Chapter 11 Gases

1. An autoclave used to sterilize equipment attains a temperature higher than 100 °C. Chapter 11 Gases 11.5 Temperature and Pressure (Gay-Lussac’s Law)

2. Gay-Lussac’s Law: P and T In Gay-Lussac’s law, • the pressure exerted by a gas is directly related to the Kelvin temperature • V and n are constant P1 = P2 T1T2

3. Learning Check Solve Gay-Lussac’s law for P2. P1 = P2 T1 T2

4. Solution Solve Gay-Lussac’s law for P2. P1 = P2 T1T2 Multiply both sides by T2 and cancel P1 x T2 = P2 x T2 T1T2 P2 = P1 x T2 T1

5. Example of Using Gay-Lussac’s Law A gas has a pressure at 2.0 atm at 18 °C. What is the new pressure when the temperature is 62 °C (V and n constant)? STEP 1Organize the data in a table of initial and final conditions. Conditions 1 Conditions 2 Know Predict P1 = 2.0 atm P2 = ? P increases T1 = 18 °C + 273 T2 = 62 °C + 273 T increases = 291 K = 335 K

6. Example of Using Gay-Lussac’s Law (continued) STEP 2Rearrange the gas law for the unknown. Solve Gay-Lussac’s law for P2. P1 = P2 T1T2 P2 = P1T2 T1 STEP 3Substitute values into the gas law to solve for the unknown. P2 = 2.0 atm x 335 K = 2.3 atm 291 K Temperature ratio increases pressure

7. Learning Check Use the gas laws to complete 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

8. Solution Use the gas laws to complete with 1) increases or 2) decreases. 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

9. Learning Check A gas has a pressure of 645 mmHg at 128 °C. What is the temperature in Celsius if the pressure increases to 1.50 atm (n and V remain constant)?

10. Solution STEP 1Organize the data in a table of initial and final conditions. P2 = 1.50 atm x 760 mmHg = 1140 mmHg 1 atm Conditions 1 Conditions 2 Know Predict P1 = 645 mmHg P2 = 1140 mmHg P increases T1 = 128 °C + 273 T2 = K – 273 T increases = 401 K = ? °C

11. Solution (continued) STEP 2Rearrange the gas law for the unknown. Solve Gay-Lussac’s law for T2. P1 = P2 T1T2 T2 = T1 x P2 P1 STEP 3Substitute values into the gas law to solve for the unknown. T2 = 401 K x 1140 mmHg = 709 K – 273 = 436 °C 645 mmHgPressure ratio increases temperature

12. Vapor Pressure and Boiling Point Vapor pressure • is the pressure above water at equilibrium in a closed container • at the boiling point is equal to the external pressure

13. Boiling Point of Water The boiling point of water • depends on the vapor pressure • is lower at higher altitudes

14. An autoclave used to sterilize equipment attains a temperature higher than 100 °C. Boiling Point of Water The boiling point of water • is increased by using an autoclave to increase external pressure

15. Learning Check In each of the following pairs of atmospheric and vapor pressure, indicate if water will or will not boil. Atmospheric Vapor Boiling Pressure Pressure Occurs? A. 760 mmHg 760 mmHg B. 960 mmHg 760 mmHg C. 520 mmHg 620 mmHg

16. Solution Atmospheric Vapor Boiling Pressure Pressure Occurs? A. 760 mmHg 760 mmHg yes B. 960 mmHg 760 mmHg no C. 520 mmHg 620 mmHg yes