2. Grahm’s Law of Effusion Effusion Equation Application

3. Review Kinetic Molecular Theory  Gases have low density  Gas molecules interact in elastic collisions  Gas molecules are always in motion  There are no forces of attraction between gas molecules  The speed of gas molecules is proportional to the temperature Pressure = Force / Area  Force - Newtons (N) ; Area - m2 m2 ; Pressure = N / m2 m2 = Pa Units of Pressure  1 atm = 760 mm of Hg = 760 torr = 101.3 kPa = 1.013 x 10 5 Pa

4. Review Partial Pressure  Dalton’s Law of Partial Pressure Total Pressure = Pressure 1 + Pressure 2 Stoichiometry of Gases  Avogadro’s Law Volume, molecules, and mols of gases are all proportional to the coefficients of the balanced chemical equation  Standard Molar Volume 22.4 L / mol of STP

5. Introduction Grahm’s law of effusion compares the rate at which two gases go through the same opening at the same temperature and pressure This relationship can be used to determine the identity of unknown gases Throughout the lecture form an answer to the following question  Why is there an inverse relationship between the molar mass of a gas and the rate at which it effuses?

6. Diffusion & Effusion Diffusion - spontaneous mixing of gas particles from a high concentration to a low concentration Effusion - a process by which gas particles pass through a tiny opening What causes the particles to go through the opening?

7. Effusion Simulation http://phet.colorado.edu/en/simulation/gas-properties http://phet.colorado.edu/en/contributions/view/3237

8. Kinetic Energy Grahm’s law can be derived from comparing the kinetic energy of two gases at the same temperature What is Kinetic energy? Energy of motion KE = 1/2 mv 2 Temperature determines the kinetic energy  At the same temperature, two gases will have the same KE

9. What did we see? Why did we only open the lid a little bit? Why did we hold the temperature constant? Which size of particle left the container more quickly?

10. Derivation KE A = KE B 1/2 m A (v A ) 2 = 1/2 m B (v B ) 2 m A (v A ) 2 =m B (v B ) 2 m B m B m A / m B (v A ) 2 = (v B ) 2 (v A ) 2 (v A ) 2 m A / m B = (v B ) 2 /(v A ) 2 √ m A / √m B = v B / v A

11. Grahm’s Law of Effusion “the rates of the effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses” VAVA VBVB = √M B √M A

12. What can we do with this formula? Compare the rates of effusion of different gases  Which will gas will leave a container first  Which gas will fill a room first Determine the identity of an unknown gas

13. Basic Steps for Effusion Problems Determine the givens and the unknowns Rearrange equation to solve for unknown Plug in known numbers Calculate square roots first Finish Solving Problem

14. Examples How much faster does Neon gas effuse than Xenon gas? Molar Mass of Neon = 20.17 g/mol Molar Mass of Xenon = 131.29 g/mol VAVA VBVB = √M B √M A V Ne V Xe = √M Xe √M Ne V Ne V Xe = √131.29 g/mol √20.17g/mol = 2.56 times

15. Examples If a sample of Helium gas effuses 6 times faster than an unknown sample of gas, what is the molar mass of the unknown gas? Molar Mass of Helium = 4.00 g/mol VAVA VBVB = √M B √M A √M B V He V unknown √M A = x 6V He V unknown √4.00g/mol = x =144.00g/mol