Ch12_Gases - BEHAVIOR OF GASES Chapter 12 1 2 3 Importance...

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Unformatted text preview: BEHAVIOR OF GASES Chapter 12 1 2 3 Importance of Gases • Airbags fill with N 2 gas in an accident. gas • Gas is generated by the decomposition of sodium azide, NaN3. azide, • 2 NaN3 ---> 2 Na + 3 N2 ---> Hot Air Balloons — How Do They Work? THREE STATES OF MATTER 4 General Properties of Gases • There is a lot of “free” space in a gas. • Gases can be expanded infinitely. • Gases occupy containers uniformly and completely. • Gases diffuse and mix rapidly. 5 6 Properties of Gases Gas properties can be modeled using math. Model depends on— • V = volume of the gas (L) • T = temperature (K) • n = amount (moles) • P = pressure (atmospheres) Page 1 7 Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643) Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down related to • Hg density • column height 8 Pressure Column height measures P of atmosphere • 1 standard atm = 760 mm Hg = 29.9 inches = about 34 feet of water SI unit is PASCAL, Pa, where 1 atm = 101.325 kPa 9 IDEAL GAS LAW 10 11 PV=nRT Brings together gas properties. Can be derived from experiment and theory. Boyle’s Law If n and T are constant, then PV = (nRT) = k (nRT) This means, for example, that P goes up as V goes down. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire. 12 Robert Boyle (1627-1691). Son of Early of Cork, Ireland. Page 2 Charles’s Law If n and P are constant, then V = (nR/P)T = kT (nR/P)T V and T are directly related. 13 14 15 Charles’s original balloon Charles’s Law Jacques Charles (17461823). Isolated boron and studied gases. Balloonist. Modern long-distance balloon Balloons immersed in liquid N 2 (at -196 ˚C) will (at shrink as the air cools (and is liquefied). 16 Charles’s Law Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. 17 18 Avogadro’s Hypothesis twice as many molecules The gases in this experiment are all measured at the same T and P. Page 3 Using PV = nRT How much N2 is req’d to fill a small room is with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC ? 19 Using PV = nRT How much N2 is req’d to fill a small room with a volume of 960 is cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? 20 21 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) liq) Decompose 1.1 g of H 2O2 in a flask with a in volume of 2.50 L. What is the pressure of O 2 oC? Of H O? at 25 2 Bombardier beetle uses decomposition of hydrogen peroxide to defend itself. R = 0.082057 L•atm/K•mol 0.082057 L•atm/K•mol R = 0.082057 L•atm/K•mol 0.082057 L• atm/K•mol Solution 1. Get all data into proper units V = 27,000 L 27,000 T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) atm/760 = 0.98 atm Solution 2. Now calc. n = PV / RT calc. n= (0.98 atm)(2.7 x 104 L) (0.0821 L• atm/K•mol)(298 K) n = 1.1 x 103 mol (or about 30 kg of gas) mol 22 23 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) liq) Decompose 1.1 g of H 2O2 in a flask with a in volume of 2.50 L. What is the pressure of O 2 at 25 oC? Of H2O? Solution Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) liq) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. in What is the pressure of O2 at 25 oC? Of H2O? at Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) liq) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. in What is the pressure of O2 at 25 oC? Of H2O? at 24 Solution Solution 1 mol 1.1 g H2 O2 • 34.0 g 0.032 mol P of O2 = n RT/V = (0.016 mol)(0.0821 L•atm/K•mol)(298 K) 2.50 L Strategy: Calculate moles of H 2O2 and then and moles of O 2 and H2O. and Finally, calc. P from n, R, T, and V. calc. 0.032 mol H2O 2 • 1 mol O2 = 0.016 mol O2 2 mol H2 O 2 P of O2 = 0.16 atm 0.16 Page 4 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) liq) What is P of H 2O? Could calculate as above. But recall Avogadro’s hypothesis. V P n at same T and P at n at same T and V at 25 26 27 Dalton’s Law of Partial Pressures 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) liq) 0.32 atm 0.16 atm What is the total pressure in the flask? Dalton’s Law Ptotal in gas mixture = P A + PB + ... in ... Therefore, Ptotal = P(H2O) + P(O 2) = 0.48 atm P(H There are 2 times as many moles of H 2O as moles of O 2. P is proportional to n. Therefore, P of H 2O is twice that of O 2. P of H2O = 0.32 atm Dalton’s Law: total P is sum of PARTIAL pressures. John Dalton 1766-1844 GAS DENSITY Screen 12.5 28 Low density GAS DENSITY Screen 12.5 29 30 USING GAS DENSITY The density of air at 15 oC and 1.00 atm is 1.23 g/L. What is the molar mass of air? 1. Calc. moles of air. Calc. V = 1.00 L P = 1.00 atm T = 288 K PV = nRT P n = V RT m P = M•V RT where M = molar mass n = PV/RT = 0.0423 mol 2. Calc. molar mass Calc. High density d= m PM = V RT d and M proportional and mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol mass/mol g/mol Page 5 ...
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