Presentation on theme: "Dalton’s Law of Partial Pressures"— Presentation transcript:
1Dalton’s Law of Partial Pressures (This is our LAST law!!)
2Background 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 don’t get “the bends” as they surface.Nitrogen and helium behave differently under pressure.
3Dalton’s Law Ptotal = P1 + P2 + P3 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.Ptotal = P1 + P2 + P3
5Merging Dalton’s + Ideal Gas laws P1 = n1RT ; P2 = n2RT ; P1 = n3RTV V VPtotal = P1 + P2 + P3 = n1RT + n2RT + n3RTV V V= n1 (RT ÷ V) + n2(RT ÷ V) + n3(RT ÷ V)= (n1 + n2 + n3)(RT ÷ V)Ptotal = ntotal (RT ÷ V)Where ntotal = the sum of the numbers of moles of the gasses in the mixture
6SO…For mixtures of ideal gasses, it is the total number of moles that is important, not the identity of the gases present.
7Sample Problem #1Mixtures of helium and oxygen are used in the air tanks of underwater divers for deep dives. For a particular dive, 12 L of O2 at 25oC and 1.0 atm and 46 L of He at 25oC and 1.0 atm were both pumped into a 5.0-L tank.Calculate the partial pressure of each gasCalculate the total pressure in the tank at 25oC (298 K).PV = nRTn = PV ÷ RTnO2 = (1.0 x 12) ÷ ( x 298) = 0.49 molnHe = (1.0 x 46) ÷ ( x 298) = 1.9 mol*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.*The tank has a volume of 5.0-L and temperature of 298 K, so we can figure out the partial pressure of each gasP = nRT ÷ VPO2 = (0.49 x x 298) ÷ 5.0 = 2.4 atmPHe = (1.9 x x 298) ÷ 5.0 = 9.3 atma)Determine the total pressure:Ptotal = PO2 + PHe = 2.4 atm atm = 11.7 atmb)
8Sample Problem #2A sample of solid potassium chlorate, KClO3, was heated in a test tube and decomposed according to the reaction:2KClO3 (s) 2KCl (s) + 3O2 (g)The oxygen produced was collected by displacement of water at 22oC. The resulting mixture of O2 and H2O vapor had a pressure of 754 torr and a volume of L. Calculate the partial pressure of O2 in the gass collected and the number of moles of O2 present. The vapor pressure of water at 22oC is 21 torr.
9Sample Problem #2 Solution We know:Ptotal = 754 torrPH2O = 21 torrNext, we solve the ideal gas law for nO2nO2 = (PO2 V) ÷ (RT)Then solve:P = atmV = LT = 22oC = = 295 KR = L atm/K molSo we can determine:Ptotal = PH2O + PO2 OR Ptotal - PH2O = PO2SO… 754 torr – 21 torr = 733 torrPO2 = 733 torrConvert pressure from torr to atm:733 torr x (1 atm/760 torr) = atmnO2 = (0.964 x 0.650)( x 295)nO2 = mol O2
10Dalton’s Law Practice Problems If a gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in 1.04-L container at 25oC, what will be the partial pressure of each gas and the total pressure in the container?How many moles of helium gas would be required to fill a 2.41-L container to a pressure of 759 mm Hg at 25oC?A sample of oxygen gas (O2) is saturated with water vapor (H2O) at 27oC. The total pressure of the mixture is 772 torr, and the vapor pressure of water is 26.7 torr at 27oC. What is the partial pressure of the oxygen gas?Suppose a gaseous mixture of 1.15 g helium and 2.91 g argon is placed in a 5.25-L container at 273oC. What pressure would exist in the container?A tank contains a mixture of 3.0 mol of N2, 2.0 mol of O2, and 1.0 mol of CO2 at 25oC and a total pressure of 10.0 atm. Calculate the partial pressure (in torr) of each gas in the mixture.