1. Gas Densities, Partial Pressures, and Kinetic-Molecular Theory Sections 10.5-10.8

2. Objectives Apply the ideal-gas equation to real gas situations. Interpret the kinetic-molecular theory of gases

3. Key Terms Partial pressures Dalton’s Law of Partial Pressures Mole fraction Kinetic-molecular theory Root-mean-square speed Effusion Graham’s Law Diffusion Mean free path

4. Gas Densities and Molar Mass Rearrange the ideal-gas equation : n = P V RT Multiply both sides by molar mass, M n M = P M V RT Product of n/V and M = density in g/L Moles x grams = grams Liter mole liter

5. Gas Densities and Molar Mass Density is expressed: d = P M RT Density depends on pressure, molar mass, and temperature

6. Example Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21 ºC and 740.0 torr.

7. Gas Mixtures and Partial Pressure Dalton’s Law of Partial Pressures: –Total pressure of a mixture equals sum of the pressures that each would exert if present alone. P t = P 1 + P 2 + P 3 + ….

8. Gas Mixtures and Partial Pressures P 1 = n 1 (RT); P 2 = n 2 (RT); P 3 = n 3 (RT);… V V V And P t = (n 1 + n 2 + n 3 + ….) RT = n t (RT) V V

9. Example 1 A gaseous mixture made from 6.00 g oxygen and 9.00 g methane is placed in a 15.0 L vessel at 0  C. What is the partial pressure of each gas, and what is the total pressure of the vessel?

10. Example 2 What is the total pressure exerted by a mixture of 2.00g hydrogen and 8.00 g nitrogen at 273 K in a 10.0 L vessel?

11. Mole Fraction, X P 1 = n 1 RT/ V = n 1 P t = n t RT/ V = n t Thus… P 1 = (n 1 /n t )P t = X 1 P t Partial press = mole frac x total press

12. Example 3 Mole fraction of N 2 in air is 0.78 (78%). If the total pressure is 760 torr, what is the partial pressure of N 2 ?

13. Homework 44, 48, and 60-68 even only

14. Sections 10.7 & 10.8 Kinetic-Molecular Theory And Effusion/Diffusion

15. Objectives Understand why gas behave as they do Apply the Kinetic-Molecular Theory to the Gas Laws Define molecular effusion and diffusion Solve problems using Graham’s Law of Effusion

16. Key Terms Kinetic-Molecular Theory Root-Mean-Square Speed Effusion Diffusion Graham’s Law of Effusion Mean Free Path

17. Kinetic-Molecular Theory Explains why gases behave as they do Developed over 100 year period Published in 1857 by Rudolf Clausius

18. Kinetic Molecular Theory * Theory of moving molecules You Must Know the 5 Postulates (page 421).

19. Five Postulates 1) Gases consist of large numbers of molecules that are in continuous, random motion. 2) The combined volume of the molecules is negligible relative to the total volume in which the gas is contained. 3) Attractive and repulsive forces between gas molecules are negligible. 4) Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as T is constant 5) The average kinetic energy of the molecules is proportional to T. At any given T, all molecules have same avg. kinetic energy

20. Root-mean-square speed, u Speed of a molecule possessing average kinetic energy Є = ½ m u 2 Є is average kinetic energy m is mass of molecule Both Є and u increase as temperature increases

21. Application to Gas Laws 1. Effect of a V increase at constant T: - Є does not change when T is constant. Thus u is unchanged. With V increase, there are fewer collisions with container walls, and pressure decreases (Boyle’s Law).

22. Application to Gas Laws 2. Effect of a T increase at constant V: - Increase T means increase of Є and u. No change in V means there will be more collisions with walls (P increase).

23. Learning Check A sample of carbon dioxide initially at STP is compressed into a smaller volume at constant temperature. How does this effect: (a) Average kinetic energy (b) rms speed (c) Total number of collisions (d) Pressure

24. Molecular Effusion & Diffusion u = 3RT M *Derived equation from the k-m theory **Less massive gas molecules have higher rms speed ***Use R in units of J/mol-K

25. Example Calculate the rms speed of a nitrogen molecule at 298K.

26. Effusion Escape of gas molecules through a tiny hole into an evacuated space

27. Diffusion Spread of one substance throughout a space or throughout a second substance

28. Graham’s Law of Effusion Effusion rate of a gas is inversely proportional to the square root of its molar mass. Rates of effusion of two gases under identical conditions*: * At same T and P in containers with identical pinholes

29. Graham’s Law of Effusion Rate directly proportional to u : u 1 = 3RT/M 1 u 2 3RT/M 2

30. Graham’s Law of Effusion

31. Diffusion and Mean Free Path Similar to Effusion (faster for lower mass molecules) BUT diffusion is slower than molecular speeds because of molecular collisions Mean Free Path : average distance traveled by a molecule between collisions –For air molecules at seal level = 6 x 10 -8 m –At about 100 km in altitude = 10 cm

32. Homework 69, 70, 73, 76, 77, 79