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;