Chapter 11 Gases

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. 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. Charles’s Law: V and T • For two conditions, Charles’s law is written V1 = V2 (P and n constant) T1T2 • Rearranging Charles's law to solve for V2 T2 x V1 = V2 x T2 T1T2 V2 = V1T2 T1

4. 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.

5. 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.

6. Learning Check Solve Charles's law expression for T2. V1 = V2 T1T2

7. Solution V1 = V2 T1T2 Cross multiply to give V1T2 = V2T1 Solve for T2 by dividing through by V1 V1T2= V2T1 V1V1 T2 = V2T1 V1

8. Example of Using Charles's Law 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 1Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 Know Predict V1 = 785 mL V2 = ? V decreases T1 = 21 °C T2 = 0 °C = 294 K = 273 K T decreases Be sure to use the Kelvin (K) temperature in gas calculations.

9. Calculations Using Charles's Law (continued) STEP 2Rearrange the gas law for the unknown. Solve Charles's law for V2 V1 = V2 T1T2 V2 = V1T2 T1 STEP 3Substitute values into the gas law to solve for the unknown. V2 = 785 mL x 273 K = 729 mL 294 K When temperature decreases, the volume decreases as predicted.

10. Learning Check 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

11. Solution STEP 1Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 Know Predict V1 = 420 mL V2 = 640 mL V increases T1 = 18 °C T2 = ? T increases = 291 K

12. Solution (continued) STEP 2Rearrange the gas law for the unknown. Solve Charles's law for T2 T2 = T1V2 V1 STEP 3Substitute values into the gas law to solve for the unknown. T2 = 291 K x 640 mL = 443 K 420 mL = 443 K - 273 K = 170 °C (2)