Dalton’s Law of Partial
Pressures
(This is our LAST law!!)
Background
• John Dalton studied the behavior of gas mixtures
– Each component of a gas mixture behaves independently
of the others.
• Example: A given amount of oxygen
exerts the same amount of pressure
whether it is alone or in the presence
of other gasses.
– Deep-sea divers use a helium-oxygen
mixture instead of a nitrogen-oxygen
(air) mixture so they don’t get “the
bends” as they surface.
– Nitrogen and helium behave differently
under pressure.
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Dalton’s Law
• For a mixture of gasses in a container, the total
pressure exerted is the sum of the partial
pressures of the gasses present.
• The Partial pressure of a gas is the pressure
that the gas would exert if it were alone in the
container.
Ptotal = P1 + P2 + P3
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Merging Dalton’s + Ideal Gas laws
P1 = n1RT ; P2 = n2RT ; P1 = n3RT
V
V
V
Ptotal = P1 + P2 + P3 = n1RT + n2RT + n3RT
V
V
V
= n1 (RT ÷ V) + n2(RT ÷ V) + n3(RT ÷ V)
= (n1 + n2 + n3)(RT ÷ V)
Ptotal = ntotal (RT ÷ V)
Where ntotal = the sum of the numbers of
moles of the gasses in the mixture
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SO…
• For mixtures of ideal gasses, it is the total
number of moles that is important, not the
identity of the gases present.
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Sample Problem #1
Mixtures of helium and oxygen are used in the air tanks of
underwater divers for deep dives. For a particular dive, 12 L of O2
at 25oC and 1.0 atm and 46 L of He at 25oC and 1.0 atm were both
pumped into a 5.0-L tank.
A)
B)
Calculate the partial pressure of each gas
Calculate the total pressure in the tank at 25oC (298 K).
PV = nRT
*Because the partial pressure of each gas depends on the total moles
present, we must first calculate n for each gas using the ideal gas law.
n = PV ÷ RT
nO2 = (1.0 x 12) ÷ (0.08206 x 298) = 0.49 mol
nHe = (1.0 x 46) ÷ (0.08206 x 298) = 1.9 mol
*The tank has a volume of 5.0-L and temperature of 298 K, so we can
figure out the partial pressure of each gas
P = nRT ÷ V
PO2 = (0.49 x 0.08206 x 298) ÷ 5.0 = 2.4 atm
PHe = (1.9 x 0.08206 x 298) ÷ 5.0 = 9.3 atm
Determine the total pressure:
Ptotal = PO2 + PHe = 2.4 atm + 9.3 atm = 11.7 atm
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Sample Problem #2
A sample of solid potassium chlorate, KClO3, was heated
in a test tube and decomposed according to the
reaction:
2KClO3 (s)  2KCl (s) + 3O2 (g)
The oxygen produced was collected by displacement of
water at 22oC. The resulting mixture of O2 and H2O
vapor had a pressure of 754 torr and a volume of 0.650L. Calculate the partial pressure of O2 in the gass
collected and the number of moles of O2 present. The
vapor pressure of water at 22oC is 21 torr.
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Sample Problem #2 Solution
We know:
• Ptotal = 754 torr
• PH2O = 21 torr
So we can determine:
Ptotal = PH2O + PO2 OR Ptotal - PH2O = PO2
SO… 754 torr – 21 torr = 733 torr
PO2 = 733 torr
Next, we solve the ideal gas law for nO2
Convert pressure from torr to atm:
nO2 = (PO2 V) ÷ (RT)
733 torr x (1 atm/760 torr) = 0.964 atm
Then solve:
P = 0.964 atm
V = 0.650 L
T = 22oC = 22 + 273 = 295 K
R = 0.08206 L atm/K mol
nO2 = (0.964 x 0.650)
(0.08206 x 295)
nO2 = 0.0259 mol O2
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Dalton’s Law Practice Problems
1.
If a gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in 1.04-L
container at 25oC, what will be the partial pressure of each gas and the
total pressure in the container?
2.
How many moles of helium gas would be required to fill a 2.41-L
container to a pressure of 759 mm Hg at 25oC?
3.
A sample of oxygen gas (O2) is saturated with water vapor (H2O) at 27oC.
The total pressure of the mixture is 772 torr, and the vapor pressure of
water is 26.7 torr at 27oC. What is the partial pressure of the oxygen
gas?
4.
Suppose a gaseous mixture of 1.15 g helium and 2.91 g argon is placed in
a 5.25-L container at 273oC. What pressure would exist in the container?
5.
A tank contains a mixture of 3.0 mol of N2, 2.0 mol of O2, and 1.0 mol of
CO2 at 25oC and a total pressure of 10.0 atm. Calculate the partial
pressure (in torr) of each gas in the mixture.
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