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Chapter 13 Gases.

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Presentation on theme: "Chapter 13 Gases."— Presentation transcript:

1 Chapter 13 Gases

2 Make a Flip Book

3 Boyle’s Gas Law How are pressure and volume related? As the pressure increases, the volume decreases proportionally.

4 Boyle’s Gas Law

5 Identify and label 2 situations
A diver blows a 0.75 liter air bubble 10m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the volume of the air in the bubble at the surface?

6 Identify and label 2 situations
A diver blows a 0.75 liter air bubble 10m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the volume of the air in the bubble at the surface? 10m underwater Surface V1 = 0.75 L V2 = ? P1 = 2.25 atm P2 = 1.03 atm

7 Solve using Boyle’s Law
P1V1 = P2 V2 2.25 0.75 = 1.03 V2 Use magic Algebra  both sides by 1.03 V2 = 2.25 0.75 1.03 V2 = 1.6 L

8 Check Answer P1V1 = P2 V2 2.25 0.75 = 1.03 V2 V2 = 1.6 L
The pressure decreased by about half, 2.25 to 1.03. So, the volume should double. Is 1.6 about twice 0.75?

9 STP 1 atm 760 torr 760 mmHg 273º K 0º C Standard Pressure
Standard Temperature 273º K 0º C

10 Charles’s Law How are temperature and volume related?
The volume of a given amount of gas is directly proportional to its kelvin temperature at constant pressure.

11 Temperature in kelvin units
Convert celsius degrees to kelvins Tk = Tc

12 Charles’s Law A helium balloon in a closed car occupies a volume of 2.23 L at 40.0C. If the car is parked on a hot day and the temperature inside rises to 75.0C, what is the new volume of the balloon, assuming the pressure remains constant.

13 Set up two situations Car with balloon T1 = 40C+ 273 = 313
V1 = 2.32 L Car parked in the heat T2 = 75.0C = 348 V2 = unknown

14 Assume the pressure and amount of gas remain constant
What volume will the gas in the balloon occupy at 250 K? A gas at 89C occupies a volume of 0.67L. At what celsius temperature will the volume increase to 1.12L? The celsius temperature of a 3.00L sample of gas is lowered from 80.0C to 30.0C. What will be the resulting volume of the gas?

15 Assume the pressure and amount of gas remain constant
What volume will the gas in the balloon occupy at 250 K? 1.8L A gas at 89C occupies a volume of 0.67L. At what celsius temperature will the volume increase to 1.12L? 330C The celsius temperature of a 3.00L sample of gas is lowered from 80.0C to 30.0C. What will be the resulting volume of the gas? 2.58L

16 Gay-Lussac’s Law How are temperature and pressure of gas related?
Remember the temperature must be in kelvins.

17 Example The pressure of the oxygen gas inside a canister is 5.00 atm at 25.0 C. The canister is located at a camp high on Mount Everest. If the temperature there falls to C , what is the new pressure inside the canister?

18 Set up 2 situations Warm T1 = 25.0 C +273 P1 = 5.00 atm Cold
P2 = unknown

19 Analyze the answer The temperature decreased, so the pressure should decrease. The unit is atm matches the pressure unit given.

20 Examples The pressure in an automobile tire is 1.88atm at 25.0 C. What will be the pressure if the temperature increases to 37.0 C. Helium gas in a 2.00L cylinder is under 1.12 atm pressure. At 36.5 C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder?

21 Examples The pressure in an automobile tire is 1.88atm at 25.0 C. What will be the pressure if the temperature increases to 37.0 C. 1.96 atm Helium gas in a 2.00L cylinder is under 1.12 atm pressure. At 36.5 C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder? C

22 The Combined Gas Law Temperature measured in kelvins.

23 Example A gas at 110 kPa and 30C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0C and the pressure increases to 440kPa, what is the new volume?

24 Set up 2 situations Initial P1 = 110 kPa T1 = 30.0 C +273 V1 = 2.00L
Raised Temp P2 = 440 kPa T2 = 80.0 C + 273 V2 = unknown


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