Ch12_Gases

Ch12_Gases - BEHAVIOR OF GASES 1 Chapter 12 2 3 Hot Air...

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Unformatted text preview: BEHAVIOR OF GASES 1 Chapter 12 2 3 Hot Air Balloons — How Do They Work? Importance of Gases • Airbags fill with N 2 gas in an accident. • Gas is generated by the decomposition of sodium azide, NaN3. • 2 NaN3 ---> 2 Na + 3 N2 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. Page 1 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) 7 Pressure 8 Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643) 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 P of Hg pushing down related to • Hg density • column height IDEAL GAS LAW PV=nRT Brings together gas properties. Can be derived from experiment and theory. 10 11 Boyle’s Law If n and T are constant, then PV = (nRT) = k This means, for example, that P goes up as V goes down. Page 2 Pressure 9 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. Robert Boyle (1627-1691). Son of Early of Cork, Ireland. 12 13 Charles’s Law If n and P are constant, then V = (nR /P)T = kT V and T are directly related. 14 Jacques Charles (17461823). Isolated boron and studied gases. Balloonist. Modern long-distance balloon 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. twice as many molecules Page 3 15 Charles’s Law Charles’s original balloon Balloons immersed in liquid N 2 (at -196 ˚C) will shrink as the air cools (and is liquefied). 17 18 Avogadro’s Hypothesis Hypothesis The gases in this experiment are all measured at the same T and P. Using PV = nRT 19 Using PV = nRT R = 0.082057 L•atm/K•mol 0.082057 Solution 1. Get all data into proper units V = 27,000 L 27,000 T = 25 oC + 273 = 298 K 25 P = 745 mm Hg (1 atm/760 mm Hg) 745 = 0.98 atm n= Strategy: Calculate moles of H 2O2 and then moles of O 2 and H2O. Finally, calc. P from n, R, T, and V. Bombardier beetle uses decomposition of hydrogen peroxide to defend itself. (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) 22 Gases and Stoichiometry Gases and Stoichiometry Solution 2. Now calc. n = PV / RT R = 0.082057 L• atm/K•mol 0.082057 21 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) Decompose 1.1 g of H 2O2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 oC? Of H2O? How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC ? 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) Decompose 1.1 g of H 2O2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 oC? Of H2O? Solution 20 23 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution 1.1 g H2 O2 • 0.032 mol H2O 2 • 1 mol 34.0 g 0.032 mol 1 mol O2 = 0.016 mol O2 2 mol H2 O 2 Page 4 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the pressure of O2 at 25 oC? Of H2O? Solution P of O2 = n RT/V = (0.016 mol)(0.0821 L•atm/K•mol)(298 K) 2.50 L P of O2 = 0.16 atm 24 Gases and Stoichiometry 25 What is the total pressure in the flask? n at same T and P at n at same T and V at Ptotal in gas mixture = P A + PB + ... Therefore, Ptotal = P(H2O) + P(O 2) = 0.48 atm 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 GAS DENSITY Screen 12.5 Dalton’s Law: total P is sum of PARTIAL pressures. 28 Low Low density density GAS DENSITY Screen 12.5 PV = nRT nRT P n = V RT m P = M•V RT where M = molar mass High High density density Dalton’s Law Dalton’s Law 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) 0.32 atm 0.16 atm What is P of H 2O? Could calculate as above. But recall Avogadro’s hypothesis. P 27 Dalton’s Law of Partial Pressures 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) V 26 d= m PM = V RT John Dalton 1766-1844 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. V = 1.00 L 1.00 P = 1.00 atm T = 288 K n = PV/RT = 0.0423 mol 2. Calc. molar mass d and M proportional and Page 5 mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol mol ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Mcquade during the Fall '08 term at FSU.

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