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UNIT 6 SOLUTIONS AND GASES 6.8 What is the relationship between P, V, and T of an ideal gas? May 4, 2011 AIM: What are the effects of pressure, volume,

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Presentation on theme: "UNIT 6 SOLUTIONS AND GASES 6.8 What is the relationship between P, V, and T of an ideal gas? May 4, 2011 AIM: What are the effects of pressure, volume,"— Presentation transcript:

1 UNIT 6 SOLUTIONS AND GASES 6.8 What is the relationship between P, V, and T of an ideal gas? May 4, 2011 AIM: What are the effects of pressure, volume, and temperature on ideal gases? DO NOW: 1.When you heat a marshmallow in the microwave, the marshmellow expands. What is the relationship between temperature and volume? (direct or indirect/inverse?) Direct: as temperature increases, volume increases 2. Why the hot-air balloon can fly. What is happening to the air molecules that cause the balloon to float in the air? The air molecules are expanding, causing the pressure in the balloon to be less than that of outside of the balloon.

2 LAB Complete the Lab. After you are done with the lab, please fill out the graphic organizer on the back of you worksheet. – If you are having difficulty filling out the graphic organizer, please access this PowerPoint (6.8) online!

3 BOYLE’S LAW PRESSURE AND VOLUME OF A GAS (at constant TEMPERATURE)

4 BOYLE’S LAW: TREND At constant temperature, as pressure increases… VOLUME DECREASES This is an INDIRECT RELATIONSHIP

5 MATHEMATICAL EQUATION P 1 V 1 = P 2 V 2

6 CHARLES’S LAW VOLUME AND TEMPERATURE OF A GAS (at constant PRESSURE)

7 CHARLES’S LAW: TREND At constant pressure, as volume increases… TEMPERATURE INCREASES This is an DIRECT RELATIONSHIP

8 MATHEMATICAL EQUATION V 1 = V 2 T 1 = T 2

9 How do we solve problems relating to gases? I. BOYLE’S LAW Example 1 A sample of gas in a syringe has a volume of 9.66 mL at a pressure of 64.4 kPa. The plunger is depressed until the pressure is 94.6 kPa. What is the new volume, assuming constant temperature? Step 1. List the givens. V 1 = V 2 = P 1 =P 2 = 9.66 ml 64.4 kPa 94.6 kPa ? Step 2. Write out the formula. P 1 V 1 = P 2 V 2

10 Step 2. Write out the formula. P 1 V 1 = P 2 V 2 Step 3. Plug in. 9.66ml x 64.4kPa= 94.6kPa x V 2 Step 1. List the givens. V 1 = V 2 = P 1 =P 2 = 9.66 ml 64.4 kPa 94.6 kPa ? 622.104ml x kPa = 94.6kPa x V 2 94.6kPa 6.576ml= V 2

11 CHARLES’S LAW Temperature must always be in KELVINS!

12 How do we solve problems relating to gases? II. CHARLES’S LAW Example 1 The volume of a sample of a gas at 0  C is 200.0 L. If the volume is decreased to 100.0 L at constant pressure, what will be the new temperature of the gas? Step 1. List the givens. V 1 = V 2 = T 1 =T 2 = 200 L 0 o C = 273K ? 100 L Step 2. Write out the formula. V 1 = V 2 T 1 T 2

13 Step 4. Plug in. 200 L = 100 L 273K T 2 0.7326 = 100L T 2 136.5K= T 2 Step 1. List the givens. V 1 = V 2 = T 1 =T 2 = 200 L 0oC0oC ? 100 L Step 3. Check the units for temperature. (Convert to Kelvins) 0 o C = 273K Step 2. Write out the formula. V 1 = V 2 T 1 T 2

14 INDEPENDENT PRACTICE Complete the Independent Practice PLEASE COMPLETE THE WORKSHEET FOR HOMEWORK


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