Dalton’s Law of Partial Pressure
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Transcript of Dalton’s Law of Partial Pressure
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 fraction PP(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 #1The air contains 0.03% CO2 (v/v). Calculate:
a) the PPCO2 on a day when the Patm = 98 kPa;
b) 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 #2A 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;