12.2 DALTON’S LAW OF
PARTIAL PRESSURES
CONTEXT
We have looked at how P, T, V are connected
 We know the ideal gas law: PV=nRT
 We haven’t really discussed mixtures of gases
 If there is a mixture, how much of the total
pressure is due to each gas?
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In a mixture of non-reacting gases
Each gas particle behaves independently
 Particles spread themselves out to occupy the space
available
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DALTON’S LAW OF PARTIAL
PRESSURES
The total pressure of a mixture of non-reacting gases
is equal to the sum of the partial pressures of the
individual gases (the sum of the pressures that each
gas would exert if it were alone).
Ptot = P1 + P2 + P3 + …
PARTIAL PRESSURES OF ATMOSPHERIC
GASES
Air consists of a mixture of gases:
N2 = 78.08% O2 = 20.95%
Ar = 0.93 % CO2 = 0.03%
If Standard atmospheric pressure is 101.325 kPa,
determine the partial pressure of each gas.
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P N2 = 79.11 kPa
P O2 = 21.23 kPa
P Ar = 0.94 kPa
P CO2 = 0.03 kPa
PT = PN2 + PO2 + Par + PC02
PT ≅ 101.325 kPa
GAS COLLECTION OVER WATER
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Gases can be collected by the
downward displacement of water
The gas collected is a mixture of
the desired gas and water.
This partial pressure of water
vapour will be a component of the
total pressure
To find the partial pressure of the
desired gas, the vapour pressure of
water must be subtracted.
The partial pressure of water
vapour depends on the
temperature of the water
GAS COLLECTION OVER WATER
Hydrogen gas from the reaction of Zinc with hydrochloric
acid was collected by the displacement of water.
Atmospheric conditions were 100.5 kPa and 20˚C.
Determine the partial pressure of hydrogen gas.
Ptot = Patm = PH2 + PH2O
PH2 = Patm - PH2O
= 100.5 kPa - 2.3 kPa
= 98.2 kPa
EXAMPLE

Oxygen gas is collected over water at 297 K and an
atmospheric pressure of 1.0 atm. The volume of gas is
128 mL. Calculate the mass of O2 obtained.
Hint: Water vapour at 297 K is 3 kPa.
T = 297 K
PT = 1.0 atm
= 101.325 kPa
PH2O = 3 kPa
V = 128 mL = 0.128 L
m=?
1. PT = PO2 + PH2O
PO2 = 101.325 kPa – 3 kPa
PO2 = 98.325 kPa
2. PV =nRT
3. m = n x M
n = PV / RT
m = (5.10 x 10-3)(32.00)
n = (98.325)(0.128)/(8.314)(297)
m = 0.16 g
-3
n = 5.10 x 10 mol
HOMEWORK
P. 460 # 22 -25
 Worksheet
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