1. The Gas Laws Learning about the special behavior of gases (section 12.3) Note Pack page 3, bottom Objective #1

2. Boyle’s Law State Boyle’s Law in words: –When temperature and the number of particles are held constant, pressure and volume are inversely related Equation: P 1 V 1 = P 2 V 2 Graph Boyle’s Law:

3. Example 1 A high-altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.) When dealing with gases, we’re going to start by using values of: Pressure Volume Temperature (K) In the problem above, please identify what you are given, and what you want.

4. Example 1 A high-altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.) P 1 V 1 = P 2 V 2  (103 kPa)(30L) = (25 kPa) (x)

5. Example 1 A high-altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.) P 1 V 1 = P 2 V 2  (103 kPa)(30L) = (25 kPa) (x) 3090 = 25x

6. Example 1 A high-altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.) P 1 V 1 = P 2 V 2  (103 kPa)(30L) = (25 kPa) (x) 3090 = 25x 25 25 123.6 = x

7. Example 1 A high-altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.) P 1 V 1 = P 2 V 2  (103 kPa)(30L) = (25 kPa) (x) 3090 = 25x 25 25 123.6 = x So, the answer is 123.6 L (we’ll use the same unit for Volume as V 1, unless otherwise instructed.

8. Example 2 The pressure on 2.5 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant? In the problem above, please identify what you are given, and what you want: Pressure Volume Temperature (K)

9. Example 2 The pressure on 2.5 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant? P 1 V 1 = P 2 V 2  (105 kPa)(2.5L) = (40.5 kPa)(X)

10. Example 2 The pressure on 2.5 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant? P 1 V 1 = P 2 V 2  (105 kPa)(2.5L) = (40.5 kPa)(X) 262.5 = 40.5x

11. Example 2 The pressure on 2.5 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant? P 1 V 1 = P 2 V 2  (105 kPa)(2.5L) = (40.5 kPa)(X) 262.5 = 40.5x 40.5 40.5 X = 6.48L

12. Example 3 A gas with a volume of 4.0 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant? In the problem above, please identify what you are given, and what you want: Pressure Volume Temperature (K)

13. Example 3 A gas with a volume of 4.0 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant? P 1 V 1 = P 2 V 2  (205 kPa)(4.0L) = (x)(12.0L) 68.3 kPa

14. Charles’s Law - State Charles’s Law in Words: –When pressure and amount of gas is held constant, Kelvin temperature and volume are directly related. Show this law in a formula: What would the graph for this look like? NOTE: Temp for all gases needs to be in Kelvin!! 

15. Example 1 A balloon inflated in a room at 24 o C has a volume of 4.0 L. The balloon is then heated to a temperature of 58 o C. What is the new volume if the pressure remains constant? Begin by changing the temp to Kelvin ( o C + 273 = K) 24 o C + 273 = 297 Kelvin 58 o C + 273 = 331 Kelvin

16. Example 1 A balloon inflated in a room at 24 o C has a volume of 4.0 L. The balloon is then heated to a temperature of 58 o C. What is the new volume if the pressure remains constant? V1 = V2 T1 = T2

17. Example 1 A balloon inflated in a room at 24 o C has a volume of 4.0 L. The balloon is then heated to a temperature of 58 o C. What is the new volume if the pressure remains constant? V1 = V2 4.0L = X T1 = T2 297 = 331

18. Example 1 A balloon inflated in a room at 24 o C has a volume of 4.0 L. The balloon is then heated to a temperature of 58 o C. What is the new volume if the pressure remains constant? V1 = V2 4.0L = X T1 = T2 297 = 331 To solve, cross multiply and then divide… X = 4.46L

19. Example 2 If a sample of gas occupies 6.8 L at 325 o C, what will be its volume at 25 o C, if the pressure does not change?

20. Example 2 If a sample of gas occupies 6.8 L at 325 o C, what will be its volume at 25 o C, if the pressure does not change? V 1 = V 2 6.8L = X T 1 = T 2 598 = 298 X = 3.39 L   To solve: Cross-multiply & divide (6.8L)(298K) = (598)(X)

21. Example 3 Exactly 5.0 L of air at - 50.0 o C is warmed to 100 o C. What is the new volume if the pressure remains constant? V1 = V2 T1 = T2 Change temp into Kelvin

22. Example 3 Exactly 5.0 L of air at - 50.0 o C is warmed to 100 o C. What is the new volume if the pressure remains constant? V1 = V2 5.0L = X. T1 = T2223K = 373K

23. Example 3 Exactly 5.0 L of air at - 50.0 o C is warmed to 100 o C. What is the new volume if the pressure remains constant? V1 = V2 5.0L = X. T1 = T2223K = 373K To solve: Cross-multiply & divide

24. Example 3 Exactly 5.0 L of air at - 50.0 o C is warmed to 100 o C. What is the new volume if the pressure remains constant? V1 = V2 5.0L = X. T1 = T2223K = 373K X = 8.36 L To solve: Cross-multiply & divide

25. Gay-Lussac’s Law State Gay-Lussac’s Law in words: When the amount of gas and the volume of the gas are held constant, Kelvin temperature and pressure are directly related. Express this in a formula: What would the graph look like?

26. Example 1 A gas left in a used aerosol can is at a pressure of 103 kPa at 25 o C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928 o C ? P1 = P2 T1 = T2

27. Example 1 A gas left in a used aerosol can is at a pressure of 103 kPa at 25 o C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928 o C ? P1 = P2103 kPa = X T1 = T2298 K = 1201K

28. Example 1 A gas left in a used aerosol can is at a pressure of 103 kPa at 25 o C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928 o C ? P1 = P2103 kPa = X T1 = T2298 K = 1201K X = 415 kPa

29. Example 2 A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? P1 = P2 T1 = T2

30. Example 2 A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? P1 = P26.58 kPa = X T1 = T2 539 K = 211K

31. Example 2 A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? P1 = P26.58 kPa = X T1 = T2 539 K = 211K X = 2.58 kPa

32. That concludes content for Obj. #1. Practice makes permanent!