Chapter 6 10/14/13. Gas Law. Volume change with temperature and pressure.

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Chapter 6 10/14/13 Gas Law 1. Properties of a Gas a. Neither definite shape nor volume i. Uniformly fills any container i Exerts pressure on surroundings Volume change with temperature and pressure. b. Mixes completely with other gases. c. Much less dense than solids, liquids. 2. Parameters Affecting Gases a. Pressure (P) b. Volume (V) c. Temperature (T) d. Number of Moles (n) e. PV= nrt 3. The Gas Phase a. The atmosphere i. Layer of gases 50 km thick i Composition is fairly consistent. Properties vary with location 1. Pressure, density 4. Pressure a. Pressure: i. Force/unit area (P = F/A Atmospheric pressure = pressure exerted due to gravity acting on air above Earth s surface. b. Units of Pressure

i. SI units = Newton/meter 2 = 1 Pascal (Pa) i 1 standard atmosphere (1 atm) = 101,325 Pa 1 atm = 760mm Hg = 760 torr. 5. Measurement of Pressure a. Barometer: measures atmospheric pressure. b. Height of Hg column based on balance of forces: i. Gravity (pulls Hg down) Atmospheric pressure (pushes Hg up into evacuated tube) 6. Measuring Pressure: Manometer a. Closed Systems: i. ΔH of Hg column is a direct measure of gas pressure (b) b. Open Systems: i. ΔH is negative if P gas < P atm (d) ΔH is positive is P gas > P atm (e) 7. Boyle s Law a. Gases are compressible i. Volume decreases are pressure increases b. Boyle s Law i. P oc 1/V (T and n fixed) Or P*V = Constant i Or, P 1V 1 = P 2V 2 8. Practice: Boyle s Law a. A balloon is filled with hydrogen to a pressure of 1.35 atm and has a volume of 2.54L. if temperature remains constant, what will the volume be when the pressure is increased to 2.50 atm? i. P 1= 1.35 atm P 2= 2.50 atm V 1 = 2.54 L 1.35atm(2.54L)/2.50atm = 1.37L

9. Charles s Law a. Charles s Law i. V oc T (P, n constant) i Or V 1/T 1 = V 2/T 2 b. Volume of a gas extrapolates to zero at absolute zero (0 Kelvin or -273 C) 10. Avogadro s Law a. Volume is directly proportional to thenumber of moles of gas, V oc n (T, P constant) b. V/n = constant c. Or, V 1/n 1 = V 2/n 2 11. Amonton s Law a. P oc T (n, V constant) b. P/T = constant c. P 1/T 1 = P 2/T 2 12. Combined Gas Law a. Boyle s Law: P * V = constant b. Charles s Law: V/T = constant c. Avogadro s Law: V/n = constant d. Combining the Gas laws: P*V/n*T = constant e. If n is constant, then PV/T = constant, and P 1V 1/T 1 = P 2V 2/T 2 13. Ideal Gas Law a. Combined Gas Law: PV/nT = Constant = R b. Which rearranges to : PV = nrt i. R = universal gas constant = 0.08206 L atm K -1 mol -1 i P = pressure ( in atm) V = Volume (in liters)

iv. n = moles 14. Practice: Ideal Gas Law v. T = temperature ( in Kelvin) a. Calculate the pressure of 1.2 mol of methane gas in a 3.3L container at 25 C. 15. Practice: Ideal Gas Law i. 1.2(.0821)(298)/3.3 = 8.9atm a. An experiment shows that a 0.495g sample of an unknown gas occupies 127 ml at 98 C and 754 torr pressure. Calculate the molar mass of the gas (hint: Mol Mass = g/n) b. PV = nrt i. n=m/m M= mrt/pv 1..495g(.0821)(371K)/(.127L)(754torr/760torr =atm) = 120g/mol 16. Reference Points for Gases. a. Standard Temperature and Pressure (STP) i. P = 1 atmosphere; T = 0 C b. Molar Volume i. For one mole of an ideal gas at STP (calculated from the ideal gas law) 1. V = 22.4 L 17. Stoichiometric Calculations Using Gases a. Stoichiometric Calculations: i. Depends on mole/mole ratios of reactants and/or products. i Moles or gas can be calculated from ideal gas law if P, V, and T are known. R = PV/RT 18. Practice: Gas Stoichiometry a. Automobile air bags inflate during a crash or sudden stop y the rapid generation of N 2(g) from sodium azide: 2NaN 2(s) 2Na(s) + 3N 2(g)

How many grams of sodium azide are needed to produce sufficient N 2(g) to fill a 45 X 45 X 25cm bag to a pressure of 1.20atm at 15 C? b. PV=nRT i. 45*45*25 = 506.25cm 3 =50.625L i N= 2.57 mol N 2 iv. 2.57mol N 2 X 2 molnan 2/3 mol N 2 X 60.01g/1mol = 102.8 = 103gNaN 2 19. Gas Density a. Can be calculated from Molar Mass and Molar volume (V/n) b. From Ideal Gas Law i. PV=nRT P/RT = n/v = m/mv M= molar mass i Density: d=m/v = PM/RT When P in atm, T in Kelvin, d = g/l 20. Buoyancy: Gas Densities a. Buoyancy depends on differences in gas densities b. Depends on: i. Molar masses 1. He(g) = 0.169 g/l* 2. N 2(g) = 1.19 g/l* Temperature c. * = at 15 C and 1 atm. 21. Practice: Gas Densities 1. Charles s Law: density decreases as temperature increases a. When HCl(aq) and NaHCO 3(aq) are mixed together, a reactiontakes place in which a gas is one of the products. The gas has a d = 1.81 g/l at 1.00 atm and 23 C. What is the molar mass and identity of the gas? 22. Dalton s Law of Partial Pressures a. For a misture of gases in a container:

i. P total = P 1+P 2+P 3+ Depends on total moles of gas, not the identity of the gas. 23. Mole Fraction & Partial Pressure a. Mole Fraction: i. Ratio of the # of moles of a given component in a mixture to the total # of moles in a mixture: 1. X 1 = n 1/n total = n 1/n 1+n 2+n 3+ /// b. Mole Fractions in Terms of Pressure: c. n 1 = P 1V/RT; when V, T are constant, P 1 oc n 1 d. By Substitution: i. X 1 = P 1/P total = n 1/n total i P 1 = X 1P total X 1 = mole fraction 24. Practice: Mole Fraction a. At 25 C, a 1.0L flask ontains 0.030 moles of nitrogen, 150.0mg of oxygen and 4X10 21 molecules of ammonia. Calculate the partial pressure and mole fraction of each gas. i. N 2 =.030 mol. i iv. 150mg x 1g/1000mg X 1 mol O 2/32.0g = 4.69X10-3 mol. 4x10 21 moleculesx 1 mol/6.022x10 23 molecules = 6.64x10-3 mol.030+4.69x10-3 +6.64x10-3 = 0.0413 mol. v..030mol/.0413 =.73 O 2=.11 NH 3=.16 =1.0 25. Collecting a Gas over Water a. 2KClO 3(g) 2KCl(s) + 3O 2(g) b. Gases Collected: i. O 2(g) and H 2O(g) c. P total = P O2 +P H2O 26. Practice: Partial Pressure of Water

a. A sample of KClO 3 is heated and decomposes to produce O 2 gas. The gas is collected by water displacement at 25 C. The total volume of the collected gas is 229mL at a pressure of 754 torr. How many moles of oxygen are formed? Table 6.4 i. P H20 @ 25 C = 23.8mm Hg= 23.8 torr i P O2 = P T- P H20 = (754-238) = 730 mmhg atm =.0090 mol. PV= nrt 27. Kinetic Molecular Theory a. Assumes that gas molecules: i. Have tiny volumes compared with their container s volume. i iv. Move randomly and constantly. Have average kinetic energy that is proportional to absolute temperature. Engage in elastic collisions with walls of container and other gas molecules. v. Act independently of other gas molecules. b. Gas Laws Involving Pressure: i. Boyle s Law: Decreasing volume increases number of collisions/area, Pressure increases (post 2,4) i Dalton s Law: Total pressure depends only on total number moles of gas, not on their identities (post 5) Avogadro s Law: increasing n increases the number of collisions, gas expands to keep pressure gonstant (post 2,4) c. Gas Laws Involving Temperature: i. Charles s Law: Increasing T increases kinetic energy; force of collisions increases and gas expands to maintain constant P (post 2,3,4) Amonton s Law: increasing T will increase force of collisions if volume is kept constant; P will increase (post 2,3,4) d. Average Kinetic Energy: KE avg = 1/2mu 2 rms i. U rms = the root-mean squared speed of the molecules; m = molecular mass. KE depends on:

1. Mass 2. Velocity oct 28. Diffusion and Effusion a. Graham s Law: i. Rate of Effusion and Diffusion: oc 1/ M Relative Rates of Effusion: (Rate) gas1 /(Rate) gas2 = M 2 / M 1 i Diffusion (Distance): (Distance) gas1/(distance) gas2 = M 2 / M 1 29. Practice: Graham s Law a. List the following gases, which are at the same temperature, in the order of increasing rate of diffusion: O 2, He, NO.