Download week-1 - OSU Chemistry

Dr. C. Weldon Mathews
Chem 122
Office: 0042 Evans Lab
Telephone: 292-1574
email: [email protected]
course web site:
http://www.chemistry.ohio-state.edu/~mathews/chem122wi07/
Chapter 10
10.1
10.2
10.3
Office hours: TR 12:30 - 2:00 pm
TR 4:00 - 5:00 pm
or by appointment
1st
Lecture
&
1st Quiz
10.4
10.5
Chapters we’ll cover in Chem 122:
10, 11, 13, 14, 15, 16, 17 (17.1-17.3)
10.6
10.0-10.6
10.7
First Week:
Second Week: 10.7-10.9
and
10.8
11.1-11.5
First Quiz: Week of Jan 8 (second week)
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
Review Chem 121, especially Chaps 8 and 9
Bloom’s Taxonomy–
[Approaches to learning Chemistry.]
Study Habits and Study Resources:
Knowledge – Simple recall of facts
a) “Lectures” and “Reading” - minimal impact by themselves
Comprehension – Translate into your own words or equations.
b) “Chemistry is not a Spectator Sport!”
Application – Apply concepts to specific situations; recognizing and solving a
problem when the equations are not given.
c) Recitation and Laboratory TAs
Analysis – Application plus recognition of important parts of problem.
d) Ask questions and seek help whenever you need it!
EXPECTATIONS
Prof. Janet Tarino, OSU Mansfield
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.
e) Web resources:
http://www.chemistry.ohio-state.edu/
http://www.chemistry.ohio-state.edu/~mathews/chem122wi07/
/~rbartosz/
/~rzellmer/
chemistry ->Undergraduate Program->Interactive Tutorials
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
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.
1
Characteristics of Gases
Characteristics of Gases
• Unlike liquids and solids, they
¾
¾
¾
¾
Expand to fill their containers.
Are highly compressible.
Have extremely low densities
Mix homogeneously with other gasses
Gases
Gases
Pressure
Pressure
F
P=
A
• Pressure is the force acting on an object per unit area:
F
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.
P=
F = ma
= (104 kg)(9.8m/s2)
= Newton
= 1 x 105 kg-m/s2
= 1 x 105 N
P=
F 1 x 105 N
=
A
1 m2
P = 1 x 105 N/m2
= 1 x 105 Pa
= 1 x 102 kPa
2
Pressure
Pressure
Atmosphere Pressure and the Barometer
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
• 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.
Notice that the top end
of the BAROMETER is
closed and that a
“Torricelli VACUUM”
exists above the mercury.
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.
Here Ph may be positive or negative!
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.
3
Let’s run an “experiment”
animation
The Gas Laws
The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law
The Pressure-Volume Relationship: Boyle’s Law
• Mathematically:
V = constant ×
1
k
=
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.
V = f(P) = ?
find VP = cnst = k
V = f(1/P)
= k(1/P)
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 !!!
4
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
• 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 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
• 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.
Ideal Gas Equation
The Quantity-Volume Relationship:
Avogadro’s Law
• 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:
V∝
nT
nT
=k
P
P
5
Ideal Gas Equation
Recall this
slide and
how to convert
grams to moles.
Ideal Gas Equation
The Quantity-Volume Relationship:
Avogadro’s Law
• 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?
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 “ / “ )
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
Fig. 10.12 shows the actual results for a number of gases. Notice
that the “ideal gas model” does a pretty good job.
6
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
has units of mol·L-1
=
V RT
nM
PM
=d =
RT And now the units are
V
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.
g·L-1
Ideal Gas Equation
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) Æ 2 Na + 3 N2 (g) ?
Volumes of Gases in Chemical Reactions
Consider now the application of these ideas to chemical reactions.
eg,
2 NaN3 (s) Æ 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.
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 + L
• Each gas obeys the ideal gas equation:
RT ⎞
Pi = ni ⎜⎛
⎟
⎝V ⎠
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.
⎛ 1 mol NaN 3 ⎞⎛ 3 mol N 2 ⎞
⎟⎟⎜⎜
⎟⎟ = 0.75 mol N 2
32.51 g NaN 3 ⎜⎜
⎝ 65.02 g NaN 3 ⎠⎝ 2 mol NaN 3 ⎠
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
• Combing the equations
RT ⎞
Ptotal = (n1 + n2 + n3 + L)⎜⎛
⎟
⎝V ⎠
Partial Pressures and Mole Fractions
• Let ni be the number of moles of gas i exerting a partial
pressure Pi, then
Pi = Χ i Ptotal
where Χi is the mole fraction (ni/nt).
7
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.
Gas Mixtures &
Partial Pressures
Collecting Gases over Water
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
• It is common to synthesize gases and collect them by
displacing a volume of water.
The total pressure is just
• To calculate the amount of gas produced, we need to
correct for the partial pressure of the water:
Ptotal = Pgas + Pwater
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
10.7 Kinetic Molecular Theory
Collecting Gases over Water
• 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.
8