rspt 1060
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RSPT 1060
MODULE C
Lesson #3
Properties of Gases
OBJECTIVES
• At the end of this module, the student will:
• Define terms associated with the properties of gases and gas mixtures.
• Identify abbreviations used during expressions of the properties of gases or gas mixtures.
• State the six principle assumptions that explain the unique properties of gases.
• State the effect of gas particle size on gas density.• State the effect of the distance between particles
of a gas on its compressibility.• State the effect of kinetic energy of a gas on
diffusion of that gas through an environment.• State the effect of molecular attraction of a gas on
its physical properties.
OBJECTIVES
• At the end of this module, the student will: • State the effect of kinetic energy of a gas on the pressure it
exerts.• State the relationship between the kinetic activity of a gas,
the temperature of the gas and the pressure it exerts. • List the major gases that comprise the atmosphere and their
fractional concentrations.• Describe the clinical significance of hyperbaric & hypobaric
conditions.• State Dalton’s Law.• Given the barometric pressure, calculate the partial pressure
of a gas.• Explain how changes in altitude will affect partial pressure
and fractional concentration of the gas.• Describe why gases need to be converted to Standard
Conditions.
OBJECTIVES
• At the end of this module, the student will: • Given a table of conversion factors, convert between the
following standard conditions: • ATPS• STPD• BTPS
• Describe how gases in the atmosphere differ from gases in the lungs and blood.
• State the partial pressure of water vapor at body temperature (37 C) with a relative humidity of 100%.
• List the normal values for various partial pressures found in the lungs and blood
• State how pressure is measured. • Express one atmosphere in units commonly encountered in
Respiratory Therapy.• Convert between units of pressure commonly encountered
in Respiratory Therapy.
OBJECTIVES
• At the end of this module, the student will:
• Explain the significance of Avogadro's Law and number.
• Calculate the gram-molecular weight of a gas.
• Calculate the density of gases and gas mixtures commonly found in Respiratory Therapy.
• Explain the relationship between density, mass and volume.
Kinetic-Molecular Theory
Six principle assumptions
that explain the
unique properties of gases.
Assumption 1: Gas Particle Size
Theory:Gas consists of tiny particles called molecules.
Gas Property:Gas density is very low. (mass/volume)
Assumption 2: Distance Between Particles
Theory:The distance between molecules of gas is very great compared to the size of the molecules.
The volume occupied by a gas mainly consists of the empty space between molecules.
Gas Property:
Gases are easily compressed.
Clinical example: Gas compressed into cylinders
Assumption 3: Gas Particle Movement
Theory:Gas molecules are in rapid motion and move in straight lines, frequently colliding with each other and with the walls of the container.
Gas Property:
Gases diffuse. One gas can move through the empty space between another gas.
Clinical example: odors moving across a room
Assumption 4: Molecular Attraction
Theory:Gas molecules do not attract each other.
Gas Property:Gases can fill any size container so they have no definite volume or shape.
Gases are easily expanded.
Gases flow.
Clinical example: gas expands as it leaves a compressed gas cylinder.
Assumption 5: Collisions and Energy Loss
Theory:When molecules of a gas collide with each other or with the walls of the container, they bounce back with no loss of energy.
Such collisions are said to be perfectly elastic.
Gas Property:Gas exerts pressure equally in all directions.
Gases do not settle out and stop moving.
Clinical examples:
Gases may layer somewhat in the atmosphere (heavy near bottom) but molecular activity and gas flow limits this to some extent.
In a gas cylinder the pressure is the same everywhere inside.
Assumption 6: Kinetic Energy and Temperature
Theory:
The average kinetic energy of the molecule is the same for all gases at the same temperature. The average kinetic energy increases as the temperature increases and decreases as the temperature decreases.
Gas Property:
Heating a gas increases the pressure it exerts on the walls of it’s container and cooling a gas decreases the pressure exerted on the walls of it’s container.
Clinical examples: Gas cylinder in a fire can rupture.
Gay-Lussac’s Law.
Question:
I. They have large spaced between molecules & atoms.
II. The have low densitiesIII. They have high densities
A. I ONLYB. II ONLYC. I and II ONLYD. II and III ONLYE. I, II, and III
Why can gases be compressed easily?
Question:
A. Molecular particles of gases are in constant motion
B. Molecules of one gas cannot move in the spaces between other gas molecules
C. Gas molecules are highly attracted to each other
D. The distance between gas molecules is very small
Which of the following is true regarding why a gas with an odor can travel across a room?
Question:
A. True
B. False
All gas molecules eventually run out of energy and settle to the lowest surface.
Question:
A. True
B. False
The pressure inside a compressed gas cylinder is greater at the bottom than it is at the top.
Question:
A. Increase the energy & increase the activity
B. Increase the energy but decrease the activity
C. Decrease the energy but increase the activity
D. Decrease the energy and decrease the activity
E. It has no affect on energy or activity
What will heating a gas do to its kinetic energy & molecular activity?
Gas in the Atmosphere
A. Definitions
B. Relationship between: Altitude Pressure Gas Concentrations
Definitions:
• Pressure – (P) The force per unit of surface area (pounds per square inch or psi). Colliding molecules exert pressure.
• Tension – Pressure of gas exerted in a liquid.
• Fractional Concentration – Percent expressed as a decimal. Percent is parts per 100 parts.
• Partial Pressure – Portion of the total pressure being exerted by one gas in a gas mixture.
Definitions:• Barometric Pressure – (PBARO) Pressure
exerted by gases in the atmosphere also known as atmospheric pressure or ambient pressure.
• One Atmosphere – (PATM) 760 mmHg pressure
• Hypobaric – Pressure below 760 mmHg
• Hyperbaric – Pressure above 760 mmHg
• Water Vapor Pressure – (PH2O) Pressure exerted by water in the gas (vapor) form.
1034 cm H2O…14.7 psi…760 torr…29.92 in Hg
Relationships
Relationship between:
• Altitude
• Pressure
• Gas Concentrations
PO2 mmHg
18
29
47
73
110
159
Altitude
a. At Sea Level Normobaric One atmosphere (760 mmHg)
b. Above Sea Level Hypobaric condition < 760 mmHg
c. Below Sea Level Hyperbaric condition >760 mmHg
Sea Level • 760 mmHg
• One atmosphere
• Composition of air:• FO2 = .2095 (21%)
• FN2 = .7809 (78%)
• FCO2 = 0.0003 (0.03%)
• Traces of Argon, Neon, Krypton, Hydrogen, Xenon. Ozone, Radon
Altitude, Pressure & Gas Concentrations
Calculating Partial Pressure
Pgas = Fractional Concentration x Barometric Pressure
• Example:
FO2 x PBaro = PO2
.2095 (21%) x 760 = 159.22 or 159 mmHg
Above Sea LevelLess then 760 mmHg
• Hypobaric
• Composition of air:• FO2 = .2095 (21%)
• FN2 = .7809 (78%)
• FCO2 = 0.0003 (0.03%)
• Traces of Argon, Neon, Krypton, Hydrogen, Xenon. Ozone, Radon
Increased Altitude and Pressure Changes
Pressure Changes With Hypobaric Conditions
Altitude (ft) PBaro (mmHg)
Hypobaric
Oxygen %
(FO2)
Oxygen pressure (PO2 mmHg)
0 760 21% 159
10,000 532 21% 110
20,000 349 21% 73
30,000 226 21% 47
40,000 141 21% 29
50,000 87 21% 18
Altitude and Pressure Changes
• As altitude increases, gravitational pull decreases.
• Gas molecules move apart.
• Density decreases.
• Molecule collisions decrease.
• Pressure decreases.
Summary: Hypobarism and Pressure Changes
• As the altitude increases the barometric pressure ___________.
• As the altitude increases the gas composition (%) _______________.
• As the altitude increases, the partial pressure exerted by each gas ________.
• _____________ occurs at high altitudes because the oxygen pressure is too low.
Why are oxygen masks needed on airplanes?
People who have traveled by air are familiar with the safety instructions given by the crew before flight. Instructions are included on how to use the oxygen masks. When and why are these masks needed?
Cruising altitude 30,000 ftPBaro outside plane = 226 mmHg
FO2 = _____________
PO2 = _____________
Why are oxygen masks needed on airplanes?
Oxygen masks provide FO2 of 0.7 (70%)
Cruising altitude 30,000 ft
PB outside plane = 226 mmHg
What PO2 will this provide? _______
Is this adequate to support life? ____
Below Sea LevelAbove 760 mmHg
• Hyperbaric
• Composition of air:• FO2 = .2095 (21%)
• FN2 = .7809 (78%)
• FCO2 = 0.0003 (0.03%)
• Traces of Argon, Neon, Krypton, Hydrogen, Xenon. Ozone, Radon
Reduced Altitude and Pressure Changes
Pressure Changes In Hyperbaric Conditions
Below Sea Level
(ft)
PBaro (mmHg)
Hypobaric
Oxygen %
(FO2)
Oxygen pressure (PO2 mmHg)
0 760 (1 ATM) 21% 159
33 1520 (2 ATM) 21% 321
66 3040 (4 ATM) 21% 638
132 6080 (8 ATM) 21% 1277
Summary: Hyperbarism and Pressure Changes
• As the altitude decreases the barometric pressure ___________.
• As the altitude decreases the gas composition (%) ________________.
• As the altitude decreases, the partial pressure exerted by each gas ________.
• Hypoxemia __________ below sea level because the oxygen pressure is _____.
Hyperbaric Medicine
• 2 – 3 Atmospheres - used to increase oxygen in the blood
• Up to 6 atmospheres - used for high pressure therapy.
Egan
Fig. 38-25
A, Fixed hyperbaric chamber.
B, Monoplace chamber.
Hyperbarism: Indications•Indications for 2 – 3 Atmospheres
• Used to increase partial pressure of oxygen in the blood
• CO Poisoning• Wound care (kills anaerobic bacteria)• Gas Gangrene• Cyanide poisoning• Extreme blood loss (anemia)
Hyperbarism: Indications
• Indications for Up to 6 atmospheres used for high pressure therapy.• Treatment for nitrogen bubbles in the blood
stream.• The bends (decompression sickness) as a
result of deep sea diving accidents.
• Air embolism
Hyperbarism: Complications
• Barotrauma of ears, gas in pleural space, gas bubbles in blood stream.
• Oxygen toxicity
• Fire hazards
• Rapid decompression problems
• Cataract worsening
• Claustrophobia
Hyperbaric Oxygen Therapy
• For additional information read:
Egan - Hyperbaric Oxygen, pages 891 – 895
Dalton’s Lawo “The total pressure of a gas mixture is
equal to the sum of the partial pressures of its individual gases.”
o PTOTAL = P1 + P2 + P3 + P4…
o Each gas exerts a pressure proportional to its concentration.
Dalton’s Law
• Partial Pressure - Partial pressure is the pressure exerted by a single gas.
• PPartial = PTotal x Fractional Concentration
• Example:760 x 0.21 = 160 mmHg
760 x 0.4 = 304 mmHg
• Example:• PAtm = PN2 + PO2 + PCO2 + PAr…..
Dalton’s Law - Practice
• What is the total pressure in a dry gas mixture given the following information:• PO2 150 mmHg
• PN2 500 mmHg
• Anesthetic gas 8 mmHg
Answer: 150 + 500 + 8 = 658 mmHg
NOTE: Dry means there is no effect from water vapor pressure….
Water Vapor Pressure
• Pressure exerted by water in the vapor (gaseous) form.
• PH2O
• In the lungs: • All gas is warmed to body temperature (37° C)
and humidified to a saturation of 100% (fully saturated).
• This results in a water vapor pressure (PH2O) of 47 mmHg.
• It also results in a content (density) of 44 mg/L.
Water Vapor in Calculations• Before working with gas mixtures
containing water vapor – remove the water vapor by subtracting the proper amount.• Inspired (I) (remove 47 mm Hg)
• Dry (subtract nothing)
• In between, the amount of water vapor will vary based on temperature.
Water Vapor - Practice• If the total pressure in a gas mixture is
730 mmHg, what is the partial pressure of the anesthetic gas if the PO2 is 150 mmHg, PH2O is 47 mmHg and PN2 is 480 mmHg?
Water Vapor Pressure
• Question:• How do I know how much pressure the water
vapor is exerting?
• Answer:• It is based on two things:
• Temperature of the gas• % saturation
• It’s easy - Look at the chart!
Water Vapor Pressures & Contents at Selected Temperatures
TABLE 6-3 Egan page 103
Dalton’s Law - Practice•Gas fully saturated with water
•Temperature 37°C
•PBaro 760 mmHg
•FiO2 0.21
•What is PO2?
•Gas fully saturated with water•Temperature 32°C•PBaro 760 mmHg•FiO2 0.21
•What is PO2?
Gas Composition Room Air vs. Alveoli
Gas % Room AirPartial
pressure
% AlveoliPartial
Pressure
Nitrogen 78% 593
mmHg
74% 562
mmHg
Oxygen 21% 160 14% 106
Carbon Dioxide
0.03% 0.2 5% 38
Argon, Neon,
Krypton…
0.9% 6.8 <1% <6
Water Vapor
0% 0 6.2% 47
Gas in the Lungs & Blood
A. Gas in the lungs
B. Gas in the blood
C. Comparison of lungs vs. blood
D. Gas conditions
O2 CO2 O2 CO2
156 0 115 35
110 0 80 40
80 40 40 45
Arterial sideArterial side Venous sideVenous side
ExhalationInhalation
Air
Lung
Blood Blood
Lung
Air
A. Gas in the lungs
• PAO2 = oxygen pressure in the lungs• Calculated value• PAO2 = (PBaro – PH20) x FiO2 – (PaCO2/0.8)• PBaro = barometric pressure • PH20 = 47 mmHg• FiO2 = fractional concentration of oxygen
• PaCO2 = CO2 in the blood• Measure by blood gas analyzer• Normal 35 – 45 mmHg
• 0.8 = Respiratory Exchange Ratio• Measured by indirect calorimetry• CO2 Production / O2 consumption• 200 mL/min./250 mL/min
Magnified view of an alveolus.
Alveolar walls, or septa, are occupied mainly by capillaries.
The interstitium contains a few interstitial fibers, composed mostly of reticular support fibers, elastic fibers, and one interstitial cell.
The small distance between blood and air makes gas exchange remarkably efficient.
PAO2
Scanning Electron Micrographs of Alveolar Air Spaces.
Note thin partitions or septa between adjacent alveoli.
Gas in the Blood
• PaO2 = oxygen pressure in the blood• Measured by blood-gas analysis (ABG)• Normal arterial level 80 – 100 mm Hg
• So how do we determine if the amount measured is OK?• A-aDO2
• PaO2/PAO2 Ratio
• PaO2/FiO2 Ratio
Alveolar-Arterial Difference
PaO2 = 80 - 100
PAO2 = 106
PH2O = 47 mmHg
PO2 = 160 mmHg
PAO2 – PaO2 = 5 – 15 mm Hg
Alveolar - arterial oxygen gradient• PAO2 – PaO2 Gradient• A-aDO2
• Used as an index of oxygenation.• 100 mm Hg – 90 mm Hg
• Normal on room air = 5 - 20 mmHg• Normal breathing 100% O2 = 25-65 mmHg
• Example: Given the following information, calculate the A-aDO2. PaO2 of 70 mm Hg PBaro: 747 mm HgFiO2: .21 PaCO2: 40 mm Hg RQ: 0.8
ANSWER: PAO2: [(747-47) x .21]-(40/.8) = 97 A-aDO2: 97 – 70 = 27 mm Hg
a/A Ratio
• PaO2/PAO2
• Evaluates the percentage of oxygen that moves from the alveolus to the artery.
• Expressed as a percentage, rather than a numerical value
• Normal: Greater than 75%
• Example: Given the following information, calculate the PaO2/PAO2. PaO2 of 70 mm Hg PBaro: 747 mm HgFiO2: .21 PaCO2: 40 mm Hg RQ: 0.8
ANSWER: PAO2: [(747-47) x .21]-(40/.8) = 97 PaO2/PAO2: 70/97 = .72 or 72%
PF Ratio
• PaO2 / FiO2
• Number used to track severity of oxygenation problem
• You DON’T need to calculate the PAO2
• Normal: 400 - 500• Severe problem < 200 (ARDS)
• Example: Given the following information, calculate the PaO2/FiO2 ratio. PaO2 of 70 mm Hg PBaro: 747 mm HgFiO2: .21 PaCO2: 40 mm Hg RQ: 0.8
ANSWER: PaO2/FiO2 = 70/.21 = 333
Gas Conditions• When comparing different gases, (density,
pressure, mass…) we have to make sure the conditions are the same for both gases.
• Conditions like:• Temperature• Barometric Pressure• Volume• Water Vapor Pressure
Gas Conditions• There are four conditions we use:
• ATPS – ambient temperature and pressure saturated
• ATPD – ambient temperature and pressure dry
• STPD- Standard temperature and pressure dry
• BTPS- Body temperature and pressure saturated
Gas Conditions• Room conditions - ATPS
• Room Temperature (given to you)
• Ambient Pressure (given to you)
• Saturated with water vapor (chart)
• Room conditions - ATPD• Room Temperature (given to you)
• Ambient Pressure (given to you)
• Dry (no water vapor pressure – PH2O = 0 mm Hg)
Gas Conditions• Standard Conditions - STPD
• Temperature = O° C• Pressure = 760 mmHg• Dry (no water vapor)
• Inside the lungs – BTPS• Body Temperature = 37°C• Ambient Pressure (given to you)
• Saturated with water vapor (chart)
Standard Conditions
Specific Conditions for GasesPage 25 - Sibberson
INITIALS TEMPERATURE (C) PRESSURE (mmHg)
Amount of water in gas
STPD 0 1 atmosphere (760 mmHg)
Dry
BTPS 37 Ambient Saturated
ATPS Room Temp. Ambient Saturated
ATPD Room Temp. Ambient Dry
Measuring Pressure andConverting
• Measuring Atmospheric Pressure• Mercury Barometer• Aneroid Barometer
• Measuring Airway and Blood Pressures• Manometer• Sphygmomanometer• Water column• Strain gauge
The major components of a mercury barometer.
Atmospheric pressure
Aneroid barometer
Atmospheric pressure
A mechanical manometer used to measure a patient's airway & blood pressure.
Sphygmomanometer
Blood pressure
Strain-gauge Blood Pressure Transducer.
A, No pressure is applied.
B, Pressure is applied to the transducer. An ammeter shows a change in electrical current proportional to the magnitude of pressure applied.
Units of Measure
Unit of measure One Atmosphere
Metric mm Hg or Torr 760
cm H2O 1034
Ft H2O 33.9
British lbs/in2 or psi 14.7
cm Hg 76
Standard International kpa 101.3
in Hg 29.9
Millibar 1014
Conversion
Given value x Unknown (atm pressure) = Unknown
1 Given (atm pressure)
Example: Convert 105 ft H2O to psi
105 ft H2O x psi = psi
1 ft H2O
105 ft H2O x 14.7 psi = 45.5 psi
1 33.9ft H2O
One atm in psi
One atm in ft H2O
ConversionConversion factor:
Pressure of one atmosphere of unknown & given values
Example: Convert 96 cmH2O to mmHg
96 cm H2O x mm Hg = mm Hg
1 cm H2O
96 cm H2O x 760 mm Hg = 70.6 mm Hg
1 1034 cmH2O
Conversion
Conversion factor:
Pressure of one atmosphere of unknown value
Pressure of one atmosphere of given value
• Where do I get the pressure of one atmosphere?• Look at the conversion chart given to you.
Units of MeasureUnit of measure One Atmosphere
Metric mm Hg or Torr 760
cm H2O 1034
Ft H2O 33.9
British lbs/in2 or psi 14.7
cm Hg 76
Standard International kpa
101.3
in Hg 29.9
Millibar 1014
Comparing Gases
A. Avogadro’s Law
B. Avogadro’s Number
C. Molar Volume
D. Moles
(Things I need to know to calculate density of gases)
Why should I care about density?•You will be administering different gases to your patients
•Air•Oxygen•Nitrogen•Helium•Carbon Dioxide•Carbon Monoxide•Nitric Oxide
•Density is one factor to consider when determining:
•which gas is given •for what reason•to which patient
Avogadro’s LawOne mole of any gas under the same conditions of temperature and pressure will contain the same number of molecules and occupy the same space.
Oxygen Air
1 atm - 0C - Dry 1 atm - 0C - Dry
STPD
http://www.bulldog.u-net.com/avogadro/avoga.html
In English Please!• One dozen of any kind on eggs under
the same conditions of temperature and pressure will contain the same number of eggs (12) and occupy the same space (an egg carton).
• So what changes between types of eggs from small to jumbo?• Weight (or mass) of eggs
Avogadro’s Law (as it applies to gases)
• The number of molecules in a mole will be Avogadro’s number:• 6.02x1023
• The volume (space) of a mole is the Molar Volume:• 22.4 liters
• What the heck is a “MOLE”?
Avogadro’s Number
• At STPD, one mole of gas always contains 6.02 x 1023 molecules and has a volume of 22.4 Liters.
6.02 x 1023
molecules
22.4Liters
STPD
Standard Temperature = 0°C
Standard Pressure = 1 atm
Dry = PH2O = 0 mmHg
Molar Volume
• Molar Volume is 22.4 L
• At STPD, one mole of gas always contains 6.02 x 1023 molecules and has a volume of 22.4 Liters.
• It is the space occupied by one mole of gas at STPD.
6.02 x 1023
molecules
Moles
• One mole of any pure substance:
• contains the same number of particles – 6.02x1023. This is Avogadro’s Number.
• Occupies the same volume 22.4 Liters. This is the Molar Volume.
One Mole of different gases under the same conditions of temperature and pressure contain the same number of molecules and occupy the same space.
The variable is the MASS of the gases.
Oxygen Air
1 atm - 0C - Dry 1 atm - 0C - Dry
STPD
Mole – a unit of measure
• Based on mass of molecules
• Each molecule has a different mass (weight).
• How do we find out what the mass is?
• Find out what elements & quantities a molecule is made of
• i.e., H2O = (2)H & (1)O
• Go to the periodic table and find the masses (amu)• Add them up for the mass of the molecule• Change the unit of measure to gram (g)
Atomic Mass & Gram Molecular Weights
SUBSTANCE MASS CALCULATION
Oxygen (O2) 2 oxygen @ 16 grams each
2 x 16 = 32 gms
Carbon Dioxide (CO2)
Potassium Chloride (KCl)
Sulfuric Acid (H2SO4)
MOLESSubstance Mass # Moles Types of
Particles
Oxygen (O) 16g 1 6.02x1023
atoms
Oxygen (O2) 32g 1 6.02x1023
molecules
CO2 ____ g 1 6.02x1023
molecules
Nitrogen (N2) ____ g 1 6.02x1023
molecules
Comparison of Moles at STPD
Oxygen (O2)
32g
22.4L
Neon (Ne)
20g
22.4L
Helium (He)
4g
22.4LSame volume &
# particles (6.02 x 1023)
Different weights and THEREFORE densities.
Practice:• Gram Molecular Weight
• Sibberson – Sample Problems First Set – Pages 13 – 14
• Use Periodic Table or chart on page 16
Calculating Density – Mass/Volume
A. Density of a single gas
B. Comparison of gases
C. Density of a gas mixture
Density of Single Gas
Gram Molecular Weight = Density Molar Volume
O2 = 32g = 1.43 g/L density @ STPD 22.4 L
CO2 = 44g = 1.96 g/L density @ STPD 22.4 L
Density of Single Gas
Gram Molecular Weight = Density
Molar Volume
He =
Comparison•Oxygen
•Volume 22.4 L•STPD•6.02 x 1023 molecules
•Weight 32g•Density 1.43 g/L
32 g/22.4 L = 1.43 g/L
•Carbon Dioxide•Volume 22.4 L•STPD•6.02 x 1023 molecules
•Weight 44g•Density 1.96 g/L
44 g/22.4 L = 1.96 g/L
Although the number of molecules and volume are the same, the weights, densities and masses are different.
Practice• Sibberson – Sample Problems Second Set –
Pages 14 – 15
• See Periodic Table or chart on page 16
Density of Gas Mixtures
• Find % of all constituent gases
• Find the GMW of all constituent gases
• Use the following formula:• (Gas #1 x %) + (Gas #2 x %)
22.4 Liters
What is the density of a 80%/20% mixture of He/Ox (Heliox)?
Practice• Sibberson – Sample Problems Third Set –
Pages 15 – 17
• See Periodic Table or chart on page 16
ASSIGNMENTS• Read Egan:
• Properties of Gases, pages 105 - 109 (skip Henry’s Law)
• Hyperbaric Oxygen, pages 891 – 895• Helium Therapy, pages 897 – 899
ASSIGNMENTS• Do Sibberson’s Math:
• Chapter 7 - Sample Problems Fourth Set: Dalton’s Law – Pgs. 84 – 86
• Chapter 7 - Practice Exercises 46 – 60 Pgs 91 – 91
• Chapter 10 – Sample Problems First, Second & Third Set: Resp. Exchange ratio – pgs 112 – 115
• Chapter 10 – Sample Problems Fourth & Fifth Set: Alveolar Air Equation & Gradient – pgs 115 – 116
• Chapter 10 – Practice Problems 1 – 45
ASSIGNMENTS• Do Sibberson’s Math:
• Chapter 2 – First Sample Set: Gram Molecular Weight, pgs. 13 -14
• Chapter 2 – Second Sample Set: Gas Density, pgs 14 -15
• Chapter 2 – Third Sample Set: Gas Mixture Density, pgs 15 -16
• Chapter 2 Practice Exercises #1 - 20, pgs.27-29
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