Gas Laws Review A Station to Station Activity. Adventures of Milkman Dan Essential Vocabulary: Imploded: A condition of internal collapse caused by an.

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Gas Laws Review A Station to Station Activity

Adventures of Milkman Dan Essential Vocabulary: Imploded: A condition of internal collapse caused by an imbalance of pressure. Too much external pressure and insufficient internal pressure cause this condition. Vacuum Tube:A tube connected to a vacuum pump. The pump can remove all gas particles from any chamber it is connected to, lowering the internal pressure of the chamber. Kitten:A disease-infested, untrainable, allergen-spreading, disloyal waste of mammalian flesh that is perceived by weak minds as something “cute.”

In the final panel, Milkman Dan says that he “imploded kittens in a vacuum tube.” How could Dan accomplish such a feat? What Gas Law was Milkman Dan bravely trying to validate, despite objections from weak minds? We love you, Milkman Dan. Your devotion to science uplifts us all…

1 What is the formula for Boyle’s Law? What do each of the variables represent? P 1 V 1 = P 2 V 2 P = pressure, V = volume 1 = initial conditions 2 = final conditions

2 What is the relationship between the pressure and the volume in Boyle’s Law? This is an inverse relationship. When pressure goes Up, volume goes down. And when pressure goes down, volume goes up

3 Charles’ Law states the temperature and volume are related when what is held constant? Pressure is held constant.

4 What is the relationship between temperature and volume based on Charles’ Law? This is a direct relationship. When volume goes up, temperature goes up. When volume goes down, temperature goes down.

5 What is the formula for Charles’ Law? V 1 = V 2OR V 1 T 2 = V 2 T 1 T 1 T 2

6 According to Charles’ Law if the temperature of a gas is increased, what would happen to the volume of that gas? The volume will increase.

7 Which gas law is demonstrated by the data above? Boyle’s Law: As P goes up to 6 books, Volume goes down to 7.5 mL

8 The gas in a 2 L piston is maintained at a pressure of 3 atm. When the piston is compressed to 1.2 L, what happens to the pressure? This is Boyle’s Law: P 1 V 1 = P 2 V 2 P 1 = 2 L V 1 = 3 atmP 2 = P 1 V 1 = 2L * 3 atm = 5 atm P 2 = ?V 2 1.2L V 2 = 1.2 L

9 Gay-Lussac’s Law states the when the pressure of a gas Increases the temperature of that gas increases. This is an example of a(n) direct relationship.

10 What number is added to a celsius temperature to convert it into Kelvin? 273

11 If a sample of gas occupies 6.80 L at 325 ˚C, what will be its volume at 25 ˚C if the pressure does not change? V 1 = 6.8L T 1 = 325 ˚C = 598 K V 2 = ? T 2 = 25 ˚C = 298K V 2 = V 1 T 2 = 6.8L * 298 K = 3.39 K T 1 598K

12 The __Collisions______________ of the particles in a gas with the walls of the container results in gas _pressure_______________. The total pressure in a mixture of gases is equal to the sum of the __pressures______________ of each gas present. This is known as Dalton’s Law of Partial Pressures.

13 A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? P 1 = 6.58 kPa T 1 = 529 K P 2 = ? T 2 = 211 K P 2 = P 1 T 2 = 6.58 kPa * 211 K = 2.62kPa T 1 529K

14 Boyle’s, Charles’, and Gay-Lussac’s Laws can be written as a single expression called the __Ideal Gas Law______________. Another expression is the ideal gas law, _PV = nRT_____________, where n = __# of moles________. The letter R is the __Gas__________ constant, and is equal to ___ L*atm/mol*K or 8.31 L*kPa/mol* K _____________.

15 The manufacturer of an aerosol deodorant packaged in a 150-mL container wishes to produce a container of the same size that will hold twice as much gas. How will the pressure of the gas in the new product compare with that of the gas in the original container? The pressure will double because the moles of gas is doubling.

16 Describe what happens to kinetic energy during gas particle collisions and as Kelvin temperature increases. The kinetic energy of the particles increases, so both the number and force of their collisions increase. As a result, the pressure increases as well.

17 When the temperature of a gas is lowered sufficiently, the gas _condenses____ into a liquid. As the temperature is lowered further, the liquid becomes a _solid____________. This reflects the Kinetic Molecular Theory. Increasing the kinetic energy allows particles to become free from each other and become gases. Decreasing kinetic energy prevents them from escaping the attractions they have for other particles.

18 Convert the following temperatures to degrees Celsius? a. 700 K – 273 = 427 o C b. 0 K – 273 = -273 o C c. 273 K – 273 = o C d. 459 K – 273 = 186 o C e K – 273 = 752 o C

19 What is the formula for the Ideal Gas Law? PV = nRT

20 An 18 liter container holds grams of oxygen gas (O 2 ) at 45 °C. What is the pressure in the container? A)Find the molar mass of oxygen. B) Convert celsius to Kelvin. C) Use the Ideal Gas Law to solve. Solution is on next slide

20 continued A)Oxygen = g O 2 = 2 x = g/mol 16 g O 2 x 1 mol O 2 = 0.5 mol O g O 2 B) 45 O C = 318K C)P = ? V = 18L n = 0.5 mol R = L*atm/mol*K T = 318 K P = nRT = 0.5 mol x L* atm/mol*K x 318K = atm V18 L

21 What does STP mean? Standard Temperature and Pressure What are the values of T and P at STP? T = 25 O C or 298K P = 1 atm

17.5 If the volume of a balloon was changed from 2 L to 6 L, what happened to the pressure if the starting pressure was 10 atm?

Practice Problems 1.A gas occupies 4.31 liters at a pressure of atm. Determine the volume if the pressure is increased to 1.25 atm L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure (as required by Charles' Law)? (0.755 atm) (4.31 liters) = (1.25 atm) (x)

Just two more! L of a gas is collected at 22.0°C and mmHg. When the temperature is changed to standard (25°C), what is the new pressure? (You must convert °C to K first.) 4. A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass in the sample. n = [ (1.00 atm) (56.2 L) ] / [ ( L atm mol¯ 1 K¯ 1 ) (273.0 K)]