1. Gas Laws Day 2

2. Gas Law Foldable Fold the left and right to the middle. Cut along solid lines (but only to the crack!)

3. Daltons Law of Partial Pressure The pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. P Total = P 1 + P 2 + P 3 ….

4. Example A balloon is filled with air (O 2, CO 2, & N 2 ) at a pressure of 1.3 atm. If P O 2 = 0.4 atm and P CO 2 = 0.3 atm, what is the partial pressure of the nitrogen gas?

5. P Total = P 1 + P 2 + P 3 …. P total = P O 2 + P CO 2 + P N 2 1.3 atm = 0.4 atm + 0.3 atm + P N 2 P N 2 = 0.6 atm

6. The Combined Gas Law Expresses a relationship between pressure, volume, and temperature (and moles) of a fixed amount of gas. It takes all three gas laws: (Boyles, Charless, & Avogadros) and combines them form one usable equation.

7. The Combined Gas Law P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

8. Example 1 A gas is cooled from 45˚C to 20˚C. The pressure changes from 103 kPa to 101.3 kPa as the volumes settles to 16.0 L. What was the initial volume? P 1 = 103 kPa V 1 = ? T 1 = 45 o C + 273 = 318 K P 2 = 101.3 kPa V 2 = 16.0 L T 2 = 20 o C + 273 = 293 K

9. (103 kPa)(V 1 ) = (101.3 kPa)(16.0 L) (n 1 ) (318 K) (n 2 ) (293 K) V 1 = 17.1 L P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

10. Example 2 (on back of foldable) A 1.5 mole sample of methane was originally 0.5 L at 25˚C at 1.1 atm. If we decreased the volume of the container to 0.25 L, increased the pressure to 2.0 atm and added 2.5 moles, what would be the new temperature in ˚C? P 1 = 1.1 atm V 1 = 0.5 L T 1 = 25 + 273= 298 K n 1 = 1.5 moles P 2 = 2.0 atm V 2 = 0.25 L T 2 = ? n 2 = 1.5+2.5 = 4 moles

11. T 2 = 101.59 K – 273 = -171.41 ˚C P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 *** easier if you solve for T 2 first THEN plug in the #s *** P 1 V 1 n 2 T 2 = P 2 V 2 n 1 T 1 P 1 V 1 n 2 P 1 V 1 n 2 = (2.0 atm)(0.25 L)(1.5 mol)(298K) (1.1 atm)(0.5 L)(4.0 mol)

12. Boyles Law: Pressure vs. Volume At a constant temperature, the volume of a fixed mass of gas varies inversely with the pressure P 1 V 1 = P 2 V 2 if T 1 = T 2

13. Inverse Indirect

14. Pressure and Volume with constant Temperature Changing the volume of the container changes the amount of space between the particles. The less space, the more the particles collide with each other and the walls.

15. Uses of Boyles Law Testing materials stability and ability to maintain their shape under force. Compressing gases for use in cooking cylinders, SCUBA tanks, and shaving cream. Used to describe density relationships between gases.

16. Example If a gas expands from a volume of 5 L to 25 L at an initial pressure of 3.5 atm, what will be its new pressure? P 1 = 3.5 atm V 1 = 5 L P 2 = ? V 2 = 25 L

17. P 1 V 1 = P 2 V 2 V 2 V 2 P 2 = (3.5 atm)(5 L) 25 L P 2 = 0.7 atm P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

18. Charles Law: Volume vs Temperature At constant pressure, the volume of a fixed mass of gas varies directly with the Kelvin temperature V 1 = V 2 if P 1 =P 2 T 1 T 2

19. Direct

20. Charles Law Graph Temp decreases Vol decreases. Temperature (K)Volume (mL) 5461092 373746 283566 274548 273546 272544 200400 50100 00

21. Temperature and Volume with constant Pressure Changing the temperature but requiring the pressure to stay the same causes the volume to increase.

22. Uses of Charles Law Describing the properties of gases, liquids, and solids at extremely low temperatures. Hot air ballooning Used to describe density relationships of gases.

23. Example If 22.4 L of oxygen is heated from 23˚C to 50 ˚C, what is its new volume? V 1 = 22.4 L T 1 = 296 K V 2 = ? T 2 = 323 K

24. V 1 = V 2 T 1 T 2 V 2 = V 1 T 2 T 1 V 2 = (22.4 L)(296 K) = 20.5 L (323 K) P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

25. Absolute Zero Temperature at which all molecular motion stops. It is defined by 0 K or -273 C. Scientists used Charles Law to extrapolate the temperature of absolute zero.

26. Avogadros Law: Volume vs. Moles At a constant temperature & pressure, the volume of a gas varies directly with the moles V 1 = V 2 n 1 n 2

27. Avogadros Law As the number of moles increases, the volume expands to make room for the additional gas

28. Example If 4.65 L of CO 2 increases from 0.8 moles to 3.75 moles, what is the new volume of the gas? V 1 = 4.65 L n 1 = 0.8 moles V 2 = ? L n 2 = 3.75 moles P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 V 1 = 4.65 L

29. V 2 = V 1 n 2 n 1 V 2 = (4.65 L)(3.75 mol) = 21.79 L (0.8 mol) (20 L) V 1 = V 2 n 1 n 2

30. Whew! My brain hurts!!!