1. Honors Chemistry, Chapter 10 Page 1 Chapter 10 – Physical Characteristics of Gases

2. Honors Chemistry, Chapter 10 Page 2 Kinetic-Molecular Theory Kinetic-molecular theory is based on the idea that particles of matter are always in motion. Ideal Gas – an imaginary gas that perfectly fits all the assumptions of the kinetic-molecular theory.

3. Honors Chemistry, Chapter 10 Page 3 Assumptions of Kinetic-Molecular Theory 1.Gases consist of large numbers of tiny particles that are far apart relative to their size. 2.Collisions between gas particles and between particles and container walls are elastic collisions. An elastic collision is one in which there is no net loss of kinetic energy. 3.Gas particles are in continuous, rapid, random motion. They therefore posses kinetic energy, which is energy of motion.

4. Honors Chemistry, Chapter 10 Page 4 Assumptions of Kinetic-Molecular Theory 4.There are no forces of attraction or repulsion between gas particles. 5.The average kinetic energy of gas particles depends on the temperature of the gas. KE = ½ mv 2 where KE is kinetic energy m is mass and v is velocity

5. Honors Chemistry, Chapter 10 Page 5 Nature of Gases Expansion – Gases have no definite shape or definite volume so they expand to fill the container. Gas particles move rapidly in all directions (assumption 3) without significant attraction or repulsion between them (assumption 4). Fluidity – Because the attractive forces between gas particles are insignificant (assumption 4), gas particles glide easily past one another.

6. Honors Chemistry, Chapter 10 Page 6 Nature of Gases Low Density – The density of a gas is about 1/1000 the density of the same substance in the liquid or solid state. Compressibility – During compression, the gas particles, which are initially very far apart (assumption 1), are crowded closer together. Under the influence of pressure, the volume of a gas can be greatly decreased.

7. Honors Chemistry, Chapter 10 Page 7 Nature of Gases Diffusion – Gases spread out and mix with one another, even without being stirred. This spontaneous mixing of particles of two substances caused by their random motion is called diffusion. Effusion is a process by which gas particles pass through a tiny opening. Rates of effusion of different gases are directly proportional to the velocities of their particles.

8. Honors Chemistry, Chapter 10 Page 8 Real Gases A real-gas is a gas that does not behave completely according to the assumptions of the kinetic-molecular theory. Deviations from ideal-gas behavior usually occur at high pressures and/or low temperatures where the attractive forces between molecules begin to play a role.

9. Honors Chemistry, Chapter 10 Page 9 Chapter 10, Section 1 Review 1.State the kinetic molecular theory of matter, and describe how it explains certain properties of matter. 2.List the five assumptions of the kinetic- molecular theory of gases. 3.Define the terms: ideal gas and real gas.

10. Honors Chemistry, Chapter 10 Page 10 Chapter 10, Section 1 Review continued 4.Describe each of the following characteristic properties of gases: expansion, density, fluidity, compressibility, diffusion, and effusion. 5.Describe the conditions under which a real gas deviates from “ideal” behavior.

11. Honors Chemistry, Chapter 10 Page 11 Pressure and Force Pressure (P) is defined as the force per unit area on a surface. pressure = force/area The SI unit for force is the newton, abbreviated N. It is the force that will increase the speed of a one kilogram mass by 1 meter per second per second that it is applied. force = mass x acceleration

12. Honors Chemistry, Chapter 10 Page 12 Example: Pressure on the Feet of a Ballet Dancer Acceleration of gravity is 9.8 m/s/s What is the force of a 51 Kg dancer on the floor? F = m x a = 51 kg x 9.8 m/s/s = 500 N What is the pressure? –Flat footed: 500 N/325 cm2 = 1.5 N/cm2 –Two feet tip toes: 500 N /13 cm2 = 38.5 N/cm2 –One foot tip toes:500 N /6.5 cm2 = 77 N/cm2

13. Honors Chemistry, Chapter 10 Page 13 Units of Pressure A common unit of pressure is millimeters of mercury, mm Hg. A pressure of 1 mm of Hg is now called 1 torr in honor of Torricelli. One atmosphere is defined as being exactly equivalent to 760 mm Hg. In SI units, pressure is expressed in derived units called pascals. One pascal (Pa) is defined as the pressure exerted by a force of one newton (1 N) acting on an area of 1 square meter. One atmosphere is 1.01325 x 10 5 Pa or 101.325 kPa.

14. Honors Chemistry, Chapter 10 Page 14 Units of Pressure UnitSymbolDefinition PascalPa1Pa = 1 N/m 2 Millimeter of mercury mm HgPressure that supports a 1 mm column of mercury torr 1 torr = 1 mm Hg 1 AtmosphereAtm760 mm Hg 101.325 kPa

15. Honors Chemistry, Chapter 10 Page 15 Standard Temperature and Pressure For purposes of comparison, scientists have agreed on standard conditions of exactly 1 atm pressure and 0 o C. These conditions are called standard temperature and pressure and are commonly abbreviated STP.

16. Honors Chemistry, Chapter 10 Page 16 Pressure Units Conversions Convert 0.830 atm to mm of Hg and kPa 0.830 atm x 760 mm of Hg/atm = 631 mm of Hg 0.830 atm x 101.325 kPa/atm = 84.1 kPa

17. Honors Chemistry, Chapter 10 Page 17 Chapter 10, Section 2 Review 1.Define pressure and relate it to force. 2.Describe how pressure is measured. 3.Convert units of pressure. 4.State the standard conditions of temperature and pressure.

18. Honors Chemistry, Chapter 10 Page 18 Gas Laws Gas laws are simple mathematical relationships between the volume, temperature, pressure, and amount of a gas. Boyle’s Law states that the volume of a fixed mass of gas varies inversely with the pressure at constant temperature.

19. Honors Chemistry, Chapter 10 Page 19 Illustration of Boyle’s Law

20. Honors Chemistry, Chapter 10 Page 20 Mathematical Expression of Boyle’s Law V = k/P or PV = k P 1 V 1 = k P 2 V 2 = k P 1 V 1 = P 2 V 2 P 1 V 1 / V 2 = P 2

21. Honors Chemistry, Chapter 10 Page 21 Volume-Pressure Data V – mLP – atmP x V 12000.5600 1.0600 3002.0600 2003.0600 1504.0600 1205.0600 1006.0600

22. Honors Chemistry, Chapter 10 Page 22 Boyle’s Law Example Problem V = 150. mL of O 2 at 0.947 atm. What is the volume at 0.987 atm (at constant temperature)? Formula: P 1 V 1 / P 2 = V 2 0.947 atm x 150. mL / 0.987 atm = 144 mL of O 2

23. Honors Chemistry, Chapter 10 Page 23 Absolute Zero Absolute zero, -273.15 o C., is the lowest temperature possible. This is assigned a value of zero on the Kelvin scale. To convert from Celsius to Kelvin: K = 273.15 + o C. To convert from Kelvin to Celsius: o C. = K – 273.15

24. Honors Chemistry, Chapter 10 Page 24 Charles’s Law Charles’s Law states that the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. V = kT or V/T = k V 1 / T 1 = V 2 / T 2

25. Honors Chemistry, Chapter 10 Page 25 Plot of Volume vs. Temperature

26. Honors Chemistry, Chapter 10 Page 26 Charles’s Law Example Problem A sample of Ne gas has a volume of 752 mL at 25 o C. What volume will the gas occupy at 50 o C. if the pressure remains constant? Convert temperatures to Kelvin: 25 o C. + 273 = 298 K. 50 o C. + 273 = 323 K.

27. Honors Chemistry, Chapter 10 Page 27 Charles’s Law Sample Problem V 2 = V 1 x T 2 / T 1 V 2 = 752 mL Ne x 323 K. / 298 K = 815 mL of Ne

28. Honors Chemistry, Chapter 10 Page 28 Gay-Lussac’s Law Gay-Lussac’s Law: The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature. P = kT or P/T = k P 1 /T 1 = P 2 /T 2

29. Honors Chemistry, Chapter 10 Page 29 Gay-Lussac’s Law Example Problem Gas in an aerosol can is 3.00 atm at 25 o C. What is the pressure at 52 o C.? P 2 = P 1 x T 2 /T 1 Convert temperatures to Kelvin: 25 o C. + 273 = 298 K. 52 o C. + 273 = 325 K.

30. Honors Chemistry, Chapter 10 Page 30 Gay-Lussac’s Law Example Problem P 2 = 3.00 atm x 325 K / 298 K = 3.27 atm

31. Honors Chemistry, Chapter 10 Page 31 Combined Gas Law The combined gas law expresses the relationship between pressure, volume, and temperature of a fixed amount of gas. PV/T = k Or P 1 V 1 /T 1 = P 2 V 2 / T 2

32. Honors Chemistry, Chapter 10 Page 32 Combined Gas Law Example Problem A helium-filled balloon has a volume of 50.0 L at 25 o C. and 1.08 atm. What volume will it have at 0.855 atm and 10 o C.? Convert the temperatures to Kelvin: 25 o C. + 273 = 298 K 10 o C. + 273 = 283 K. V 2 = P 1 V 1 T 2 /( P 2 T 1 )

33. Honors Chemistry, Chapter 10 Page 33 Combined Gas Law Example Problem V 2 = P 1 V 1 T 2 /( P 2 T 1 ) V 2 = 1.08 atm x50 L Hex283 K/(0.855atm x298 K) = 60.0 L

34. Honors Chemistry, Chapter 10 Page 34 Dalton’s Law of Partial Pressures Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. P T = P 1 + P 2 + P 3 + …

35. Honors Chemistry, Chapter 10 Page 35 Partial Pressure Example Oxygen gas is collected over water. The barometric pressure was 731 torr and the temperature was 20.0 o C. What was the partial pressure of the oxygen collected? P T = 731 torr P water = 17.5 torr at 20.0 o C.

36. Honors Chemistry, Chapter 10 Page 36 Partial Pressure Example P oxygen = P T – P water P oxygen = 731 torr – 17.5 torr = 713.5 torr

37. Honors Chemistry, Chapter 10 Page 37 Chapter 10, Section 3, Review 1.Use the kinetic-molecular theory to explain the relationship between gas volume, temperature, and pressure. 2.Use Boyle’s law to calculate volume- pressure changes at constant temperature. 3.Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.

38. Honors Chemistry, Chapter 10 Page 38 Chapter 10, Section 3, Review continued 4.Use the combined gas law to calculate volume-temperature-pressure changes. 5.Use Dalton’s law of partial pressures to calculate the partial pressures and total pressures.