1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 11 Gases 11.4 Temperature and Volume (Charles’s Law) As the gas in the hot-air balloon is heated, it expands.

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Charles’s Law In Charles’s law, the Kelvin temperature of a gas is directly related to the volume P and n are constant when the temperature of a gas increases, its volume increases

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 For two conditions, Charles’s law is written V 1 = V 2 (P and n constant) T 1 T 2 Rearranging Charles's law to solve for V 2 T 2 xV 1 = V 2 x T 2 T 1 T 2 V 2 = V 1 T 2 T 1 Charles’s Law: V and T

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Learning Check Use the gas laws to complete the following statements with 1) increases or 2) decreases. A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12 L to 24 L. D. Volume _______ when T changes from 15 °C to 45 °C. 4

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. Solution A. Pressure (1) increases when V decreases. B. When T decreases, V (2) decreases. C. Pressure (2) decreases when V changes from 12 L to 24 L. D. Volume (1) increases when T changes from 15 °C to 45 °C. 5

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 Learning Check Solve Charles's law expression for T 2. V 1 = V 2 T 1 T 2

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 Solution V 1 = V 2 T 1 T 2 Cross multiply to give V 1 T 2 =V 2 T 1 Solve for T 2 by dividing through by V 1 V 1 T 2 =V 2 T 1 V 1 T 2 =V 2 T 1 V 1

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 A balloon has a volume of 785 mL at 21°C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)? STEP 1 Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 Know Predict V 1 = 785 mL V 2 = ? V decreases T 1 = 21 °C T 2 = 0 °C = 294 K = 273 K T decreases Be sure to use the Kelvin (K) temperature in gas calculations. Example of Using Charles's Law

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 Calculations Using Charles's Law (continued) STEP 2 Rearrange the gas law for the unknown. Solve Charles's law for V 2 V 1 = V 2 T 1 T 2 V 2 = V 1 T 2 T 1 STEP 3 Substitute values into the gas law to solve for the unknown. V 2 = 785 mL x 273 K = 729 mL 294 K When temperature decreases, the volume decreases as predicted.

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 A sample of oxygen gas has a volume of 420 mL at a temperature of 18 °C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443 °C 2) 170 °C 3) –82 °C Learning Check

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 STEP 1 Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 Know Predict V 1 = 420 mL V 2 = 640 mL V increases T 1 = 18 °C T 2 = ? T increases = 291 K Solution

12. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 STEP 2 Rearrange the gas law for the unknown. Solve Charles's law for T 2 T 2 = T 1 V 2 V 1 STEP 3 Substitute values into the gas law to solve for the unknown. T 2 = 291 K x 640 mL = 443 K 420 mL = 443 K - 273 K = 170 °C (2) Solution (continued)