
Given:
R = 0.08206 L∙ atm ∙ mol-1 ∙ K-1 (…from reference tables)
V = 65.0 L
( …Liters works for the constant)
P = 1.09 atm
( …atm works for the constant)
n= ?
(…this is mol but you need this value to figure out the # of molecules)
T = 25.0°C
(screech to a halt .... you need the absolute temperature … convert!)
Step 2: Plug and chug:
Equation: PV = nRT
solving for n gives: * n = PV = * (1.09 atm)(65.0 L)
RT (0.08206 L•atm/ K•mol)(298K)
* n = 2.90 mol of dinitrogen gas
Step 3: use unit cancellation to convert moles to molecules
*molecules = 2.90 mol | 6.02 x 1023 molecules | = 1.75 x 1024 molecules of N2
1 mol
166
2) The O2(g) in an oxygen tent has a volume of 5.56 Liters at 21.0°C & a pressure of 1,140 torr.
a) How many kilograms of oxygen gas are in the oxygen tent?
b) How many molecules of oxygen gas (dioxygen) are in the oxygen tent?
c) Assume the average resting person inhales approximately 288,887 dioxygen
molecules with each breath. How many “breaths” of dioxygen are technically
available in the oxygen tent?
Step 1: Take stock of what you are given and what you need. Make any necessary conversions.
Given:
R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
V = 5.56 L
(… is Liters is okay?)
P = 1,140 torr
(…is torr okay?)
n= ?
T = 21.0°C
(…is °C okay?) Do you have all of the correct units required by the Universal Gas
Constant? … Do you need to make any conversions …or are you good to go?
Step 2: Plug and chug to find moles (n):
PV = nRT
rearranged becomes
* n = PV =
RT
* (1.50 atm) (5.56 L)
(0.08206 L•atm/ K•mol)(294K)
* n = 0.346 mol of dioxygen gas.
Step 3:
a) *kilograms = 0.346 mol | 32 grams O2 | 1 kilogram | = 0.0111 kg of O2
1 mol O2
1,000 grams
Ans: 0.0111 kg
b)*molecules of O2 = 0.346 mol O2| 6.02 x 1023 molecules | = 2.08 x 1023 molecules
1 mol
Ans: 2.08 x 1023 molecules
c)*breaths
= 2.08 x 1023molecules | 1 breath
| = 7.20 x 1027 breaths
288,887 molecules
Ans: 7.20 x 1017 breaths
167
3) 2.00 moles of methane gas are placed in a rigid 5.00 Liter container and heated to 100.°C. What
pressure, in atmospheres, will be exerted by the methane?
Given:
*R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
V = 5.00 L
P=?
n = 2.00 mol
T = 100°C
*P = nRT/V
P = (2.00 mol) 0.08206 L∙atm ∙ mol-1 ∙ K-1)(373 K)
5.00 L
Ans: 12.2 atm
a) What is the pressure in kPa?
* kPa = 12.2 atm | 101.3 kPa
1 atm
| = 1,235.86 kPa
Ans: = 1240 kPa
b) Assuming all the same conditions, estimate the pressure, in atmospheres, exerted
by 6.00 mol of methane.
*P = nRT/V
*P = (6.00 mol) 0.082057 L∙atm ∙ mol-1 ∙ K-1)(373 K)
5.00 L
Ans: approx 36.6 atm (36.7)
4) A gas sample containing 1.32 grams of helium at a pressure of 900. torr is cooled to 288 Kelvin.
What is the volume, in Liters, occupied by the gas, at these conditions?
Given:
*R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
V= ?
P = 900 torr
n = ? but you have 1.32 grams of He
T = 288 K
*V = nRT
P
❶
molHe = 1.32 grams| 1 mol | = 0.330 mol
4 grams
❷
atm = 900. torr | 1 atm | = 1.18
760 torr
*V = (0.330 mol)(0.08206 L∙atm ∙ mol-1 ∙ K-1)(288)
1.18 atm
Ans: 6.61 L
168
5) An engineer pumps 140.grams of carbon monoxide gas into a cylinder that has a capacity of
20.0 L. What is the pressure in atm of carbon monoxide (CO(g)) inside the cylinder at 25.0°C?
Ans: 6.11 atm
* Given:
R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
P=?
V= 20.0 L
n=?
T = 298 K
n = ? but you have 140 g of CO
* PV = nRT
*moles = 140 grams CO| 1 mol| = 5.00 mol
28 g
or P = nRT
V
*P = (5.00 mol) (0.08206 L∙atm ∙ mol-1 ∙ K-1)(298 K)
20.0 L
6) Which of the following changes to a system containing one mole of an ideal-gas in a balloon, at STP
causes the pressure to increase, based on the kinetic molecular theory of gases and the ideal gas law?
1)
2)
3)
4)
Increasing the size, but not the number of the gas particles
Increasing the average kinetic energy of the gas particles
Increasing the polarity (dipole moment) of the gas particles
Increasing the molar mass of each pas particle by 5%.
ans: *2
V) Moving On: Molar Mass and an Application of the Ideal Gas Law
The ideal gas law can be used to determine the molar mass (M) of a gas, when you know the pressure,
absolute temperature, mass of gas and volume of gas.
A) The key is to understand that the number of moles of gas, n can be expressed as grams per mole mass
or rather: n = grams
this essentially allows you to substitute in grams/M in for n
M
1) PV = nRT
becomes
PV = grams RT
M
If you grasp that substitution then it should be a short hop to rearranging the equation to
solve for molar mass:
M = *gRT
PV
169
TRY THIS: 20.0 grams of an unknown gas occupy 2.00 Liters at STP. Calculate the molar mass of the gas.
* Given:
R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
P = 1 atm
V= 2.00 L
T = 273 K
n = ? but you have 20 g of gas
M = *(20.0)(0.08206 L∙atm ∙ mol-1 ∙ K-1)(273)
(1 atm) (2.00 L)
Ans: 224 g/mol
B) Practice Determining the Molar Mass:
1) When 0.900 g of an unidentified gas occupies a volume of 0.500 L at 101 °C and
0.970 atm pressure, what is its molar mass?
*Given:
R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
P = 0.970 atm
V= 0.500 L
n = ? but you have a 0.900 gram mass
T = 101 °C
n = 5.00 mol
moles = 140 grams CO| 1 mol|
PV = nRT or with n substitution: PV = gRT
M
rearrange solving for M
*M = gRT
PV
*M = (0.900 g) (0.08206)(374)
(0.970) (0.500)
Ans: 56.95 g = 57.0 g
170
2) Determine the molar mass of an unknown gas that has a volume of 72.5 mL at
a temperature of 68.0°C, a pressure of 0.980 atm, and a mass of 0.207 g
*Given:
R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
P = 0.990 atm
V= 0.0725 L
n = ? but you have a 0.207 gram mass
T = 341 K
*M = gRT
PV
*M = (0.207 g)(0.08206)(341)
(0.980) (0.0725)
Ans: 81.5 g/mol
3) A sample of an unknown gas has a mass of 0.116 g. It occupies a volume of
25.0 mL at a temperature of 127°C and has a pressure of 155.3 kPa. Calculate
the molar mass of the gas.
*Given:
R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
P = 0.155.3 kPa … change to 1.533 atm
V= 0.0250 L
n = ? but you have a 0.116 gram mass
T = 400 K
PV = gRT
M
rearrange solving for M
*M = gRT
PV
*M = (0.116g) 0.08206 L∙atm ∙ mol-1 ∙ K-1)(400.K)
(1.533 atm)(0.025L)
Ans: 99.3 g/mol
3a) Using the following mole masses, which of the following molecular
formulae could represent the gas?
C = 12.0g/mol
H = 1.0 g/mol
i) C4H10Cl2
S= 32.0 g/mol
ii) C5H4Cl
O= 15.9 g/mol Cl = 35.4 g/mol
iii) SO2Cl2
iv) SO3
Ans: ii
171
4) 13.9 grams of a gas at 1.11 atm and a temperature of 31.0 C occupies a volume
of 7.10 L. Calculate the molar mass of the gas.
*Given:
R = 0.08206 L∙atm ∙ mol-1 ∙ K-1
P = 1.11 atm
V= 7.10 L
n = ? but you have a 0.13.9 gram mass
T = 304 K
*M = (13.9) 0.082057 L∙atm ∙ mol-1 ∙ K-1)(304K)
(1.11 atm)(7.10 L)
Ans. 43.99 g/mol = 44.0 g/mol
4a) Using the following mole masses, which of the following molecular
formulae could represent the gas?
C = 12.0g/mol H = 1.0 g/mol S= 32.0 g/mol O= 15.9 g/mol Cl = 35.4 g/mol
i) SO2
ii) CCl4
iii) C6H6
iv) CO2
Ans = iv
Practice, Practice, Practice … Keep pushing forward….
1) Refer to the reading regarding Cannizzaro’s refinement process for determining molecular masses, found
in this packet. He used Avogadro’s work, but according to the reading, how did Cannizzaro know when
the pressures of his gas samples were equivalent to each other?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
172
2) What mass, in kilograms, of CO2 is needed to fill an 80.0 L tank to a pressure of 150.0 atm at 27.0°C?
Hint 1: * You have volume, pressure, temp, the identity
and the need to find mass OF A GAS … try the ideal
gas law.
Hint 2: *PV = nRT
(150.0)(80.0) = n(0.08206)(300.15)
Hint 3: * solve for n (which is the # of moles)…
Convert
the number of moles to grams (or directly into
kilograms), using the mole mass of carbon dioxide …
ans:* mass = 21.5 kg
3) How many molecules of gas would you have if you had a volume of 38.0 L under a pressure of 1,432 mmHg
at standard temperature?
Hint 1: * You have volume, pressure, temp, of a gas!
You are asked for molecules, which is related to moles
… Try the ideal gas law, solving for “n” and then
convert to molecules using Avogadro’s constant of
6.092 x 10 2
ans * 1.93 x 1023 molecules
4) At what Celsius temperature would 2.10 moles of N2 gas have a pressure of 1.25 atm and a volume
of 25.0 L?
ans -92.0 °C
5) 185 grams of an unknown gas exert a pressure of 9.42 atmospheres, in a 15.0 L cylinder at 25.0oC.
Calculate the molar mass of the gas.
ans: 32.0 g/mol
173
6) Which of the following molecular formulae could represent the substance in question 5?
1) HCl
2) O2
3) Cl2
4) HCN
7) A 50.0 L cylinder is filled with argon gas to a pressure of 10,130.0 kPa at 30.0oC. What mass of argon gas
is in the cylinder?
8) Under what conditions will a real gas most closely approximate the predictions of the ideal gas law?
1) high pressure, high temperature
2) low pressure, low temperature
3) high pressure, low temperature
4) low pressure, high temperature
9) Which of the following gases should best approximate the behaviors of an ideal gas?
1) CO2
2) H2
3) C2H6
4) SO3
10) Several commercial drain cleaners (e.g. Crystal DRANO) contain NaOH(s) and small amounts of Al(s)
When one of these cleaners is added to water, a reaction occurs resulting in the formation of hydrogen gas
bubbles.
2 Al(s) + 2 (OH)-1(aq) + H2O(l) → 3 H2(g) + 2 AlO2-(aq)
The purpose of the H2(g) bubbles is to agitate the solution and thereby increase the cleansing action.
According to the balanced reaction equation, how many liters of hydrogen gas, are released when
0.200 grams of Al are consumed in an excess of (OH)-1 at 25 C and 1 atm?
Hint 1:* I get the sense that ultimately use of
the ideal gas law is appropriate, since this is
a gas problem, at non-STP conditions ….but
we’ll need moles of hydrogen gas.
Hint 2:* Think of stoichiometry to calculate
the number of moles of hydrogen gas
…using 0.200 grams of Al as a given.
Hint 3: *Use the ideal gas law to convert the
moles of H2 to Liters.
Ans: 6) 2 7) 8,040 grams 8) 4 9) 2 Ideal gas behavior is best approximated by real gases which are nonpolar molecules and
extremely small with very few e-, thus limiting the effect of IMF. Hydrogen gas (dihydrogen) best approximates these properties
(characteristics)… While CO2 and C2H6 are both nonpolar molecules, they are much larger in mass and thus have a far greater number
of electrons…creating a greater IMF between the molecules. 10) 0.271 L
174
V) Moving On … An application of the Ideal Gas Law and density
The ideal gas law can be used to determine the density of a gas … Attack the Trivedi Flash at sections
5.15 and 5.16. Please note that the equations re: density via the re-arrangement of the ideal gas law is
NOT on your tables. I would memorize this equation, or grasp the rather elegant derivation
through which the narration on the flash guides you. Essentially d = m/v … and the following equation,
simply starts with the ideal gas law … It solves for m/v and calls that solution d (for density) … It’s
d = (M)(P) …to determine the density of a gas at a specific
RT
absolute temperature and pressure.
d = density, M = molar mass. *At const. P&T, d∝ M
Clearly this equation can be used to determine the molar mass, given the density … Rearrange the above
equation to solve for molar mass:
* M = dRT
P
Assignment : Prep-Read p. B & L: 417-418
VI) Mixtures of Gases, partial pressure…and mole fractions
A) John Dalton is our hero on this section … Back in 1803 John Dalton built on his work on color
blindness (Daltonism), atomic structure, and his soon to be enunciated Law of Definite Proportions.
1) Dalton’s Law of Partial Pressures is the first to address the behavior of gases, in a mixture.
a) The total pressure of a mixture of gases equals the sum of the pressures that each
would exert, were that gas present alone. Also, the total pressure is proportional
to the total number of moles of gas. The pressure exerted by a specific
gaseous component of a gaseous mixture is called a partial pressure.
b) Ptotal = P1 + P2 + P3 + … each gas of a mixture exerts its own (partial) pressure.
Each gas of a mixture diffuses due to its own partial
pressure and does so independently of the other gases
(and/or total pressure) of the mixture.
Now, the biologist in me is totally in love with this idea. Picture it
An alveolus of a mammalian lung, where gas exchange between
CO2 and O2 may occur…. Dalton’s Law applies here. In spite of the
equality of total pressure in the alveolus and blood, the CO2 and O2
act independently of each other … even though their partial pressures
comprise the total pressure.
175
For instance, the total pressure in the alveolus is 144 mmHg as is the
total pressure of the gases in the blood … You might think then, that
there would be no exchange of gas … the total pressures are equivalent.
However, Dalton’s Law of Partial Pressures expresses the idea that the
O2 will act independently of the total pressure and will move from the
higher partial pressure of O2 in the alveolus (at 105 mmHg) to the lower
pressure of O2 in the blood. So, utterly elegant …
http://faculty.stcc.edu/AandP/AP/AP2pages/Units21to23/respiration/alveolar.htm
2) The ideal gas law, again, comes to our theoretical rescue, when trying to dig
down to the basic assumptions and relationships. As written in your text (p. 416)...
Ptotal = P1 + P2 + P3 + … is a wonderful application of the ideal gas law.
P1 = n1 (RT);
V
P2 = n2 (RT);
V
P3 = n3 (RT);
V … where n = moles
The number of moles of each gas may be different … but each gas will expand
to occupy the volume of the container (Thus, V is an equivalent value) … and
due to the relatively low specific heat of gases, the entire sample will reach an
equivalent temperature (Thus, T is the same for the entire sample).
Notice that the
number of moles
of gas is related
to total pressure.
So: Pt = (n1 + n2 + n3 + ….)(RT) OR
V
Pt = nt (RT)
V
At a constant temperature and volume, the total pressure of a gas mixture
is really determined by the total number of moles of gas present … (This is
accurate, whether it is a mixture or a single gaseous compound).
176
3) This sum of partial pressures plays an important role in such issues as finding
the pressure of a “dry” gas, when using displacement of water as a collection
means in lab.
Total pressure is
due to the collected
gas and to water
vapor … Wherever
there is liquid water,
there too, is its
vapor. Don’t forget
this.
Gas collection via water displacement:
Dalton’s Law
provides us a way of
determining the
http://www.docbrown.info/page13/ChemicalTests/GasPreparation.htm “dry” gas that was
collected.
For instance, consider a human lung as a relatively closed system. The
water of the lung will vaporize to a partial pressure of about 47 torr. Assuming
the total pressure inside the lung is 760 torr, calculate the pressure of “dry air”.
ans: 713 torr
Pressure of Water Vapor (kPa) at various Temperatures
Temp
oC
Pressure
kPa
Temp
o
C
Pressure
kPa
Temp
o
C
Pressure
kPa
0
0.6
20
2.3
30
4.2
3
0.8
21
2.5
32
4.8
5
0.9
22
2.6
35
5.6
8
1.1
23
2.8
40
7.4
10
1.2
24
3.0
50
12.3
12
1.4
25
3.2
60
19.9
14
1.6
26
3.4
70
31.2
16
1.8
27
3.6
80
47.3
18
2.1
28
3.8
90
70.1
19
2.2
29
4.0
100
101.3
Try This:
a) A student collected 35.0 mL of H2(g) via water displacement, at a temperature of 22C. If the
barometric pressure of the room were 98.0 kPa, calculate the partial pressure of the dry hydrogen
gas.
ans: 95.4 kPa
*PT = Pwater vapor + Phydrogen Or:
98.0 kPa = 2.6 kPa + Phydrogen
177
b) A student collected 25.0 mL of H2(g) via water displacement, at a temperature of 20C. If the
barometric pressure of the room were 106.0 kPa, calculate the partial pressure of the dry hydrogen
gas.
ans: 103.7 kPa
c) A student collected 40.0 mL of H2(g) via water displacement, at a temperature of 30C. If the
barometric pressure of the room were 103.0 kPa, calculate the partial pressure of the dry hydrogen
gas.
ans: 98.8 kPa
d) A mixture of oxygen, hydrogen and nitrogen gases exerts a total pressure of 278 kPa. If the partial
pressures of the oxygen and the hydrogen are 112 kPa and 101 kPa respectively, what would be the
partial pressure exerted by the nitrogen. (http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson74.htm) ans: 65 kPa
e) A mixture of neon and argon gases exerts a total pressure of 2.39 atm. The partial pressure of the
neon alone is 1.84 atm, what is the partial pressure of the argon?
ans: 0.55 atm
(http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson74.htm)
TN: Hexane Demo
4) To carry this conversation, further, the concept of mole fraction becomes important.
a) The ideal gas law, can be used to determine the mole fraction … and to solve
for the independent pressure of a gas relative to the total mixture:
P1 = n1RT/V
Pt = ntRT/V
= n1 …. n1/nt is the mole fraction
nt
The mole fraction is the ratio of the number of moles of one component
gas of a mixture of gases to the total number of moles in the mixture.
When we multiply the total number of moles by the mole fraction,
we can determine the percent of a specific gas in a mixture.
i) The above arithmetic relationship is more commonly written as:
X1 = moles of X
total moles of gas
where X1 is the mole fraction
Then, this relationship can be used by applying it to
P1 = (Xi)(Ptotal) …. but memorization is
important, only if you don’t yet see that this is essentially
a “part” vs. “whole” problem, akin to finding a percentage.
178
b) e.g. A 1.00 Liter container at 25.0 °C contains 0.0200 mol of N2(g) and
0.0300 mol of NH3(g) [ammonia]. The total pressure of the gaseous
mixture is 1.22 atmospheres.
-What is the mole fraction of ammonia gas?
*XNH3 = 0.0300 mol
XNH3 = 0.600 [no units]
0.0500 mol
- What is the partial pressure of ammonia gas?
*PNH3 = (0.600)(1.22 atm)
PNH3 = 0.732 atm
Try This: A gas mixture has a total pressure of 810.0 torr and a temperature of
290 K. It is comprised of 0.5000 mol He(g), 1.500 mol H2(g).
Calculate the partial pressures due to helium and hydrogen gas.
Solution: There are 2.000 moles of gas total. Hence, 1 mole of gas exerts
* 405.0
torr of pressure … Now … think…. each mol exerts
this pressure …so….
*810 torr/2 moles = 405.0 torr per mole
Since the helium is only 0.500 mol: (0.5000 mol) 405.0 torr/mol)
partial pressureHe = 202.5 torr
Since the hydrogen gas is 1.500 mol: (1.500 mol)(405.0 torr/mol)
partial pressureH2 = 607.5 torr.
*Check the work: PT = PHe + PH2 and
*810 torr = 202.5 torr + 607.5 torr
Essentially this is the concept of mole fraction. The total number of moles is
2.000 mol of which 0.5000 mol is helium gas… That means that ¼ of the mixture is
due to helium (or 25% of the mixture is due to helium)
c) e.g. A 1.00 Liter container at 25.0 °C contains 0.0200 mol of N2(g) and
0.0300 mol of NH3(g) [ammonia]. The total pressure of the gaseous
mixture is 1.22 atmospheres.
179
-What is the mole fraction of ammonia gas?
*XNH3 = 0.0300 mol
XNH3 = 0.600 [no units]
0.0500 mol
- What is the partial pressure of ammonia gas?
*PNH3 = (0.600)(1.22 atm)
PNH3 = 0.732 atm
I wrote that this is really akin to a percentage …for instance in the above
example 3/5ths or 60% of the mixture molecules are N2 and of course,
the partial pressure (0.732 atm) is 60% of the total pressure.
d) What if you wished to calculate the total pressure of a gas mixture?
Engineers and research chemists will need to calculate total pressures
as gaseous mixtures are evolved … not reacting … but collecting in
a vessel … and thus exerting a pressure.
A 0.50 Liter container holds 0.25 mol N2(g) and 0.15 mol He(g) at 25°C.
Calculate the total pressure exerted by the mixture of gases.
i) Get with a neighbor … any thoughts as to approach? I have 2 different
ways of solution … a 2 step and a 1 step …
*Use the ideal gas law to calculate the partial pressure of each gas
2 step
*Use Dalton’s Law of Partial Pressures
*OR take a different tact: PT = (nN2 + nHe) RT
V
Solution: * PN2 = nN2RT
V
and PHe = nHeRT
V
*PT = PN2 + PHe
ans: approx. 2.0 x 101 atm
180
VII) Kinetic Molecular Theory and the Distributions of Molecular Speed
A) As second year students it is probably no surprise that the molecules in a gaseous sample share
the same average kinetic energy (which is directly proportional to the absolute [Kelvin] temperature)
1) Given the variety of interactions between the gas molecules and the container walls, various
speeds are exhibited, but there is a distribution of the molecular speeds which may prove to be
of value.
2) The distribution will follow along the lines of our old friend, the Maxwell-Boltzmann graph
a) The peak (ump) is the most likely speed and the graph will expand and flatten as
the temperature increases, indicating that the range of molecular speeds increases
with an increase in temperature. Whereas the root-mean-square speed (rms) is the
speed of a molecule possessing a kinetic energy identical to the average kinetic
energy of the sample.
The following diagram from your text (B & L) p. 419 summarizes the distribution issues.
181
It is interesting that the authors at:
http://chempaths.chemeddl.org/services/chempaths/?q=book/General%20Chemistry%20Textbook/Gases/1
395/kinetic-theory-gases-distribution-molecular-speeds mention:
There are many things in nature which depend on the average (rms) velocity of
molecules. A mercury thermometer is one, and the pressure of a gas is another. Many
other things, however, are influenced by the number of very fast molecules rather than
by the average velocity. One example is the human finger—in a sample of H2(g) at
300 K it feels pleasantly warm, but at 373 K it will blister. This important difference
in behavior is not caused by an 11-percent increase in the average velocity of the
molecules. It is caused by a dramatic increase in the number of very energetic
molecules, which occurs when the temperature is raised from 300 to 373 K.
3) Calculating the Root Mean Square Velocity:
a) Equation urms = √3RT
√𝑀
(Memorize this… It is not on your tables)
Due to units: R = 8.3145 kg m2/s2/K•mol
T must be in Kelvin
M in in kilograms !!!!!!!!!!!!!
e.g) What is the average velocity (the root mean square velocity) of a molecule in a sample of
oxygen at 0.00 °C?
ans: 461.3 m/s
Did you convert the mole mass of O2 to kg?
urms = √3RT
√𝑀
=
Here is where that visualization of a gas sample in a cylinder with a moveable piston, comes into play…
B) Now … focus upon the simple gas laws … and learn how a gas sample will change
1) Assume a 1-Liter container with a moveable piston at constant temperature:
a) double the pressure: What happens to the volume? *it is reduced by ½
b) reduce the volume to ¼: What happened to the pressure? *it was quadrupled
c) double the pressure: What happened to the density? *it is doubled
d) triple the pressure: What happens to the average kinetic energy? * RTS temperature
is constant…. so average KE is constant.
2) Assume a 1-Liter container with a moveable piston at constant pressure:
a) double the absolute temperature: What happens to the volume? * it doubles
b) reduce the volume by 1/2 : What happens to the Kelvin temperature? *cut to ½
c) double the absolute temperature: What happens to the density? *it is cut to ½
182
3)
Assume a 1-Liter container with a constant volume:
a) double the absolute temperature: What happens to the pressure? *it doubles
Work at Sample Exercise 10.12 (p. 420) in the B&L text. This is a rather important skill.
Correct your work and bring in your questions.
What are you learning … at what are you getting better?
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183
VIII) Graham’s Law of Diffusion and Effusion
Fact1: The rates of effusion and diffusion of gaseous molecules is based upon the velocity of the
molecules
Fact2: The velocity of a gas molecule varies inversely with its mass (Lighter travels faster than
heavier given the same temperature)
Fact3: Kinetic Energy = ½ mv2
Fact4: rate of diffusiongas A =
rate of diffusiongas B
where v = urms
& this equation calculates the
translational kinetic energy
molecular mass Gas B or √𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐺𝑎𝑠 𝐵
√𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐺𝑎𝑠 𝐴
molecular mass Gas A
Since it is a ratio, it does not matter whether we use molecular mass or molar mass
A) The equation is rooted (if you will pardon the pun) in the root mean square velocity equation.
That equation essentially is: urms = √𝟑𝑹𝑻
√𝑴
Now: Diffusion rate for gas 1 = urms for gas 1 =
Diffusion rate for gas 2 = urms for gas 2
A nice little derivation of concept….
√3𝑅𝑇
√𝑀1 = √𝑀2
√3𝑅𝑇
√𝑀1
√𝑀2
TN: See http://physics.stackexchange.com/questions/54316/root-mean-square-speed-of-gas for the origin of the “3” in rms calculation
B) Effusion: *the escape or passage of gas molecules through a tiny hole, or along a pre-determine
pathway, into an evacuated chamber
“Classically” in class, we speak of effusion occurring through a pinhole
Through this pinhole gas molecules move from a high pressure into an
area of vacuum or of lower pressure, at least. Helium moving through the
very small pores of a latex rubber balloon is a nice example of effusion.
1) At constant temperature, the rate of effusion of a gas is inversely proportional to the square
root of its molar mass. The rate of effusion * increases as the square root of the temperature.
184
C) Diffusion: *the spread of one substance through a space or throughout a second substance.
1) Graham’s Law essentially says that at the same temperature and pressure, the gas of lesser
molecular mass will effuse or diffuse at a greater rate or speed, when compared to gas
molecules of greater molecular mass. Smaller/Lighter molecules travel faster…than
larger/heavier molecules assuming the same conditions
In diffusion the molecules of one substance can spread into the region occupied by the
molecules of another substance, in a series of random steps, undergoing collisions as
they move. This is, essentially, how pheromones or perfumes move out into a space.
The rate of diffusion is based on a number of factors … molecular mass,
temperature and the other gas(es). Perfume vapors, for instance, do not spread out
through a room as quickly as Graham’s Law predicts. The interactions with other
molecules (the mean free path) (and, I submit, the rate of oxidation of the vapors in O2)
may or can affect the rate of diffusion. This is a story of approximations and general
ideas ….
Check out: http://www.youtube.com/watch?v=H7QsDs8ZRMI#t=104
2) What Graham noticed that the rate of diffusion of gases was inversely proportional to the
square roots of their densities.
rate of diffusion ∝ 1
√𝑑𝑒𝑛𝑠𝑖𝑡𝑦
And, noting Avogadro’s Hypothesis that per liter of gas, at the same temperature and
pressure, the number of molecules is the same …. The relationship between the
rate of diffusion and density boiled down to molecular mass… since the density is
directly proportional to its molar mass at the same temperature and pressure.
https://www.chem.tamu.edu/class/majors/tutorialnotefiles/graham.htm
Hence: rate of diffusion ∝ 1
√𝑴
…. elegant
3) Based upon your understanding of Graham’s Law, which gas diffuses faster
at the same temperature and pressure ... ammonia (NH3) or argon (Ar)?
Answer *ammonia,
because * it is a smaller molecule, of lesser molecular mass
a) Prove it! How many times faster does *NH3
*rate of NH3
rate of Ar
diffuse/effuse faster than * Ar ?
= √39.48 𝑔𝑟𝑎𝑚𝑠 = 1.523 Or… The rate of diffusion of ammonia is
approximately 1.5 times greater than that
√17.03 𝑔𝑟𝑎𝑚𝑠
of argon.
Note that the gas in question is your numerator … That is; since we were seeking the rate of
ammonia vs. argon … I set the ratio
ammonia … or rather I put “x” in the numerator
argon
185
D) Examples of Problem Types re: Diffusion and Effusion:
1) Find the ratio: Comparing the rate of diffusion or effusion of 2 different gases:
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.
*Rate of effusion H2 = √𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑈𝐹6 =
Rate of effusion UF6
√𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐻2
√352.06 = *13.2
√2.016
* ans H2 effuses at a rate 13.2 times greater than that of UF6
2) Calculate the molecular mass (or the molar mass) of a gas, given the diffusion rates:
Calculate the molecular mass of a gas, which diffuses at a rate 1/50 of hydrogen’s rate.
Think … should this gas, be lighter or heavier in terms of molecular mass, than H2
given that it diffuses at a mere fraction of hydrogen’s rate?
Think: H2 diffuses 50 times faster … so reflect as to how that appears in the equation
*Rate of diffusion X = √𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 =
Rate of diffusion H2
√𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑋
*
1 = √2.016 𝑔/𝑚𝑜𝑙
50
√𝑥
=
1 = 2.016 g/mol =
2500
x
ans: 5040 g/mol
3) Comparison of effusion rates
Two porous containers are filled with hydrogen and neon respectively. Under identical
conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half the
neon to escape? …. For this problem, I know Ne is more massive than H2, so I am going to set Ne as rate 1, even
though it is the gas, ultimately in question … only because I want to avoid the interpretation of a
fractional rate of effusion…
rate of effusionhydrogen =
rate of effusionneon
√20.18 𝜇
√2.016 𝜇
=
*H2 effuses 3.16 times faster.
*Since the H2 escapes 3.16 times as fast as Ne, we can calculate the amount of Ne leaving in 6 hours:
* Assume 1 gram of the gases. So if we assume Ne is 1, then H2 diffuses 3.16 times faster. If we assume
1 gram of each gas then 2/3rds or 0.667 grams of hydrogen will diffuse or… 0.667 = x
3.16
1
or
0.667 = 0.211 of the mass of Ne escapes in 6 hours
3.16
*Now, to calculate the time required for ½ of the Ne to escape … knowing 0.211 of the mass
escapes in 6 hours …. 0.211 = 0.500
0.211 x = 3 x = 14.2 h
6h
x
ans = 14.2 h
186
4) Comparing densities, using rates of effusion:
When the density of hydrogen is 0.090 g/L and its rate of diffusion is 5.93 times that of
an unknown gas, what is the density of the unknown gas at STP? This will combine our
work on molar volume with Graham’s Law
Think: You really want to find the molar density of the unknown gas … That is you want to
get to Density = molar mass …. In order to get to that equation you really need to
22.4 L
know the molar mass (or at least, the molecular mass)
of the unknown ….
rate of x =
rate of H2
√𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐻2
√𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑥
*= 1
=
5.93
√2.016 𝑔/𝑚𝑜𝑙
√𝑥
1
= 2.016 g/mol
35.16
x
* x = 70.9 g/mol
Now that you know the molecular mass … and thus the molar mass … you can
determine the density:
* densisty = 70.9 g
22.4 L
= 3.17 g/L
ans: 3.17 g/L
5) Finding a distance The key here is to assume that any distance given is in 1 second, so that
you can create a rate…when a rate is not given….
Assume a molecule of CH4 diffuses a distance of 0.530 m from a point source, calculate the
distance (in meters) that a molecule of N2 would diffuse under the same conditions for the
same period of time.
* rate of N2
= √𝑚𝑒𝑡ℎ𝑎𝑛𝑒
rate of CH4
√𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛
x
= √16
0.530
√28
=
=
x
= 0.756
0.530
ans: 0.400 m/s
187
Sheer Goofiness … and a (very simple) Graham’s Law problem….
Austin Powers has been imprisoned in a room equipped with four gas vents! (Muhahahaha!!!)
Now, each gas has the potential of killing him by directly poisoning him or by suffocating him.
From vent 1 the ever stinky H2S will be emitted. From vent 2, intoxicating N2O will flow
(laughing gas). From vent 3, flammable CH4 will pour and from vent 4, deadly HCN will diffuse
Austin has but one chance ... if he could use the periodic table left
hanging fortunately in his cell and plug up the vents with four
gas impermeable cloths fortuitously left in the room - he can survive.
In what order should he plug up the vents, using Graham’s
Law of Diffusion (Thank goodness he studied this while a
student and happened to review the gas law just before going to
bed the night before this all happened... Phew!!)
Austin should plug up vent *3
first, then *4 , then *1 and lastly *2
E) Additional work on effusion:
http://www.businessesgrow.com/wp-content/uploads/2011/04/austin-powers.jpg
Fortunately, each outlet has been clearly labeled. Even more fortunately, each gas will
enter the room based solely upon Graham’s Law of effusion. And most fortunately of all,
Dr. Evil has explained this all to Austin.
It’s a real gas, Yeah Baby!
Recall: At constant temperature, the rate of effusion of a gas is inversely proportional to the square
root of its molar mass. The rate of effusion increases as the square root of the temperature.
This allows us to see the above equation of Fact4 a bit differently, in terms of time
The times it takes for the same amount of two substances to effuse through a small
hole are inversely proportional to the rates at which they effuse … hence THE
EQUATION CHANGES ….
a)
time for A to diffuse
time for B to diffuse
=√𝑀𝑎
𝑀𝑏
notice that it is Ma/Mb … the inverse of
Graham’s law….
This relationship can be used to estimate the molar mass of a substance by comparing the
time required for a given amount of an unknown substance to effuse with that required
for the same amount of a substance with a known molar mass …Let’s go to the races!!!
188
TRY THIS: It takes 30. mL of argon 40.s to effuse through a porous barrier. The same volume
of vapor of a volatile compound extracted from Caribbean sponges takes 120.s to
effuse through the same barrier under the same conditions of temperature and pressure.
Stop for a second: Think: 40.s for the argon …. 120. s for the unknown.
Before doing the math … Which molecule do you think is travelling
at a greater rate? * Ar
Graham’s Law essentially says … smaller is faster … so make a
prediction about the unknown’s molar mass …. *it is larger
time for A to diffuse
time for B to diffuse
=
𝑀𝑎
√
𝑀𝑏
* 120.s = √𝑥
40. s √39.95
3.0 = √𝑥
or
√39.95 g/mol
9.0 = x
39.95 g/mol
ans: *x = 359.55 g/mol or … 360 g/mol with sig figs.
Does the answer agree with your thinking?
b) and while not on the AP… What if there were a need to compare effusion rates at
different temperatures? … Well as temperature increases, the rate of effusion increases
and for any given gas, the rate of effusion increases as the square root of the
temperature.
Rate of effusion at Temp2 = √𝑇2
Rate of effusion at Temp1
√𝑇1
Taking this further … From Atkins (p. 152)
Because the rate of effusion is proportional to the average speed of the molecules,
it may be inferred that the average speed of molecules in a gas, is proportional to the
square root of the temperature : Average speed ∝ √𝑇
This very important relation begins to reveal the significance of one of the more
elusive concepts in science …the nature of temperature. ….When referring to a gas
the temperature is an indication of the average speed of the molecules, and the higher
the temperature, the higher the average speed of the molecules.
The two relations can be combined ….to
Average speed ∝ √𝑻
√𝑴
…This then leads us to the idea that the average speed of molecules in a gas is
directly proportional to the square root of the temperature and inversely proportional to
square root of the molar mass!!!
Not bad for a guy who postulated “Graham’s Law” back in 1848 based upon his
experimental results!
189
PRACTICE: Because Graham’s Law is such a crazy quilt of applications, here are a few more practice
problems on Graham’s Law. I’ve tried to work out the problems and the answers are given (or at least
allueded to) in each of the “hint” text boxes. The answers are also given at the end of the problem set.
1) A compound composed of carbon, hydrogen, and chlorine diffuses through a pinhole 0.411 times as fast as
neon. Select the correct molecular formula for the compound:
1) CHCl3
2) CH2Cl2
3) C2H2Cl2
4) C2H3Cl
Hint 1: *
rate of effusionx = √20.18 𝑔/𝑚𝑜𝑙 = 0.411 times
rate of effusionneon √𝑥𝑔/𝑚𝑜𝑙
as fast as Ne
Hint 2:* square both sides and solve for
“X” by cross multiplying and dividing 20.18 by
0.168921 (which is the square of 0.411)
ans:* choice 1 is has molecular mass of approx.
119.35 µ
2) How much faster does hydrogen escape through a porous container than sulfur dioxide?
1) 2.44 times faster
2) 3.27 times faster
3) 3.80 times faster
4) 5.64 times faster
Hint 1:
*rateH2 = √64.06 𝜇
rate SO2 √2.016 µ
Hint 2: *Take the square root of each
value and divide
3) Calculate how many times faster carbon dioxide diffuses relative to ozone (O3) at the same temperature.
1) 1.15 times faster
2) 1.04 times faster
3) 1.68 times faster
4) 1.92 times faster
Hint 1: *rateCO2 = √48 𝜇
rate O3 √44 µ
Hint 2: *Take the square root of each value
and divide
4) 2.278 x 10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 95.70 s. Under identical
conditions, 1.738 x 10-4 mol of argon gas takes 81.60 s to effuse. What is the molar mass of the unidentified
substance?
1) 28.02 g/mol
2) 30.68 g/mol
3) 31.79 g/mol
4) 34.73 g/ mol
Hint 1:*I know, you want to use the time vs.
M … but think… You have a number of
moles PER time … So, calculate a rate for
the unknown and argon.
Hint 2:* ratex = √𝑀𝐴𝑟
rateAr = √𝑀𝑥
Hint 3:* 0.000002380 = √39.95𝜇
0.000002129
√𝑥
190
5) It takes 10.0 minutes for 1.00 mol of Br2(g) to effuse through a
membrane. How long would it take the same number of moles of
Ar(g) to effuse through the same membrane?
1) 4.22 min
2) 5.00 min
3) 2.31 min
4) 2.50 min
Two ways to solve:
Hint 1:*Use time a = √𝑀𝑎
time b √𝑀𝑏
Hint 2: *
x
= √39.95µ
10 min
√159.8𝜇
Hint 3: *square both sides
x2/100 = 39.95/159.8
and solve for x
*******OR Use * rate Ar = √159.8𝜇
rate Br2 √39.95
Hint: * rate Ar =
0.100 mol/min
ans: *5 min (choice 2)
√159.8𝜇
√39.95𝜇
6) In an effusion experiment, it was determined that nitrogen gas, N2, effused at a rate 1.812 times faster than
an unknown gas. What is the molar mass of the unknown gas? (molecular mass of N2 = 28.014 µ/molecule)
1)
2)
3)
4)
89.03 g/mol
90.99 g/mol
91.98 g/mol
93.46 g/mol
Hint 1: *The rate of effusion of Nitrogen to
the unknown is 1.812 times faster … and rate
= the inverse of the square roots of the
molecular masses….
Hint 2: * 1.1812 = √𝑥
√28.014µ
ans:* choice 3)
7) What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate
of 3.62 mol/min (I’ve worked out the problem, on the next page, under the “answers”)
1)
2)
3)
4)
3.01 mol/min
2.56 mol/min
2.04 mol/min
5.82 mol/min
191
8) Why are the rates of diffusion of nitrogen gas and carbon monoxide almost identical at the same
temperature?
* The rate of diffusion is inversely proportional to the square root of the density of gas, which can be
replaced by the the molecular mass of the gas, per Avogadro’s Hypothesis. Thus, the rate of diffusion
is based upon the molecular mass. When comparing nitrogen (dinitrogen) to carbon monoxide,
one can note that their molecular masses are very, very similar. Thus, using Graham’s Law one
should expect the rates of diffusion to be very, very similar. Nitrogen has a molecular mass
of 28.02 µ/molecule (to 4 sig figs), while carbon monoxide has a molecular mass of
28.01 µ/molecule (to 4 sig figs).
Answers:
1) 1 check out the exemplar re: finding a molecular mass
2) 4 just run a “normal” Graham’s Law problem … essentially dividing the square roots of the molecular masses to find out how
many times greater the diffusion rate . Notice that I put the heavier sulfur dioxide in the numerator when finding the square root
… I do this because, I know hydrogen is going to have a greater rate of diffusion, due to its lesser molecular mass … and all the
choices are greater than 1 …
3) 2
4) 3 Use the values of mol and seconds to find a rate … (divide the two values for each gas). Once you have
rates, you can set up your Graham’s Law equation.
5) 2
6) 3
7) 2 rate 1 is 3.62 mol/min … rate 2 = x … Now, you know that the second gas (at rate 2) is twice as heavy,
so, don’t worry about the numbers … just use 2 vs. 1 or 2 vs 4 …. or 3 vs. 6 … who cares as long as one is twice as large
as the other.
3.62 mol/minligther = √4
square both sides ….
13.1044 = 4
xheavier gas
x2
2
√2
be sure you work with x2 thus:
4x2 = 26.2
x2 = 6.55
x =2.56 mol/min and this makes sense (hopefully)
because the molecules of this gas are supposed to be larger
than the other gas molecules, so these larger molecules
should diffuses at a smaller (lower, lesser) rate … and
2.56 mol/min is a lesser rate than 3.62 mol/min
*********************************
192
X) Real Gases and Deviations from the “Ideal” and Kinetic Molecular Theory
The Trivedi Flash does a very nice job… but this selection from Purdue University
captures the issue nicely:
The behavior of real gases usually agrees with the predictions of the ideal gas equation
to within 5% at normal temperatures and pressures.

At low temperatures or high pressures, real gases deviate significantly from
ideal gas behavior.

In 1873, while searching for a way to link the behavior of liquids and gases, the
Dutch physicist Johannes van der Waals developed an explanation for these
deviations and an equation that was able to fit the behavior of real gases over a
much wider range of pressures.
Van der Waals realized that two of the assumptions of the kinetic molecular
theory were questionable.

The kinetic theory assumes that gas particles occupy a negligible fraction of the
total volume of the gas.

It also assumes that the force of attraction between gas molecules is zero.
The first assumption works at pressures close to 1 atm. But something happens
to the validity of this assumption as the gas is compressed. Imagine for the moment
that the atoms or molecules in a gas were all clustered in one corner of a cylinder, as
shown in the figure below. At normal pressures, the volume occupied by these particles
is a negligibly small fraction of the total volume of the gas. But at high pressures, this
is no longer true. As a result, real gases are not as compressible at high pressures
as an ideal gas.

The volume of a real gas is therefore larger than expected from the ideal gas
equation at high pressures. In the van der Waals equation:
P + a n2 (V – nb) = nRT
V2
When the pressure is relatively small, and the volume is reasonably large, the nb term
is too small to make any difference in the calculation. But at high pressures, when the
volume of the gas is small, the nb term corrects for the fact that the volume of a real
gas is larger than expected from the ideal gas equation.
The assumption that there is no force of attraction between gas particles cannot be
true. If it were, gases would never condense to form liquids. In reality, there is a
small force of attraction between gas molecules that tends to hold the molecules
to each other.
193
This intermolecular force of attraction has two consequences:
(1) gases condense to form liquids at low temperatures and
(2) the pressure of a real gas is sometimes smaller than expected for an ideal gas.
To correct for the fact that the pressure of a real gas is smaller than expected from the
ideal gas equation, van der Waals added a term to the pressure in this equation. This
term contained a second constant (a) and has the form: an2/V2.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/deviation.php
As a summary: When gases are polar, massive, at high pressure and low temperature, they
do not behave like ideal gases. They may even condense into liquids or freeze into solids
under these conditions.
Question: Why does being a polar molecule explain the deviation from ideal behavior and KMT?
*Polar molecules exhibit stronger IMF, thus, they are more likely to attract each other, interfering
with the ability of the gas molecules to expand away from each other. Thus a gas of polar
molecules has a far greater chance of failing to obey the KMT.
194
TAKE A GANDER AT THE FOLLOWING HANDY-DANDY REFERENCE CHART.
Beyond the obvious issues, try to see if the following table helps direct your work.
When I write “obvious issues” I mean the variables of the gas law … For instance,
you should consider the Ideal Gas Law when asked to find volume, or temp,
pressure etc…. But what if …



I want to find …
the problem asks for …
the problem gives me…
number of molecules of gas
molar mass (of formula mass)
density of a gas
rate of effusion or diffusion
ratio between two diffusing gases
time for a gas to effuse or diffuse
distance a gas travelled
rate of effusion at different temps
root mean square velocity
total pressure of a mixture of gases
partial pressure of a gas
mass of gas(es) in a gas mixture
pressure of “dry gas”
temperature or volume at constant
pressure
pressure or volume at constant temp
pressure or temp at constant volume
then, consider using ….
Ideal Gas Law
Ideal Gas Law OR Graham’s Law
Ideal Gas Law OR Graham’s Law OR Molar Density
Graham’s Law
Graham’s Law
Graham’s Law
Graham’s Law
Graham’s Law (but not on AP test)
urms equation
Dalton’s Law of Partial Pressure
Dalton’s Law of Partial Pressure OR Ideal Gas Law …& you may
need to consider using mol fractions (see below)
Dalton’s Law of Partial Pressure: (mol fraction)
 determine the moles of each gas
 add to find the total number of moles in the mixture
 divide to find the mole fraction of the gas and then
 multiply the mole fraction by the total pressure
Dalton’s Law of Partial Pressure
Combined Gas Law worked down to Charles’ Law
Combined Gas Law worked down to Boyle’s Law
Combined Gas Law worked down to Gay-Lussac’s
PRACTICE … The answers to 1 -3 will vary. Answers for the vocabulary building section of questions are
on the last page. Most of the other answers are given directly or implied in the “hint” textboxes.
1) Identify one tenet of the Kinetic Molecular Theory van der Waals disputed when it came to the
behavior of real gases.
____________________________________________________________________________
2) Assume van der Waals is accurate in that the tenet of the KMT you selected does affect the physical
behavior of real gases. Identify how the failure of the KMT assumption affects the behavior of real gases.
_____________________________________________________________________________
195
3) Very often teachers tell their students, that H2 and He are the real gases which best approximate ideal gas
behavior. Gases such as butane (C4H8) or sulfur dioxide (SO2) tend to deviate from ideal gas behavior far
more significantly than hydrogen gas or helium.
In a short response, identify one characteristic of either helium or hydrogen gas (or butane and sulfur
dioxide) which account for these facts.
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
___4) Under what conditions of temperature and pressure should a real gas, such as He or H2 best approximate
ideal gas behavior?
1) high temperature and high pressure
2) low temperature and low pressure
3) low temperature and high pressure
4) high temperature and low pressure
___5) This term is the ratio of the number of moles of one component gas of a mixture of gases to the
total number of moles in the mixture.
1) partial pressure
2) mole fraction
3) root mean square speed
4) molar density
___6) The pressure exerted by a specific gaseous component of a gaseous mixture is called
1) partial pressure
2) mole fraction
3) root mean square speed
4) molar density
___7) The simple gas law which states that the pressure is inversely proportional to the volume of a gas
sample at constant temperature
1) Avogadro’s Law
2) Boyle’s Law
___8) An absolute temperature scale is
1) Celsius
2) Centigrade
3) Charles’ Law
3) Kelvin
4) Gay-Lussac’s Law
4) Fahrenheit
196
___9) A mixture consisting of 4.90 grams CO and 8.50 grams SO2, two atmospheric pollutants, exerts a
pressure of 0.761 atm when placed in a sealed container. What is the partial pressure of the SO2 in
this mixture?
Hint 1: *This is a Dalton’s Law which deals with mole
fractions.
1) 0.33 atm
2) 0.18 atm
3) 0.13 atm
4) 0.43 atm
Hint 2:* Convert grams to moles. Add the two mol
values together to get a total number of moles.
4.90g| 1 mol| vs
28 g
8.50 g | 1 mol |
64 g
* 0.175 mol CO + 0.133 mol SO2 = 0.308 mol TOTAL
Gas
Hint 3: *Find mol fraction of SO2: 0.133/0.308 = 0.432
Hint 4: * The partial pressure of SO2 =
(mol fraction)(Total pressure)
(0.432) (0.761 atm) ans* 0.33 atm
___10) A sample of an unknown gas effuses in 8.2 minutes. An equal volume of krypton in the same apparatus
under the same conditions effuses in 4.0 minutes. What is the likely identity of the unknown gas?
1) UF6
2) Br2
3) PCl5
4) Ne
Hint 1:* If you could figure out the molecular mass
(formula mass … mole mass… GFM ….whatever you
hope to call it) …. you can figure out the identify of the
substance.
Hint 2: * Use Graham’s Law, with TIME (not rate…) in
the ratio. This causes us to use a slightly different version:
time 1 =
√𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 1
time 2
√𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 2
Hint 3: * 8.2 minutes = √𝑥
4.0 minutes √83.80𝜇
consider squaring
both sides….
Hint 4: *ans: 1) or UF6
11) Air is a source of reactants for many chemical processes. To determine how much air is needed for
these reactions, it is useful to know the partial pressures of the components. A sample of dry air with a
total mass of 1.00 grams consists almost entirely of 0.76 grams of N2 and 0.24 grams of O2. Calculate
the partial pressure of each gas, when the total pressure is 0.87 atm. (Atkins p 151)
Hint 1: * This is a mole fraction problem … so
find the # of moles and ….
Hint 2: * N2 = 0.027 mol vs. 0.0075 mol for a
total number of 0.0345 mol.
Hint 3: * This means that the mole fraction of
N2 = 0.7826
Hint 4: *Find the partial pressure of N2 :
(0.7826)(0.87) = 0.681 atm and the O2 = 0.189
atm
197
12) A baby with a severe bronchial infection is in respiratory distress. The anesthetist administers, heliox,
a mixture of helium and oxygen with 92.3% by mass, O2. What would be the partial pressure of oxygen
being administered to the baby if the atmospheric pressure were 730 torr? (Atkins p 151)
Hint 1: *This is a Dalton’s Law which deals with
mole fractions.
Hint 2:* Assume 100 grams of gas of which 92.3% are
O2 and thus 7.7% is He by mass. This means that
92.3 grams of the gas are due to oxygen and 7.7 grams
are due to He.
Hint 3: * Convert grams to moles. Add the two mol
values together to get a total number of moles of gas
in the heliox mixture.
92.3g| 1 mol| vs
7.7 g | 1 mol |
32 g
4g
* 2.88 mol O2 + 1.9 mol He = 4.8 mol TOTAL
Gas
Hint 4: * Find the mol fraction of O2 : 2.88/4.8 = 0.60
Hint 5: * The partial pressure of oxygen =
(mol fraction)(Total pressure)
(0.60) (730 Torr) ans* 438 or 440 Torr
13) A hydrogen gas thermometer is very similar to a manometer. Essentially it measures temperature by
observing the changes in volume of H2 gas kept at a constant pressure. A bulb filled with H2 gas is
submerged into the test substance. Changes in the volume of the hydrogen gas, at constant pressure are
noted.
Essentially, the temperature of the test substance can be determined based upon the ideal gas law:
PV = nRT, where P, n, and R, are constant. Thus the absolute temperature is proportional to changes in
the volume of the hydrogen gas.
(keep going)
This sort of hydrogen thermometer can help determine temperatures ranging from -200 °C to 500 °C.
A hydrogen gas thermometer was found to have a volume of 100.0 cm3 when placed in an ice-water bath at
0°C. When the same gas thermometer is immersed in boiling liquid chlorine, the volume of hydrogen at the
same pressure is found to be 87.2 cm3. What is the temperature of the boiling point of chlorine?
Hint 1: * P,n and R are constant …
only V and T vary.
ans: *238 K
198
14) The following three figures represent different mixtures of gases in containers of the same temperature
and volume. The unshaded spheres represent particles of a gas with a smaller molar mass, than the
particles of a gas represented by the shaded spheres. How do the total pressures of the gases compare?
Explain your reasoning. [Hint …count the spheres ….think Dalton’s or Ideal Gas Law]
a
b
c
*The pressures in a, b, and c are the same. The volume and temperature are the same, as is the total
number of
moles. Part of Dalton’s Law states that the total pressure is proportional to the total number of moles of
gas. The
partial pressures of each gas changes from diagram to diagram, but since there are 8 spheres (total),
Dalton’s also
tells us that the sum of the partial pressures must add up to the total. Given that the number of spheres is
equal in
each situation, we can analogize this to a constant total number of moles. Hence, V, T, n and R are all
constant,
thus, according to the ideal gas law, P must be constant.
15) A mixture of three gases, Ar, Ne and He has a total pressure of 1000 torr in a 1 Liter container at 25°C.
Given the following representation of the mixture, what is the partial pressure of Argon? Explain your
reasoning.
Key: Argon
Neon
Helium
ans: * 400 torr
*your explanation should include
a statement about the mole
fraction of Ar = 0.40 and a
reference to Dalton’s Law about
the total pressure being proportional
to the total number of moles
199
16) The following figure represents the condition of two gases effusing through a pinhole from container 1 to
container 2 after 10 minutes. Container 2 initially had no particles of gas. Which gas has a lesser
(smaller) molar mass (or formula mass)? Explain…
= gas A
= gas B
*Gas B must have the smaller or lesser formula mass. According to Grahams’ law of effusion, the
rate of effusion of a gas is inversely proportional to the square root of its molar mass. The gas with
the smaller formula mass (molar mass) will escape or pass from the original container (1) through
the pinhole, into container 2 more rapidly than the gas with a larger formula mass. More molecules
of gas B have effused through the pinhole after 10 minutes than gas A, thus molecules of gas B have
have effused at a faster rate. It can be inferred that they must have the smaller formula mass.
Assignment: I could keep writing and writing … but get to the Trivedi flash, Chapter 5. Take one of the
quizzes at the end of the chapter. Grade yourself. How are you doing?
Answers : 4) 4 5) 2 6) 1 7) 2 8) 3
200