The Kinetic Theory Chapter 13

The Kinetic Theory Chapter 13

Today… Turn in: Larry the Lawn Chair Guy Reading Our Plan:
Grab a WS Packet & Calendar Notes – Kinetic Theory Worksheet #1 Homework (Write in Planner): WS #1 – Due next class

Kinetic Theory The Kinetic Molecular Theory of Matter is a concept that basically states that atoms and molecules possess an energy of motion (kinetic energy) that we perceive as temperature. In other words, atoms and molecules are constantly in motion, and we measure the energy of these movements as the temperature of that substance. This means if there is an increase in temperature, the atoms and molecules will gain more energy (kinetic energy) and move even faster.

The Kinetic Theory of Gases
Gases consist of hard, small spheres Gas particles have insignificant volume because they are so small and so spread out.

Explains… Compressibility – measure of how much the volume of matter decreases under pressure

Kinetic Theory No attractive or repulsive forces exist between particles

As a result… Gases are free to move in their containers

Kinetic Theory Gas particles move rapidly in constant random motion

Affected by… Temperature and Volume

The effect of volume on temperature…
Smaller Volume = Faster Movement of Particles (more collisions) = Increase in Temperature (HOT!)

The effect of volume on temperature…
Larger Volume = Slower Movement of Particles (fewer collisions) = Decrease in Temperature (COLD!)

Real Life Example Smoke out of a chimney on a cold winter day hangs in the air because the gas particles are moving so slowly (in the summer smoke travels very quickly through the air)

Kinetic Theory All collisions are perfectly ELASTIC (the total kinetic energy remain CONSTANT)

Video Demonstration

Quick Check Summarize the kinetic theory in the space provided.
Share your responses with a shoulder partner. Partner A shares for 30 seconds and then Partner B shares for 30 seconds. Partner A is the student with the most pets.

Gas Pressure Gas pressure results from the collisions of particles with the walls of a container If there are no particles, there cannot be collisions, which means there would be no pressure. This explains why there is no pressure in a vacuum.

Gas Pressure Air exerts pressure on Earth because the gases in our atmosphere push down (collide with) the Earth’s surface. This is called atmospheric pressure. Atmospheric pressure decreases as you climb a mountain because there is less air pressing down on you.

Quick Check Think of a pressure in your life. How does it relate to pressure as we’ve discussed here?

Think About it… What is the relationship between kinetic energy and temperature?

Absolute Zero The temperature of an object depends on how fast the particles in it are moving

Absolute Zero Absolute Zero (0 K) is the lowest possible temperature that can be reached – it is when motion of particles STOPS!

Absolute Zero Temperature and Particle Movement Web Demonstration

Boiling (Vaporization/Evaporation)
Changes of State Change of State Phases Example Boiling (Vaporization/Evaporation) Liquid → Gas Melting Solid → Liquid Condensation Gas → Liquid Freezing Liquid → Solid Sublimation Solid → Gas Deposition Gas → Solid

Phase Diagram A phase diagram tells you the state of matter at each temperature and pressure for a substance. Triple point – when all three states of matter occur simultaneously in a container.

Phase Diagram

Phase Diagrams Two phases exist on the lines Water is in a single phase in each of the colored regions. At the triple point, water is a solid, liquid, and a gas at the same time!

Phase Diagram

Real Life Applications
Cooking at high altitudes. Demo – boiling water in a syringe

Quick Check Quick Check – Look at the phase diagram for CO2. Label which state of matter is occurring at each point. A = B = C = Which change of state occurs when you go from point C to point B? Which change of state occurs when you go from point A to point B?

Today… Turn in: Finish Lab Test – Weigh & Calculate Our Plan:
Grab a WS Packet & Calendar Notes – Kinetic Theory Worksheet #1 Online Inquiry Lab Homework (Write in Planner): WS #1 – Due next class

Variables to describe gases
The Behavior of Gases Variables to describe gases

Variables Used to Describe Gases
Pressure Measured in atm, torr, mm Hg or Pascals Barometer

Temperature Measured in Kelvin, Fahrenheit, or Celsius Thermometer

Volume Measured in Liters or m3

Amount of Gas Measured in Moles

STP Stands for Standard Temperature and Pressure
Standard Conditions of 1 atm pressure and 0 degrees Celsius

Temperature Conversion Factors
K = C + 273 F = 9/5(C) + 32 C = 5/9 (F – 32)

Other Conversion Factors
1 cm3 = 1 mL 1 dm3 = 1 L 1000 Pa = 1 kPa 1000 mL = 1 L 1 atm = 760 mm Hg 1 atm = 101 kPa 1 torr = 1 mm Hg 1000 L = 1 kL

Conversion Example 1 760 mm Hg 87 atm X = 66120 mm Hg 1 atm
How many mm Hg are in 87 atm? 760 mm Hg 87 atm X = 66120 mm Hg 1 atm

Conversion Example 2 101.325 kPa = X 0.22 kPa 760 mm Hg 1.65 mm Hg
How many kPa are in 1.65 mm Hg? kPa = 1.65 mm Hg X 0.22 kPa 760 mm Hg

Have to convert to Celsius first!
Conversion Example 3 Convert 36oF into Kelvin. Have to convert to Celsius first! 5/9 (36 – 32) = 2.22oC K = = K

Try it Out! Convert 23oC into Kelvin. (296 K) How many Pa is 3.5 atm? (354,637.5 Pa)

ASSIGNMENT Complete Worksheet #1 – Conversions!

Today… Turn in: Get WS#1 out to Check Grab your chromebook Our Plan:
Online Inquiry Labs Notes Worksheet #2 Clicker Review - Scenarios Homework (Write in Planner): WS #2 – Due next class

Online Inquiry Lab Go to Mrs. C’s website, Unit 9, Resources

Find 4 “Dates” Fill out your date card on p. 8 of your booklet. People at your table are your family and we don’t date family!

Gas Laws Challenge! What happens to a bag of potato chips when it is placed in the freezer? What happens when it is left in a hot car?

Gas Laws Challenge! Aerosol cans have a warning on them, indicating not to incinerate them or to store them above a certain temperature. Explain why this is the case and what will happen if they are exposed to extreme temperatures.

Boyle’s Law

Boyle’s Law Named after its founder Robert Boyle English

The Original Experiment
Robert Boyle – 1662 Used a manometer – J-shaped piece of tubing with one end closed Sealed a certain volume of air in the closed end of the tube Varied the pressure (mm Hg) and watched the change in volume

The Relationship As pressure increases, volume decreases OR As pressure decreases, volume increases

Variables that are Constant
The amount of gas (moles) AND The temperature of the gas

P1V1 = P2V2 The Equation Where P = Pressure and V = Volume
Where 1 = Initially (1st) and 2 = Finally (2nd)

The Graph

The Graph This is called an inverse relationship

Practice Problem A sample of gas has a volume of 12.0 L and a pressure of 1.00 atm. If the pressure of gas is increased to 2.00 atm, what is the new volume of the gas? MATH ALERT!

Practice Problem

Practice Problem Step 1: List the variables that you know. P1 = 1 atm V1 = 12.0 L P2 = 2 atm V2 = ?

Step 2: Plug the numbers into the equation
Practice Problem Step 2: Plug the numbers into the equation P1V1 = P2V2 (1 atm)(12.0L) = (2 atm)(V2)

Practice Problem Step 3: Use algebra to solve for the unknown variable
(1 atm)(12.0 L) = V2 (2 atm) V2 = 6.0 L

Try it Out! A sample of gas has a pressure of 3.00 atm and a volume of 4.6 L. If the volume of gas is decreased to 3.2 L, what is the new pressure of the gas? Enter in Clicker!

Try it Out! Step 1: List the variables that you know. P1 = 3 atm V1 = 4.6 L P2 = ? V2 = 3.2 L

Step 2: Plug the numbers into the equation
Try it Out! Step 2: Plug the numbers into the equation P1V1 = P2V2 (3 atm)(4.6L) = (x)(3.2L)

Try it Out! Step 3: Use algebra to solve for the unknown variable
(3 atm)(4.6 L) = P2 (3.2 L) P2 = 4.3 atm

Quick Check Come up with at least 1 real-life application for this gas law and write it in the appropriate spot. Find your 3:00 date and share your answers.

Real-Life Applications
SYRINGES To draw fluids in: the volume is increased in the syringe, which causes the pressure inside to be less than that outside, so the liquid is forced in To force fluid out: the volume is decreased, so the pressure inside is greater than outside, so the fluid is forced out

Real-life Applications
LUNGS & DIAPHRAGM When we breathe in (inspire), the diaphragm is lowered and the chest wall is expanded, increasing the volume for the chest cavity. Outside air enters the lungs because it is at a higher pressure than the air in the chest cavity. When we breathe out (expire), the diaphragm rises and the chest wall contracts, decreasing the volume of the chest cavity. The pressure is increased, and some air is forced out.

Tire Pump Air-filled Automotive Shock Absorbers Tire Pressure Gauge

Real Life Applications
Airplanes Automobile Pistons Meteorology Straws

Scuba Diving – “The Bends”

Diving – “The Bends” As a diver descends, the water exerts greater pressure. More gas pressure is required to keep the lungs expanded. This increased gas pressure causes more nitrogen to dissolve in the divers blood.

Diving – “The Bends” Below a depth of about 30 meters, dissolved nitrogen interferes with the transmission of nerve impulses. The effects are similar to those of alcohol and include dizziness, slowed reaction time, and an inability to think clearly.

Diving – “The Bends” As the diver returns to the surface, pressure decreases and dissolved nitrogen is released from the blood. Bubbles of nitrogen can block small blood vessels and reduce the supply of oxygen to cells, causing severe pain in the joints, dizziness, vomiting, or even death.

Question Why are gases transported under great pressure?

Charles’ Law

The Inventor Named after Jacques Charles, but actually discovered by Joseph Louis Gay-Lussac when he was only 23 years old. Gay-Lussac named it after Charles because of his previous work with gases.

The Inventor Hot air balloons were very popular at the time, so the two men decided to do studies on the relationship between volume and temperature.

Used a manometer – immersed the J-shaped tube in a water bath. By adjusting the temperature of the water they changed the temperature of the gas in the tube. Charles and Gay-Lussac watched what happened to the volume of the gas when they changed the temperature

The Relationship As temperature is increased, volume increases. OR
As temperature is decreased, volume decreases.

Variables that are constant
Pressure AND Amount of gas (moles)

The Equation V1 / T1 = V2 / T2 Where V = volume and T = Temperature

The Graph

The Graph This is called a direct relationship

Practice Problem The temperature of a 4.00 L sample of gas is changed from 10.0 degrees Celsius to 20.0 degrees Celsius. What will the volume of this gas be at the new temperature if the pressure is held constant?

IMPORTANT! When solving gas law problems, all temperatures must be in KELVIN!

Practice Problem

Practice Problem Step 1: List the variables that you know.
V1 = 4.00 L T1 = 10 ◦C = 283K V2 = ? T2 = 20 ◦C =293K

V1 / T1 = V2 / T2 Practice Problem
Step 2: Plug the numbers into the equation V1 / T1 = V2 / T2 (4.00 L/ 283 K) = (V2/293 K)

Practice Problem Step 3: Use algebra to solve for the unknown variable – Cross multiply 1172=283x X=4.14 L

Try it Out! The volume of a tire at 32○C is 2 L. What is the temperature if the volume of the tire is 5.4 L? Enter in Clicker!

Try it Out! Step 1: List the variables that you know.
V1 = 2 L T1 = 32 ◦C = 305 K V2 = 5.4 L T2 = ?

Try it Out! Step 2: Plug the numbers into the equation V1 / T1 = V2 / T2 (2 L/ 305 K) = (5.4 L/x)

Try it Out! Step 3: Use algebra to solve for the unknown variable – Cross multiply 2x = 1647 x = K

Quick Check Come up with at lest 1 real-life application for this gas law and write it in the appropriate spot. Find your 6:00 date and share your answers.

Hot Air Balloons A propane heater is used to heat the air in the balloon. As the air is heated, the gas expands, becoming less dense. Because the density of the air inside the balloon is less than the density outside the balloon, the balloon is buoyant.

Hot Air Balloons As the air cools, it contracts and becomes more dense and less buoyant. Thus the balloonist can control altitude by heating or cooling the gas to increase or decrease buoyancy.

Tire Volume Bridges Food in a freezer

Gay-Lussac’s Law

The Experiment Discovered by Joseph Louis Gay-Lussac in 1802
He did experimentation on hot air balloons

The Relationship As the temperature increases, the pressure increases
OR As the temperature decreases, the pressure decreases

The Relationship To understand the relationship, think about the kinetic theory – if you increase the temperature, the particles move faster and thus hit the side of the container more often. This causes the pressure to increase. If the container is solid, the volume doesn’t change

Variables that are constant
Volume AND Number of Moles

Where P = Pressure and T = Temperature
The Equation P1/T1 = P2/T2 Where P = Pressure and T = Temperature

The Graph

The Graph This is a direct relationship

Practice Problem A gas has a pressure of atm at 50 degrees Celsius. What is the pressure at standard temperature?

Practice Problem Remember to convert all temperatures to Kelvin!

Practice Problem

P1 = atm T1 = 50 ◦C P2 = ? T2 = 0 ◦C

P1 / T1 = P2 / T2 Practice Problem
Step 2: Plug the numbers into the equation P1 / T1 = P2 / T2 (0.370atm/ 323 K) = (P2/273 K)

= 323x x = atm

Try it Out A gas has a pressure of atm at 120 K. What is the temperature at atm? Enter in Clicker!

Try it Out! Step 1: List the variables that you know.
P1 = atm T1 = 120 K P2 = atm T2 = ?

Try it Out! Step 2: Plug the numbers into the equation P1 / T1 = P2 / T2 (0.891atm/120 K) = (0.428 atm/x)

Try it Out! Step 3: Use algebra to solve for the unknown variable 0.891x = 51.36 P2 = K

Quick Check Come up with at lest 1 real-life application for this gas law and write it in the appropriate spot. Find your 12:00 date and share your answers.

Aerosol Cans Autoclave – machine that sterilizes medical equipment Tire Pressure

Quick Check Which gas law describes each situation on p. 11 of your notebooklet? Share your answers with your 9:00 partner.

STOP! Worksheet Time! Complete Worksheet #2 by next class

Wrap Up Clicker Review

Today… Our Plan: Turn in: Get WS#2 out to check
Magic Squares Review Activity Notes WS #3 Gases Reading Homework (Write in Planner): WS #3 & Reading – Due next class

Gas Laws Review Boyle’s Law Charles’ Law Gay-Lussac’s Law

Helpful Reminders! All Temperatures must be in KELVIN!
Measurements of the same variable must be in the same units Can’t have 1 mL and 2 L Can’t have 123 kPa and 0.9 atm CONVERT to one or the other STP (Standard Temperature & Pressure) Standard Temp = 0 Celsius Standard Pressure = 1 atm

Review Time Complete the Review Worksheet with a partner. Be sure to each use 2 different colored writing utensils and take turns writing!

The Combined Gas Law P1V1 = P2V2 T1 T2
Put all of the Gas Law Equations together to form one! P1V1 = P2V2 T T2

Practice Problem If a balloon containing 1000L of gas at 50 degrees Celsius and 101 kPa rises to an altitude where the pressure is 50.5 kPa and the temperature is 10 degrees Celsius, what would the volume of the balloon be under these new conditions?

Practice Problem

P1 = 101 kPa T1 = 50 ◦C V1 = 1000 L P2 = 50.5 kPa T2 = 10 ◦C V2 = ?

Practice Problem Step 2: Plug the numbers into the equation
P1V1 = P2V2 T T2 (1000 L x 101 kPa)/323 K = (V2 x 50.5 kPa)/283 K

Practice Problem Step 3: Use algebra to solve for the unknown variable = (V2 x 50.5 kPa)/283 K V2 = L

Practice Problem 2 A container of krypton occupies a volume of 15.0 L at a pressure of 210 kPa and a temperature of 110 K. Find the new temperature when the volume is 25.0 L and the pressure is 790 kPa.

Practice Problem

P1 = 210 kPa T1 = 110 K V1 = 15 L P2 = 790 kPa T2 = ? V2 = 25 L

P1V1 = P2V2 T T2 (15 L x 210 kPa)/110 K = (25L x 790 kPa)/x

28.64 = (25L x 790 kPa)/x CROSS MULTIPLY! 28.64x = 19750 X= K

Try it Out! A sample of gas at 47○C and 1.03 atm occupies a volume of 2.20 L. What volume would this gas occupy at 107○C and atm? (3.41 L)

Try it Out! A gas has a volume of 1.75 L at -23○C and 150 kPa. At what temperature would the gas occupy 1.30 L at 210. kPa? (260 K)

STOP! Complete Worksheet #3 by next class. When you finish choose a gas reading to complete by next class.

Today… Homework (Write in Planner): Turn in: Get out WS#3 to check
Turn Reading in to Basket Our Plan: Review Problem Quiz Lab Homework (Write in Planner): Nothing

Review Problem If 10.0 liters of oxygen at STP are heated to 512 °C, what will be the new volume of gas if the pressure is also increased to mm Hg? 14.4 L

Today… Turn in: Nothing – grab goggles and apron and have pencil for lab (clear off everything else) Our Plan: Penny Mystery Lab Finish Gas Laws Lab – DUE TODAY Homework (Write in Planner): Labs if you don’t finish in class!

Today… Turn in: Labs (if you haven’t already) Our Plan:
Password Vocabulary Review Notes – Ideal Gas Law Worksheet #4 Clicker Review Homework (Write in Planner): WS #4 – Due next class

The Ideal Gas Law

Ideal Gas Law Definition
Mathematical relationship between pressure, volume, temperature, and the number of moles of gas.

The effect of changing moles on volume

The effect of changing moles on pressure

The Ideal Gas Law Constant
R = L x atm mol x K The numbers from the problem must be in those units to solve! Volume in Liters Pressure in atm Temperature in Kelvin

The Equation PV = nRT Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

Practice Problem What is the volume, in Liters, of mole of oxygen gas at 20oC and atm of pressure?

Practice Problem

Practice Problem Step 1: Identify the variables. V= ? K x mol
P= atm R= L x atm V= ? K x mol n= moles T= 20oC

Practice Problem Step 2: Check the units on all variables
Temperature should be in Kelvin, so convert it! K= 20oC + 273 K= 293

(0.974 atm) (V) = (0.250 moles)( Lx atm) (293K) K x mol

Practice Problem V = 6.0138 L x atm 0.974 atm V = 6.17 L
Step 4: Use algebra to solve for the unknown variable. V = L x atm 0.974 atm V = 6.17 L

Practice Problem What is the pressure, in atmospheres, of 12 grams of neon gas at 32○C that fills a 2L gas tube?

Practice Problem

Practice Problem Step 1: Identify the variables. V= 2 L K x mol n=
P= ? R= L x atm V= 2 L K x mol n= T= 32oC 12 g x 1 mole = 0.6 moles 20 g

Practice Problem Step 2: Check the units on all variables
Temperature should be in Kelvin, so convert it! K= 32oC + 273 K= 305

Practice Problem (P)(2) = (0.6)(0.0821)(305K)
Step 3: Plug the numbers into the equation (P)(2) = (0.6)(0.0821)(305K)

Practice Problem P = 15.02 2 P = 7.51 atm
Step 4: Use algebra to solve for the unknown variable. P = 15.02 2 P = 7.51 atm

Try It Out! What is the mass of 3 L of nitrogen gas at 40oC and atm of pressure?

Try It Out Step 1: Identify the variables. V= ? K x mol
P= atm R= L x atm V= ? K x mol n= 0.3 moles T= 40oC

Try It Out Step 2: Check the units on all variables
Temperature should be in Kelvin, so convert it! K= 40oC + 273 K= 313 K

Try It Out Step 3: Plug the numbers into the equation
(0.564 atm) (V) = (0.30 moles)( Lx atm) (313K) K x mol

Try it Out V = 7.709 L x atm 0.564 atm V = 13.67 L
Step 4: Use algebra to solve for the unknown variable. V = L x atm 0.564 atm V = L

STOP! Begin working on WS #4

Today… Turn in: Get out WS#4 to Check Our Plan: Diffusion Demo Notes
Worksheet #5 & #6 Test Review Homework (Write in Planner): Worksheet #5/#6 Test Next Class!

Demo Time…

Ideal Gases vs. Real Gases
An ideal gas is one that follows the gas laws at all conditions of pressure and temperature and conforms precisely to the assumption of the kinetic theory. Ideal gases do not exist Real gases: Do have volume There are attractions between particles

Ideal Gas Law & Stoichiometry
Last unit we calculated volumes of gas produced in chemical reactions using the molar volume of a gas (22.4 L). Remember, this only works if the reaction is carried out at STP (0 degrees C and 1 atm pressure) Under any other conditions, you must use stoichiometry in combination with the ideal gas law.

Example 1 An air sample containing H2S at atm and 29 degrees C is treated with a catalyst to promote the reaction, H2S (g) + O2 (g) → H2O (g) + S(s). If 3.2 g of solid S was collected, calculate the volume of H2S in the original sample.

Example 2 How much NaN3 is needed to inflate a 50.0 L air bag containing N2 to 1.15 atm at 25.0 ºC given the following chemical reaction? 2 NaN3 (s) → 2 Na (s) + 3 N2 (g)

Try It Out! In the chemical reaction used in automotive air-bag safety systems, N2(g) is produced by the decomposition of sodium azide, NaN3(s), at a somewhat elevated temperature: 2 NaN3(s) --> 2 Na(l) N2(g) What volume of N2(g), measured at 25 °C and atm, is produced by the decomposition of 62.5 g NaN3? 35.9 L

Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures
Partial Pressure: the pressure of each gas in a mixture Dalton’s Law: the total pressure of a mixture of gases is equal to the sum of the partial pressures of each component gas

Dalton’s Equation PT = P1 + P2 + P3 + ….

Practice Problem A mixture of three gases, A, B, and C, is at a total pressure of 6.11 atm. The partial pressure of gas A is 1.68 atm; that of gas B is 3.89 atm. What is the partial pressure of gas C?

Practice Problem 6.11 = Pc Pc = 0.54 atm

Graham’s Law

Diffusion Spontaneous mixing of the particles of 2 substances caused by their random motion Quick Check – Think of examples Examples: Smells Steam/Smoke Others?

Effusion Process by which gas particles pass through a tiny opening
Video

Graham’s Law of Effusion
The rate of effusion of a gas depends on its size. The more massive a molecule, the slower it effuses. Can also be applied to the diffusion of gases. Video

Quick Check Think of examples of effusion.

STOP! Complete WS #5 Begin Test Review

Today… Turn in: WS#5 Get Test Review out to Check Our Plan:
Quiz, Quiz, Trade Questions on Test Review Test Read the Airbag Lab (p. 1 & 2) and answer questions on p. 3 Homework (Write in Planner): Enjoy your Spring Break

Boyle’s Law P1V1 = P2V2 Charles’ Law V1/T1 = V2/T2 Gay-Lussac Law P1/T1 = P2/T2 Combined Gas Law (P1V1)/T1 = (P2V2)/T2 Ideal Gas Law PV = nRT Dalton’s Law PT = P1 + P2 + P3 +… Gases Collected Over Water Patm = Pgas + PH2O Graham’s Law rate = √M2/M1

When to Convert… Kelvin ALWAYS PV=nRT
Otherwise, as long as the variables are in the same unit you’re fine.

How to do the problems on the test….
A gas occupies 11.2 L at atm. What is the pressure if the volume becomes 15,000 mL? (0.64 atm)

A sample of hydrogen gas has a volume of 65.0 mL at a pressure of atm and a temperature of 16.0 ○C. What volume will the hydrogen occupy at mm Hg and 25○C? (67.6 mL)

An engineer pumps 5.00 mol of carbon monoxide gas into a cylinder that has a capacity of 20.0 L. What is the pressure in kPa of CO inside the cylinder at 25.0○C? (618 kPa)

The Kinetic Theory Chapter 13

Similar presentations

Presentation on theme: "The Kinetic Theory Chapter 13"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

The Kinetic Theory Chapter 13

Similar presentations

Presentation on theme: "The Kinetic Theory Chapter 13"— Presentation transcript:

Similar presentations

About project

Feedback