Graham's law

1. Graham's Law Diffusion and Effusion Din | Ramil | Santos | Luna | Inton | Bunayog

2. Motivation

3. ! To be able to understand the basic principles of Graham's Law. To analyse the difference of diffusion from effusion To define diffusion and effusion To identify the applications of Graham's Law checklist (Objectives)

4. Let’s start the discussion Gases (Short) Kinetic Molecular Theory Diffusion Effusion Thomas Graham Graham’s Law

5. Before we start Before we start We would like to point out that we strongly disagree with this* and you should too. *Only from this point and until this report ends :D

6. gases Topic: Just Gas Topic: Just Gas The entire space in which molecules of a gas move represents the volume of the gas. » M oles of gases are always in rapid, random, straight line motion, endlessly bumping against each other and hitting the walls of the container . This is discussed in the Kinetic Molecular Theory .

7. Topic: Kinetic molecular Theory Topic: Kinetic molecular Theory Kinetic Molecular Theory Gases are very far from each other. There are no interaction (attraction) between the molecules. They collide with each other and with the walls of the container in a perfectly elastic manner.

8. Diffusion Topic: Diffusion » The rate at which two gases mix ; movement of gases from high concentration to low concentration ; mixing of gases due to molecular motion Topic: Diffusion

9. Topic: E ffusion » movement of gases from high pressure to low pressure Topic: E ffusion Effusion

10. Topic: Diffusion and Effusion » Diffusion is the process of slowly mixing two gases together. Topic: Diffusion and Effusion Diffusion % E ffusion » Effusion is the process that occurs when a gas is permitted to escape its container through a small opening.

11. Topic: Diffusion and Effusion » Diffusion movement of gases from high concentration to low concentration Topic: Diffusion and Effusion Diffusion % E ffusion » Effusion movement of gases from high pressure to low pressure

12. Topic: Thomas Graham Topic: Thomas Graham Thomas graham He formulated the law of diffusion . Also called Graham’s Law in 1846. » 1805 -1869 » Scottish chemist

13. Topic: Graham’s Law Topic: Graham’s Law Graham’s law of diffusion "The velocity of a gas at a certain temperature is inversely proportional to the square root of its molecular mass."

14. Topic: Graham’s Law Topic: Graham’s Law Graham’s law of diffusion Graham's law is a gas law which relates the rate of diffusion or effusion of a gas to its molar mass. This means light gasses effuse or diffuse quickly and heavier gases effuse or diffuse slowly .

15. Topic: Graham’s Law Topic: Graham’s Law Graham’s law of diffusion This means light gasses effuse or diffuse quickly and heavier gases effuse or diffuse slowly . Hydrogen Helium Helium is lighter than Hydrogen

16. Topic: Graham’s Law Topic: Graham’s Law Graham’s law of diffusion This gas law is derived directly from the kinetic molecular theory. Consider two gas molecules: Each gas molecule has its corresponding mass, m , and volume, v . If they are at the same absolute temperature, then their kinetic energies are equal. a B

17. Topic: Graham’s Law Topic: Graham’s Law Graham’s law of diffusion Each gas molecule has its corresponding mass, m , and volume, v . If they are at the same absolute temperature, then their kinetic energies are equal.

18. Topic: Graham’s Law Topic: Graham’s Law Graham’s law of diffusion The mass can be related to the molecular weight of the gas "The velocity of a gas at a certain temperature is inversely proportional to the square root of its molecular mass."

19. Topic: Graham’s Law Topic: Graham’s Law Graham’s law of diffusion The mass can be related to the molecular weight of the gas Gases with high molecular weight tend to diffuse more slowly than gases with low molecular weight Higher molecular weight : Numerator Lower molecular weight : Denominator Higher velocity: Numerator Lower velocity: Denominator

20. let’s try it out ?

21. Sample Problem How much faster does O 2 escape through a porous container than SO 2 ? (Use the periodic table to get the molecular weight of gas) Sample problem MW O2 = 32.0 g/ mol MW SO2 = 64.0 g/ mol Solution: Substituting these values in the Equation for Graham’s law of diffusion. MW SO 2 MW O 2 V SO 2 V O 2 V SO 2 V O 2 64 g/ mol 32 g/ mol = = 8 5.66 V SO 2 V O 2 Given:

22. Sample Problem How much faster does O 2 escape through a porous container than SO 2 ? (Use the periodic table to get the molecular weight of gas) Sample problem Final answer: This means that O 2 diffuses 1.41 times as fast as SO 2 . = 8 5.66 V SO 2 1.41 = V SO 2 V O 2 V O 2 = 1.41 Solution: V SO 2 V O 2 V O 2 = 1.41 V SO 2

23. applications

24. APPLICATIONS APPLICATIONS 1. When someone sautés meat with garlic and onion, the volatile substances responsible for the aroma of the spices vaporize and mix with the gases in the air.

25. APPLICATIONS APPLICATIONS 2. A balloon filled with air and a boy allows the gas to escape, the gas molecules will diffuse among the molecules of air and mixture of gases will occupy the whole room. Thus, we say that a gas spreads throughout the space available to it.

26. Quiz Ready?

27. quiz Quiz 1. How much faster does O2 escape through a porous medium than CO2 given the same conditions? 2 . Determine how much faster He would escape through a porous medium than N 2 under the same conditions.

28. Answers Ready?

29. V C O 2 Answers Answers 1. How much faster does O 2 escape through a porous medium than CO 2 given the same conditions? MW C O 2 MW O 2 V C O 2 V O 2 V C O 2 V O 2 44 g/ mol 32 g/ mol = = 6.63 (6.6332) 5.66 (5.6568) V C O 2 V O 2 1.17 = V C O 2 V O 2 = V O 2 = 1.17 MW O2 = 32.0 g/ mol MW C O2 = 44.0 g/ mol Solution: Given: Final answer:

30. V C O 2 Answers Answers 2 . Determine how much faster He would escape through a porous medium than N 2 under the same conditions. MW N 2 MW He V N 2 V He V N 2 V He 28 g/ mol 4 g/ mol = = 5.29 2 V N 2 V He 2.65 V N2 = V N 2 V He = 2.65 V N2 V He MW N 2 = 28.0 g/ mol MW C O2 = 4 .0 g/ mol Solution: Given: Final answer: