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V1 = 4.6 L V2 = 330 L P1 = 1.18 atm P2 = ? atm P1V1 = P2V2

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Presentation on theme: "V1 = 4.6 L V2 = 330 L P1 = 1.18 atm P2 = ? atm P1V1 = P2V2"— Presentation transcript:

1 V1 = 4.6 L V2 = 330 L P1 = 1.18 atm P2 = ? atm P1V1 = P2V2
The pressure of air on a 4.6 L container is 1.18 atm. What is the new pressure if the sample is transferred to a 330 L container? V1 = 4.6 L V2 = 330 L P1 = 1.18 atm P2 = ? atm P1V1 = P2V2 1.18 x 4.6 = P2 x 330 1.18 atm x 4.6 L = P2= atm 330 L

2 V1 = 5.3 L V2 = ? L T1 = 65oC = 338 K T2 = 42oC = 315 K V1 = V2 T1 T2
A balloon has a volume of 5.3 L and is filled at a temperature of 65°C. What volume will the balloon occupy at a temperature of 42°C? P? V1 = 5.3 L V2 = ? L T1 = 65oC = 338 K T2 = 42oC = 315 K V1 = V2 T1 T2 = V2 338 K K 5.3 L x 315 K = V2= 4.9 L 338 K

3 T1= 25oC = 298 K T2= -10.0oC = 263 K P1 = 5.00 atm P2 = ? atm P1 = P2
The pressure of the oxygen gas inside a canister is 5.00 atm at 25°C. The canister is located at a camp high on Mount Everest. If the temperature falls to -10.0°C, what is the new pressure inside the canister? T1= 25oC = 298 K T2= -10.0oC = 263 K P1 = 5.00 atm P2 = ? atm P1 = P2 T1 T2 = P2 298 K K 5.00 atm x 263 K = P2= 4.41 atm 298 K

4 A gas at 110 kPa and 30.0°C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0°C and the pressure increases to 440 kPa, what is the new volume? V1 = 2.00 L V2 = ? L T1= 30.0oC = 303 K T2 = 80.0oC = 353 K P1 = 110 kPa P2 = 440 kPa P1V1 = P2V2 T T2 110 x 2.00 = 440 x V2 303K K 110 kPa x 2.00 L x 353 K = V2= 0.58 L 303 K x 440 kPa

5 V1 = 50.0 mL V2 = ? mL T1= 23.0oC = 296.0 K T2 = 273 K P2 = 1 atm
A gas has a volume of 50.0 ml at 0.90 atm and 23.0°C. How much is the gas volume at STP? V1 = 50.0 mL V2 = ? mL T1= 23.0oC = K T2 = 273 K P1 = 0.90 atm P2 = 1 atm P1V1 = P2V2 T T2 0.90 x 50.0 = 1 x V2 296.0 K 273 K 0.90 atm x 50.0 mL x 273 K = V2= 41.5 mL 296.0 K x 1 atm


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