dr. c. weldon mathews chem 122 office: 0042 evans lab telephone: 292-1574 email:...

Dr. C. Weldon Mathews Chem 122Office: 0042 Evans LabTelephone: 292-1574email: [email protected] 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

mailto:[email protected]

Chapter 10 Gases

10.1 Characteristics of Gases 39810.2 Pressure 400

Atmosopheric Pressure and the Barometer10.3 The Gas Laws 404

The Pressure-Volume Relationships: Boyle's LawThe Temperature-volume Relationship: Charles's LawThe Quantity-Volume Relationship: Avogadro's Law

10.4 The Ideal-Gas Equation 408Relating the Ideal-Gas Equation and the Gas Laws

10.5 Further Applications of the Ideal-Gas Equation 413Gas Densities and Molar MassVolumes of Gases in Chemical Reactions

10.6 Gas Mixtures and Partial Pressures 417Partial Pressures and Mole FractionsCollecting Gases over Water

10.7 Kinetic-Molecular Theory 420Application to the Gas Laws

10.8 Molecular Effusion and Diffusion 423Graham's Law of EffusionDiffusion and Mean Free Path

10.9 Real Gases: Deviations from Ideal Behavior 427The van der Waals Equation

1st Lecture&

1st Quiz

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.

Evaluation – Judge the value of materials in terms of internal and external criteria.

This is a grossly abbreviated adaptation from Bloom, B. S. (Ed.) (1956) Taxonomy of educational objectives: The classification of educational goals: Handbook I, cognitive domain. New York; Toronto: Longmans, 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

Expectedin this

course.

Expectedin

Research.

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 ityour 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.

Gases

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

• Pressure is the force acting on an object per unit area:

• 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.

AF

P

Pressure

AF

P

F = ma = (104 kg)(9.8m/s2)

= 1 x 105 kg-m/s2

= 1 x 105 N

2

5

1

101

m

Nx

A

FP

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


Notice that the top endof the BAROMETER isclosed and that a “Torricelli VACUUM”exists above the mercury.

Pressure


• 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.

Pressure

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.

The Gas Laws

In our discussions aboutgases, we’ll often usethe gas cylinder with amovable piston as ahelpful analogy.

You also may think of a bicycle pump as anexample.

Let’s run an “experiment”

animation

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:

• 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.

P

k

PV

1constant

kPV

constant

PV

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:

The Gas LawsThe Gas Laws

kT

TV

constant kconstant

T

V

These are typical ofobservations you mightmake in the lab.

Notice that on this plotthe equation is of theform y = a + b x andthe volume does NOT go to 0 at x = 0 !!!

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 -273.15C. • We define absolute zero, 0 K = -273.15C.• 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 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.



• 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.



• Mathematically:

• 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.

nk

nV

constant

The Gas Laws


The Gas Laws

• Consider the three gas laws.

• We can combine these into a general gas law:

), (constant 1

TnP

V

), (constant PnTV

),(constant TPnV

• Boyle’s Law:

• Charles’s Law:

• Avogadro’s Law:

P

nTk

P

nTV

Ideal Gas Equation

• If the k is now defined as R, the proportionality constant of (called the gas constant), then

• The ideal gas equation is:

• R = 0.08206 L·atm/mol·K = 8.314 J/mol·K

PnT

RV

nRTPV

But how can you derive the value of R?

Ideal Gas Equation


Recall thisslide andhow to convertgrams to moles.

Ideal Gas Equation

• We define STP (standard temperature and pressure) = 0C, 273.15 K, 1 atm.

• And now we see the volume of 1 mol of gas at STP is:

L 41.22

atm 000.1K 15.273KL·atm/mol· 0.08206mol 1

PnRT

V

nRTPV

Ideal Gas Equation

( 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.

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

22

22

11

11TnVP

TnVP

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

RTP

dV

nRTP

Vn

nRTPV

MM

has units of mol·L-1

And now the units are g·L-1

Applications of The Ideal Gas Equation

Gas Densities and Molar Mass• The molar mass of a gas can be determined as follows:

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.

PdRTM

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 ofnitrogen gas generated may be calculated.

From this, the volume may be calculated at a given temperature andpressure.

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) 2 Na + 3 N2 (g) ?

In order to answer this question, we need to know how many molesof nitrogen will be generated (see Chem 121). Then we apply theideal gas law.

23

2

3

33 75.0

2

3

02.65

151.32 Nmol

NaNmol

Nmol

NaNg

NaNmolNaNg

gasNofL

atm

KKmolatmLNmolVor

P

nRTVequationtheyieldsnRTPV

22 81.16

1

273/0821.075.0

• 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:

• Each gas obeys the ideal gas equation:

321total PPPP

VRT

nP ii

Gas Mixtures & Partial Pressures

• Combing the equations

Partial Pressures and Mole Fractions

• Let ni be the number of moles of gas i exerting a partial pressure Pi, then

where i is the mole fraction (ni/nt).

VRT

nnnP 321total

totalPP ii


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 Hesimilarly, 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 = 0.538/0.956Likewise, X2 = 0.329 and X3 = 0.108

note that ∑ Xi = 1.00 always (within error limits): 0.563 + 0.329 + 0.108 = 1.00

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:

watergastotal PPP


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.

10.7 Kinetic Molecular Theory10.7 Kinetic Molecular Theory

dr. c. weldon mathews chem 122 office: 0042 evans lab telephone: 292-1574 email:...

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