CHAPTER 9 GASES: THEIR PROPERTIES AND BEHAVIOUR The Air Around Us Measuring Gases Any sample of gas has mass, volume and density. Unlike solids and liquids, gases expand to fill all of the space available. Gas mixture are always homogeneous at equilibrium.They are also very much more compressible than solids and liquids. Under the same conditions carbon dioxide, CO2(g) o, is heavier than air (higher density) and hydrogen, H2 (g)mis lighter (lower density) than air. So, in pouring carbon dioxide it must be poured down and hydrogen would have to be poured up! Most of a sample of a gas at atmospheric pressure is empty space. 9.1 Amount of Gas There are various properties that we can measure for a sample of gas: Mass m Volume V Pressure P Temperature T To be useful in our calculations we need to be able to relate these measurements to the number of moles present in the sample. 9.2 Measurement of Pressure Torricelli (1608 - 1647) - Mercury Barometer The atmospheric pressure varies from day to day but can normally support a column of mercury about 760 mmHg high. Consider a column 760 mm tall x 1 cm2 cross-section. What pressure does it exert in Pascal the SI pressure unit? The densityof mercury, Hg = 13.534 g cm-3.. Mass of the column = 76.0 cm x 1 cm2 x 13.534 g cm-3 = 1028.84 g = 1.028584 kg So, Force = mass x acceleration -2 = 1.029 kg x 2.81 m s = 10.09 kg m s-2 The pressure is force per unit area so now we need to convert to the SI base unit Since this is the force per cm2 we must convert to the force per m2. 1 cm2 = ? m2 1 2 1 2 -4 2 = ( 100 m) = 1000 m = 10 m 10.09 kg m s-2 5 -1 -2 5 -2 Pressure = 10-4 m2 = 1.009 x 10 kg m s or 1.009 x 10 N m = 101 kPa (since 1 N m-2 = 1 Pa pascal) The Standard Atmosphere The standard atmosphere is defined as = 760 mmHg = 101.3 kPa = 1 atm. 1 standard atm = 101.3 kPa = 760 mmHg. Since Hg expands with temperature, correction must be applied (density changes!) Since mercury is recognized as a poisonous substance (remember the Mad Hatter!) mercury barometer are not often seen today. One might ask why water is not used instead of mercury. However, because the density of water is only 1.0 g cm-3 a water barometer would have to be about ~ 33.7 feet or ~10 metres tall not ideal for the dining room! 9.3 Note: Standard pressure for gas measurements has been redefined to be 1 bar, or 100.000 kPa rater than 101.3 kPa (1 atm) . Most household barometers are aneroid barometers like the one in the picture opposite. Measuring Gas Pressure The Gas Laws 9.4 Early Study 1660 Robert Boyle - Boyle’s Law How does the volume of a gas change as the pressure changes (keeping the temperature and the number of moles constant)? This can be studied quite easily using something a simple as a bicycle pump and give results similar to the following graphs: Volume is inversely proportional to pressure 1 1 V } P or V = constant x P so PV = constant at constant n and T Amount of substance Amount of Substance/Volume V varies directly with n so V = k1n where V k1 is a constant. P & T constant Double moles and the volume doubles (if P and T constant)! n, number of moles Charles’ Law V directly proportional to T absolute V Charles, (1746 - 1823) and (Gay-Lussac 1778 - 1850) If we use the Kelvin Scale - add 273.1 to -273 the Celsius temp. - get new expression o O t C V is directly proportional to T V = Vo + k t 9.5 V = kT So we have V = k n and V = k T The laws can be combined PV = k (amount of substance constant) PV = k n T m The new k is a constant for any sample of gas and this we call R the gas const. Ideal Gas Law PV = nRT The law fits behaviour of well-behaved gases quite well but there are significant deviations for most real gases especially near the boiling point. Avogadro Law (1776 - 1856) Equal volumes of gases at the same temperature and pressure contain equal numbers of mole Consider the reaction N2(g) + 3 H2(g) ® 2 NH3(g) Say we have 1 volume of N2(g) and 5 volumes of H2(g) and the reaction goes to completion. What will the final volume be at the same temperature and pressure? N2(g) + 3 H2(g) ® 2 NH3(g) Total volume Initial 1 volume 5 volumes 0 volumes 6 volumes Final 0 volumes 2 volumes 2 volumes 4 volumes (limiting) (excess) 9.6 Standard Temperature & Pressure STP Standard Pressure 1 atm 760 mmHg (760 torr) 101.3 kPa Standard Temperature 273.15 K (0 ºC) At STP, the molar volume for a well behaved (ideal) gas is about 22.4 L mol–1 Using these values, we get R by rearranging PV = nRT PV 760 mm Hg x 22.4 L R = nT = 1 mol x 273.15 K R = 62.3 L mmHg K–1 mol–1 or if the pressure is measured in atmospheres 1 atm x 22.4 L R = 1 mol x 273.15 K R = 0.0820 L atm K-1 mol-1 or if the pressure is measured in kilopascals, kPa 101.3 kPa x 22.4 L R = 1 mol x 273.15 K R = 8.31 L kPa K-1 mol-1 but 1 L kPa = 1 N m = 1 J so R = 8.314 J K-1 mol-1 9.7 Simple Calculations Example If 3.66 L of a gas is collected at 0.886 atm and 100 ºC, what would the volume be at 1.000 atm pressure? Keeping n & T constant. PV PV = nRT but at constant n and T we can say T = constant P1V1 P2V2 P1V1 = constant = P2V2 or = T 1 T 2 P1V1 V2 = will expect the volume to decrease! P2 0.886 atm x 3.66 L = 1.000 atm Ans: 3.24 L Example A rigid car tyre is pumped to 28.0 lbs in-2 at –10.0 ºC. What is the pressure in the tyre at 20.0 ºC? Here we have constant V and n P V P V P P Again PV = nRT and 1 1 = 2 2 but V is constant 1 = 2 T 1 T 2 T1 T2 This time the pressure should increase.... We must convert the temperatures to Kelvin T1 = –10.0 + 273.15 K = 263.15 K and T2 = 20.0 + 273.15 K = 293.15 K P1T2 293.15 K P = = 28.0 lbs in-2 x = 31.2 lbs in-2. 2 T1 263.15 K This is enough to be a problem, so be careful if you plan to drive fast! Example: Making O2 in the Laboratory What volume of “dry” O2 collected at 25.0ºC and 1.0 atm when 50.0 mL of 5.0% H2O2 is decomposed (density 1.0 g mL-1) 9.8 2 H2O2 t 2 H2O + O2(g) 1 1 1/2 -1 5.0% Mass H2O2 = 50.0 mL x 1.0 g mL x 100% = 2.50 g 2.50 g available moles of H2O2 = = 0.07350 mol 34.01 g. mol-1 mol O2 = ½ x 0.07350 mol = 0.0368 mol nRT 0.0368 mol x 0.0821 L atm K-1 mol-1 x 298.1K V = P = 1.0 atm = 0.900 L Ans: The volume of dry O2 = 0.90 L Stoichiometry in Gas Reactions Once we can calculate moles of gases, we can use gas volumes in stoichiometry calculations. Consider the reaction: N2(g) + 3 H2(g) ® 2 NH3(g) What volume of hydrogen, H2, measured at 3.00 atm and 28.0 ºC is required to make 1.00 kg of ammonia, NH3? 1.00 kg x 1000 g kg-1 Moles NH3 required = 17.03 g mol-1 = 58.719 mol. Divide by 2 to focus on the ammonia: 1 3 2 N2(g) + 2 H2(g) ® 1 NH3(g) 3 Moles of H2 required = 2 x 58.719 mol = 88.079 mol. Using the ideal gas law PV = nRT nRT and rearranging gives V = P and substituting 88.079 mol x 0.0821 L atm K-1 mol-1 x 301.15 K V = 3.00 atm = 726 L Ans: 726 L 9.9 250 mL of dry HCl(g) at 25.0ºC and 790 mmHg is bubbled into 50.0 mL of ice cold water so that it all dissolves. The solution is then made up to 100.0 mL. What is the concentration of HCl (molarity) in the final solution? mol PV CHCl = V b PV = nRT n = RT 790 mmHg x 0.250 L 760 mmHg atm-1 n = 0.0821 L atm -1 K-1 mol-1 x 298.15 K =0.0106145 mol, so 0.0106145 mol -1 cHCl = 100.0 x 10-3 L = 0.106 mol L Ans: 0.106 mol L-1 Notice the concentration of the HCl could also be found by titration What volume of the CO2(g) + H2O(g) gases produced when 1.00 g of octane, C8H18 , burns - measured at, say, 130 ºC at a car exhaust at 100.0 kPa? So first write the balanced equation (complete combustion of octane gives only carbon dioxide and water) C8H18(g) + 12½ O2(g) ® 8 CO2(g) + 9 H2O(g) 1.00 g -3 Moles of octane burned = 114.22 g mol-1= 8.755 x 10 mol -3 so total moles of gas formed = 17 x 8.755 x 10 mol = 0.1488 mol nRT 0.1488 mol x 8.314 L kPa K-1 mol-1 x 403.15 K now, V = P = 100.0 kPa = 4.99 L Ans: 4.99 L Mixtures of Gases - Dalton’s Law of Partial Pressures Recall - Equal volumes of gases at the same T & P contain equal moles.