Dr. C. Weldon Mathews
Chem 122
Office: 0042 Evans Lab
Telephone: 292-1574
email: mathews.6@osu.edu
course web site:
http://www.chemistry.ohio-state.edu/~mathews/chem122wi07/
Office hours: TR 12:30 - 2:00 pm
TR 4:00 - 5:00 pm
or by appointment
Chapters we’ll cover in Chem 122:
10, 11, 13, 14, 15, 16, 17 (17.1-17.3)
First Week:
10.0-10.6
Second Week: 10.7-10.9
and
11.1-11.5
First Quiz: Week of Jan 8 (second week)
Review Chem 121, especially Chaps 8 and 9
Chapte r 10
10.1
10.2
10.3
1st Lecture
&
1st Quiz
10.4
10.5
10.6
10.7
10.8
10.9
Gases
Characteristics of Gases
398
Pressure
400
Atmosopheric Pressure and the Barometer
The Gas Laws
404
The Pressure-Volume Relationships: Boyle's Law
The Temperature-volume Relationship: Charles's Law
The Quantity-Volume Relationship: Avogadro's Law
The Ideal-Gas Equation
408
Relating the Ideal-Gas Equation and the Gas Laws
Further Applications of the Ideal-Gas Equation
413
Gas Densities and Molar Mass
Volumes of Gases in Chemical Reactions
Gas Mixtures and Partial Pressures 417
Partial Pressures and Mole Fractions
Collecting Gases over Water
Kinetic-Molecular Theory
420
Application to the Gas Laws
Molecular Effusion and Diffusion 423
Graham's Law of Effusion
Diffusion and Mean Free Path
Real Gases: Deviations from Ideal Behavior
427
The van der Waals Equation
Bloom’s Taxonomy–
[Approaches to learning Chemistry.]
Knowledge – Simple recall of facts
Comprehension – Translate into your own words or equations.
Application – Apply concepts to specific situations; recognizing and solving a
problem when the equations are not given.
Analysis – Application plus recognition of important parts of problem.
Synthesis – Assemble components into a form new to them,
i.e. design a research plan or devise a synthetic scheme.
Expected
in this
course.
Evaluation – Judge the value of materials in terms of internal and external
criteria.
Expected
This is a grossly abbreviated adaptation from Bloom, B. S. (Ed.) (1956) Taxonomy of educational
in objectives:
The classification of educational goals: Handbook I, cognitive domain. New York; Toronto: Longmans,
Research.
Green. Use Google to find other references.
see also http://www.coun.uvic.ca/learn/program/hndouts/bloom.html
http://www.officeport.com/edu/blooms.htm
http://www.kurwongbss.eq.edu.au/thinking/Bloom/blooms.htm
Study Habits and Study Resources:
a) “Lectures” and “Reading” - minimal impact by themselves
b) “Chemistry is not a Spectator Sport!”
Prof. Janet Tarino, OSU Mansfield
c) Recitation and Laboratory TAs
d) Ask questions and seek help whenever you need it!
EXPECTATIONS
e) Web resources:
http://www.chemistry.ohio-state.edu/
http://www.chemistry.ohio-state.edu/~mathews/chem122wi07/
/~rbartosz/
/~rzellmer/
chemistry ->Undergraduate Program->Interactive Tutorials
First Lab Experiment: Calorimetry and Hess’s Law
You’ll need to review material from Chem 121:
Calorimetry, Chap 5, pp 182-187
Hess’s Law, Chap 5, pp 187-191
Enthalpies of Formation, pp 191-196
(and of Reactions)
Some other things you need to know:
Your Carmen Student ID. If you can’t find it
your TA will be able to help in lab or recitation.
Call Number for your section (eg 04499-3)
and your Chemistry Dept. Section Number
(eg 109), as shown on one of the following slides.
Characteristics of Gases
• Unlike liquids and solids, they




Expand to fill their containers.
Are highly compressible.
Have extremely low densities
Mix homogeneously with other gasses
Gases
Characteristics of Gases
Gases
Pressure
• Pressure is the force acting on an object per unit area:
F
P
A
• Gravity exerts a force on the earth’s atmosphere
• A column of air 1 m2 in cross section exerts a force of 105
N, with a mass of about 104 kg or 2.2 x 104 lbs.
• The pressure of a 1 m2 column of air is about 100 kPa.
Pressure
F
P
A
F = ma
= (104 kg)(9.8m/s2)
= 1 x 105 kg-m/s2
= 1 x 105 N
F 1 x 105 N
P 
A
1 m2
P = 1 x 105 N/m2
= 1 x 105 Pa
= 1 x 102 kPa
= Newton
Pressure
Atmosphere Pressure and the Barometer
• SI Units: 1 N = 1 kg.m/s2; 1 Pa = 1 N/m2.
• Atmospheric pressure is measured with a barometer.
• Standard atmospheric pressure is the pressure required to
support 760 mm of Hg in a column.
• Units: 1 atm = 760 mmHg = 760 torr = 1.01325  105 Pa
= 101.325 kPa.
Pressure
Atmosphere Pressure and the
Barometer
Notice that the top end
of the BAROMETER is
closed and that a
“Torricelli VACUUM”
exists above the mercury.
Pressure
Atmosphere Pressure and the
Barometer
• The pressures of gases not open
to the atmosphere are measured
in manometers.
• A MANOMETER consists of a
bulb of gas attached to a U-tube
containing Hg or other liquids:
– If Pgas < Patm then Pgas + Ph2 = Patm.
– If Pgas > Patm then Pgas = Patm + Ph2.
Here Ph may be positive or negative!
Pressure
Notice again the various units that may be used
to measure Pressure:
Easiest to refer to the ‘Standard Atmospheric Pressure’ which is
defined as 1 atm
1 atm = 101.325 kPa = 1.01325 x 105 Pa =
= 760 mmHg = 760 torr
Memorize these relations!
They will be useful and help you with conversion factors
between the various units.
The Gas Laws
•
•
•
•
•
The Pressure-Volume Relationship:
Boyle’s Law
Weather balloons are used as a practical consequence to
the relationship between pressure and volume of a gas.
As the weather balloon gets further from the earth’s
surface, the atmospheric pressure decreases.
As a consequence, the volume of the balloon increases.
Boyle’s Law: the volume of a fixed quantity of gas is
inversely proportional to its pressure.
Boyle used a manometer to carry out the experiment.
In our discussions about
gases, we’ll often use
the gas cylinder with a
movable piston as a
helpful analogy.
You also may think of a
bicycle pump as an
example.
Let’s run an “experiment”
animation
The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law
V = f(P) = ?
find VP = cnst = k
V = f(1/P)
= k(1/P)
The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law
• Mathematically:
1
k
V  constant 

P
P
PV  constant
PV  k
• A plot of V versus P is a hyperbola.
• Similarly, a plot of V versus 1/P must be a straight line
passing through the origin.
The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
• We know that hot air balloons expand when they are
heated.
• Charles’s Law: the volume of a fixed quantity of gas at
constant pressure increases as the temperature increases.
• Mathematically:
V  constant  T
 kT
V
 constant  k
T
These are typical of
observations you might
make in the lab.
Notice that on this plot
the equation is of the
form y = a + b x and
the volume does NOT
go to 0 at x = 0 !!!
The Gas Laws
•
•
•
•
•
The Temperature-Volume Relationship:
Charles’s Law
A plot of V versus T is a straight line.
When T is measured in C, the intercept on the
temperature axis is -273.15C.
We define absolute zero, 0 K = -273.15C.
Note the value of the constant reflects the assumptions: of
a constant amount of gas and pressure.
And now the equation is of the form y = bx , i.e. a = 0.
The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
• Gay-Lussac’s Law of combining volumes: at a given
temperature and pressure, the volumes of gases which
react are ratios of small whole numbers.
The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
• Avogadro’s Hypothesis: equal volumes of gas at the same
temperature and pressure will contain the same number
of molecules.
• Avogadro’s Law: the volume of gas at a given
temperature and pressure is directly proportional to the
number of moles of gas.
The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
• Mathematically:
V  constant  n
 kn
• We will show that 22.4 L of any gas at 0 C and 1 atm
contain 6.02  1023 gas molecules = 1 mole of molecules.
The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
Ideal Gas Equation
• Consider the three gas laws.
1
V  (constant n, T )
• Boyle’s Law:
P
• Charles’s Law:
V  T (constant n, P)
• Avogadro’s Law:
V  n (constant P, T )
• We can combine these into a general gas law:
nT
nT
V
k
P
P
Ideal Gas Equation
• If the k is now defined as R, the proportionality constant
of (called the gas constant), then
nT 

V  R 
 P 
• The ideal gas equation is:
PV  nRT
• R = 0.08206 L·atm/mol·K = 8.314 J/mol·K
But how can you derive the value of R?
Recall this
slide and
how to convert
grams to moles.
Ideal Gas Equation
The Quantity-Volume Relationship:
Avogadro’s Law
Ideal Gas Equation
• We define STP (standard temperature and pressure) =
0C, 273.15 K, 1 atm.
• And now we see the volume of 1 mol of gas at STP is:
PV  nRT
nRT 1 mol 0.08206 L·atm/mol·K 273.15 K 
V

 22.41 L
P
1.000 atm
( notice the “>” should be a “ / “ )
Fig. 10.12 shows the actual results for a number of gases. Notice
that the “ideal gas model” does a pretty good job.
Ideal Gas Equation
The Ideal-Gas Equation and the Gas Laws
• If PV = nRT and n and T are constant, then PV = constant
and we have Boyle’s law.
• Other laws can be generated similarly.
• In general, if we have a gas under two sets of conditions,
then
P1V1 P2V2

n1T1 n2T2
Applications of The
Ideal Gas Equation
Gas Densities and Molar Mass
• Density has units of mass over volume.
• Rearranging the ideal-gas equation with M as molar mass
we get
PV  nRT
n
P

V RT
nM
PM
d 
V
RT
has units of mol·L-1
And now the units are
g·L-1
Ideal Gas Equation
Gas Densities and Molar Mass
• The molar mass of a gas can be determined as follows:
dRT
M
P
Volumes of Gases in Chemical Reactions
• The ideal-gas equation relates P, V, and T to number of
moles of gas.
• The n can then be used in stoichiometric calculations.
Ideal Gas Equation
Volumes of Gases in Chemical Reactions
Consider now the application of these ideas to chemical reactions.
eg,
2 NaN3 (s)  2 Na + 3 N2 (g)
Given the mass of sodium azide that reacts, the number of moles of
nitrogen gas generated may be calculated.
From this, the volume may be calculated at a given temperature and
pressure.
What volume of gas at 760 torr and 0 oC would be generated from
32.51 g of sodium azide which decomposes according to the equation
2 NaN3 (s)  2 Na + 3 N2 (g) ?
In order to answer this question, we need to know how many moles
of nitrogen will be generated (see Chem 121). Then we apply the
ideal gas law.
 1mol NaN3  3 mol N 2 

  0.75 mol N 2
32.51 g NaN3 
 65.02 g NaN3  2 mol NaN3 
nRT
P
0.75 mol N 2 0.0821 L  atm / mol  K 273 K   16.81 L of N gas
or V 
2
1 atm
PV  nRT yields the equation V 
Gas Mixtures &
Partial Pressures
• Since gas molecules are so far apart, we can assume they
behave independently.
• Dalton’s Law: in a gas mixture the total pressure is given
by the sum of partial pressures of each component:
Ptotal  P1  P2  P3  
• Each gas obeys the ideal gas equation:
RT 

Pi  ni 

V 
Gas Mixtures &
Partial Pressures
• Combing the equations
RT 

Ptotal  n1  n2  n3  

V 
Partial Pressures and Mole Fractions
• Let ni be the number of moles of gas i exerting a partial
pressure Pi, then
Pi  i Ptotal
where i is the mole fraction (ni/nt).
A mixture of gases containing 0.538 mol He(gas 1), 0.315 mol Ne (gas 2),
and 0.103 mol Ar (gas 3) is confined in a 7.00-L vessel at 25 0C.
(a) Calculate the partial pressure of each gas. (b) Calculate the total
(b) pressure in the vessel. (c) Calculate the mole fraction of each gas.
P1 = n1 RT / V = (0.538 mol)(0.0821 L-atm/mol-K)(298 K) / (7.00 L)
= 1.88 atm of He
similarly, P2 = 1.10 atm of Ne, and P3 = 0.360 atm Ar
The total pressure is just
PT = P1 + P2 + P3 = ∑ Pi
= 1.88 + 1.10 + 0.360 = 3.34 atm
The mole fraction of He is X1 = P1 / PT = 1.88/3.34 = 0.563
(This also could be obtained from X1 = n1 / nT = 0.538/0.956
Likewise, X2 = 0.329 and X3 = 0.108
note that ∑ Xi = 1.00 always (within error limits):
0.563 + 0.329 + 0.108 = 1.00
Gas Mixtures &
Partial Pressures
Collecting Gases over Water
• It is common to synthesize gases and collect them by
displacing a volume of water.
• To calculate the amount of gas produced, we need to
correct for the partial pressure of the water:
Ptotal  Pgas  Pwater
Gas Mixtures &
Partial Pressures
Collecting Gases over Water
10.7 Kinetic Molecular Theory
• Theory developed to explain gas behavior.
• Theory based on properties at the molecular level.
• Assumptions:
– Gases consist of a large number of molecules in constant
random motion.
– Volume of individual molecules negligible compared to volume
of container.
– Intermolecular forces (forces between gas molecules)
negligible.