1. WHAT FORMULA IS USED FOR SOLVING BOYLE’S LAW? FOR CHARLES’ LAW? Quick Write p128:

2. Gay-Lussac's Law

3. Joseph-Louis Gay-Lussac In the early 1800’s, Gay-Lussac ascended to a height of approximately 23,000 ft in a hot air balloon to study variations in the Earth's electro-magnetic intensity relative to altitude. On this flight, he experienced the effects of oxygen deprivation but still managed to collect air samples at over 20,000 ft, study the variation of pressure and temperature, and continue his observations on electro- magnetism.

4. What is Gay-Lussac’s Law? Gay-Lussac's Law states that the pressure of a sample of gas at constant volume, is directly proportional to its temperature in Kelvin.

5. Mathematically Gay-Lussac’s Law can be expressed as P 1 = P 2 T 1 T 2 P 1 T 2 = P 2 T 1 P 1 is the initial pressure T 1 is the initial temperature (in Kelvin) P 2 is the final pressure T 2 is the final temperature (in Kelvin)

6. Gay-Lussacs’s Law Practice Problems 1.A tank of gas has a pressure of 2.75 atm at 20°C. What will be the new pressure in the tank if its temperature is increased to 100°C? P 1 T 2 = P 2 T 1 P 1: T 1 : P 2 : T 2 :

7. 2.A gas system has initial pressure of 793 kPa and temperature of 34.5°C If the pressure changes to 446 kPa, what will the resultant temperature be in K? P 1 T 2 = P 2 T 1 P 1 : T 1 : P 2 : T 2 :

8. COMBINED GAS LAW The combined gas law is a combination of Boyle's Law and Charles's Law; hence its name the combined gas law. In the combined gas law, the volume of gas is directly proportional to the absolute temperature and inversely proportional to the pressure.

9. Mathematically, Combined Gas Law can be expressed as P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 P 1 is the initial pressure V 1 is the initial volume T 1 is the initial temperature (in Kelvin) P 2 is the final pressure V 2 is the final volume T 2 is the final temperature (in Kelvin)

10. Combined Gas Law Practice Problems 1.A sample of gas occupies a volume of 75 mL at a pressure of 725 mmHg and a temperature of 18°C. What volume would this sample of gas occupy at 800 mmHg and 298K? P 1 V 1 T 2 = P 2 V 2 T 1 P 1 : V 1 : T 1 : P 2 : V 2 : T 2 :

11. 2.A sample of gas occupies a volume of 3.75 L at a pressure of 1.15 atm and a temperature of 25°C. At what temperature will the sample of gas occupy 4 L at a pressure of 1 atm? P 1 V 1 T 2 = P 2 V 2 T 1 P 1 : V 1 : T 1 : P 2 : V 2 : T 2 :

12. Abbreviations atm – atmosphere mm Hg - millimeters of mercury torr - another name for mm Hg Pa - Pascal (kPa = kilo Pascal) K - Kelvin °C - degrees Celsius Problems: 1. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure? 2.500.0 liters of a gas are prepared at 700.0 mm Hg and 200.0 °C. The gas is placed into a tank under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the volume of the gas? 3. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 °C to 30.0 °C and a pressure change from 760.0 mm Hg to 360.0 mm Hg? 4.What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300.0 L at 0.250 atm and 400.0 K. 5.At conditions of 785.0 torr of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 745.0 torr and 30.0 °C? 6.A gas occupies a volume of 34.2 mL at a temperature of 15.0 °C and a pressure of 800.0 torr. What will be the volume of this gas at standard conditions? 7.The volume of a gas originally at standard temperature and pressure was recorded as 488.8 mL. What volume would the same gas occupy when subjected to a pressure of 100.0 atm and temperature of minus 245.0 °C? 8.At a pressure of 780.0 mm Hg and 24.2 °C, a certain gas has a volume of 350.0 mL. What will be the volume of this gas under STP Conversions K = °C + 273 1 cm 3 (cubic centimeter) = 1 mL (milliliter) 1 dm 3 (cubic decimeter) = 1 L (liter) = 1000 mL Standard Conditions 0.00 °C = 273 K 1.00 atm = 760.0 mm Hg = 101.325 kPa = 101,325 Pa Combined Gas Law Worksheet: ISN p 127