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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.1 Chapter 7 Gases 7.6 The Combined Gas Law.

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Presentation on theme: "General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.1 Chapter 7 Gases 7.6 The Combined Gas Law."— Presentation transcript:

1 General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.1 Chapter 7 Gases 7.6 The Combined Gas Law

2 General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. Summary of Gas Laws The gas laws can be summarized as follows: 2

3 General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.3 The combined gas law uses Boyles Law, Charless Law, and Gay-Lussacs Law (n is constant). P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law

4 General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.4 A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? Step 1 Set up data table: Conditions 1Conditions 2 P 1 = atm P 2 = 3.20 atm V 1 = L (180 mL) V 2 = 90.0 mL T 1 = 29 °C = 302 KT 2 = ? Combined Gas Law Calculation

5 General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.5 STEP 2 Solve for T 2 P 1 V 1 =P 2 V 2 T 1 T 2 T 2 = T 1 x P 2 x V 2 P 1 V 1 STEP 3 Substitute values to solve for unknown. T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K atm 180. mL T 2 = 604 K 273 = 331 °C Combined Gas Law Calculation (continued)

6 General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.6 A gas has a volume of 675 mL at 35 °C and atm pressure. What is the volume (mL) of the gas at -95 °C and a pressure of 802 mmHg (n constant)? Learning Check

7 General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc.7 STEP 1 Set up data table Conditions 1Conditions 2 T 1 = 308 K T 2 = -95 °C = 178 K V 1 = 675 mL V 2 = ? P 1 = 646 mmHg P 2 = 802 mmHg STEP 2 Solve for V 2 V 2 =V 1 x P 1 x T 2 P 2 T 1 STEP 3 Substitute values to solve for unknown. V 2 = 675 mL x 646 mmHg x 178 K = 314 mL 802 mmHg x 308 K Solution


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