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Warm Up2-7-14 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.

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Presentation on theme: "Warm Up2-7-14 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."— Presentation transcript:

1 Warm Up 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. Agenda -Pick up binders (40pts) -Notes Chp 12-3 (gas laws) -WS The gas law Homework Quiz next class

2 12-3 The Gas Laws Mathematical relationships between volume, temperature, pressure and amount of gas

3 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!!!

4 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

5 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

6 Boyles Law: Pressure & Volume As pressure increases, volume decreases P V P 1 V 1 = P 2 V 2

7 Boyles 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

8 Boyles 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

9 Boyles 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

10 Boyles 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

11 Boyles 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

12 Charless Law: Volume & Temperature When temperature increases, volume increases T V V1V1 = V2V2 T1T1 T2T2 All temperatures must be in Kelvin!!!

13 Charless 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

14 Charless 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

15 Charless 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

16 Charless 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

17 Charless 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

18 Gay-Lussacs Law: Pressure & Temperature When temperature increases, pressure increases T P P1P1 = P2P2 T1T1 T2T2 Remember: all temperatures must be in Kelvin!!!

19 Gay-Lussacs 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

20 Gay-Lussacs 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

21 Gay-Lussacs 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

22 Gay-Lussacs 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

23 Gay-Lussacs 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

24 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


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