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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.
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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.
Consider air. What are the constituents of air and their % by volume? N2(g) @ 78% (v/v) O2(g) @ 21% (v/v) Ar(g) @ 1 % (v/v) + other gases such as CO2 . . . What is % of each gas in air by mole and by (partial) pressure?
Partial Pressures of Component Gases in Air Gas% (v/v)mol %mol fractionPP(atm)PP(kPa) N2 78 78 0.78 0.78 79 O2 21 21 0.21 0.21 21 Ar 1 1 0.01 0.01 1.0 Note: PP = partial pressure mole fraction of gas 1 = Χ1. (Χ = Greek letter chi = mol fraction)
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, PT, on a diver? PT = 4 atm = 1 atm (due to air) + 3 atm (due to water) What is the PPO2 at a depth of 30 m? PPO2 = (0.21) * 4 atm = 0.84 atm
So what’s Dalton’s Law of Partial Pressure (PP)? PT = P1 + P2 + P3 + . . . where P1 is the PP of gas 1; etc and P1 = Χ1*PT P2 = Χ2*PT , etc
sample problem #1 The air contains 0.03% CO2 (v/v). Calculate: • the PPCO2 on a day when the Patm = 98 kPa; • The mole fraction of CO2 in air. Solution: a) PPCO2 = 0.03/100 * 98 kPa = 0.03 kPa. b) ΧCO2 = 0.03/100 * 1 mol = 3 x 10-4.
sample problem #2 A gas contains a mixture of 72.3% (v/v) methane, CH4 and 27.7% ethane, C2H6. If the PPCH4 is 250 kPa, calculate the PT and the PPC2H6. Solution: PPCH4 = (72.3/100) * PT = 250 kPa. PT = 250/0.723 = 346 kPa PC2H6 = 346 – 250 = 96 kPa
Application to Mountaineering . . . Consider Crescent’s Outreach trip to Tanzania Visit Amani Home go on safari climb Kilimanjaro
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% O2 at these altitudes.
But Patm = 350 mmHg or 0.46 atm or 47 kPa By Dalton’s Law of PP, PP of O2 = 21% of 47 kPa = 9.8 kPa cf. PP of O2 at sea level is 21 kPa. Ouch!!!
homework p 557 LC 7 – 12;