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LecturePLUS Timberlake1 Chapter 7 Gases The Combined Gas Law Volume and Moles (Avogadros Law) Partial Pressures

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LecturePLUS Timberlake2 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Rearrange the combined gas law to solve for V 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 = P 1 V 1 T 2 P 2 T 1

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LecturePLUS Timberlake3 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Isolate V 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 = P 1 V 1 T 2 P 2 T 1

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LecturePLUS Timberlake4 Learning Check C1 Solve the combined gas laws for T 2.

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LecturePLUS Timberlake5 Solution C1 Solve the combined gas law for T 2. (Hint: cross-multiply first.) P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 T 2 = P 2 V 2 T 1 P 1 V 1

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LecturePLUS Timberlake6 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?

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LecturePLUS Timberlake7 Data Table Set up Data Table P 1 = 0.800 atm V 1 = 0.180 L T 1 = 302 K P 2 = 3.20 atm V 2 = 90.0 mL T 2 = ?? ??

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LecturePLUS Timberlake8 Solution Solve for T 2 Enter data T 2 = 302 K x atm x mL = K atm mL T 2 = K - 273 = °C

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LecturePLUS Timberlake9 Calculation Solve for T 2 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T 2 = 604 K - 273 = 331 °C

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LecturePLUS Timberlake10 Learning Check C2 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?

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LecturePLUS Timberlake11 Solution G9 T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = 0.315 L = 315 mL P 1 = 0.850 atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 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

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LecturePLUS Timberlake12 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?

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LecturePLUS Timberlake13 Learning Check C3 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.

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LecturePLUS Timberlake14 Solution C3 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.

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LecturePLUS Timberlake15 Avogadros Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V 1 = V 2 n 1 n 2 initial final

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LecturePLUS Timberlake16 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)

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LecturePLUS Timberlake17 Learning Check C4 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? P 1 = V 1 = T 1 = K P 2 = V 2 = ?? T 2 = K V 2 = 15 L x atm x K = 6.8 L atm K

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LecturePLUS Timberlake18 Solution C4 P 1 = 1.0 atm V 1 = 15 L T 1 = 273 K P 2 = 2.0 atm V 2 = ?? T 2 = 248 K V 2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K

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LecturePLUS Timberlake19 Molar Volume At STP 4.0 g He 16.0 g CH 4 44.0 g CO 2 1 mole 1 mole1mole (STP) (STP)(STP) V = 22.4 L V = 22.4 L V = 22.4 L

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LecturePLUS Timberlake20 Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L

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LecturePLUS Timberlake21 Learning Check C5 A.What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g2) 0.357 g3) 1.43 g

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LecturePLUS Timberlake22 Solution C5 A.What is the volume at STP of 4.00 g of CH 4 ? 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 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

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LecturePLUS Timberlake23 Daltons Law of Partial Pressures Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. P T = P 1 + P 2 + P 3 +.....

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LecturePLUS Timberlake24 Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N 2 593.4 mmHg 20.95% O 2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO 2 0.2 mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg 2 2 2 Total Pressure760 mm Hg

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LecturePLUS Timberlake25 Learning Check C6 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.143) 0.109

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LecturePLUS Timberlake26 Solution C6 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557

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LecturePLUS Timberlake27 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 0.5 mole O 2 + 0.3 mole He + 0.2 mole Ar 1 mole H 2

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LecturePLUS Timberlake28 Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 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 O 2 in scuba tanks used for deep descents.

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LecturePLUS Timberlake29 Learning Check C7 A 5.00 L scuba tank contains 1.05 mole of O 2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

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LecturePLUS Timberlake30 Solution C7 P = nRT P T = P O + P He V 2 P T = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L(K mol) =7.19 atm

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