2. Wednesday, Jan. 16 th : “A” Day Thursday, Jan. 17 th : “B” Day Agenda  Homework Questions/Collect  Quiz over sec. 12.1: “Characteristics of Gases”  Sec. 12.2: “The Gas Laws” Boyle’s Law, Charles’s Law, Gay-Lussac’s Law, Avogadro’s Law, Combined Gas Law  Homework: Sec. 12.2 review, pg. 432: #1-12  Concept Review: “The Gas Laws” Be prepared for a quiz covering this section next time….

3. Homework Pg. 422: #1-12  Questions/Problems on the homework?  Hand In

4. Section 12.1 Quiz “Characteristics of Gases”  You should be OK to complete this on your own using your notes and your book. Have Fun!

5. Section 12.2: “The Gas Laws”  In this section, the relationship between the measurable properties of gases will be studied using the following variables: P = pressure exerted by the gas T = temperature in KELVINS of the gas V = total volume occupied by the gas n= number of moles of the gas

6. Pressure-Volume  In 1662, English scientist Robert Boyle found that as the pressure on a gas is increased in a closed container, the volume of the gas decreases. Pressure Volume

7. Gas molecules in a car-engine cylinder Pressure-Volume

8. Boyle’s Law P 1 V 1 = P 2 V 2

9. Pressure is inversely related to volume at constant temperature P V

10. Sample Problem B, Pg. 425 Solving Pressure-Volume Problems A given sample of gas occupies 523 mL at 1.00 atm. The pressure is increased to 1.97 atm, while the temperature remains the same. What is the new volume of the gas? Use Boyle’s Law: P 1 V 1 = P 2 V 2 P 1 = 1.00 atm V 1 = 523 mL P 2 = 1.97 atm Plug in variables and SOLVE for V 2 V 2 = 265 mL (3 sig figs)

11. Additional Practice If 650 mL of hydrogen is stored in a cylinder with a moveable piston at 225 kPa and the pressure is increased to 545 kPa at constant temperature, what is the new volume? Use Boyles Law: P 1 V 1 = P 2 V 2 P 1 = 225 kPa V 1 = 650 mL P 2 = 545 kPa Plug variables in and SOLVE for V 2 V 2 = 268 mL (3 sig figs)

12. Temperature-Volume Relationships  In 1787, French physicist Jacques Charles discovered that a gas’s volume is directly proportional to the temperature on the Kelvin scale if the pressure remains constant. Temperature Volume

13. Temperature-Volume

14. Charles’s Law V 1 = V 2 T 1 T 2 Remember: Temperature is in Kelvins!

15. Volume is directly proportional to Kelvin temperature at constant pressure T V

16. Sample Problem C, Pg. 428 Solving Volume-Temperature Problems A balloon is inflated to 665 mL volume at 27°C. It is immersed in a dry-ice bath at −78.5°C. What is its volume, assuming the pressure remains constant? Use Charles’s Law: V 1 = V 2 T 1 = T 2 V 1 = 665 mL T 1 = 27˚C + 273 = 300 K T 2 = -78.5˚C + 273 = 194.5 K Plug in variables and solve for V 2. V 2 = 431 mL (3 sig figs)

17. Additional Practice If the original temperature of a 62.2 L sample of a gas is 150˚C, what is the final temperature of the gas (in degrees C) if the new volume of gas is 24.4 L and the pressure remains constant? Use Charles’s Law: V 1 = V 2 T 1 = T 2 V 1 = 62.2 L T 1 = 150˚C + 273 = 423 K V 2 = 24.4 L Plug in variables and solve for T 2. Then change to ˚C. T 2 = -107°C (3 sig figs)

18. Temperature-Pressure Relationships  In 1802, French scientist Joseph Gay-Lussac discovered that if the temperature of a gas is doubled in a closed container of fixed volume, the pressure will double as well. Temperature Pressure

19. Gay-Lussac’s Law P 1 = P 2 T 1 T 2 Remember: Temperature is in Kelvins!

20. Pressure is directly proportional to Kelvin temperature, at constant volume T P

21. Sample Problem D, pg. 430 Solving Pressure-Temperature Problems An aerosol can containing gas at 101 kPa and 22°C is heated to 55°C. Calculate the pressure in the heated can. Use Gay-Lussac’s Law: P 1 = P 2 T 1 T 2  P 1 = 101 kPa  T 1 = 22˚C + 273 = 295 K  T 2 = 55˚C + 273 = 328 K  Plug in variables and solve for P 2 112 kPa

22. Additional Practice The pressure in a bottle of soda pop is 505 kPa at 20.0˚C. What is the new pressure if someone warms the sealed bottle to 65.0˚C? Use Gay-Lussac’s Law: P 1 = P 2 T 1 T 2  P 1 = 505 kPa  T 1 = 20.0˚C + 273 = 293 K  T 2 = 65.0˚C + 273 = 338 K  Plug in variables and solve for P 2 583 kPa

23. Volume-Molar Relationships  In 1811, Italian scientist Amadeo Avogadro proposed the idea that equal volumes of ALL gases, under the same conditions, have the same number of particles. Quite the looker, eh?

24. Avogadro’s Law  Gas volume is directly proportional to the number of moles of gas at the same temperature and pressure. V = kn or V 1 = V 2 n 1 n 2 n = number of moles of gas k = proportionality constant

25. Volume-Molar Relationships

26. Avogadro’s Law  At STP (0˚C and 1 atm pressure) the volume of 1 mole of ANY gas is 22.41 L.  The MASS of 22.41 L of a gas at STP will be equal to the gas’s molar mass. (molar mass = the mass, in grams, of 1 mole)

27. Combined Gas Law  The combined gas law is a combination of Boyles’ law and Charles’ law. + P 1 V 1 = P 2 V 2 T 1 T 2 Remember: Temperature is in Kelvins!

28. Combined Gas Law Example When 500 mL of O 2 gas at 25˚C and 1.045 atm is cooled to -40˚C and the pressure is increased to 2.00 atm, what is the new volume of the gas?  Use the combined gas law: P 1 V 1 = P 2 V 2 T 1 T 2  P 1 = 1.045 atm  V 1 = 500 mL  T 1 = 25˚C + 273 = 298 K  P 2 = 2.00 atm  T 2 = -40˚C + 273 = 233 K V 2 = 204 mL

29. Additional Example A gas at STP occupies 28 cm 3 of space. If the pressure changes to 3.8 atm and the temperature increases to 203°C, find the new volume.  Use the combined gas law: P 1 V 1 = P 2 V 2 T 1 T 2  P 1 = 1.00 atm  V 1 = 28.0 cm 3  T 1 = 0°C + 273 = 273 K  P 2 = 3.8 atm  T 2 = 203°C + 273 = 476 K V 2 = 12.8 cm 3

30. How do you know which gas law to use?  Look at the data given in the problem.  What do you know from the problem? Temperature, volume, pressure, etc.  Choose the formula that uses the variables given and solve for the missing variable…

31. Homework  Sec. 12.2 review, pg. 432: #1-12  Concept Review: “The Gas Laws” Next Time:  Quiz over Sec. 12.2  Lab Write-up  Gas Laws and Drinking Straws Activity