Chapter 14 “The Behavior of Gases”

Section 14.1 The Properties of Gases OBJECTIVES: Explain why gases are easier to compress than solids or liquids are. Describe the three factors that affect gas pressure. HW: 14.1 Cornell notes and section assessment questions due Fri

Compressibility (only vocab word new to this unit) Gases EXPAND to fill containers, unlike solids or liquids The reverse is also true: Gases are easily compressed Compressibility: measure of how much the volume of matter decreases under pressure

Compressibility This is why “air bags” are in cars In an accident, the air compresses more than a solid (steering wheel, dash, glass) Impact can force the gas particles closer together, because there is a lot of empty space between them

Compressibility At room temperature, the distance between particles is about 10x the diameter of the particle  How does the volume of the particles in a gas compare to the overall volume of the gas?

4 Variables that describe a Gas The four variables and their common units: 1. pressure (P) in kilopascals 2. volume (V) in Liters 3. temperature (T) in Kelvin 4. amount (n) in moles The amount of gas, volume, and temperature are factors that affect gas pressure.

1. Amount of Gas When we inflate a balloon, we are adding gas molecules. Increasing the number of gas particles increases the number of collisions  pressure increases (Increased gas  increased pressure) If temperature is constant, doubling the # of particles doubles pressure

Number of molecules and pressure are directly related More molecules means more collisions, and… Fewer molecules means fewer collisions… What happens if you have an area with high pressure and lower pressure?

2. Volume of Gas In a smaller container, particles hit sides of the container more often. As volume decreases, pressure increases. (think of a syringe)

2. Volume of Gas Volume and pressure are inversely related to each other ( Decrease volume  Increased pressure)

3. Temperature of Gas Raising the temperature of a gas increases the pressure, if the volume is held constant. (Temp. and Pres. are directly related) The molecules hit the walls harder, and more frequently! Fig. 14.7, page 417 Should you throw an aerosol can into a fire? What could happen? When should your automobile tire pressure be checked?

What’s going to happen to these gas particles as the balloon is heated?

Graphing activity You’re creating 3 graphs, one to represent each of the 3 gas laws Boyle’s law Charles’ law Gay-Lussac’s law Things to remember: Every graph needs a title, labeled axes, and units How do you figure out the scale for a graph?

Discussion review of 14.1 What is compressibility, and why is it only possible to compress one state of matter? What happens to the particles in a gas when temperature is increased? When decreased? What would happen if the amount of a gas in a container increased? What happens to the pressure in a container if temperature stays the same, amount of gas stays the same, but volume is decreased? What do the following variables indicate? V, P, T, n What units are used for each?

Section 14.2 The Gas Laws OBJECTIVES: Describe the relationships among the temperature, pressure, and volume of a gas. Use gas laws to solve problems.

5.3.11 Bellringer ( (BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure? Homework 14.2 w/section assessment questions due TOMORROW GRAPH due TOMORROW (WE’LL START IN CLASS) 14.3 w/section assessment questions due FRIDAY

5.3.11 Bellringer (BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure? Knowns Unknown V1 = 3.0 L P2 = ? V2 = 1.5 L P1 = 100 kPa

5.3.11 Bellringer (BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure? Knowns Unknown V1 = 3.0 L P2 = ? V2 = 1.5 L If… P1V1=P2V2 P1 = 100 kPa

5.3.11 Bellringer (BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure? Knowns Unknown V1 = 3.0 L P2 = ? V2 = 1.5 L Then… P1V1=P2 P1 = 100 kPa V2

5.3.11 Bellringer (BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure? Knowns Unknown V1 = 3.0 L P2 = ? V2 = 1.5 L Then… P1V1=P2 P1 = 100 kPa V2 So, P2 = 100 kPa x 3.0 L 1.5 L

5.3.11 Bellringer (BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure? Knowns Unknown V1 = 3.0 L P2 = ? V2 = 1.5 L Then… P1V1=P2 P1 = 100 kPa V2 So, P2 = 100 kPa x 3.0 L = 200 kPa 1.5 L

The Gas Laws are mathematical The gas laws describe HOW gases behave  gas behavior The amount of change can be calculated with mathematical equations. You need to know both!

Robert Boyle (1627-1691) Boyle was born into an aristocratic Irish family Became interested in medicine and the new science of Galileo and studied chemistry.  A founder and an influential fellow of the Royal Society of London Wrote extensively on science, philosophy, and theology.

#1. Boyle’s Law - 1662 Gas pressure is inversely proportional to the volume, when temperature is held constant. Pressure x Volume = a constant Equation: P1V1 = P2V2 (T = constant)

Graph of Boyle’s Law – page 418 Boyle’s Law says the pressure is inverse to the volume. Note that when the volume goes up, the pressure goes down

- Page 419

Jacques Charles (1746-1823) French Physicist Part of a scientific balloon flight on Dec. 1, 1783 – was one of three passengers in the second balloon ascension that carried humans This is how his interest in gases started It was a hydrogen filled balloon – good thing they were careful!

#2. Charles’s Law - 1787 The volume of a fixed mass of gas is directly proportional to the Kelvin temperature, when pressure is held constant.

Converting Celsius to Kelvin Gas law problems involving temperature will always require that the temperature be in Kelvin. (Remember that no degree sign is shown with the kelvin scale.) Reason? There will never be a zero volume, since we have never reached absolute zero. Kelvin = C + 273 °C = Kelvin - 273 and

- Page 421

Joseph Louis Gay-Lussac (1778 – 1850) French chemist and physicist Known for his studies on the physical properties of gases. In 1804 he made balloon ascensions to study magnetic forces and to observe the composition and temperature of the air at different altitudes.

#3. Gay-Lussac’s Law - 1802 The pressure and Kelvin temperature of a gas are directly proportional, provided that the volume remains constant. How does a pressure cooker affect the time needed to cook food? (Note page 422) Sample Problem 14.3, page 423

PRACTICE PROBLEMS P. 419 #8 (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 constant? Knowns Unknown 4.00 L 205 kPa 12.0 L

PRACTICE PROBLEMS P. 419 #8 (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 constant? Knowns Unknown V1 = 4.00 L P2 = ? (Look at BR) P1 = 205 kPa P2 = P1V1 V2 = 12.0 L V2

PRACTICE PROBLEMS P. 419 #8 (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 constant? Knowns Unknown V1 = 4.00 L P2 = ? (Look at BR) P1 = 205 kPa P2 = P1V1 V2 = 12.0 L V2 P2 = 205 kPa x 4.00 L 12.00 L

PRACTICE PROBLEMS P. 421 #10 (CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to 100.0 degrees Celsius. What is the new volume if the pressure remains constant? Knowns Unknown

PRACTICE PROBLEMS P. 421 #10 (CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to 100.0 degrees Celsius. What is the new volume if the pressure remains constant? Knowns Unknown V1 = 5.00 L V2 = ? T1 = -50 °C T2 = 100.0 °C

PRACTICE PROBLEMS P. 421 #10 (CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to 100.0 degrees Celsius. What is the new volume if the pressure remains constant? Knowns Unknown V1 = 5.00 L V2 = ? T1 = -50 °C + 273 = 223K T2 = 100.0 °C + 273 = 373K

PRACTICE PROBLEMS P. 421 #10 (CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to 100.0 degrees Celsius. What is the new volume if the pressure remains constant? Knowns Unknown V1 = 5.00 L V2 = ? T1 = -50 °C + 273 = 223K T2 = 100.0 °C + 273 = 373K V1 = V2 T1 T2

PRACTICE PROBLEMS P. 421 #10 (CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to 100.0 degrees Celsius. What is the new volume if the pressure remains constant? Knowns Unknown V1 = 5.00 L V2 = V1T2 T1 = -50 °C + 273 = 223K T1 T2 = 100.0 °C + 273 = 373K V1 = V2 T1 T2

PRACTICE PROBLEMS P. 423 #12 (Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. 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. Knowns Unknown P1 = 198 kPa T2 = ? T1 = 27°C P2 = 225 KPa

PRACTICE PROBLEMS P. 423 #12 (Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. 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. Knowns Unknown P1 = 198 kPa T2 = ? T1 = 27°C P2 = 225 Kpa P1 = P2 T1 T2

PRACTICE PROBLEMS P. 423 #12 (Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. 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. Knowns Unknown P1 = 198 kPa T2 = ? T1 = 27°C T1 = T2 P2 = 225 Kpa P1 P2

PRACTICE PROBLEMS P. 423 #12 (Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. 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. Knowns Unknown P1 = 198 kPa T2 = ? T1 = 27°C P2 T1 = T2 P2 = 225 Kpa P1

PRACTICE PROBLEMS P. 423 #12 (Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. 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. Knowns Unknown P1 = 198 kPa T2 = ? T1 = 27°C +273=300K P2 T1 = T2 P2 = 225 Kpa P1

#4. The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas. Sample Problem 14.4, page 424

The combined gas law contains all the other gas laws! If the temperature remains constant... P1 x V1 P2 x V2 = T1 T2 Boyle’s Law

P1 x V1 P2 x V2 = T1 T2 Charles’s Law The combined gas law contains all the other gas laws! If the pressure remains constant... P1 x V1 P2 x V2 = T1 T2 Charles’s Law

P1 x V1 P2 x V2 = T1 T2 Gay-Lussac’s Law The combined gas law contains all the other gas laws! If the volume remains constant... P1 x V1 P2 x V2 = T1 T2 Gay-Lussac’s Law

Compute the value of an unknown using the ideal gas law. Section 14.3 Ideal Gases OBJECTIVES: Compute the value of an unknown using the ideal gas law.

PRACTICE PROBLEMS P. 424 #13 (Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume? Knowns Unknown P1 = V2 = ? T1 = How do we rearrange? V1 = P2 = T2 =

PRACTICE PROBLEMS P. 424 #13 (Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume? Knowns Unknown P1 = V2 = ? T1 = How do we rearrange? V1 = P2 = T2 =

PRACTICE PROBLEMS P. 424 #13 (Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume? Knowns Unknown P1 = 155 kPa V2 = ? T1 = 25°C How do we rearrange? V1 = 1.00 L P2 = 605 kPa T2 = 125 °C

PRACTICE PROBLEMS P. 424 #13 (Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume? Knowns Unknown P1 = 155 kPa V2 = ? T1 =25°C= 298K How do we rearrange? V1 = 1.00 L P2 = 605 kPa T2 =125°C=398K

PRACTICE PROBLEMS P. 424 #14 (Combined gas law) A 5.00L air sample has a pressure of 107 kPa at a temperature of -50 °C. If the temperature is raised to 102°C and the volume expands to 7.00 L, what will the new pressure be? Knowns Unknown P1 = 107 kPa P2 = ? T1 = -50°C V1 = 1.00 L How will we rearrange? T2 = 102°C V2 = 7.00L

Compare and contrast real an ideal gases. Section 14.3 Ideal Gases OBJECTIVES: Compare and contrast real an ideal gases.

5. The Ideal Gas Law #1 R = 8.31 (L x kPa) / (mol x K) Equation: P x V = n x R x T Pressure times Volume equals the number of moles (n) times the Ideal Gas Constant (R) times the Temperature in Kelvin. R = 8.31 (L x kPa) / (mol x K) The other units must match the value of the constant, in order to cancel out. The value of R could change, if other units of measurement are used for the other values (namely pressure changes)

Ideal Gases We are going to assume the gases behave “ideally”- in other words, they obey the Gas Laws under all conditions of temperature and pressure An ideal gas does not really exist, but it makes the math easier and is a close approximation. Particles have no volume? Wrong! No attractive forces? Wrong!

Ideal Gases There are no gases for which this is true (acting “ideal”); however, Real gases behave this way at a) high temperature, and b) low pressure. Because at these conditions, a gas will stay a gas! Sample Problem 14.5, page 427

Real Gases and Ideal Gases

Ideal Gases don’t exist, because: Molecules do take up space There are attractive forces between particles - otherwise there would be no liquids formed

Real Gases behave like Ideal Gases... When the molecules are far apart. The molecules do not take up as big a percentage of the space We can ignore the particle volume. This is at low pressure

Real Gases behave like Ideal Gases… When molecules are moving fast This is at high temperature Collisions are harder and faster. Molecules are not next to each other very long. Attractive forces can’t play a role.

Section 14.4 Gases: Mixtures and Movements OBJECTIVES: Relate the total pressure of a mixture of gases to the partial pressures of the component gases.

Section 14.4 Gases: Mixtures and Movements OBJECTIVES: Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses.

#7 Dalton’s Law of Partial Pressures For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . P1 represents the “partial pressure”, or the contribution by that gas. Dalton’s Law is particularly useful in calculating the pressure of gases collected over water.

Connected to gas generator Collecting a gas over water – one of the experiments in Chapter 14 involves this.

= 6 atm Sample Problem 14.6, page 434 If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3: 2 atm + 1 atm + 3 atm = 6 atm 1 2 3 4 Sample Problem 14.6, page 434

Diffusion is: Molecules moving from areas of high concentration to low concentration. Example: perfume molecules spreading across the room. Effusion: Gas escaping through a tiny hole in a container. Both of these depend on the molar mass of the particle, which determines the speed.

Diffusion: describes the mixing of gases Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Molecules move from areas of high concentration to low concentration. Fig. 14.18, p. 435

Effusion: a gas escapes through a tiny hole in its container -Think of a nail in your car tire… Diffusion and effusion are explained by the next gas law: Graham’s

End of Chapter 14