2. Dalton’s Law of Partial Pressure

3.  John Dalton responsible for atomic theory responsible for atomic theory also studied gas mixtures also studied gas mixtures  the P of gas mixture is the sum of the individual pressures of each gas alone  the P that each gas exerts in the mixture is independent of the P that are exerted by other gases

4. Dalton’s Law of Partial Pressure  The total pressure in a container is the sum of the partial pressures of all the gases in the container.  P T = P 1 + P 2 + P 3 + …P n Where P 1, P 2, and P n are the partial pressures of the gases involved  Partial Pressure- P of each gas in mixture

5. Why?  the particles of each gas in a mixture have an equal chance to hit the walls  so each gas exerts P independent of that exerted by other gases  total P is result of the total # of collisions per unit of wall area

6.  Ex. A mixture of oxygen, hydrogen and nitrogen gases exerts a total pressure of 278 kPa. If the partial pressures of the oxygen and the hydrogen are 112 kPa and 101 kPa respectively, what would be the partial pressure exerted by the nitrogen. P T = 278 kPa P O = 112 kPa P H = 101 kPa P N = ? P T = P O + P H + P N 278 kPa = 112 kPa + 101 kPa + P N P N = 65.0 kPa

7. Vapour Pressure Defined  Vapour pressure is the pressure exerted by a vapour. Ex. the H 2 O(g) in a sealed container. Yet, molecules both leave and join the surface, so vapour pressure also pushes molecules up. Eventually the air above the water is filled with vapour pushing down. As temperature , more molecules fill the air, and vapour pressure . To measure vapour pressure we can heat a sample of liquid on top of a column of Hg and see the pressure it exerts at different °C.

8. Gas Collected Over Water

9. Production of oxygen by thermal decomposition of KCIO3.

10. Water Displacement  gas produced is less dense than water so it replaces the water in the bottle  gas collected is not pure because it contains vapor from the water P T = P gas + P water equal to atmospheric pressure set for a certain T

11. Example  Oxygen gas from decomposition of KClO 3 was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 mm Hg and 20.0 ° C respectively. If the partial pressure of water vapor is 17.5 mm Hg at 20.0 ° C. What was the partial pressure of oxygen collected? P T = P DG + P H2O 731.0 mm Hg = P O2 + 17.5 mm Hg P O2 = 713.5 mm Hg

12. Example  Find the partial pressure by 2 gases (A and B) mixed if the overall pressure is 790 mmHg. The percent by volume is A: 20% and B: 80%.  P T = P A + P B = 790 mmHg  A: 0.20 x 790 = 158 mmHg  B: 0.80 x 790 = 632 mmHg

13. P T = P DG + P H2O P T = 113.0 kPa P H2O = 1.6 kPa P DG = ? P DG = 113.0 kPa – 1.6 kPa P DG P DG = 111 kPa Ex. A sample of hydrogen gas is collected over water at 14.0 o C. The pressure of the resultant mixture is 113.0 kPa. What is the pressure that is exerted by the dry hydrogen alone?

14. If the above gas is 2.3 L, what is the new volume at standard pressure? P 1 = 111.4 kPa V 1 = 2.3 L P 2 = 101.325 kPa V 2 = ? P 1 V 1 = P 2 V 2 V 2 = 2.53 L