1. © 2014 Pearson Education, Inc. Chapter 11 Gases 11.4 Temperature and Volume (Charles’s Law) Learning Goal Use the temperature–volume relationship (Charles’s law) to determine the final temperature or volume when the pressure and amount of gas are constant. Chapter 11 Lecture Basic Chemistry Fourth Edition

2. © 2014 Pearson Education, Inc. Charles’s law states that the temperature (K) of a gas is directly related to its volume (V) when P and n are constant if the temperature (K) increases, then the volume (V) also increases Charles’s Law Figure 11.5 Charles’s law: The Kelvin temperature of a gas is directly related to the volume of the gas when there is no change in the pressure and amount of gas. When the temperature increases, making the molecules move faster, the volume must increase to maintain constant pressure.

3. © 2014 Pearson Education, Inc. Charles’s law is a direct relationship between volume and temperature. A direct relationship between variables means the variables increase or decrease together. All temperatures used in gas law calculations must be converted to Kelvin (K) temperature. Charles’s Law No change in number of moles and pressure

4. © 2014 Pearson Education, Inc. The equation for Charles’s law can be rearranged to solve for any factor. To obtain V 2, multiply both sides by T 2. Charles’s Law Equation

5. © 2014 Pearson Education, Inc. Use the gas laws to complete the following statements with increases or decreases. A. Temperature _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12 L to 4 L. D. Volume _______ when T changes from 15 o C to 45 o C. Learning Check

6. © 2014 Pearson Education, Inc. Use the gas laws to complete the following statements with increases or decreases. A. Temperature decreases when V decreases. B. When T decreases, V decreases. C. Temperature decreases when V changes from 12 L to 4 L. D. Volume increases when T changes from 15 o C to 45 o C. Solution

7. © 2014 Pearson Education, Inc. Solve Charles’s law expression for T 2. Learning Check

8. © 2014 Pearson Education, Inc. Solve Charles’s law expression for T 2. Cross multiply to give: V 1 T 2 = V 2 T 1 Solve for T 2 by dividing through by V 1 Solution

9. © 2014 Pearson Education, Inc. Learning Check A balloon has a volume of 785 mL at 21 o C. If the temperature drops to 0 o C, what is the new volume of the balloon (n, P constant)?

10. © 2014 Pearson Education, Inc. A balloon has a volume of 785 mL at 21 o C. If the temperature drops to 0 o C, what is the new volume of the balloon (n, P constant)? Step 1 Organize the data in a table of initial and final conditions. We can predict that as T decreases, V decreases. Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? T 1 = 21 °C = 294 KT 2 = 0 °C = 273 K Learning Check

11. © 2014 Pearson Education, Inc. A balloon has a volume of 785 mL at 21 o C. If the temperature drops to 0 o C, what is the new volume of the balloon (n, P constant)? Step 2 Rearrange the gas law equation to solve for the unknown quantity. Solve Charles’s law for V 2. Decreasing temperature decreases the volume when moles and pressure are constant. Solution

12. © 2014 Pearson Education, Inc. A balloon has a volume of 785 mL at 21 o C. If the temperature drops to 0 o C, what is the new volume of the balloon (n, P constant)? Step 2 Rearrange the gas law equation to solve for the unknown quantity. Solution

13. © 2014 Pearson Education, Inc. A balloon has a volume of 785 mL at 21 o C. If the temperature drops to 0 o C, what is the new volume of the balloon (n, P constant)? Step 3 Substitute values into the gas law equation and calculate. When temperature decreases, the volume decreases as predicted. Solution