P1V1 = P2V2 (99.0 kPa) (300.0 mL) = (188 kPa) V2

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P1V1 = P2V2 (99.0 kPa) (300.0 mL) = (188 kPa) V2
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Presentation transcript:

P1V1 = P2V2 (99.0 kPa) (300.0 mL) = (188 kPa) V2 1. The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what will be the new volume? P1V1 = P2V2 (99.0 kPa) (300.0 mL) = (188 kPa) V2 (99.0 kPa) (300.0 mL) = V2 (188 kPa) V2 = 158 mL

2. The pressure of a sample of Helium in a 1.00 L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00 L container? P1V1 = P2V2 (0.988 atm) (1.00 L) = P2 (2.00 L) (0.988 atm) (1.00 L) = P2 (2.00 L) P2 = 0.494 atm

4. What volume will the gas in the balloon at the right occupy at 250 K? V1 = V2 T1 T2 4.3 L = V2 V2 = 3.1 L 350 K 250 K

5. A gas at 89 oC occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? V1 = V2 T1 T2 0.67 L = 1.12 L 362 K T2 T2 = 605 K = 330 oC

6. The Celsius temperature of a 3.00 L sample of gas is lowered from 80.0 oC to 30.0 oC. What will be the resulting volume of this gas? V1 = V2 T1 T2 3.00 L = V2 V2 = 2.58 L 353 K 303 K

8. The pressure in an automobile tire is 1.88 atm at 25.0 oC. What is the pressure if the temperature increases to 37.0 oC? P1 = P2 T1 T2 1.88 atm = P2 P2 = 1.96 atm 298 K 310. K

P1 = P2 T1 T2 1.12 atm = 2.56 atm T1 = 136 K T1 310. K = - 137 oC 9. Helium gas in a 2.00 L cylinder is under 1.12 atm of pressure. At 36.5 oC that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder? P1 = P2 T1 T2 1.12 atm = 2.56 atm T1 = 136 K T1 310. K = - 137 oC

P1V1 = P2V2 T1 T2 1.02 atm * V1 = 1.23 atm * 0.224 ml 295 K 373 K 11. A sample of air in a syringe exerts a pressure of 1.02 atm at 22.0 oC. The syringe is placed in a boiling-water bath at 100.0 oC. The pressure is increased to 1.23 atm by pushing the plunger in, which reduces the volume to 0.224 ml. What was the initial volume? P1V1 = P2V2 T1 T2 1.02 atm * V1 = 1.23 atm * 0.224 ml 295 K 373 K V1 = 0.214 ml

P1V1 = P2V2 T1 T2 1.30 atm * 146.0 mL = 2.60 atm * V2 278 K 275 K 12. A balloon contains 146.0 mL of gas confined at a pressure of 1.30 atm and a temperature of 5.0 oC. If the pressure doubles and the temperature decreases to 2.0 oC, what will be the volume of gas in the balloon? P1V1 = P2V2 T1 T2 1.30 atm * 146.0 mL = 2.60 atm * V2 278 K 275 K V2 = 72.2 mL