P.O.D What is the standard temperature and pressure (STP)? 2. How much space does 1 mole of hydrogen gas occupy at STP? 3. Convert 30 Celsius to Kelvin. Agenda --Notes Chapter 12-3 (Gas Laws) -WS The Combined Gas Laws

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

Boyle’s Law: Pressure & Volume As pressure increases, volume decreases ↑P ↓V ↓P ↑V P 1 V 1 = P 2 V 2

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 atm. What will the volume of the gas be at a pressure of atm? Given: V 1 = 150 mL P 1 = atm P 2 = atm Find: V2V2V2V2

Boyle’s Law: Example 1 Given: V 1 = 150 mL P 1 = atm P 2 = atm Find: V2V2V2V2 Plan: P1V1P1V1P1V1P1V1= P2V2P2V2P2V2P2V2 P1V1P1V1P1V1P1V1= V2V2V2V2 P2P2P2P2 Solve: V2V2V2V2= (0.947atm)(150 mL) atm V2V2V2V2= 144 mL ↑P ↓V ? yes 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: P 1 = 1.26 atm V 1 = 7.40 L V 2 = 2.93 L Find: P2P2P2P2

Boyle’s Law: Example 2 Given: P 1 = 1.26 atm V 1 = 7.40 L V 2 = 2.93 L Find: P2P2P2P2 Plan: P1V1P1V1P1V1P1V1= P2V2P2V2P2V2P2V2 P1V1P1V1P1V1P1V1= P2P2P2P2 V2V2V2V2 Solve: P2P2P2P2= (1.26atm)(7.40 L) 2.93 L P2P2P2P2= 3.18 atm Check: ↓V ↑P ? yes

Charles’s Law: Volume & Temperature When temperature increases, volume increases ↑T ↑V ↓T ↓ V V1V1 = V2V2 T1T1 T2T2 All temperatures must be in Kelvin!!!

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: V 1 = 752 mL T 1 = 25 ºC = 298 K T 2 = 50 ºC = 323 K Find: V2V2V2V2

Charles’s Law: Example 1 Given: V 1 = 752 mL T 1 = 25 ºC = 298 K T 2 = 50 ºC = 323 K Find: V2V2V2V2 Plan: V1V1V1V1= V2V2V2V2 T1T1T1T1 T2T2T2T2 V 1 T 2 = V2V2V2V2 T1T1T1T1 Solve: V2V2V2V2= (752 mL)(323 K) 298 K V2V2V2V2= 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: V 1 = 2.75 L T 1 = 20 ºC = 293 K V 2 = 2.46 L Find: T2T2T2T2

Charles’s Law: Example 2 Given: V 1 = 2.75 L T 1 = 20 ºC = 293 K V 2 = 2.46 L Find: T2T2T2T2 Plan: V1V1V1V1= V2V2V2V2 T1T1T1T1 T2T2T2T2 V 2 T 1 = T2T2T2T2 V1V1V1V1 Solve: T2T2T2T2= (2.46 L)(293 K) 2.75 L T2T2T2T2= 262 K – 273 = -11 ºC Check: ↓V ↓↓ ↓↓ T ? yes

Gay-Lussac’s Law: Pressure & Temperature When temperature increases, pressure increases ↑T ↑P ↓T ↓P P1P1 = P2P2 T1T1 T2T2 Remember: all temperatures must be in Kelvin!!!

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: P 1 = 3.00 atm T 1 = 25 ºC = 298 K T 2 = 52 ºC = 325 K Find: P2P2P2P2

Gay-Lussac’s Law : Example 1 Given: P 1 = 3.00 atm T 1 = 25 ºC = 298 K T 2 = 52 ºC = 325 K Find: P2P2P2P2 Plan: P1P1P1P1= P2P2P2P2 T1T1T1T1 T2T2T2T2 P 1 T 2 = P2P2P2P2 T1T1T1T1 Solve: P2P2P2P2= (3.00 atm)(325 K) 298 K P2P2P2P2= 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: P 1 = 1.8 atm T 1 = 20 ºC = 293 K P 2 = 1.9 atm Find: T2T2T2T2

Gay-Lussac’s Law : Example 2 Given: P 1 = 1.8 atm T 1 = 20 ºC = 293 K P 2 = 1.9 atm Find: T2T2T2T2 Plan: P1P1P1P1= P2P2P2P2 T1T1T1T1 T2T2T2T2 P 2 T 1 = T2T2T2T2 P1P1P1P1 Solve: T2T2T2T2= (1.9 atm)(293 K) 1.8 atm T2T2T2T2= 310 K – 273 = 37 ºC Check: ↑P ↑↑ ↑↑ T ? yes

Combined Gas Law: Pressure, Volume &Temperature Combines all three gas laws into one equation!! Memorize this equation!!! P1V1P1V1 = P2V2P2V2 T1T1 T2T2 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 atm and 10 ºC? Given: V 1 = 50.0 L P 1 = 1.08 atm T 1 = 25 ºC = 298 K P 2 = atm T 2 = 10 ºC = 283 K V 2= Find: V2V2V2V2

Combined Law: Example 2 A sample of air has a volume of mL at 67 ºC under 2 atm. At what temperature will its volume be 50.0 mL under 2 atm? Given: V 1 = L T 1 = 67 ºC = 340 K V 2 = 50.0 mL P 1 = P 2 Find: T2T2T2T2

Quiz Topics Next Class 1.KMT 2.Properties of Gases 3.Combined gas law calculation 4.P, V, T relationship 5.Converting Celsius to Kelvin