Warm Up 2-7-14 1.What is STP? 2. How much space does 1 mole of hydrogen gas occupy at STP? 3. How many torr is in 5 atm? 4. Convert 30 Celsius to Kelvin. Standard temperature pressure 1. 0Celsius 1 atm 3. 3800torr Agenda -Pick up binders (40pts) -Notes Chp 12-3 (gas laws) -WS The gas law Homework Quiz next class

12-3 The Gas Laws Mathematical relationships between volume, temperature, pressure and amount of gas

Combined Gas Law: Pressure, Volume &Temperature Combines all three gas laws into one equation!! Memorize this equation!!! P1V1 = P2V2 T1 T2 Remember: all temperatures must be in Kelvin!!!

Combined Law: Example 1 A helium filled balloon has a volume of 50.0 L at 25 ºC and 1.08 atm. What volume will it have at 0.855 atm and 10 ºC? Given: V1 = 50.0 L P1 = 1.08 atm T1 = 25 ºC + 273 = 298 K P2 = 0.855 atm T2 = 10 ºC + 273 = 283 K V2= Find: V2

Combined Law: Example 2 A sample of air has a volume of 140.0 mL at 67 ºC under 2 atm. At what temperature will its volume be 50.0 mL under 2 atm? Given: V1 = 140.0 L T1 = 67 ºC + 273 = 340 K V2 = 50.0 mL P1 = P2 Find: T2

Boyle’s Law: Pressure & Volume As pressure increases, volume decreases ↑P ↓V ↓P ↑V P1V1 = P2V2

Boyle’s Law Think about why this is true: Pressure is caused by gas molecules hitting the container If the volume of the container is decreased, the same number of gas molecules are moving in a much smaller area & will hit the container more often

Boyle’s Law: Example 1 A sample of oxygen gas has a volume of 150 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm? Given: V1 = 150 mL P1 = 0.947 atm P2 = 0.987 atm Find: V2

Boyle’s Law: Example 1 Given: V1 = 150 mL P1 = 0.947 atm Find: V2 Plan: P1V1 = P2V2 V2 P2 Solve: V2 = (0.947atm)(150 mL) 0.987 atm 144 mL Check: ↑P ↓V ? yes

Boyle’s Law: Example 2 A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be? Given: P1 = 1.26 atm V1 = 7.40 L V2 = 2.93 L Find: P2

Boyle’s Law: Example 2 Given: P1 = 1.26 atm V1 = 7.40 L V2 = 2.93 L Find: P2 Plan: P1V1 = P2V2 P2 V2 Solve: P2 = (1.26atm)(7.40 L) 2.93 L 3.18 atm Check: ↓V ↑P ? yes

Charles’s Law: Volume & Temperature When temperature increases, volume increases ↑T ↑V ↓T ↓ V All temperatures must be in Kelvin!!! V1 = V2 T1 T2

Charles’s Law Think about why this is true: Increase in temp. causes molecules to move faster Faster molecules  more collisions More collisions  more pressure inside container More pressure  bigger volume

Charles’s Law: Example 1 A sample of neon gas occupies a volume of 752 mL at 25 ºC. What volume will the gas occupy at 50 ºC? Given: V1 = 752 mL T1 = 25 ºC + 273 = 298 K T2 = 50 ºC + 273 = 323 K Find: V2

Charles’s Law: Example 1 Given: V1 = 752 mL T1 = 25 ºC + 273 = 298 K T2 = 50 ºC + 273 = 323 K Find: V2 Plan: V1 = V2 T1 T2 V1 T2 Solve: V2 = (752 mL)(323 K) 298 K 815 mL Check: ↑T ↑ V ? yes

Charles’s Law: Example 2 A helium balloon has a volume of 2.75 L at 20 ºC. If you bring it outside on a cold day, the volume decreases to 2.46 L. What is the outside temperature in ºC? Given: V1 = 2.75 L T1 = 20 ºC + 273 = 293 K V2 = 2.46 L Find: T2

Charles’s Law: Example 2 Given: V1 = 2.75 L T1 = 20 ºC + 273 = 293 K V2 = 2.46 L Find: T2 Plan: V1 = V2 T1 T2 V2 T1 Solve: T2 = (2.46 L)(293 K) 2.75 L 262 K – 273 = -11 ºC Check: ↓ V ↓ T ? yes

Gay-Lussac’s Law: Pressure & Temperature When temperature increases, pressure increases ↑T ↑P ↓T ↓P Remember: all temperatures must be in Kelvin!!! P1 = P2 T1 T2

Gay-Lussac’s Law Think about why this is true: Increase in temp. causes molecules to move faster Faster molecules  more collisions More collisions  more pressure inside container

Gay-Lussac’s Law: Example 1 Gas in an aerosol can is at a pressure of 3.00 atm 25 ºC. What would the pressure be in the can at 52 ºC? Given: P1 = 3.00 atm T1 = 25 ºC + 273 = 298 K T2 = 52 ºC + 273 = 325 K Find: P2

Gay-Lussac’s Law : Example 1 Given: P1 = 3.00 atm T1 = 25 ºC + 273 = 298 K T2 = 52 ºC + 273 = 325 K Find: P2 Plan: P1 = P2 T1 T2 P1 T2 Solve: P2 = (3.00 atm)(325 K) 298 K 3.27 atm Check: ↑T ↑P ? yes

Gay-Lussac’s Law: Example 2 Before you leave on a road trip, the pressure in your car tires is 1.8 atm at 20 ºC. After driving all day, the pressure gauge read 1.9 atm. What temperature (in ºC) are your tires? Given: P1 = 1.8 atm T1 = 20 ºC + 273 = 293 K P2 = 1.9 atm Find: T2

Gay-Lussac’s Law : Example 2 Given: P1 = 1.8 atm T1 = 20 ºC + 273 = 293 K P2 = 1.9 atm Find: T2 Plan: P1 = P2 T1 T2 P2 T1 Solve: T2 = (1.9 atm)(293 K) 1.8 atm 310 K – 273 = 37 ºC Check: ↑P ↑ T ? yes

Quiz Topics Next Class KMT Properties of Gases Combined gas law calculation P, V, T relationship Converting Celsius to Kelvin