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Daltons Law of Partial Pressures (This is our LAST law!!)

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Background John Dalton studied the behavior of gas mixtures – Each component of a gas mixture behaves independently of the others. Example: A given amount of oxygen exerts the same amount of pressure whether it is alone or in the presence of other gasses. – Deep-sea divers use a helium-oxygen mixture instead of a nitrogen-oxygen (air) mixture so they dont get the bends as they surface. – Nitrogen and helium behave differently under pressure. 2

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Daltons Law For a mixture of gasses in a container, the total pressure exerted is the sum of the partial pressures of the gasses present. The Partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. P total = P 1 + P 2 + P 3 3

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Merging Daltons + Ideal Gas laws 5 P 1 = n 1 RT ; P 2 = n 2 RT ; P 1 = n 3 RT V V V P total = P 1 + P 2 + P 3 = n 1 RT + n 2 RT + n 3 RT V V V = n 1 (RT ÷ V) + n 2 (RT ÷ V) + n 3 (RT ÷ V) = (n 1 + n 2 + n 3 )(RT ÷ V) P total = n total (RT ÷ V) Where n total = the sum of the numbers of moles of the gasses in the mixture

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SO… For mixtures of ideal gasses, it is the total number of moles that is important, not the identity of the gases present. 6

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Sample Problem #1 Mixtures of helium and oxygen are used in the air tanks of underwater divers for deep dives. For a particular dive, 12 L of O 2 at 25 o C and 1.0 atm and 46 L of He at 25 o C and 1.0 atm were both pumped into a 5.0-L tank. A)Calculate the partial pressure of each gas B)Calculate the total pressure in the tank at 25 o C (298 K). 7 *Because the partial pressure of each gas depends on the total moles present, we must first calculate n for each gas using the ideal gas law. PV = nRT n = PV ÷ RT n O 2 = (1.0 x 12) ÷ ( x 298) = 0.49 mol n He = (1.0 x 46) ÷ ( x 298) = 1.9 mol P = nRT ÷ V P O 2 = (0.49 x x 298) ÷ 5.0 = 2.4 atm P He = (1.9 x x 298) ÷ 5.0 = 9.3 atm *The tank has a volume of 5.0-L and temperature of 298 K, so we can figure out the partial pressure of each gas Determine the total pressure: P total = P O 2 + P He = 2.4 atm atm = 11.7 atm

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Sample Problem #2 A sample of solid potassium chlorate, KClO 3, was heated in a test tube and decomposed according to the reaction: 2KClO 3 (s) 2KCl (s) + 3O 2 (g) The oxygen produced was collected by displacement of water at 22 o C. The resulting mixture of O 2 and H 2 O vapor had a pressure of 754 torr and a volume of L. Calculate the partial pressure of O 2 in the gass collected and the number of moles of O 2 present. The vapor pressure of water at 22 o C is 21 torr. 8

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Sample Problem #2 Solution We know: P total = 754 torr P H 2 O = 21 torr Next, we solve the ideal gas law for n O 2 n O 2 = (P O 2 V) ÷ (RT) Then solve: P = atm V = L T = 22 o C = = 295 K R = L atm/K mol 9 So we can determine: P total = P H 2 O + P O 2 OR P total - P H 2 O = P O 2 SO… 754 torr – 21 torr = 733 torr P O 2 = 733 torr Convert pressure from torr to atm: 733 torr x (1 atm/760 torr) = atm n O 2 = (0.964 x 0.650) ( x 295) n O 2 = mol O 2

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Daltons Law Practice Problems 1.If a gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in 1.04-L container at 25 o C, what will be the partial pressure of each gas and the total pressure in the container? 2.How many moles of helium gas would be required to fill a 2.41-L container to a pressure of 759 mm Hg at 25 o C? 3.A sample of oxygen gas (O 2 ) is saturated with water vapor (H 2 O) at 27 o C. The total pressure of the mixture is 772 torr, and the vapor pressure of water is 26.7 torr at 27 o C. What is the partial pressure of the oxygen gas? 4.Suppose a gaseous mixture of 1.15 g helium and 2.91 g argon is placed in a 5.25-L container at 273 o C. What pressure would exist in the container? 5.A tank contains a mixture of 3.0 mol of N2, 2.0 mol of O2, and 1.0 mol of CO2 at 25 o C and a total pressure of 10.0 atm. Calculate the partial pressure (in torr) of each gas in the mixture. 10

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