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Chapter 12 Gases

Chapter 12 Gases. The Combined Gas Law Volume and Moles (Avogadro’s Law) Partial Pressures. Combined Gas Law. Combines Boyles Law, Charles’ Law and Gay Lussac’s Law. Combined Gas Law Problem.

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Chapter 12 Gases

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  1. Chapter 12Gases The Combined Gas Law Volume and Moles (Avogadro’s Law) Partial Pressures LecturePLUS Timberlake

  2. Combined Gas Law Combines Boyles Law, Charles’ Law and Gay Lussac’s Law LecturePLUS Timberlake

  3. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? LecturePLUS Timberlake

  4. Data Table Set up Data Table P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L V2= 90.0 mL T1 = 302 K T2 = ?? Fill in with increase or decrease P change will _____________ T. V change will _____________ T. LecturePLUS Timberlake

  5. Solution Enter data for T and P, then T and V T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C LecturePLUS Timberlake

  6. Calculation Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL (inc. T) (dec. T) T2 = 604 K - 273 = 331 °C LecturePLUS Timberlake

  7. Learning Check C1 A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg? LecturePLUS Timberlake

  8. Solution C1 T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K - 273 = - 95°C LecturePLUS Timberlake

  9. Volume and Moles How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume? LecturePLUS Timberlake

  10. Learning Check C2 True (1) or False(2) 1.___The P exerted by a gas at constant V is not affected by the T of the gas. 2.___ At constant P, the V of a gas is directly proportional to the absolute T 3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V. LecturePLUS Timberlake

  11. Solution C2 True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V. LecturePLUS Timberlake

  12. Avogadro’s Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V1 = V2 n1 n2 initial final LecturePLUS Timberlake

  13. STP The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg) LecturePLUS Timberlake

  14. Learning Check C3 A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C? P1 = V1 = T1 = K P2 = V2 = ?? T2 = K V2 = 15 L x atm x K = 6.8 L atmK LecturePLUS Timberlake

  15. Solution C3 P1 = 1.0 atm V1 = 15 L T1 = 273 K P2 = 2.0 atm V2 = ?? T2 = 248 K V2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm273 K LecturePLUS Timberlake

  16. Molar Volume At STP 4.0 g He 16.0 g CH4 44.0 g CO2 1 mole 1 mole 1mole (STP) (STP) (STP) V = 22.4 L V = 22.4 L V = 22.4 L LecturePLUS Timberlake

  17. Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L LecturePLUS Timberlake

  18. Learning Check C4 A.What is the volume at STP of 4.00 g of CH4? 1) 5.60 L 2) 11.2 L 3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g LecturePLUS Timberlake

  19. Solution C4 A.What is the volume at STP of 4.00 g of CH4? 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He LecturePLUS Timberlake

  20. Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. P = 1.00 atm P = 1.00 atm 1 mole H2 0.5 mole O2 + 0.3 mole He + 0.2 mole Ar LecturePLUS Timberlake

  21. Health Note When a scuba diver goes under water, the high pressure of the water causes more N2 (g)to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents. LecturePLUS Timberlake

  22. Learning Check C6 A scuba tank contains O2 with a pressure of 0.540 atm and He at 855 mm Hg? What is the total pressure in mm Hg in the tank? LecturePLUS Timberlake

  23. Solution C6 0.450 atm x 760 mm Hg = 342 mm Hg 1 atm 342 mm Hg + 855 mm Hg = 1197 mm Hg LecturePLUS Timberlake

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