Sample Problem : Boyle’s Law

1. Sample Problem: Boyle’s Law The gas in a balloon has a volume of 4 L at 100 kPa. The balloon is released into the atmosphere, and the gas in it expands to a volume of 8 L. What is the pressure on the balloon at the new volume?Show Work: V1 = _____ V2 = _____ P1 = ______ P2 = ______ Eqn= ________________ 8 L 4 L ? 100 kPa 100 kPa x 4 L = P2 x 8 L P1V1 = P2V2 • 100 kPa x 4 L = P2 = 50 kPa • 8 L

2. Sample Problem: Charles’ Law • A sample of gas occupies 24 m3 at 100 K. What volume would the gas occupy at • 400 K? Show Work: • V1 = _____ V2 = _____ • T1 = ______ T2 = ______ • Eqn= ___ ___ 24 m3 ? 400 K 100 K V1=V2 T1 T2 24 m3= V2 100 K 400 K • 400 K x 24 m3= V2 = 96 m3 • 100 K

3. Sample Problem: Gay-Lussac Law Helium gas is under 1.12 atm pressure. At 36.5 oC, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas? Show Work: T1 = _____ T2 = _____ P1 = ______ P2 = ______ Eqn= ___ ___ ? 36.5 oC 2.56 atm 1.12 atm P1=P2 T1 T2 1.12 atm= 2.56 atm T1 309.5 K • 1.12 atm x 309.5 K = T1 = 135.4K • 2.56 atm

4. 1. Chlorine gas occupies 250 mL when its pressure is 720 mm Hg. What volume will it occupy when its pressure is 750 mm Hg? V2 = ? L V1 = 250 mL P2 = 750 mm Hg P1 = 720 mm Hg P1V1 = P2V2 720 mm Hg x 250 mL = 750 mm Hg x V2 720 mm Hg x 250 mL = V2= 240 mL 750 mm Hg

5. 2. Argon gas has a volume of 240 mL at 700 mm Hg. What pressure is needed to reduce its volume to 60 mL? V2 = 60 mL V1 = 240 mL P2 = ? mm Hg P1 = 700 mm Hg P1V1 = P2V2 700 mm Hg x 240 mL = P2 x 60 mL 700 mm Hg x 240 mL = P2 = 2800 mm Hg 60 mm Hg

6. 3. A sample of water vapor is collected at 672 mm Hg. If the pressure is increased to 842 mm Hg, the volume changes to 78.9 ml. What was the original volume of the water vapor? V2 = 78.9 mL V1 = ? mL P1 = 672mm Hg P2 = 842 mm Hg P1V1 = P2V2 672 mm Hg x V1= 842 mm Hg x 78.9 mL 842 mm Hg x 78.9 mL = V1 = 98.9 mL 672 mm Hg

7. Converting from ˚C to K Converting from K to ˚C (Remember K = ˚C + 273)(Remember ˚C = K – 273) • 37˚C = __________ K 465 K = __________ ˚C • 2˚C = ___________ K 273 K = ___________ ˚C • 150˚C = _________ K 487 K = ___________ ˚C • 84˚C = __________K 182 K = ___________ ˚C • 534˚C = _________ K 65 K = ____________ ˚C 192 310 275 0 423 214 357 -91 -208 807

8. 1) If 90.0 mL of Hydrogen is collected at 37 oC. At what temperature will it occupy 180.0 ml? V1 = 90.0 mL V2 = 180.0 mL T1 = 37oC = 310 K T2 = ? V1=V2 T1 T2 90.0 mL = 180.0 mL 310 K ? K 310 K x 180.0 mL = T2 = 620 K 90.0 mL (347oC )

9. 2) If 25.6 mL of Nitrogen is collected at 62 oC, what volume will it occupy if it is cooled to 31 oC? V1 = 25.6 mL V2 = ? mL T1 = 62oC = 335 K T2 = 31oC = 304 K V1=V2 T1 T2 25.6 mL = ? mL 335 K 304 K 304 K x 25.6 mL = V2 = 23.2 mL 335 K

10. 3) If Carbon monoxide has a volume of 45.6 mL at -40 oC, at what temperature will the volume be 138.9 ml? V1 = 45.6 mL V2 = 138.9 mL T1 = -40oC = 233 K T2 = ? V1=V2 T1 T2 45.6 mL = 138.9 mL 233 K ? K 233 K x 138.9 mL = T2 = 709 K 45.6 mL (436oC)

11. 1. Oxygen gas is at 42°C and a pressure of 6.4 atm. What will the pressure be if the temperature lowers to 25°C? P1 = 6.4 atm P2 = ? atm T1 = 42oC = 315 K T2 = 25oC = 298 K P1=P2 T1 T2 6.4 atm = ? atm 315 K 298 K 298 K x 6.4 atm = P2 = 6.05 atm 315 K

12. 2. Hydrogen gas is under 2.5 atm pressure. At 55°C, the pressure has changed to 6.2 atm. What is the initial temperature of the gas? P1 = 2.5 atm P2 = 6.2 atm T1 = ? K T2 = 55oC = 328 K P1=P2 T1 T2 2.5 atm = 6.2 atm ? K 328 K 328 K x 2.5 atm = T1 = 132 K 6.2 atm

13. 3. The temperature in a greenhouse is 95°C. The door is left open and the temperature decreases to 75°C and the pressure is at 1.2 atm. What was the initial pressure inside the greenhouse? P1 = ? atm P2 = 1.2 atm T1 = 95oC = 368 K T2 = 75oC = 348 K P1=P2 T1 T2 ? atm = 1.2 atm 368 K 348 K 368 K x 1.2 atm = P2 = 1.27 atm 348 K (1.3 atm)