Bell-Ringer DATE: Why does a collision with an inflated air bag cause much less damage than a collision with a steering wheel or dashboard? Hint: Gases.

Name: Bell-Ringer DATE: Why does a collision with an inflated air bag cause much less damage than a collision with a steering wheel or dashboard? Hint: Gases.
Uploaded: 2017-09-07T23:36:13+00:00
Duration: PTM27S40
Channel: Agatha Watts
Description: Bell-Ringer DATE: Why does a collision with an inflated air bag cause much less damage than a collision with a steering wheel or dashboard? Hint: Gases.

Bell-Ringer DATE: Why does a collision with an inflated air bag cause much less damage than a collision with a steering wheel or dashboard? Hint: Gases are compressible, unlike solids. The ability of a gas to be compressed allows your forward motion; momentum/energy to be transferred to the gas particles, thus slowing your motion down

P T V Properties of Gases Compressibility
Section 14.1 P T V Properties of Gases Compressibility Factors Affecting Gas Pressure Amount of gas Volume Temperature

1. Compressibility Compressibility is a measure of how much the volume of matter decreases under pressure.

Gases are easily compressed because of the space between the particles in a gas.
The volume of the particles in a gas is small compared to the overall volume of the gas. The distance between particles in a gas is much greater than the distance between particles in a liquid or solid. Under increased pressure, the particles in a gas are forced closer together.

This model shows identical air samples in two different containers
This model shows identical air samples in two different containers. Each container has 8 nitrogen molecules and 2 oxygen molecules. In the larger container, the molecules are farther apart. In the smaller container, the air sample is compressed, and the molecules are closer together.

Explain why air is easily compressed, but wood is not easily compressed.
The volume of the particles in a gas, such as air, is small compared to the overall volume of the gas. So, the distance between particles in air is much greater than the distance between particles in a solid, such as wood. Under pressure, the particles in a gas can be forced closer together, or compressed.

2. Factors Affecting gas pressure
Unit of pressure: kPa Amount of gas Number of moles (n) Volume of gas (V) in liters Temperature (T) in Kelvins

14-1 Gas Laws Factors Affecting Gas Pressure 1.) Amount of Gas You can use kinetic theory to predict and explain how gases will respond to a change of conditions. If you inflate an air raft, for example, the pressure inside the raft will increase. Collisions of gas particles with the inside walls of the raft result in the pressure that is exerted by the enclosed gas. Increasing the number of particles increases the number of collisions which explains why the gas pressure increases.

Factors Affecting Gas Pressure
14-1 Gas Laws Factors Affecting Gas Pressure When a gas is pumped into a closed rigid container, the pressure increases as more particles are added. If the number of particles is doubled, the pressure will double. Once the pressure exceeds the strength of the container, the container will burst.

14-1 Gas Laws Factors Affecting Gas Pressure If the pressure of the gas in a sealed container is lower than the outside air pressure, air will rush into the container when the container is opened. When the pressure of a gas in a sealed container is higher than the outside pressure, the gas will flow out of the container when the container is unsealed.

The operation of an aerosol can depends on the movement of a gas from a region of high pressure to a region of lower pressure.

14-1 Gas Laws Factors Affecting Gas Pressure 2.) Volume You can raise the pressure exerted by a contained gas by reducing its volume. The more the gas is compressed, the more pressure the gas exerts inside the container. Increasing the volume of the contained gas has the opposite effect. If the volume is doubled, the particles can expand into a volume that is twice the original volume. With the same number of particles in twice the volume, the pressure of the gas is cut in half.

14-1 Gas Laws Factors Affecting Gas Pressure A piston can be used to force a gas in a cylinder into a smaller volume. This slide models the relationship between volume and pressure. When the volume is increased, the pressure the gas exerts is decreased. When the volume is decreased, the pressure the gas exerts is increased.

14-1 Gas laws Factors Affecting Gas Pressure 3.) Temperature The temperature increases and the average kinetic energy of the particles in the gas increases. Faster-moving particles strike the walls of their container with more energy.

An increase in temperature causes an increase in the pressure of an enclosed gas.
The container can explode if there is too great an increase in the gas pressure.

How does the pressure of a contained gas change when the volume of the gas is increased?
A. The pressure increases. B. The pressure decreases. C. The pressure does not change.

A properly inflated soccer ball will travel farther.
CHEMISTRY & YOU Which do you think would travel farther if kicked with the same amount of force: a properly inflated soccer ball or an underinflated soccer ball? A properly inflated soccer ball will travel farther.

If the pressure is too high, the ball may burst when it is kicked.
CHEMISTRY & YOU What might happen to an overinflated soccer ball if you kicked it too hard? If the pressure is too high, the ball may burst when it is kicked.

P T V Bell Ringer Make your PTV card:
DATE: 10/14/2014 Bell Ringer P T V Make your PTV card: Consider what will happen when a helium-filled balloon is released into the sky. Assume the temperature remains constant. Write down your thoughts and be prepared to share with the class.

The Gas laws Boyle’s Law 𝑷 𝟏 ×𝑽 𝟏 = 𝑷 𝟐 × 𝑽 𝟐 Charles’s Law
Gay-Lussac’s Law Combined Gas Law 𝑷 𝟏 ×𝑽 𝟏 = 𝑷 𝟐 × 𝑽 𝟐 𝑽 𝟏 𝑻 𝟏 = 𝑽 𝟐 𝑻 𝟐 𝑷 𝟏 𝑻 𝟏 = 𝑷 𝟐 𝑻 𝟐 𝑷 𝟏 ×𝑽 𝟏 𝑻 𝟏 = 𝑷 𝟐 ×𝑽 𝟐 𝑻 𝟐

14-2 Gas laws Boyle’s law 𝑷 𝟏 ×𝑽 𝟏 = 𝑷 𝟐 × 𝑽 𝟐 If the temperature is constant, as the pressure of a gas increases, the volume decreases. Boyle’s law states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure.

Interpret Graphs As the volume decreases from 1.0 L to 0.5 L, the pressure increases from 100 kPa to 200 kPa.

CHEMISTRY & YOU Inhale, hold, and exhale. Think about Boyle's law and how it applies to breathing. How does Boyle's law explain why air enters your lungs when you inhale and leaves when you exhale? 1.) When you inhale, lung volume increases. Pressure decreases, and air moves in. When you exhale, lung volume decreases. Pressure increases and air moves out. 2.) As a diver ascends, pressure decreases. A decrease in pressure means an increase in volume. If a diver holds her breath while ascending, air volume in the lungs will increase. Why do you think scuba divers are taught never to hold their breath as they ascend from deep water?

14-2 Boyle’s Law Sample Problem 14.1 A balloon contains 30.0 L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.) 𝑷 𝟏 ×𝑽 𝟏 = 𝑷 𝟐 × 𝑽 𝟐 V2 = 1.24  102 L

14-2 Boyle’s Law Nitrous oxide (N2O) is used as an anesthetic. The pressure on L of N2O changes from 105 kPa to 40.5 kPa. If the temperature does not change, what will the new volume be? 𝑷 𝟏 ×𝑽 𝟏 = 𝑷 𝟐 × 𝑽 𝟐 V2 = 6.48 L

14-2 Boyle’s Law A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains the same? 𝑷 𝟏 ×𝑽 𝟏 = 𝑷 𝟐 × 𝑽 𝟐 P2 = 68.3 kPa

14-2 Boyle’s Law The volume of a gas at 99.6 kPa and 24◦C is 4.23 L. What volume will it occupy at 93.3 kPa and 24◦C? V2 = 4.52 L

14-2 Boyle’s Law A sample of neon gas occupies a volume of 677 mL at kPa. What is the pressure of the sample if the volume is decreased to 642 mL? P2 = 141 kPa

14-2 Gas laws Charles’s Law 𝑽 𝟏 𝑻 𝟏 = 𝑽 𝟐 𝑻 𝟐 As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant. Charles’s law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant.

Interpret Graphs

CHEMISTRY & YOU A hot air balloon contains a propane burner onboard to heat the air inside the balloon. What happens to the volume of the balloon as the air is heated? According to Charles’s law, as the temperature of the air increases, the volume of the balloon also increases.

14-2 Charles’S Law Sample Problem 14.2 A balloon inflated in a room at 24oC has a volume of L. The balloon is then heated to a temperature of 58oC. What is the new volume if the pressure remains constant? NOTE: Convert oC  K K = oC + 273 𝑽 𝟏 𝑻 𝟏 = 𝑽 𝟐 𝑻 𝟐 V2 = 4.46 L

14-2 Charles’S Law If a sample of gas occupies 6.80 L at 325oC, what will its volume be at 25oC if the pressure does not change? NOTE: Convert oC  K K = oC + 273 𝑽 𝟏 𝑻 𝟏 = 𝑽 𝟐 𝑻 𝟐 V2 = 3.39 L

14-2 Charles’S Law Exactly 5.00 L of air at -50.0oC, is warmed to 100.0oC. What is the new volume if the pressure remains constant? NOTE: Convert oC  K 𝑽 𝟏 𝑻 𝟏 = 𝑽 𝟐 𝑻 𝟐 V2 = 8.36 L

14-2 Charles’S Law The volume of a gas is 0.80 L at kpa and 0oC. What volume will it occupy at kPa and 24oC. V2 = 0.87 L

14-2 Charles’S Law What is the temperature of a 2.3 L balloon if it shrinks to a volume of L when it is dipped into liquid nitrogen at a temperature of 77 K? T2 = 276 K

Gay- Lussac’s Law 𝑷 𝟏 𝑻 𝟏 = 𝑷 𝟐 𝑻 𝟐
14-2 Gas laws Gay- Lussac’s Law 𝑷 𝟏 𝑻 𝟏 = 𝑷 𝟐 𝑻 𝟐 As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant. Gay-Lussac’s law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant.

14-2 gay-lussac’s Law Gay-Lussac’s law can be applied to reduce the time it takes to cook food. In a pressure cooker, food cooks faster than in an ordinary pot because trapped steam becomes hotter than it would under normal atmospheric pressure.

𝑷 𝟏 𝑻 𝟏 = 𝑷 𝟐 𝑻 𝟐 14-2 gay-lussac’s Law
Sample Problem 14.3 Aerosol cans carry labels warning not to incinerate (burn) the cans or store them above a certain temperature. This problem will show why it is dangerous to dispose of aerosol cans in a fire. The gas in a used aerosol can is at a pressure of 103 kPa at 25oC. If the can is thrown onto a fire, what will the pressure be when the temperature reaches 928oC? 𝑷 𝟏 𝑻 𝟏 = 𝑷 𝟐 𝑻 𝟐 NOTE: Convert oC  K P2 = 4.15 × 102 kPa

The pressure in a sealed plastic container is 108 kPa at 41oC. What is the pressure when the temperature drops to 22oC? Assume that the volume has not changed. NOTE: Convert oC  K 𝑷 𝟏 𝑻 𝟏 = 𝑷 𝟐 𝑻 𝟐 P2 = 101 kPa

The pressure in a car tire is 198 kPa at 27oC. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant. NOTE: Convert oC  K 𝑷 𝟏 𝑻 𝟏 = 𝑷 𝟐 𝑻 𝟐 T2 = 341 K (68 oC )

14-2 gay-lussac’s Law A sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at 211 K? P2 = 2.58 kPa

14-2 gay-lussac’s Law A pressure cooker containing kale and some water starts at 298 K and 101 kPa. The cooker is heated, and the pressure increases to 136 kPa. What is the final temperature inside the cooker? T2 = 400 K

Combined Gas Law 𝑷 𝟏 ×𝑽 𝟏 𝑻 𝟏 = 𝑷 𝟐 ×𝑽 𝟐 𝑻 𝟐
14-2 Gas laws Combined Gas Law 𝑷 𝟏 ×𝑽 𝟏 𝑻 𝟏 = 𝑷 𝟐 ×𝑽 𝟐 𝑻 𝟐 When only the amount of gas is constant, the combined gas law describes the relationship among pressure, volume, and temperature. Combined gas law is a single expression that combines Boyle’s law, Charles’s law, and Gay-Lussac’s law.

Suppose you hold the temperature constant (T1 = T2).
You can derive the other laws from the combined gas law by holding one variable constant. Suppose you hold the temperature constant (T1 = T2). Suppose you hold the volume constant (V1 = V2). P1  V1 = T2 P2  V2  T1 P1  V1 = P2  V2 Apply this to the other laws as well. P1  V1 = T2 P2  V2  T1

14-2 Combined gas Law Sample Problem 14.4 The volume of a gas-filled balloon is 30.0 L at 313 K and kPa pressure. What would the volume be at standard temperature and pressure (STP)? STP = 273 K/ kPa 𝑷 𝟏 ×𝑽 𝟏 𝑻 𝟏 = 𝑷 𝟐 ×𝑽 𝟐 𝑻 𝟐 V2 = 39.5 L

14-2 Combined gas Law A gas at 155 kPa and 25oC has an initial volume of L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125oC. What is the new volume? NOTE: Convert oC  K 𝑷 𝟏 ×𝑽 𝟏 𝑻 𝟏 = 𝑷 𝟐 ×𝑽 𝟐 𝑻 𝟐 V2 = L

14-2 Combined gas Law A 5.00 L air sample has a pressure of 107 kPa at a temperature of -50.0oC. If the temperature is raised to102oC and the volume expands to 7.00 L, what will the new pressure be? P2 = 1.29 × kPa

14-2 Combined gas Law The volume of a gas at 26oC and 75 kPa is 10.5 L. The pressure is increased to 116 kPa. What final temperature would be required to reduce the volume to 9.5 L? T2 = 418 K or 145oC

Which of the following equations could be used to correctly calculate the final temperature of a gas? T2 = V1  P2 V2  P1  T1 T2 = P2  T1 V1  P1  V2 A. C. T2 = V2  P2 V1  P1  T1 T2 = V1  P1 V2  P2  T1 D. B.

Bell-Ringer Review for quiz:
DATE: 10/20/2014 Review for quiz: Complete multiple choice questions #65-91 pg19-25 in your Topic 1 Matter and Energy note packet. Anything not done will become homework. You will be given 30 minutes to take your quiz. You will begin the quiz at 2:05 p.m.

Bell-Ringer Check Topic 1 Matter and Energy Notepacket:
DATE: 10/21/2014 Check Topic 1 Matter and Energy Notepacket: 65.) ) ) 0.64 atm 82.) 3.2L 88.) 5.39 L 66.) ) ) ) ) 2 67.) ) ) ) K 90.) 1 68.) ) 267 mL 79.) ) 0.1 atm 91.) 1 69.) ) L 80.) ) 118 kPa 75.) kPa 81.) ) 4

IDEAL GAS Real GAS 14-3 IDeal gas Law
An “ideal” gas is one that follows all the gas laws and precisely conforms to the KMT: Particles would have no volume No attraction between particles in the gas No such “ideal” gases exist A “real” gas: Particles do have volume Attractive forces exist between particles in the gas Because of these properties, gases have the ability to condense or solidify when compressed or cooled. (particle speed slows down) KMT = Kinetic Molecular Theory

14-3 IDeal gas Law Real gases differ most from an ideal gas at low temperatures and high pressures. (Think about your PTV card.) Pressure and temperature are directly proportional. Thus, real gases behave vey much like an ideal gas at many conditions of temperature and pressure.

What are the 3 variables of gases that you have already learned about?
14-3 IDeal gas Law What are the 3 variables of gases that you have already learned about? Pressure Temperature Volume 4th variable: the amount of gas (n) measured in moles

14-3 IDeal gas Law If you know the values for P, V, T, and n for one set of conditions, you can calculate a value for the ideal gas constant (R). Recall that 1 mol of every gas occupies 22.4 L at STP (101.3 kPa and 273 K). Insert the values of P, V, T, and n into (P  V)/(T  n). P  V T  n R = = 101.3 kPa  22.4 L 273 K  1 mol = 1 atm ×22.4 L 273 K ×1 mol R = 8.31 (L·kPa)/(K·mol) R = (L·atm)/(K·mol)

PV=nRT Ideal gas law P = Pressure in kilopascals (kPa)
V = Volume in liters (L) n = # of moles of gas (mol) R = constant 8.31 (L∙kPa)/(K∙mol) OR constant (L∙atm)/(K∙mol) T = Temperature in Kelvin (K)

14-3 IDeal gas Law Sample Problem 14.5 At 34oC, the pressure inside a nitrogen-filled tennis ball with a volume of L is 212 kPa. How many moles of nitrogen gas are in the tennis ball? n = 1.23  10–2 mol N2

14-3 IDeal gas Law When the temperature of a rigid hollow sphere containing L of helium gas is held to 621 K, the pressure of the gas is 1.89 x 103 kPa. How many moles of helium does the sphere contain? n = 2.51 x 102 mol

14-3 IDeal gas Law What pressure will be exerted by mol of a gas at 25◦C if it is contained in a L vessel? P = 1.71 x 103 kPa

Where you paying attention?
How would you rearrange the ideal gas law to isolate the temperature, T? PV nR T = A. nR PV T = C. PR nV T = B. RV nP T = D.

Bell-Ringer DATE: 10/22/2014 Please check last night’s homework with the answers below. Please make a note next to any of the questions you would like to see solved. Ideal Gas Law Worksheet: 1.) 205 K 7.) 19.9 L 2.) 1.26 mol 8.) K 3.) atm 9.) 1.62 atm 4.) L 10.) mol 5.) 5.34 atm 11.) L 6.) mol 12.) mol When you are done checking your homework, please come up and grab The Ideal and Combined Gas Laws, to work on for the class period. Anything not completed in class will become homework.

Agenda DATE: 10/27/2014 Today’s in class assignment is to make Unit 1 Interactive Achievement Test corrections. The test corrections must be completed on a separate sheet of paper and the following information must be provided for every question wrong: Rewrite the question. Identify the Correct answer. Science and theory behind the correct answer. Page number in your textbook where the science and theory for the correct answer can be found. Submit test corrections with original grade score report.

Bell-Ringer DATE: Why does a collision with an inflated air bag cause much less damage than a collision with a steering wheel or dashboard? Hint: Gases.

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Bell-Ringer DATE: Why does a collision with an inflated air bag cause much less damage than a collision with a steering wheel or dashboard? Hint: Gases.

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