The Gas Laws

1. The Gas Laws

2. Gases - 4 min

3. The Gas Laws Gas laws are concerned with the volume of gases

4. Volume Gas volume can be expressed in any of the metric units

5. Volume 1 liter L = 3 dm 1 cubic decimeter = 1000 milliliters ml = 3 1000 cubic centimeters cm

6. The Gas Laws Gas volume is determined by temperature & pressure

7. Temperature Changes in temperature cause changes in particle motion

8. Temperature Changes in particle motion cause changes in volume

9. Temperature Standard temperature is zero degrees Celsius o C 0

10. Temperature All gas calculations must be done using Kelvin temperatures

11. Temperature o C 273 K + =

12. Pressure Standard atmospheric pressure is one atmosphere at sea level

13. Pressure 1 atm 760 mm Hg = = 101 kPa 101,300 Pascles = = 1013 mb 29.92 in Hg =

14. STP Standard Temperature and Pressure

15. Boyle's Law Charles' Law Combined Gas Law Dalton's Law

16. Boyle's Law At a constant temperature, the pressure exerted by a gas varies inversely with its volume.

17. Boyle's Law Used when only pressure changes. Whatever pressure does, volume does the opposite.

18. Boyle's Law

19. Boyle's Law V2 = V1 P1 P2

20. Boyle's Law Practice Problem

21. Boyle's Law A gas occupies 15.5 dm3 at a pressure of 30 mm Hg. What is its volume when the pressure is increased to 50 mm Hg. Assume temp does not change.

22. Boyle's Law V2 = V1 P1 Write the proper equation. P2

23. Boyle's Law V2 = 15.5 dm3 30 mm Hg Substitute the given numbers. 50 mm Hg

24. Boyle's Law V2 = 15.5 dm3 30 mm Hg 50 mm Hg Since pressure increases, what will happen to volume?

25. Boyle's Law V2 = 15.5 dm3 30 mm Hg 50 mm Hg DECREASE V2 will be less than V1.

26. Boyle's Law Do the math. V2 = 15.5 dm3 30 mm Hg 50 mm Hg V2 = 9.3 dm3

27. V2 = V1 P1 All Boyle's Law problems should look like this: P2 = 15.5 dm3 30 mm Hg 50 mm Hg = 9.3 dm3 stop

28. Charles' Law At a constant pressure, the volume of a gas varies directly with the Kelvin temperature.

29. Charles' Law Used only when temperature changes. Whatever temperature does, volume does the same.

30. Charles' Law K e l v i n t e m p e r a t u r e K = o C + 2 7 3 V2 = V1 T2 T1

31. Practice Problem Charles' Law

32. Charles' Law 500 ml of air at 20 oC is heated to 60 oC with no pressure change. What is the new gas volume?

33. Charles' Law V2 = V1 T2 Write the proper equation. T1

34. Charles' Law V2 = 500 ml 60 oC Substitute the given numbers. 20 oC

35. Charles' Law V2 = 500 ml 333 K Temps must be in Kelvins. 293 K Add 273 to Celsius temps.

36. Charles' Law V2 = 500 ml 333 K 293 K Since temperature increases, what will happen to volume?

37. Charles' Law V2 = 500 ml 333 K 293 K INCREASE V2 will be greater than V1.

38. Charles' Law Do the math. V2 = 500 ml 333 K 293 K V2 = 600 ml

39. V2 = V1 T2 All Charles' Law problems should look like this: T1 = 500 ml 333 K 293 K = 600 ml stop

40. In the real world, pressure and temperature BOTH will change.

41. What do we do then??

42. Combined Gas Law Used when both temperature and pressure change

43. Combined Gas Law V2 = V1 P1 T 2 P2 T1

44. Practice Problem Combined Gas Law

45. Combined Gas Law A gas has a volume of 810 ml at 44 oC and 325 mm Hg. What would be the gas volume at 227 oC and 650 mm Hg?

46. Combined Gas Law V2 = V1 P1 T 2 Write the proper equation. P2 T1

47. Combined Gas Law V2 = 810 ml 325 mm Hg500 K Substitute the given numbers. 650 mm Hg 317 K

48. Combined Gas Law V2 = 810 ml 325 mm Hg500 K Do the math. 650 mm Hg 317 K = 640 ml

49. Calculate the volume of a gas at STP if 5.05 dm3 of the gas are collected at 27.5 oC and 95.0 kPa.

50. V1 P1 T2 V2 = P2 T1 5.05 dm3 95.0 kPa 273 K = 101 kPa 300.5 K = 4.32 dm3