1. Gases Chapter 10 H2H2H2H2 Paris 1783 Gas Bag N2N2N2N2

2. Why gases are studied separately Many common compounds exist as gases Many common compounds exist as gases Gases transport matter and energy across the globe (i.e., weather) Gases transport matter and energy across the globe (i.e., weather) Compared with those of liquids and solids, the behavior of gases is easiest to model. Compared with those of liquids and solids, the behavior of gases is easiest to model.

3. Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. (No solubility rules!) Gases have much lower densities than liquids and solids. Density of a gas given in g/L (vice g/mL for liquids) Physical Characteristics of Gases NO 2

4. Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 101,325 Pa = 101.325 kPa 1 atm = 760 mm Hg = 760 torr Pressure = Force Area (force = mass x acceleration) Height of column ∝ 1/density Fig. 10.2 mass 760 mm Hg

5. Fig. 10.3 For P ≈ 1 atm Using a Manometer to Measure Gas Pressure

6.  Pressure-Volume Relationship: Boyle’s Law P 1 V 1 = P 2 V 2  Temperature-Volume Relationship: Charles’s Law V ∝ T  Volume-Amount Relationship: Avogadro’s Law V ∝ n  The Ideal Gas Law:  The Ideal Gas Law: PV = nRT

7. As P (h) increases V decreases Fig. 10.6 Add Hg

8. P · V = constant P 1 · V 1 = P 2 · V 2 Boyle’s Law Constant temperature Constant amount of gas Fig 10.7

9. A sample of argon gas occupies a volume of 500. mL at a pressure of 626 mm Hg. What is the volume of the gas (in mm Hg) if the pressure is reduced at constant temperature to 355 mm Hg? P 1 · V 1 = P 2 · V 2 P 1 = 626 mm Hg V 1 = 500. mL P 2 = 356 mm Hg V 2 = ? V 2 = P1 · V1P1 · V1 P2P2 (626 mm Hg) (500. mL) 355 mm Hg = = 882 mL

10. As T increasesV increases

11. Variation of gas volume with temperature at constant pressure: Charles’ Law V ∝ TV ∝ T V = constant · T T (K) = t (°C) + 273.15 Temperature must be in Kelvin Constant pressure Constant amount of gas Fig 10.8

12. A sample of hydrogen gas occupies 13.20 L at 22.8 °C. At what temperature will the gas occupy half that volume if the pressure remains constant? V 1 = 13.20 L T 1 = 296.0 K V 2 = 1.60 L T 2 = ? T 2 = V2 · T1V2 · T1 V1V1 6.60 L · 296.0 K 13.20 L = = 148 K T 1 = 22.8 ( ° C) + 273.15 (K) = 296.0 K 148 K - 273.15 = -125 °C

13. Avogadro’s Law V ∝ number of moles (n) V = constant · n Constant pressure Constant temperature Fig 10.10

14. Fig 10.11 Avogadro’s Hypothesis Molar Volume

15. Propane burns in oxygen to form carbon dioxide and water vapor. How many volumes of carbon dioxide are obtained from one volume of propane at the same temperature and pressure? C 3 H 8 + 5O 2 3CO 2 + 4H 2 O 1 mole C 3 H 8 3 mole CO 2 At constant T and P 1 volume C 3 H 8 3 volumes CO 2

16. PV = nRT The conditions 0 ° C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L:

17. Fig 10.13 Comparison of Molar Volumes at STP One mole of an ideal gas occupies 22.41 L STP One mole of various real gases at STP occupy:

18. What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 ° C = 273.15 K P = 1 atm n = (49.8 g) (1 mol HCl) (36.45 g HCl) = 1.37 mol V = 1 atm (1.37 mol)(0.08216 ) (273.15 K) Latm molK V = 30.6 L

19. PV = nRT useful when P, V, n, and T do not change Modify equation when P, V, and/or T change: Initial state (1) of gas: Final state (2) of gas: Eqn [10.8] Combined Gas Law

20. Density (d) Calculations d = m V = P(MM) RT m is the mass of the gas in g MM is the molar mass of the gas Molar Mass (MM) of a Gaseous Substance dRT P MM = d is the density of the gas in g/L What happens to the density of a gas if: (a) it is heated at constant volume? (b) It is compressed at constant temperature? (b) Additional gas is added at constant volume?

21. Volumes of Gases in Chemical Reations What is the volume of CO 2 produced at 37.0 ° C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.1867 mol CO 2 V = nRT P (0.1867 mol) (0.0821 ) (310.15 K) Latm molK 1.00 atm = = 4.75 L

22. Gas Mixtures and Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2 Dalton’s Law of Partial Pressures

23. Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = nini nTnT Dalton’s Law of Partial Pressures

24. 2KClO 3 (s) 2KCl (s) + 3O 2 (g) P T = P O + P H O 22 Fig. 10.16 Collecting a water-insoluble gas over water Bottle full of oxygen gas and water vapor

25. 2KClO 3 (s) 2KCl (s) + 3O 2 (g) A 0.811 g sample of KClO 3 is partially decomposed to produce oxygen gas over water. The volume of gas collected is 0.250 L at 26 °C and 765 torr total pressure. How many grams of O 2 are collected? Sample Exercise 10.12 p 413 P T = P O + P H O 22 P O = - P H O 2 2 PTPT PV = nRT

26. Pressure of water vapor vs temperature App B p 1111

27. Kinetic Molecular Theory of Gases 1.A gas is composed of widely-separated molecules. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2.Gas molecules are in constant random motion. 3.Collisions among molecules are perfectly elastic. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. KE ∝ T

28. Fig 10.18 The Effect of Temperature on Molecular Speeds u rms ≡ root-mean-square speed Hot molecules are fast, cold molecules are slow.

29. The distribution of speeds of three different gases at the same temperature u rms = 3RT (MM)  R = 8.314 J/(mol K) Fig 10.19 The Effect of Molecular Mass on Molecular Speeds Heavy molecules are slow, light molecules are fast.

30. An “Ideal Gas” Assumptions: Gas molecules do not exert any force (attractive or repulsive) on each other i.e., collisions are perfectly elastic Volume of molecules themselves is negligible compared to volume of container i.e., the molecules are considered to be points An ideal gas “obeys” PV = nRT i.e., calculated value ≈ experimental value

31. Deviations from Ideal Behavior Assumptions made in the kinetic-molecular model:  negligible volume of gas molecules themselves  no attractive forces between gas molecules These breakdown at high pressure and/or low temperature.