Gas Laws For CHM1020

Similar documents
To convert to millimeters of mercury, we derive a unit factor related to the equivalent relationship 29.9 in. Hg = 760 mm Hg.

Name: Chapter 13: Gases

Chapter 13 Gases and Pressure. Pressure and Force. Pressure is the force per unit area on a surface. Force Area. Pressure =

Notes: Gas Laws (text Ch. 11)

Gas Law Worksheets - WS: Boyle s and Charles Law

Unit 8: Gases and States of Matter

Gases. Edward Wen, PhD

Lab Dates. CRHS Academic Chemistry Unit 11 Gas Laws Notes

Chapter 12. The Gaseous State of Matter

Worksheet 1.7: Gas Laws. Charles Law. Guy-Lassac's Law. Standard Conditions. Abbreviations. Conversions. Gas Law s Equation Symbols

CHAPTER 11: THE GASEOUS STATE

CHAPTER 14. The Behavior of Gases Properties of Gases. Factors Affecting Gas Pressure

Kinetic Molecular Theory imaginary Assumptions of Kinetic Molecular Theory: Problems with KMT:

Unit 9 Packet: Gas Laws Introduction to Gas Laws Notes:

Name Chemistry Pre-AP

2. Pressure Conversions (Add to your Conversion Sheet

Kinetic-Molecular Theory of Matter

Kinetic-Molecular Theory

Honors Chemistry Unit 7 Gas Laws Notes

Chapter 14-Gases. Dr. Walker

The Kinetic-Molecular Theory of Gases based on the idea that particles are always in motion

4.) There are no forces of attraction or repulsion between gas particles. This means that

Chemistry HP Unit 6 Gases. Learning Targets (Your exam at the end of Unit 6 will assess the following:) 6. Gases

Gases. Properties of Gases Gas Pressure

2. Convert these pressures to atm: 1 atm! Or to mm Hg, 760 mm Hg! 760 mm Hg! 1 atm. 800 mm Hg 380 mm Hg 0.75 atm 0.25 atm

Chapter 5: Gases 5.1 Pressure Why study gases? An understanding of real world phenomena. An understanding of how science works.

A. What are the three states of matter chemists work with?

Chapter 11. Recall: States of Matter. Properties of Gases. Gases

Gas Laws Chapter 14. Complete the following pressure conversion. Be sure to show how units cancel.

Gases. Unit 10. How do gases behave?

THE BEHAVIOR OF GASES

Chapter 11 The Behavior of Gases

Section 8.1 Properties of Gases Goal: Describe the kinetic molecular theory of gases and the units of measurement used for gases.

Ch. 14 The Behavior of Gases

temperature and pressure unchanging

CP Chapter 13/14 Notes The Property of Gases Kinetic Molecular Theory

In the name of Allah

Name Unit 9 Notes: Gas Laws Period. Complete throughout unit. Due on test day!

Example: 25 C = ( ) K = 298 K. Pressure Symbol: p Units: force per area 1Pa (Pascal) = 1 N/m 2

8.1 Properties of Gases. Goal: Describe the Kinetic Molecular Theory of Gases and the units of measurement used for gases.

Gases Chapter 8. Chapter 8

Gases. Chapter 8. Chapter 8. Gases Properties of Gases. We are surrounded by gases, but we are often

NOTES: Behavior of Gases

CHM 111 Unit 5 Sample Questions

Pressure of the atmosphere varies with elevation and weather conditions. Barometer- device used to measure atmospheric pressure.

CP Chapter 13/14 Notes The Property of Gases Kinetic Molecular Theory

CHEMISTRY - CLUTCH CH.5 - GASES.

General Properties of Gases

Chem 110 General Principles of Chemistry

Name /74. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Unit 14 Gas Laws Funsheets

POGIL EXERCISE 18 All You Need to Know About Gas Laws

PROPERTIES OF GASES. [MH5; Ch 5, (only)]

Unit 8: Kinetic Theory Homework Packet (90 points)

SCH3U7 Quantitative Chemistry

PSI Chemistry: Gases Multiple Choice Review

Gas Laws. Directions: Describe what contribution each of the Scientist below made to the Gas Laws and include there gas law equation.

Chapter 13. Gases. Copyright Cengage Learning. All rights reserved 1

Kinetic Molecular Theory Gases. Behavior of gases. Postulate two. Postulate one. Postulate three. Postulate four

Under ideal conditions, the rates at which different gases diffuse (spread out) are proportional to their molar masses.

13.1!"#$#%"&'%()$*+%,+-.$+/*$#

Chemistry 51 Chapter 7 PROPERTIES OF GASES. Gases are the least dense and most mobile of the three phases of matter.

States of Matter Review

AP TOPIC 6: Gases. Revised August General properties and kinetic theory

Simple Gas Laws. To facilitate comparison of gases, the following standards are used: STP: O C (273 K) and kpa. SATP: 25 C (298 K) and 101.

You should be able to: Describe Equipment Barometer Manometer. 5.1 Pressure Read and outline 5.1 Define Barometer

Behavior of Gases Chapter 12 Assignment & Problem Set

Chemistry Chapter 12. Characteristics of Gases. Characteristics of Gases 1/31/2012. Gases and Liquids

Unit 10: Gas Laws. Monday Tuesday Wednesday Thursday Friday. 10 Review for Cumulative Retest. 17 Chem Think Gas Laws Tutorial- Computer Lab-

Gases. Name: Class: Date: Matching

Gilbert Kirss Foster. Chapter 10. Properties of Gases The Air We Breathe

Behavior of Gases. Gases are mostly The molecules in a gas are separate, very small and very

Section 8: Gases. The following maps the videos in this section to the Texas Essential Knowledge and Skills for Science TAC (c).

Problem Solving. Gas Laws

Temperature Temperature

THE GAS STATE. Unit 4. CHAPTER KEY TERMS HOME WORK 9.1 Kinetic Molecular Theory States of Matter Solid, Liquid, gas.

2. Calculate the ratio of diffusion rates for carbon monoxide (CO) and carbon dioxide (CO2). υa = MB = 44 = 1.25

Expand to fill their containers, are highly compressible, have extremely low densities.

9A Gas volume and pressure are indirectly proportional.

(Test ID: ins700075) Created with INSPECT and the Online Assessment Reporting System (OARS) For Authorized Use Only

Section 10-1: The Kinetic-Molecular Theory of Matter. 1) How does the word kinetic apply to particles of matter?

World of Chemistry Notes for Students [Chapter 13, page 1] Chapter 13 Gases

Gas volume and pressure are indirectly proportional.

Gases. Chapter 5: Gas Laws Demonstration. September 10, Chapter 5 Gasses.notebook. Dec 18 10:23 AM. Jan 1 4:11 PM. Crushing 55 gallon drum

Chapter 9 Gases: Their Properties and Behavior

Chemistry 20 Unit 2 Gases FITB Notes. Topic A Characteristics of Gases

Properties of any sample of gas. Unit 5: Gases. * All gases behave according to the Kinetic Molecular Theory pg 421

Chapter 5. Pressure. Atmospheric Pressure. Gases. Force Pressure = Area

Name: Class: Date: SHORT ANSWER Answer the following questions in the space provided.

Gas Pressure. Pressure is the force exerted per unit area by gas molecules as they strike the surfaces around them.

Gases and Pressure. Main Ideas

Name: Period: Date: CHAPTER 10 NOTES 10.3: The Gas Laws

Name Hour. The Behavior of Gases. Practice B

Gas Laws V 1 V 2 T 1. Gas Laws.notebook. May 05, T = k P 1 V 1 = P 2 V 2. = 70 kpa. P. V = k. k = 1 atm = kpa

Section 5.1 Pressure. Why study gases? An understanding of real world phenomena. An understanding of how science works.

Chapter 10. Physical Characteristics of Gases

Chapter 13: The Behavior of Gases

Gases Chapter 11 (and 10)

Basic Concepts of Chemistry Notes for Students [Chapter 10, page 1] D J Weinkauff - Nerinx Hall High School. Chapter 10 Gases

Chapter 11: Gases: Homework: Read Chapter 11. Keep up with MasteringChemistry and workshops

Transcription:

Gas Laws For CHM1020 PROPERTIES OF GASES 1. Variable shape and volume (same shape and volume as container) 2. Expand uniformly (as container increases in volume, gas expands and distributes uniformly in container) 3. Compress uniformly (as container decreases in volume, gas compresses, molecules are closer together - if volume is reduced sufficiently a gas will liquefy) 4. Low density (density of air is 1 g/l, density of water is 1g/mL so water is 1000x more dense than air) 5. Mix uniformly with other gases ATMOSPHERIC PRESSURE Gas pressure is the result of constantly moving molecules striking the inside surface of their container. 1. If molecules collide more often, pressure 2. If molecules collide with more energy, pressure 3. If temp, molecules move faster and collide more frequently with more energy, therefore pressure temp = pressure temp = pressure UNIT Standard Pressure Units of Gas Pressure atmosphere 1 atm (exactly) inches of mercury 29.9 in Hg centimeters of mercury 76 cm Hg (exactly) millimeters of mercury 760 mmhg (exactly) torr 760 torr (exactly) pounds per square inch kilopascal 101 kpa 1

EXAMPLES (gas pressure unit conversion) 1. Given that air pressure inside an automobile tire is 34.0 psi, express pressure in each of the following units: a. atm 34.0 psi 1 atm = 2.31 atm b. cm Hg 76 cm Hg 34.0 psi = 176 cm Hg c. torr 760 torr 34.0 psi = 1758 torr d. kpa 101 kpa 34.0 psi = 234 kpa 2. Which of the following is the greatest gas pressure: 1 atm, 1 in Hg, 1 torr, 1 cm Hg, 1 mmhg, or 1 psi? The atm is the largest unit, so 1 atm is the greatest gas pressure from this list. You can check this my converting all units to atm, and see that all others are smaller than 1 atm. GAS PRESSURE CHANGES 1. Pressure is inversely related to the volume volume = pressure volume = pressure 2. temp = pressure (faster movement) temp = pressure (slower movement) Can be explained by collision and movement 3. Pressure is directly related to the # of moles of gas or number of gas molecules # gas molecules, then pressure EXAMPLES (gas pressure changes) 1. State whether pressure of a gas in a sealed container increases or decreases with each change: a. Volume 250mL to 500mL (decreases according to #1) b. Temp 20 C to 80 C (decreases according to #2) c. Moles of gas change from 1 mol to 1.5 mol (increases according to #3) 2. Which of the following decreases gas pressure? a. increase volume (yes, according to #1) b. increase temp (no, according to #2, gas pressure would increase) c. increase molecules (no, according to #3, gas pressure would increase) 2

BOYLES LAW (Pressure-Volume Relationships) Pressure, Volume P1V1 = K = P2V2 EXAMPLES (using Boyle s Law) 1. 5.00 L of propane gas is compressed and pressure increases from 1.00 atm to 1.50 atm. What is the new volume? Step 1: Define your variables from the given information. P1 = 1.00 atm, P2 = 1.50 atm, V1 = 5.00 L, V2 =? Step 2: Rearrange P1V1=P2V2 to solve for V2. Divide both sides by P2 to obtain: V 2 = P 1V 1 P 2 Step 3: Substitute givens into the equation for the appropriate variables and solve. V 2 = 1 atm 5.0 L 1.5 atm = 3.33 L 2. Ethane gas has a volume of 125mL at 20 C and 725 torr. What is the volume of gas when pressure decreases to 475 torr? Step 1: Define your variables from the given information. P1 = 725 torr, P2 = 475 torr, V1 = 125 ml, V2 =? Step 2: Rearrange P1V1=P2V2 to solve for V2. Divide both sides by P2 to obtain: V 2 = P 1V 1 P 2 Step 3: Substitute givens into the equation for the appropriate variables and solve. V 2 = 725 torr 125 ml 475 torr = 191 ml 3

CHARLES S LAW (Volume-Temperature Relationships) Volume can be listed in any unit, temperature MUST be in Kelvin. Volume of gas is directly proportional to the Kelvin temperature if pressure remains constant. Celsius to Kelvin Conversion: C + 273 V 1 = V 2 EXAMPLE (using Charles s Law) A krypton balloon has a volume of 555mL at 21 C. If the balloon is cooled and the volume decreases to 475mL, what is the final temperature? (Pressure remains constant). Step 1: Define your variables using the given information. V1 = 555 ml, T1 = 21 C, V2 = 475 ml, T2 =? Step 2: Change temperature units to Kelvin. T1 = 21 C + 273 = 294 K Step 3: Rearrange V 1 = V 2 to solve for T2. Crossmultiply to get: V 1 = V 2 Divide both sides by V1: = V 2 V 1 Step 4: Substitute givens into the equation for the appropriate variables and solve. = 475mL 294 K 555 ml = 252 K Step 5: Convert back to original temperature units ( C). T2 = 252 K - 273 = -21 C Kelvin to Celcius Conversion: K - 273 4

GAY- LUSSAC S LAW (Pressure-Temperature Relationships) Volume constant, pressure temperature P 1 = P 2 *Temperature must be in Kelvin EXAMPLE (using Guy-Lussac s Law) A copper container has a volume of 555mL and is filled with air at 25 C. The container is then immersed in dry ice and the pressure drops from 761 torr to 495 torr. What is the final temp of the air in the copper container? Step 1: Define your variables using the given information. P1 = 761 torr, T1 = 25 C, P2 = 495 torr, T2 =? Step 2: Change temperature units to Kelvin. T1 = 25 C + 273 = 298 K Step 3: Rearrange P 1 = P 2 to solve for T2. Crossmultiply to get: P 1 = P 2 Divide both sides by P1: = P 2 P 1 Step 4: Substitute givens into the equation for the appropriate variables and solve. = 495 torr 298 K 761 torr = 194 K Step 5: Convert back to original temperature units ( C). T2 = 194 K - 273 = -79 C 5

COMBINED GAS LAW P 1 V 1 = P 2V 2 STP = 273 K/0 C, 1 atm/760mm Hg/760 torr/76cm Hg *Temperature must be in Kelvin EXAMPLE (using Combined Gas Law) An oxygen gas sample occupies 50 ml at 27 C and 765 mm Hg. What is the final temperature if the gas is cooled to a volume of 35.5 ml and a pressure of 455 mm Hg? Step 1: Fill in the table to help solve. Pressure Volume Temperature Initial 765 mm Hg 50 ml 27 C = 300 K Final 455 mm Hg 35.5 ml? Step 2: Rearrange the equation to solve for T2. = P 2 V 2 P 1 V 1 Step 3: Substitute the values from the table into the rearranged equation. 300 K 455 mm Hg 35.5 ml = = 127 K = 146 765 mm Hg 50 ml DALTON S LAW OF PARTIAL PRESSURES Total pressure of gas mixture = sum of the individual pressures of each gas Pressure of each gas in a mixture is partial pressure Sum of individual partial pressures = Total atmospheric pressure EXAMPLE (using Dalton s Law of Partial Pressures) An atmospheric sample contains nitrogen, oxygen, argon, and trace gases. If the partial pressure of nitrogen is 592 mm Hg, oxygen is 160 mm Hg, argon is 7 mm Hg and trace gas is 1 mm Hg, what is the atmospheric pressure? Ptotal = Pnitrogen + Poxygen + Pargon + Ptrace = 592 + 160 + 7 + 1 = 760 mm Hg 6

IDEAL GAS LAW PV = nrt Temperature (Kelvin) Pressure (atm) Ideal gas constant, 0.0821 L*atm/mol K Moles of gas Volume (L) Summarizes the behavior of gases EXAMPLE (using Ideal Gas Law) What is the temperature of 0.250 mol of chlorine gas at 655 torr if the volume is equal to 3.5 L? Step 1: Identify givens. P = 655 torr (need in atm to use formula), 655 torr 1 atm 760 torr =.862 atm V = 3.5 L n = 0.250 mol R = 0.0821 L*atm/mol K T =? Step 2: Rewrite formula to solve for unknown. T = PV nr Step 3: Plug in givens to rewritten formula and solve.. 862 atm 3.5 L T = = 147 K L atm 0.250 mol 0.0821 mol K 7