Transcript P 2 - Macomb
RSPT 1060
MODULE C Lesson #4
GAS LAWS
OBJECTIVES
• • • • • • • • • •
At the end of this module, the student should be able to…
Define terms associated with gas laws.
Define Boyle’s Law.
Describe the relationship between volume, pressure, mass and temperature.
Describe how Boyle’s Law can be used to explain normal ventilation.
Given appropriate information, use the mathematical formula for Boyle’s Law to solve for an unknown.
Define Charles’s Law.
Describe the relationship between volume, pressure, mass and temperature.
State a clinical example of when Charles Law is applied in respiratory therapy.
Given appropriate information, use the mathematical formula for Charles’s Law to solve for an unknown.
OBJECTIVES
• • • • • • • •
At the end of this module, the student should be able to…
Define Gay Lussac’s Law.
Describe the relationship between volume, pressure, mass and temperature.
Give a clinical example of when Gay-Lussac's Law is applied in respiratory therapy Given appropriate information, use the mathematical formula for Gay-Lussac's Law to solve for an unknown.
State the combined gas law.
Given appropriate information, use the mathematical formula for the Combined Gas Law to solve for an unknown.
State the Universal (Ideal) Gas Law.
SUPPORTIVE READINGS
•
Egan
: Gas Behavior Under Changing Conditions, pgs. 109 – 111 •
Sibberson’s Math for RC:
• • • • • Chapter 2 – Fourth Set.
Chapter 2 – Fifth Set.
Boyle’s Law
, pgs. 17
Charles’s Law
– 19, Sample Problems , pgs. 19 – 20, Sample Problems Chapter 2 – Sixth Set.
Gay Lussac’s Law
, pgs. 20 – 21, Sample Problems Chapter 2 –
Combined Gas Law
, pgs. 22 – 26, Sample Problems Seventh & Eighth Set.
Chapter 2 – Practice Exercises, pgs. 29 – 32, #21 – 60.
Web Sites
•
http://www.grc.nasa.gov/WWW/K 12/airplane/boyle.html
Gas Laws
• Laws describing the behavior of gases.
• Supported by the Kinetic Molecular Theory • Six assumptions • Applies to most situations.
• Exceptions to these laws may occur when there is… • Extremely high pressures • Extremely low temperatures
Physical Properties Compared in the Gas Laws
• Mass – “amount of matter” • Pressure – “The force per unit of surface area” (pounds per square inch or psi). Results from molecular collisions. • Temperature – measurement of the degree of molecular activity • Volume – “space occupied by matter” For a gas it is the volume of the container because gases will always fill the container.
Gas Law - Summary Table
Boyle’s Charles’s
Law: Formula:
P1xV1=P2xV2
Constant: Temperature
& Mass Relationship: Inverse: P = V
Rearranged:
V2=P1xV1 P2 V1 = V2 T1 T2 Pressure & Mass Direct: T = V V2=V1xT2 T1 Gay Lussac’s P1 = P2 T1 T2 Volume & Mass Direct: T = P P2=P1xT2 T1 P2=P1xV1 V2 Combined P1xV1 = P2xV2 T1 T2 Mass Variable V2=V1xP1xT2 P2xT1 P2=V1xP1xT2 V2xT1 T2=P2xV2xT1 P1xV1
Memory Game
“What remains constant?” • Boyle’s = “Boiling” • (Temperature constant) • Charles’s = Charlie watches TV.
• (Pressure constant) • Gay Lussac’s =
GV
• (Volume constant) Gas volume
Boyle’s Law
• Constants: Mass & Temperature •
Measured under Isothermic conditions
• • Constant temperature Difficult to accomplish P 1 x V 1 = P 2 x V 2 •
Opposite of Isothermic: “Adiabatic”
• means a varying temperature
Boyle’s Law
•
Inverse
relationship: P 1 x V 1 = P 2 x V 2 • As pressure exerted on a gas is increased, volume will decrease.
• As pressure exerted on a gas is decreased, volume will increase.
Boyle’s Law
• Solve for the unknown : P 1 x V 1 = P 2 x V 2 P 1 x V 1 = P 2 x V 2 (both known) (one known & one unknown)
NBRC Question
• You have 3 liters of gas at 770 mmHg. The volume is changed to 2.5 liters. (The mass and temperature are constant.) Which of the following statements would be true concerning this situation ?
I.
II.
III.
IV.
V.
The pressure has increased The pressure has decreased The pressure is now 924 mmHg The pressure is now 641.7
The pressure has not changed
Calculation
• P 1 • V 1 x V 1 = P 2 x V 2 = 3 liters of gas • • P 1 V 2 = 770 mmHg = 2.5 liters (volume decrease) • P 2 = ______________? Rearrange the formula
increase or decrease?
P 2 = P 1 xV 1 V 2 =
Choose an answer
• • • • • I The pressure will increase II The pressure will decrease III The pressure will be 924 mmHg IV The pressure will be 641.7
V The pressure will not change
a.
I & II
b.
I & III
c.
II & III
d.
II & IV
e.
V only
Practice
• Sibberson’s Practical Math for RC: • Chapter 2: Boyle’s Law , pgs, 17-19, Sample Problems Fourth Set.
Problem #1
• P 1 x V 1 = P 2 x V 2 • • • • V 1 =6.4L
P 1 = 720 mmHg V 2 P 2 = 4.75L
= __________? increase or decrease?
Rearrange the formula P 2 = P 1 xV 1 V 2 =
Examples
• Closed syringe • Close off the end of a syringe and pull back. • Vacuum is formed with pressure decrease and volume increase.
• Normal breathing • Muscle contraction and inspiration causes decreased pressure in the pleural space and lungs.
• Decreased pressure yields volume movement into the lungs and increase in volume.
Temp constant Pressure change = Volume change
Charles’s Law
•
V
1
T
1
V
2
T
2
Constants: Mass & Pressure
Direct Relationship
– as temperature increases, volume increases
Charles’s Law
• Rearrange the formula to find V 2
V T
1 1
V
2
T
2
NBRC Question
• If you have 2 liters of a gas at 37°C and 752 mmHg and you change the temperature to 68°C without changing the pressure (constant pressure is Charles’s Law), what is the new volume of gas? Do you expect the volume to increase or decrease?
• • • • • A 2.2 Liters B 1.8 liters C 2 liters D 2.4 liters E 1.6 liters
Temperature Scales
• When working with gas laws – always convert temperature to Kelvin.
• ° Celsius (C) + 273 ° = ° Kelvin (K)
Calculation
V
1
T V
1 1
V
2
T
2
2 liters of gas T 1
37
C
273
K
310
K T
2
V
2
68
C
273
K
UNKNOWN
341
K
increase/d ecrease
Practice
• Sibberson’s Practical Math for RC: • Chapter 2: Charles’s Law , pgs, 19 - 20, Sample Problems Fifth Set.
Examples
• • Balloon filled with air • Put in refrigerator and it shrinks • Put by heater and it expands Pulmonary Function Testing • • • Patient exhales warm gas (37 ° spirometer (room temp).
C) into cold Measured gas volume will be less than actual volume in the lungs.
Measured volume must be corrected from ATPS to BTPS.
Gay-Lussac’s Law
• Constants: Mass & Volume
Direct Relationship
– as temperature increases, pressure increases
P
1
T
1
P
2
T
2
Gay-Lussac’s Law
P
1
T
1
• Rearrange the formula to find P 2
P T
2 2
NBRC Question
• You have 1.5 liters of a gas at 40° C and 750 mmHg pressure. The temperature of the gas is changed to 25° C without changing the volume (constant volume is Gay-Lussac’s Law), what is the new pressure of gas? Do you expect the pressure to increase or decrease?
• • • • • A The new pressure will increase by a factor of 25 B The new pressure will increase by a factor of 15 C 417 mmHg will be the new pressure D 714 mmHg will be the new pressure E There will not be a pressure change
Temperature Scales
• When working with gas laws – always convert temperature to Kelvin.
• ° Celsius (C) + 273 ° = ° Kelvin (K)
Calculation
P
1
T P
1 1
P
2
T
2
750
mmHg
T 1
40
C
273
K
313
K T
2
P
2
25
C
273
K
UNKNOWN
298
K
increase/d ecrease
P
2
P
1
T
2
T
1
Calculation
Choose an answer
A. The new pressure will increase by a factor of 25 B. The new pressure will increase by a factor of 15 C. 417 mmHg will be the new pressure D. 714 mmHg will be the new pressure E. There will not be a pressure change
Practice
• Sibberson’s Practical Math for RC: • Chapter 2: Gay Lussac’s Law Sample Problems Sixth Set.
, pgs, 20 - 21,
Examples
• • • Gas cylinder • Exposure to increased temperatures will cause the pressure in the cylinder to rise Bicycle tires • On a hot day the tire pressure will be higher than the pressure on a cold day Automobile tires • After driving a car for a while the tire pressure will increase as the tires heat up.
P
1
V
1
T
1
Combined Gas Law
P
2
V
2
T
2
•
Mass is the only constant
•
This formula can replace all previous.
Temperature & Pressure & Volume Relationships
Guidelines
• Before doing any calculations • Must correct temperature to Kelvin • Before doing any calculations • • Must subtract water vapor Use the “ Temperature & Humidity Chart ”
Sibberson Math Book – page 24
Sibberson Math Book – page 25
Set up a chart before you set up your formula
• P 1 – P H 2 O = ___ • P 2 – P H 2 O = ___ • V 1 = ___ • V 2 =___ • T 1 (convert to °K) = ____ Fill in chart then set up formula : • T 2 (convert to ° K) = ____
P
1
T V
1 1
P
2
V
2
T
2
NBRC Question
• A gas is at 42 ° C and 760 mmHg pressure. It occupies a volume of 2.5 liters. The temperature is decreased to 37 ° C and the volume decreases to 2 liters. What is the new pressure? (decreased or increased?) • • • • • Set up the table Change temp. to Kelvin Subtract P H 2 O (if indicated) Rearrange formula Solve for unknown
Set up a chart before you set up your formula
• P 1 – P H 2 O = ___ • P 2 – P H 2 O = ___ • V 1 = ___ • V 2 =___ • T 1 (convert to °K) = ____ Fill in chart then set up formula : • T 2 (convert to ° K) = ____
P
1
T V
1 1
P
2
T V
2 2
P
1
V
1
T
2
V
2
T
1
P
2
Practice
• Sibberson’s Practical Math for RC: • Chapter 2: Combined Law , pgs, 22 - 26, Sample Problems Seventh & Eighth Set.
• Seventh set is dry gases • Eighth set is gases with water vapor
Problem #1
page 26 (vapor) On hand is a gas volume of 5.8 L, at a temp. of 32C, and an atmospheric pressure of 722 mmHg, ATPS. Find the new volume if the gas was measured at STPD.
• P 1 – P H 2 O = ___ • P 2 – P H 2 O = ___ • V 1 = ___
P
1
V
1
T
1
P
2
V
2
T
2
• T 1 (convert to °K) = ____ • V 2 =___ • T 2 (convert to ° K) = ____
5. Universal Gas Law
• • • • P 1 xV 1 • = nRT n = Gram molecular weight (mass) • R = 22.4 L (molar volume) All parameters can vary • Pressure • Temperature • • Volume Mass Used in situations where mass is varying Not used in Respiratory Therapy
Gas Law - Summary Table
Boyle’s Charles’s
Law: Formula:
P 1 xV 1 =P 2 xV 2
Constant:
Temperature & Mass Relationship: Inverse: P = V
Rearranged: V 2
=P 1 xV 1 P 2 V 1 T 1 = V T 2 2 Pressure & Mass Direct: T = V
V 2
=V 1 xT 2 T 1
P 2
=P 1 xV 1 V 2 Gay Lussac’s P 1 T 1 = P T 2 2 Volume & Mass Direct: T = P
P 2
=P 1 xT 2 T 1 Combined P 1 xV 1 T 1 Mass = P 2 xV 2 T 2 Variable
V 2
=V 1 xP 1 xT 2 P 2 xT 1
P 2
=V 1 xP 1 xT 2 V 2 xT 1
T 2
=P 2 xV 2 xT 1 P 1 xV 1