1 / 28

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

adonica
Télécharger la présentation

Chapter 9a

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related