Transcript Slide 1
PowerPoint to accompany Chapter 9a Gases Characteristics of Gases Like liquids and solids, they: Still have mass and momentum Have heat capacities & thermal conductivities Can be chemically reactive Unlike liquids and solids, they: Expand to fill their containers Are highly compressible Have extremely low densities Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Pressure Pressure is the amount of force applied to an area. F P= A Atmospheric pressure is the weight of air per unit of area. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Units of Pressure Pascals Bar 1 bar = 105 Pa = 100 kPa mm Hg or torr 1 Pa = 1 N/m2 = 1 Kg/ms2 The difference in the heights in mm (h) of two connected columns of mercury Atmosphere 1.00 atm = 760 torr Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Standard Pressure Normal atmospheric pressure at sea level. This is ambient pressure, equal all around us, so not particularly felt It is equal to: 1.00 atm 760 torr (760 mm Hg) 101.325 kPa 14.696 psi Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Manometer Used to measure the difference between atmospheric pressure & that of a gas in a vessel. Figure 9.2 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Manometer calculation: Patm= 102 kPa Pgas> Patm Pgas= Patm + h Pgas= 102 kPa + (134.6mm – 103.8mm)Hg Pgas= 102 kPa + (32.6mm Hg) Note 1mmHg = 133.322 Pa Pgas= 102 kPa + (32.6x133.322x10-3 kPa) Pgas= 106 kPa = 1.06 bar =1.06/1.013 atm Pgas= 1.046 atm Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure: V ~ 1/P Figure 9.4 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia As P and V are inversely proportional A plot of V versus P results in a curve. Since PV = k V = k (1/P) This means a plot of V versus 1/P will be a straight line. Figure 9.5 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Charles’s Law: initial warm balloon, T big, V big Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Charles’s Law: final cold balloon T & V little Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Charles’s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. V =k T V = kT Figure 9.7 A plot of V versus T will be a straight line. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Avogadro’s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means: V = kn Figure 9.8 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ideal-Gas Equation So far we’ve seen that: V 1/P (Boyle’s law) V T (Charles’s law) V n (Avogadro’s law) Combining these, we get, if we call the proportionality constant, R: nT V=R P i.e. PV = nRT (where R = 8.314 kPa L/mol K) Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia P V = k (1/P) nT V=R P V Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ideal Gas Law Calculations Heat Gas in Cylinder from 298 K to 360 K at fixed piston position. (T incr. so P increases) Same volume, same moles but greater P P1 = (nR/V1)T1 or P1/T1 = P2 / T2 P 2 = P 1 T 2 / T1 P2 = P1(T2 /T1 ) = P1(360/298) = 1.2081P1 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ideal Gas Law Calculations Move the piston to reduce the Volume from 1 dm3 to 0.5 dm3. (V decreases so P increases) Same moles, same T, but greater P P1V1 = (nRT1) = P2V2 P2 = P1V1 / V2 = P1 (V1 / V2) P2 = P1(1.0/0.5) = P1(2.0) Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ideal Gas Law Calculations Inject more gas at fixed piston position & T. the n increases so P increases) Same volume, more moles but greater P P1 = n1(RT1/V1) or P1/n1 = P2 / n2 P2 = P1 n2 / n1 for n2 = 2 n1 P2 = P1(2) Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ideal Gas Law Calculations R = (PV / nT) When kPa and dm3 are used as is common R = 8.314 kPa dm3 / mol K or J/mol K It is useful to write the ideal gas law this way so that R is the constant and other variables can change from initial to final conditions This again reduces the gas law to a simple arithmetic ratio calculation The only tricks are to ensure all units are consistent or converted as needed. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia V = (nRT/P) = 1mol (8.314kPa dm3 /molK)(273.15K)/100 kPa V/mol = 22.71 dm3 = 22.71L Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Gas Densities and Molar Mass If we divide both sides of the ideal-gas equation by V and by RT, we get: n P = V RT Most of our gas law calculations involve proportionalities of these basic commodities. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Gas Densities and Molar Mass We know that moles molar mass = mass n=m So multiplying both sides by the molar mass ( ) gives: m P = V RT P i.e. = RT Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Molecular Mass We can manipulate the density equation to enable us to find the molar mass of a gas: P = RT Becomes RT = P Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia End of Part 9a Ideal Gas Law Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia