Gas Densities, Partial Pressures, and Kinetic-Molecular Theory

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 nM = PM 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 = PM 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. Pt = P1 + P2 + P3 + ….

8. Gas Mixtures and Partial Pressures P1 = n1 (RT); P2 = n2 (RT); P3 = n3 (RT);… V V V • And Pt = (n1 + n2 + n3 + ….) RT = nt (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 P1 = n1 RT/ V = n1 Pt = nt RT/ V = nt Thus… P1 = (n1/nt)Pt = X1Pt Partial press = mole frac x total press

12. Example 3 • Mole fraction of N2 in air is 0.78 (78%). If the total pressure is 760 torr, what is the partial pressure of N2?

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 • Gases consist of large numbers of molecules that are in continuous, random motion. • The combined volume of the molecules is negligible relative to the total volume in which the gas is contained. • Attractive and repulsive forces between gas molecules are negligible. • 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 • 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 mu2 Є is average kinetic energy m is mass of molecule • Both Є and u increase as temperature increases

21. Application to Gas Laws • 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: • Average kinetic energy • rms speed • Total number of collisions • 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: u1 =3RT/M1 u2 3RT/M2

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