1. Gases and Moles The Ideal Gas Equation

2. What factors affect the pressure of a confined gas? 1. Number of molecules 2. Temperature 3. Volume of the container Think in terms of the number of collisions.

3. Number of molecules Increasing the number of molecules increases the number of collisions … P  n Where n is the number of moles of molecules … which increases the pressure.

4. Temperature Increasing the temperature makes the molecules move faster, increasing the number of collisions … P  T Where T is the absolute temperature … which increases the pressure.

5. Volume Increasing the volume of the container decreases the number of collisions … P  Where V is the volume 1V … which decreases the pressure.

6. Sooooo… P  n P  T P  1V n V T

7. nTR Make it into an equation P  n V T P  V

8. The Ideal Gas Equation n TR P  V n TR PV  … is usually written as …

9. The Ideal Gas Equation nTR P  V R = 0.0821 L atm mol K R is the “gas constant”

10. The Ideal Gas Equation nTR P  V Can R be in units other than L atm mol K ?

11. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

12. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

13. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

14. The Ideal Gas Equation nTR P  V R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

15. Ideal Gas Equation The ideal gas equation relates pressure, volume, temperature and the number of moles of a quantity of gas. PV = nRT

16. Ideal Gas Equation Use the ideal gas equation whenever the problem gives you mass or moles, or asks for a mass or a number of moles. PV = nRT

17. Ideal gas equation problem: Some ammonia gas (NH 3 ) is contained in a 2.50 L flask at a temperature of 20.0 C. If there are 0.0931 moles of the gas, what is its pressure?

18. Solution PV = nRT P = (nRT)/V = (0.0931 mol ) P = 0.896 atm 2.50 L )(293 K ) (0.0821 L atm mol K

19. Here’s another one Find the volume of 1.00 mole of nitrogen gas (N 2 ) at 0.0 C and 1.00 atm of pressure.

20. Solution V = (1.00 mol ) V = (1.00 mol ) 1.00 atm )(273 K ) (0.0821 L atm mol K PV = nRT V = (nRT)/P V = 22.4 L

21. Ideal gas equation problem: How many grams of sulfur trioxide are in an 855 mL container at a pressure of 1585 torr and a temperature of 434 C? The answer is 2.46 g SO 3

22. The Ideal Gas Equation can be used to derive the Combined Gas Law

23. The Combined Gas Law Start with the ideal gas equation: PV = nRT

24. The Combined Gas Law P 1 V 1 = nRT 1 P 2 V 2 = nRT 2 and Suppose the volume, pressure and temperature change to give a new pressure, volume and temperature.

25. The Combined Gas Law Now, solve for what doesn’t change, the constants n and R: P 1 V 1 = nRT 1 P 2 V 2 = nRT 2 and

26. The Combined Gas Law P1V1P1V1P1V1P1V1 = nR T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2 Now, solve for what doesn’t change, the constants n and R: and

27. The Combined Gas Law Since both are equal to nR, we can make a new equation. and P1V1P1V1P1V1P1V1 = nR T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

28. The Combined Gas Law Since both are equal to nR, we can make a new equation. = P1V1P1V1P1V1P1V1 T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

29. The Combined Gas Law This is the Combined Gas Law = P1V1P1V1P1V1P1V1 T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

30. The Combined Gas Law It can be derived from the laws of Boyle, Amonton and Charles, or the Ideal Gas Equation the Ideal Gas Equation = P1V1P1V1P1V1P1V1 T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

31. The Ideal Gas Equation and Density

32. Density calculations And an equation for “moles”: Start with the equation for density: D = m V n = mM Where m = mass and M = molar mass

33. Density calculations into the ideal gas equation … PV = nRT Now substitute n = mM PV = mRT M and get and get

34. Density calculations to get to get Now rearrange PV = mRTM P = mRT VMVMVMVM

35. Density calculations Recall that Recall that D = mV P = DRT M P = mRT VMVMVMVM

36. Density calculations Solving for density, Solving for density, P = DRT M becomes: becomes: D =D =D =D = PMPMPMPMRT

37. Density calculations The density of a gas depends on the molar mass and the pressure and temperature. D = PMPMPMPMRT

38. Density Problem (a) at STP (b) at a pressure of 695 torr and a temperature of 40.0 C. 1. Determine the density of nitrogen, N 2, gas The answers are 1.25 g/L, and 0.996 g/L. and 0.996 g/L.

39. 2. Determine the molar mass of a gas which has a density of 8.53 g/L at a pressure of 2.50 atm and a temperature of 500.0 K? Another Problem The answer is 140. g/mol