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Chapter 9a

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

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Chapter 9a

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  1. Chapter 9a Gases

  2. 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. F A P = Pressure • Pressure is the amount of force applied to an area. • Atmospheric pressure is the weight of air per unit of area.

  4. Units of Pressure • Pascals • 1 Pa = 1 N/m2 = 1 Kg/ms2 • Bar • 1 bar = 105 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

  5. 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

  6. Manometer Used to measure the difference between atmospheric pressure & that of a gas in a vessel. Figure 9.2

  7. 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

  8. 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

  9. PV = k Since V = k (1/P) This means a plot of V versus 1/P will be a straight line. As P and V are inversely proportional A plot of V versus P results in a curve. Figure 9.5

  10. Charles’s Law: initial warm balloon, T big, V big

  11. Charles’s Law: final cold balloon T & V little

  12. V T = k Charles’s Law • The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. Figure 9.7 V = kT A plot of V versus T will be a straight line.

  13. Mathematically, this means: V = kn Avogadro’s Law • The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Figure 9.8

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

  15. nT P V= R P V = k (1/P) V

  16. 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 • P2 = P1T2 / T1 • P2 = P1(T2 /T1 ) = P1(360/298) = 1.2081P1

  17. 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)

  18. 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)

  19. 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.

  20. V = (nRT/P) = 1mol (8.314kPa dm3 /molK)(273.15K)/100 kPa V/mol = 22.71 dm3 = 22.71L

  21. n V P RT = Gas Densities and Molar Mass If we divide both sides of the ideal-gas equation by V and by RT, we get: Most of our gas law calculations involve proportionalities of these basic commodities.

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

  23. P RT RT P Becomes  =  = Molecular Mass We can manipulate the density equation to enable us to find the molar mass of a gas:

  24. End of Part 9a  Ideal Gas Law

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