1. Two cartoons in honor of my husband’s birthday…

3. Friday, Jan. 25 th : “A” Day Monday, Jan. 28 th : “B” Day Agenda  “Charles’ Law” lab – finish/collect  Begin Section 12.3: “Molecular Composition of Gases” Ideal gas law, diffusion, effusion, Graham’s law of diffusion  Homework: “Chemistry: Practice Problems for the Gas Laws” Concept Review – the 1 st 2 sections should be done… We will finish section 12.3 next time…

4. “Charles’ Law: The Effect of Temperature on Volume”  By now, you should have completed the data table and graphed your data.  Make sure your graph has a title and the axis’ are labeled, including units.  With your lab partner, take about 15 – 20 minutes to complete the conclusion questions before the lab will be due.  Remember to use complete sentences when answering all lab questions.

5. Review of Basic Gas Laws

6. Ideal Gas  Ideal gas: an imaginary gas whose particles are infinitely small and do not interact with each other. An ideal gas: 1.Does not condense to a liquid at low temperatures 2.Does not have forces of attraction or repulsion between the particles 3.Is composed of particles that have no volume

7. Ideal Gas Law  Ideal gas law: the law that states the mathematical relationship of pressure (P), volume (V), temperature (T), the gas constant (R), and the number of moles of a gas (n). PV = nRT “Pivnert” Remember: temperature is in Kelvins!

8. Ideal Gas Law  R is a proportionality constant  The value of R used in calculations depends on the units used for pressure. For Pressure in kPa: R = 8.314 L∙kPa mol∙K For Pressure in atm: R = 0.0821 L∙atm mol∙K

9. Ideal Gas Law  Real gases deviate somewhat from an ideal gas and more so at very high pressures.

10. Sample Problem E Pg. 435 How many moles of gas are contained in 22.41 liters at 101.325 kPa and 0°C? Use the ideal gas law: PV = nRT  P = 101.325 kPa  V = 22.41 liters  n = ?  R = 8.314 L∙kPa mol∙K  T = 0˚C + 273 = 273 K n = 1.00 mole

11. Additional Practice What is the volume of 4.35 moles of a gas at a pressure of 85.6 kPa and 26.0˚C? Use the ideal gas law: PV = nRT  P = 85.6 kPa  V = ?  n = 4.35 moles  R = 8.314 L∙kPa mol∙K  T = 26.0˚C + 273 = 299 K V = 126 Liters

12. Diffusion  When you open a bottle of household ammonia, the odor of ammonia gas doesn’t take long to fill the room. Why?  Diffusion: the movement of particles from regions of higher density to regions of lower density.  The process of diffusion involves an increase in entropy (S).

13. Effusion  Effusion: The passage of a gas under pressure through a tiny opening.  Scottish scientist Thomas Graham found that at constant temperature and pressure, the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass, M.

14. Graham’s law of diffusion V A = M B V B M A V = velocity (molecular speed) of gas A and B M = molar mass of gas A and B  Particles with lower molar mass travel faster than heavier particles.

15. Sample Problem F, pg 438 Oxygen molecules have an average speed of about 480 m/s at room temperature. At the same temperature, what is the average speed of molecules of sulfur hexafluoride, SF 6 ?  Gas A = SF 6 V A = ? M A : 146 g/mol  Gas B = Oxygen V B = 480 m/s M B : 32 g/mol V A__ = 32 - do division under and 480 146 - cross multiply to find V A V A = 225 m/s

16. Additional Practice The average velocity of CO 2 molecules at room temperature is 409 m/s. What is the molar mass of a gas whose molecules have an average velocity of 322 m/s under the same conditions? Gas A = CO 2 V A = 409 m/s M A = 44 g/mol Gas B = X V B = 322 m/s M B = ? 409 m/s = M B 1.27 = M B 322 m/s 44 g/mol 44 g/mol - Square each side to solve for M B M B = 71 g/mol

17. Homework  “Chemistry: Practice Problems for the Gas Laws” worksheet  DO NOT do #4 on the “Ideal Gas Law” side Of the worksheet… We will finish this section next time…