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Daltons Law of Partial Pressures. Dalton’s Law of Partial Pressures The partial pressure of a gas is the pressure of that gas in a mixture. Dalton’s Law.

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Presentation on theme: "Daltons Law of Partial Pressures. Dalton’s Law of Partial Pressures The partial pressure of a gas is the pressure of that gas in a mixture. Dalton’s Law."— Presentation transcript:

1 Daltons Law of Partial Pressures

2 Dalton’s Law of Partial Pressures The partial pressure of a gas is the pressure of that gas in a mixture. Dalton’s Law of Partial Pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. John Dalton, the English chemist who proposed the atomic theory, also studied gas mixtures. He found that in the absence of a chemical reaction, the pressure of a gas mixture is the sum of the individual pressures of each gas alone.

3 This box shows the molecules of a gas at a pressure of 0.12 atm, and at 0 degrees Celsius. This box shows another type of gas at the pressure of 0.12 atm and at 0 degrees Celsius. Finally, this box shows the combined gases at a pressure of 0.24 atm, and at 0 degrees Celsius. The rapidly moving particles of each gas have an equal chance to collide with the container walls. Therefore, each gas exerts a pressure independent of that exerted by the other gases present. The total pressure is the result of the total number of collisions per unit of wall area in a given time. Dalton’s Law of Partial Pressures

4 Gay-Lussac’s law can be expressed by the following mathematical expression: PTPT =P1P1 + P T is the total pressure of the mixture, P 1, P 2, P 3, etc. are the partial pressures of component gases 1, 2, 3, and so on. Dalton’s Law of Partial Pressures + P3P3 P2P2 + …

5 Combined Gas Law

6 The Combined Gas Law The Combined Gas Law expresses the relationship between pressure, volume, and temperature of a fixed amount of gas. The Combined Gas Law is used when TWO or THREE variables are being changed for a fixed amount of gas. For example, if we change both the pressure AND the temperature, what will happen to the volume? Because of the complication of two or more changes, we have no way to visually predict what will happen to the gas – we must go straight to the mathematical expression.

7 The Combined Gas Law can be shown using the following mathematical expression: PV T = k In the equation, k is constant and depends on the amount of gas. The Combined Gas Law

8 If k is always the same, then the PV/T ratio will remain the same before and after you make a change. Therefore, you can set the initial PV/T ratio to the changed PV/T ratio. P 1 V 1 T 1 = P 2 V 2 T 2 The Combined Gas Law

9 Find the first pressure (P 1 ). P1P1 = P 2 V 2 T 1 T 2 V 1 The Combined Gas Law Find the first Volume (V 1 ). V1V1 = P 2 V 2 T 1 T 2 P 1

10 Find the first Temperature (T 1 ). T1T1 = P 2 V 2__ T 2 P 1 V 1 The Combined Gas Law Find the second pressure (P 2 ). P2P2 = P 1 V 1 T 2 T 1 V 2

11 Find the second volume (V 2 ). V2V2 = P 1 V 1 T 2 T 1 P 2 The Combined Gas Law Find the second temperature (T 2 ). T2T2 = P 1 V 1__ T 1 P 1 V 2

12 The first volume of a Helium is 50.0 L. The first temperature of Helium is 25 degrees C + 273 = 298K. The second temperature of Helium is 10 degrees C + 273 = 283 K. The first pressure of Helium is 1.08 atm. The second pressure of Helium is 0.855 atm. Find the second volume (V 2 ) of the gas. V2V2 = First, rearrange the equation so the (V 2 ) is isolated. P 1 V 1 T 2 T 1 P 2 The Combined Gas Law

13 Now, plug in the information that you have, “P 1 ”, “V 1 ” and “T 1 ”, P 2, & T 2. V2V2 = (283 K)(1.08 atm) The “K” on the bottom cancels out the “K” on the top The “atm” on the cancels out the “atm” on the top.. Now, plug the numbers into the calculator, and you have your answer. 60.0 L (50.0 L) (0.855 atm)(298 K) The Combined Gas Law

14 The Ideal Gas Law

15 Ideal Gas Law PV = nRT Brings together gas properties. Can be derived from experiment and theory.

16 Ideal Gas Equation P V = n R T Universal Gas Constant Volume No. of moles Temperature Pressure R = 0.0821 atm L / mol K R = 8.314 kPa L / mol K Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 366

17 PV = nRT P = pressure V = volume T = temperature (Kelvin) n = number of moles R = gas constant Standard Temperature and Pressure (STP) T = 0 o C or 273 K P = 1 atm = 101.3 kPa = 760 mm Hg Solve for constant (R) PV nT = R Substitute values: (1 atm) (22.4 L) (1 mole)(273 K) R = 0.0821 atm L / mol K or R = 8.31 kPa L / mol K R = 0.0821 atm L mol K (101.3 kPa) ( 1 atm) = 8.31 kPa L mol K 1 mol = 22.4 L @ STP = R

18 Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300 o C and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine T = 300 o C P = 740 mm Hg R = 0.0821 atm. L / mol. K Step 2) Equation: V= nRT P V (500 g)(0.0821 atm. L / mol. K)(300 o C) 740 mm Hg = Step 3) Solve for variable Step 4) Substitute in numbers and solve PV = nRT

19 GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

20 GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3

21 Example Problems Calculate the molecular weight of a gas if 35.44 g of the gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at a constant temperature of 35.5 °C 1.089 g of a gas occupies 4.50 L at 20.5 °C and 0.890 atm. What is its molar mass? How many moles of gas are contained in a 50.0 L cylinder at a pressure of 100.0 atm and a temperature of 35.0 °C?

22 Solutions 1. 1.99 g/mol 2. 6.56 g/mol 3. 197.73 mol

23 Gas Stoichiometry https://www.youtube.com/watch?v=qvOVXg24Npo

24 Find vol. hydrogen gas made when 38.2 g zinc react w /excess hydrochloric acid. Pres. = 107.3 kPa; temp.= 88 o C. Gas Stoichiometry 16.3 L At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP. Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g) 38.2 g excess X L P = 107.3 kPa T = 88 o C x L H 2 = 38.2 g Zn 65.4 g Zn = 13.1 L H 2 1 mol Zn 1 mol H 2 1 mol Zn 22.4 L O 2 1 mol H 2 x mol H 2 = 38.2 g Zn 65.4 g Zn = 0.584 mol H 2 1 mol Zn 1 mol H 2 1 mol Zn P V = n R T V = n R T P = 0.584 mol (8.314 L. kPa/mol. K)(361 K) 107.3 kPa = 88 o C + 273 = 361 K

25 At standard temp and pressure. Calcium carbonate decomposes at high temperatures to form carbon dioxide and calcium oxide: CaCO3(s) ‡ CO2(g) + CaO(s) How many grams of calcium carbonate will I need to form 3.45 liters of carbon dioxide? 14.1 grams

26 At STP, Ethylene burns in oxygen to form carbon dioxide and water vapor: C2H4(g) + 3 O2(g) ‡ 2 CO2(g) + 2 H2O(g) How many liters of water can be formed if 1.25 liters of ethylene are consumed in this reaction? 2.50 liters

27 At 1 atm and 298K: WHen chlorine is added to acetylene, 1,1,2,2-tetrachloroethane is formed: 2 Cl2(g) + C2H2(g) ‡ C2H2Cl4(l) How many liters of chlorine will be needed to make 75.0 grams of C2H2Cl4? 21.8 L

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