1. Daniel L. Reger Scott R. Goode David W. Ball http://academic.cengage.com/chemistry/reger Lecture 06 (Chapter 6) The Gaseous State

2. Physical States of Matter Matter can exist as:  Solid  Liquid  Gas Temperature Dependent States http://www.uni.edu/~iowawet/H2OProperties.htmlhttp://www.uni.edu/~iowawet/H2OProperties.html; http://en.wikipedia.org/;http://en.wikipedia.org/

3. A solid has fixed shape and volume. Solid Phase Solid Br 2 at low temperature

4. A liquid has fixed volume but no definite shape. The density of a solid or a liquid is given in g/mL. Liquid Phase Liquid Br 2

5. A gas has no fixed volume or definite shape. The density of a gas is given in g/L whereas liquids and solids are in g/mL. Gas Phase Gaseous Br 2

6. Summary of Gas Phase Characteristics Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011; http://www.chemistry.wustl.edu/~edudev/LabTutorials/Airbags/airbags.html Characteristics of gases: Particles in constant random motion Particles far apart, travel in straight lines, collide frequently Low Density (particles widely separated) Indefinite Shape (little cohesion, particles expand to shape of container) Large Compressibility (gas is mostly empty space) Moderate Thermal Expansion (increase in temperature causes particles to collide with more energy, increases volume)

7. Pressure is the force per unit area exerted on a surface. The pressure of the atmosphere is measured with a barometer. Pressure of a Gas

8. Both open and closed end manometers measure pressure differences. In Figs. A and B (open manometers), h represents difference in P between a gas and the atmosphere. In Fig. C (closed manometer), h represents P of the gas. Manometers AB C

9. Units of Pressure One atmosphere of pressure (1 atm) is the normal pressure at sea level. The SI unit of pressure is the pascal (Pa), but is a very small unit and is not used frequently by chemists. 1 atm = 760 mm Hg1 atm = 101.3 kPa 1 atm = 760 torr 1 atm = 1.01 bar 1 torr = 133.3 Pa 1 atm = 29.9 in Hg 1 atm = 14.7 psi

10. Gas Laws Describe behavior of gases when mixed, subjected to pressure or temperature changes, or allowed to diffuse Laws describes relationships between temperature (T), volume (V), pressure (P) and moles (n) The physical properties of all gases behave in the same manner, regardless of the identity of the gas Boyle’s law Charles’s law Combined gas law Avogadro’s law Ideal gas law

11. A constant relationship exists between pressure (P) and volume (V) at constant temperature (T) If pressure increases, volume occupied by the gas decreases If volume increases, pressure created by the gas decreases Boyle’s Law

12. A plot of volume versus 1/P is a straight line. Boyle’s Law Formula for straight line

13. A sample of a gas occupies 5.00 L at 0.974 atm. Calculate the volume of the gas at 1.00 atm, when the temperature held is constant. Example: Boyle’s Law How do you solve this problem? What do you know? If P 1 V 1 = k 1, and k 1 is constant (at constant T), then P 1 V 1 = P 2 V 2 Check: P 1 V 1 = (5.00L)(0.974 atm) = 4.87 L·atm = k 1 V= (k 1 /P) = (4.87 L·atm)/(1.00 atm) = 4.87 L Or P 1 V 1 = P 2 V 2 rearranges to V 2 = (P 1 V 1 )/P 2 = (0.974 atm * 5.00 L)/1.00 atm = 4.87 L

14. At constant pressure, the volume of a gas sample is directly proportional to the temperature (expressed in kelvins) If temperature increases, volume increases at constant pressure Charles’s Law Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

15. A plot of volume versus temperature is a straight line. Extrapolation to zero volume yields absolute zero in temperature (which isn’t going to happen!!): -273 o C. V = k 2 x T, where T is given in units of kelvin. Charles’s Law

16. A balloon filled with oxygen gas at 25 ° C occupies 2.1 L. At constant pressure, what is the volume at 100 ° C? Example: Charles’s Law How do you solve this problem? What do you know? First, convert T value(s) from ° C to K. Ti in K = 25 + 273 = 298 K Tf in K = 100 + 273 = 373 K If then

17. P, V, T Relationships: Combined Gas Law Boyle’s law and Charles’s law can be combined to relate P, V and T, when mass of gas remains constant. Because k’’ is a constant, we can use this equation to evaluate changes in these variables over time (between some initial state and a final state). Any problem that can be solved with either Boyle’s law or Charles’s law, can also be solved using the Combined Gas Law.

18. Combined Gas Law: Examples 2500 liters of oxygen gas is produced at 1.00 atm of pressure. It is to be compressed and stored in a 20.0 liter cylinder. If temperature is constant, calculate the pressure of the oxygen in the cylinder. A 3.2 liter sample of gas is at 40°C and 1.0 atmosphere of pressure. If the temperature decreases to 20°C and the pressure decreases to 0.60 atmospheres, what is the new volume in liters?

19. Combined Gas Law: Examples A sample of a gas occupies 4.0 L at 25 o C and 2.0 atm of pressure. Calculate the volume at STP (T = 0 o C, P = 1 atm). A sample of a gas occupies 200 mL at 100 o C. If the pressure is held constant, calculate the volume of the gas at 0 o C.

20. Avogadro’s Hypothesis Two different gases of equal volume measured at same T and P contain equal numbers of molecules Mass would not be identical due to different MW’s

21. A plot of the volume of all gas samples, at constant T and P, vs. the number of moles (n) of gas is a straight line. V = k 3 x n Avogadro’s Law

22. Ideal Gas Law P = Pressure V = Volume n = number of moles T = Temperature R = Universal Gas Constant Also, because m = mass MW = molecular weight We can also express the ideal gas law as STP (Standard Temperature and Pressure) T = 0 °C P = 1.0 atm V of 1 mol ideal gas (any gas) = 22.4 L at STP

23. Ideal Gas Law Calculation Calculate the number of moles of argon gas in a 30 L container at a pressure of 10 atm and temperature of 298 K. How do we solve this problem?

24. Ideal Gas Law Calculation Calculate the number of moles of argon gas in a 30 L container at a pressure of 10 atm and temperature of 298 K. PV = nRT n = n = = 12 mol

25. Molar Mass and Density The ideal gas law can be used to calculate density (mass/volume) and molar mass (mass/moles) of a gas. At constant pressure and temperature the density of a gas is proportional to its molar mass, so the higher the molar mass, the greater the density of the gas.

26. Example: Molar Mass Calculate the molar mass of a gas if a 1.02 g sample occupies 220 mL at 95 o C and a pressure of 750 torr.

27. Example: Molar Mass Calculate the molar mass of a gas if a 1.02 g sample occupies 220 mL at 95 o C and a pressure of 750 torr.

28. Gases and Chemical Equations The ideal gas law can be used to determine the number of moles, n, for use in problems involving reactions. The ideal gas law relates n to the volume of gas just as molar mass is used with masses of solids and molarity is used with volumes of solutions.

29. Example: Gases with Equations Calculate the volume of O 2 gas formed in the decomposition of 2.21 g of KClO 3 at STP. 2KClO 3 (s)  2KCl(s) + 3O 2 (g)

30. Gas Volumes in Reactions Equal volumes of gases at same P and T contain same number of moles. In chemical rxns under same conditions, gas volumes combine in same proportions as coefficients of equation. Therefore, 3 L of hydrogen gas, combined with 2 L of nitrogen gas, forms 2 L of ammonia.

31. Calculate the volume of NH 3 gas produced in the reaction of 4.23 L of H 2 with excess N 2 gas. Assume the volumes are measured at the same temperature and pressure. Example: Gas Volumes in Reactions

32. The pressure exerted by each gas in a mixture is called its partial pressure. For a mixture of two gases A and B, the total pressure, P T, is P T = P A + P B Dalton’s Law of Partial Pressure

33. Example: Partial Pressures Calculate the pressure in a container that contains O 2 gas at a pressure of 3.22 atm and N 2 gas at a pressure of 1.29 atm. P T = P A + P B = 3.22 atm + 1.29 atm = 4.51 atm

34. Example: Partial Pressures A gas sample in a 1.2 L container holds 0.22 mol N 2 and 0.13 mol O 2. Calculate the partial pressure of each gas, and the total pressure at 50 ° C. We will use Ideal Gas law to calculate P for each of the 2 gases.

35. Mole Fraction

36. Mole fraction of the yellow gas is 3/12 = 0.25 and the mole fraction of the red gas is 9/12 = 0.75 Mole Fraction

37. Example: Partial Pressure Calculate the partial pressure of Ar gas in a container that contains 2.3 mol of Ar and 1.1 mol of Ne and is at a total pressure of 1.4 atm.

38. The volume of gas generated in a rxn can be determined by water displacement. The collected gas contains both O 2 gas from the rxn and water vapor, and each contribute to total P. Collecting Gases over Water

39. Example: Collecting Gases Sodium metal is added to excess water, and H 2 gas produced in the reaction is collected over water with the gas volume of 1.2 L. If the pressure is 745 torr and the temperature 26 o C, what was the mass of the sodium? The vapor pressure of water at 26 o C is 25 torr. 2Na(s) + 2H 2 O( l )  H 2 (g) + 2NaOH(aq)

40. Describes the behavior of (ideal) gas particles at molecular level, based on 4 postulates. 1. Gases consist of small particles that are in constant and random motion. 2. Gas particles are very small compared to the average distance that separates them. 3. Collisions of gas particles with each other and the walls of the container are elastic (i.e., no loss of kinetic energy). 4. The average kinetic energy of gas particles is proportional to the temperature on the Kelvin scale. Kinetic Molecular Theory of Gases

41. How does this theory explain the observed behavior of gases? 1.Because gases consist of small particles that are in constant and random motion and they are constantly colliding with each other and with the walls of the container, and because these collisions involve no loss of average KE, the result is increased pressure. 2. Because gas particles are very small compared to the average distance that separates them, at constant T, they can be compressed into a smaller volume (P increases), or they can expand to fill a larger volume (P decreases). 3. Because the average kinetic energy of gas particles is proportional to the temperature (on the Kelvin scale), KE of the molecules increases with increasing T resulting in more collisions over time involving greater force; thus, at constant P, V would have to increase (Charles’s law). Kinetic Molecular Theory of Gases

42. Although gas particles move at different velocities, we can calculate an average velocity (see previous slide), which is also called the root mean square (rms) speed, u rms, and is the square root of the average squared speed. This number can be derived mathematically and represented by the associated curve. Maxwell-Boltzmann distribution curves Average Speed of a Gas R = 8.314 J/mol. K; molar mass in kilograms per mole

43. From this plot and the equation, you can see that velocity, or root mean square speed, of molecules decreases with increasing molar mass (at constant T). Average Speed of a Gas

44. The Kinetic Molecular Theory also correctly described both effusion and diffusion: Effusion - the passage of a gas through a small hole into an evacuated space. Gases with low molar masses effuse more rapidly (e.g., heavier gases move more slowly). Diffusion is the mixing of particles due to motion. Effusion (Graham’s law) and Diffusion

45. Graham’s Law is frequently used to compare rates of effusion of 2 gases. Effusion (Graham’s law)

46. Effusion (Graham’s law) Example Calculate the molar mass of a gas if equal volumes of nitrogen and the unknown gas take 2.2 and 4.1 minutes, respectively, to effuse through the same small hole (at constant T and P).

47. Gases deviate from the ideal gas law at high pressures in 2 ways. Deviations from Ideal Behavior

48. The assumption that gas particles are small compared to the distances separating them fails at high pressures. The observed value of PV/nRT will be greater than 1 under these conditions. Deviations from Ideal Behavior

49. The forces of attraction between closely spaced gas molecules reduce the impact of wall collisions. These attractive forces cause the observed value of PV/nRT to decrease below the expected value of 1 at moderate pressures. Forces of Attraction in Gases

50. A gas (O 2 below) deviates from ideal gas behavior at low temperatures (near the condensation point ) and high pressures. In other words, removing heat removes energy applied to molecules, so they slow down, and in doing so, are more susceptible to effects caused by force of attraction. Ideal Gases

51. The van der Waals equation corrects for attractive forces and the volume occupied by the gas molecules. a is a constant related to the strength of the attractive forces. b is a constant that depends on the size of the gas particles. a and b are determined experimentally for each gas. (Review Example 6.17 on your own) van der Waals Equation