1. PowerPoint to accompany Chapter 9a Gases

2. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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

3. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

4. Pressure  Pressure is the amount of force applied to an area.  Atmospheric pressure is the weight of air per unit of area. P = FAFA

5. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Units of Pressure  Pascals  1 Pa = 1 N/m 2 = 1 Kg/ms 2  Bar  1 bar = 10 5 Pa = 100 kPa  mm Hg or torr  The difference in the heights in mm ( h ) of two connected columns of mercury  Atmosphere  1.00 atm = 760 torr

6. 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

7. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Manometer Figure 9.2 Used to measure the difference between atmospheric pressure & that of a gas in a vessel.

8. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Manometer calculation:  P atm = 102 kPa  P gas > P atm  P gas = P atm + h  P gas = 102 kPa + (134.6mm – 103.8mm)Hg  P gas = 102 kPa + (32.6mm Hg)  Note 1mmHg = 133.322 Pa  P gas = 102 kPa + (32.6x133.322x10 -3 kPa)  P gas = 106 kPa = 1.06 bar =1.06/1.013 atm  P gas = 1.046 atm

9. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

10. 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

11. 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 V = k (1/P) This means a plot of V versus 1/P will be a straight line. PV = k Figure 9.5

12. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Charles’s Law: initial warm balloon, T big, V big

13. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Charles’s Law: final cold balloon T & V little

14. 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. VTVT = k A plot of V versus T will be a straight line. Figure 9.7 V = kT

15. 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

16. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

17. 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 P V = R i.e. PV = nRT (where R = 8.314 kPa L/mol K)

18. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia V = k (1/ P ) nT P V = R V P

19. 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  P 1 = (nR/V 1 )T 1 or P 1 /T 1 = P 2 / T 2  P 2 = P 1 T 2 / T 1  P 2 = P 1 (T 2 /T 1 ) = P 1 (360/298) = 1.2081P 1

20. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ideal Gas Law Calculations  Move the piston to reduce the Volume from 1 dm 3 to 0.5 dm 3. (V decreases so P increases)  Same moles, same T, but greater P  P 1 V 1 = (nRT 1 ) = P 2 V 2  P 2 = P 1 V 1 / V 2 = P 1 (V 1 / V 2 )  P 2 = P 1 (1.0/0.5) = P 1 (2.0)

21. 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  P 1 = n 1 (RT 1 /V 1 ) or P 1 /n 1 = P 2 / n 2  P 2 = P 1 n 2 / n 1 for n 2 = 2 n 1  P 2 = P 1 (2)

22. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ideal Gas Law Calculations  R = (PV / nT)  When kPa and dm 3 are used as is common  R = 8.314 kPa dm 3 / 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.

23. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia V = (nRT/P) = 1mol (8.314kPa dm 3 /molK)(273.15K)/100 kPa V/mol = 22.71 dm 3 = 22.71L

24. 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: nVnV P RT = Most of our gas law calculations involve proportionalities of these basic commodities.

25. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

26. Gas Densities and Molar Mass  We know that  moles  molar mass = mass  So multiplying both sides by the molar mass (  ) gives: n   = m P  RT mVmV = i.e.  = P  RT

27. 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: Becomes P  RT  =  RT P  =

28. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia End of Part 9a  Ideal Gas Law