1. Dalton’s Law of Partial Pressure Recall the Ideal Gas Law PV = nRT This is true for all gases, except at low T and high P, conditions under which gases liquefy. Is PV = nRT true for (non-reactive) mixtures of gases? Yes.

2. Consider air. What are the constituents of air and their % by volume? N 2 (g) @ 78% (v/v) O 2 (g) @ 21% (v/v) Ar(g) @ 1 % (v/v) + other gases such as CO 2... What is % of each gas in air by mole and by (partial) pressure?

3. Partial Pressures of Component Gases in Air Gas% (v/v)mol % mol fraction PP(atm) PP(kPa) N 2 78 780.78 0.78 79 O 2 21 210.21 0.21 21 Ar 1 1 0.01 0.01 1.0 Note: PP = partial pressure mole fraction of gas 1 = Χ 1. (Χ = Greek letter chi = mol fraction)

4. Aside: For SCUBA divers Every 10 m below the surface of water = 1 atm of additional pressure. At ca. 30 m (or about 99 feet) below the surface, what will be the total pressure, P T, on a diver? P T = 4 atm = 1 atm (due to air) + 3 atm (due to water) What is the PP O 2 at a depth of 30 m? PP O 2 = (0.21) * 4 atm = 0.84 atm

6. So what’s Dalton’s Law of Partial Pressure (PP)? P T = P 1 + P 2 + P 3 +... where P 1 is the PP of gas 1; etc andP 1 = Χ 1 *P T P 2 = Χ 2 *P T, etc

7. sample problem #1 The air contains 0.03% CO 2 (v/v). Calculate: a)the PP CO 2 on a day when the P atm = 98 kPa; b)The mole fraction of CO 2 in air. Solution: a) PP CO 2 = 0.03/100 * 98 kPa = 0.03 kPa. b) Χ CO2 = 0.03/100 * 1 mol = 3 x 10 -4.

8. sample problem #2 A gas contains a mixture of 72.3% (v/v) methane, CH 4 and 27.7% ethane, C 2 H 6. If the PP CH 4 is 250 kPa, calculate the P T and the PP C 2 H 6. Solution: PP CH 4 = (72.3/100) * P T = 250 kPa. P T = 250/0.723 = 346 kPa P C 2 H 6 = 346 – 250 = 96 kPa

9. Application to Mountaineering... Consider Crescent’s Outreach trip to Tanzania Visit Amani Home go on safari climb Kilimanjaro

10. The hike up Kibo isn’t a technical climb—no ropes, crampons, etc required. So why is it so difficult? The air is “thin” up there (alt > 19,000 feet) Final assault from Kibo Hut. Air is still 21% O 2 at these altitudes.

11. But P atm = 350 mmHg or 0.46 atm or 47 kPa By Dalton’s Law of PP, PP of O 2 = 21% of 47 kPa = 9.8 kPa cf. PP of O 2 at sea level is 21 kPa. Ouch!!!

12. homework p 557LC 7 – 12;