Ch. 11: Gases

1. Ch. 11: Gases Dr. Namphol Sinkaset Chem 152: Introduction to General Chemistry

2. I. Chapter Outline • Introduction • Kinetic-Molecular Theory of Gases • Pressure • Individual Gas Laws • The Combined Gas Law • The Ideal Gas Law • Partial Pressures • Gases and Stoichiometry

3. I. The Unique Gas Phase • Physical properties of a gas are nearly independent of its chemical identity! • Gas behavior is markedly different than solid or liquid behavior. • We look at a theory that explains why gas behavior is universal and then at the origin of equations that allow us to do gas calculations.

4. II. Kinetic Molecular Theory • This simple theory is very successful in explaining the physical behavior of most gases under “normal” conditions. • Kinetic molecular theory can be summarized into 4 statements.

5. II. Kinetic Molecular Theory • A gas is a collection of particles in constant, straight line motion. • Gas particles do not attract nor repel one another. They collide with each other and the walls of the container. • There is a lot of space between gas particles. • The average KE is proportional to the kelvin temperature of the gas.

6. II. Kinetic Molecular Theory

7. II. Application of KM Theory • KM theory predicts properties of gases well. For example: • Compressibility. Gases can be compressed because of the amount of space between particles. • Assume shape and volume of container. Gas particles are in constant motion and have no interaction with each other.

8. III. Pressure • Pressure is simply a force exerted over a surface area.

9. III. Origin of Gas Pressure • Gas pressure is the result of the cumulative force of many collisions between gas particles and container walls.

10. III. Pressure Imbalance

11. III. Atmospheric Pressure • Patm is simply the weight of the earth’s atmosphere pulled down by gravity. • Barometers are used to monitor daily changes in Patm. • Torricelli barometer was invented in 1643.

12. III. Units of Pressure • For historic reasons, we have units such as torr and mm Hg. (Why?) • The derived SI unit for pressure is N/m2, known as the pascal (Pa). • Note that 1 atm = 760 mm Hg = 760 torr = 101.325 kPa. • Pounds per square inch, psi, is an everyday unit. 1 atm = 14.7 psi.

13. III. Sample Problem • Perform the pressure unit conversions below. • Convert 575 torr to atm. • Convert 2.17 atm to mm Hg.

14. IV. Gas Laws • A sample of gas can be physically described by its pressure (P), temperature (T), volume (V), and amount of moles (n). • If you know any 3 of these variables, you know the 4th (via calculation). • We look at the history of how the ideal gas law was formulated.

15. IV. Pressure and Volume

16. IV. Boyle’s Law • At constant temperature and constant amount of gas, the volume of a gas and its pressure are inversely proportional.

17. IV. Boyle’s Law and KM Theory

18. IV. Volume and Temperature

19. IV. Absolute Zero • The graph shows an extrapolation to zero volume for the gas. • Of course, zero volume is impossible, so the corresponding temperature is known as absolute zero. • Absolute zero is the coldest possible temperature; 0 K = -273.15 °C.

20. IV. Charles’s Law • For constant pressure and constant moles of gas, the volume of a gas and its kelvin temperature are directly proportional.

21. IV. Charles’s Law and KM Theory

22. V. The Combined Gas Law • Boyle’s and Charles’s Laws can be combined into a convenient form. • The equation holds only when amount of gas remains constant.

23. V. Sample Problem • What’s the final pressure of a sample of N2 with a volume of 952 m3 at 745 torr and 25 °C if it’s heated to 62 °C with a final volume of 1150 m3?

24. V. Sample Problem • A sample of N2 has a volume of 880 mL and a pressure of 740 torr. What pressure will change the volume to 870 mL at the same temperature?

25. VI. Combined Gas Law to Ideal Gas Law • The combined gas law is actual very close to the ideal gas law. • The only quantity missing is the moles of gas. • We need one more gas law derive the ideal gas law from the combined gas law.

26. VI. Volume and Moles

27. VI. Avogadro’s Law • At constant temperature and pressure, the volume of a gas and the amount of moles of gas are directly proportional.

28. VI. Avogadro’s Law and KM Theory

29. R = 0.082058 L atm/K mole VI. The Ideal Gas Law • The ideal gas law is a combination of the combined gas law and Avogadro’s Law.

30. VI. Sample Problem • What volume, in mL, does a 0.245 g sample of N2 occupy at 21 °C and 750 torr?

31. VI. What Is An Ideal Gas? • The ideal gas law works best when gases are acting ideally. • To be an ideal gas, (1) the volume of the gas particles must be small relative to space between them and (2) the forces between the gas particles are not significant. • Gases behave nonideally at low temperature and high pressure.

32. VI. Ideal Vs. Nonideal

33. VII. Mixtures of Gases • According to KM theory, each gas in a mixture of gases acts independently of the others. • Each individual gas pressure is called a partial pressure. • Dalton’s Law of Partial Pressures: the sum of all partial pressures equals the total pressure. • Ptotal = P1 + P2 + P3 + … + Pn

34. VII. Gas Collection Over Water • When gas is collected over water, the total pressure is a sum of Pgas and PH2O. • Dalton’s Law of Partial Pressure is used to calculate the pressure of the gas by itself. • Ptotal = Pgas + PH2O • Partial pressures of water are tabulated as vapor pressures.

35. VII. Collecting H2 Over Water

36. VII. Water Vapor Pressures

37. VIII. Gases and Stoichiometry • Gases can be either products or reactants in a reaction, so they can be involved in stoichiometry problems. • Mole relationships allow gas calculations via the ideal gas equation. • Note that at STP (0 °C and 1 atm) 1 mole of gas occupies 22.4 L.

38. VIII. Sample Problem • How many mL of HCl(g) forms at 725 mm Hg and 32.3 °C when 0.117 kg of NaCl reacts with excess H2SO4? H2SO4(aq) + 2NaCl(s) Na2SO4(aq) + 2HCl(g)

39. VIII. Sample Problem • How many liters of oxygen at STP are needed to form 100.0 g of water according to the reaction below? 2H2(g) + O2(g) 2H2O(g)