Dalton's Law of Partial Pressures assumes each gas in the mixture is behaving like an ideal gas.

Partial Pressure

A container of fixed volume at constant temperature holds a mixture of gas a and gas b at a total pressure of 4atm.

The total pressure in the container is proportional to the number of gas particles.

More gas particles = greater pressure.
Less gas particles = lower pressure.

If each dot represents 1 mole of gas particles, then there are 48 moles of gas particles in this container exerting a total pressure of 4atm.

Imagine the container with no particles of gas b.

Only particles of gas a are present in the same container at the same temperature.

Now the container holds only 12 moles of gas particles instead of the 48 moles of gas particles it originally contained.

Since pressure is proportional to the number of gas particles,
the pressure exerted by gas a = 12mol ÷ 48mol x 4atm = 1atm

Imagine the container with no particles of gas a.

Only particles of gas b are present in the same container at the same temperature.

Now the container holds only 36 moles of gas particles instead of the 48 moles of gas particles it originally contained.

Since pressure is proportional to the number of gas particles,
the pressure exerted by gas b = 36mol ÷ 48mol x 4atm = 3atm

Dalton's Law of Partial Pressures

The total pressure in a gas mixture is the sum of the partial pressures of each individual gas.

P_{total}

=

P_{gas a}

+

P_{gas b}

=

+

Examples

10 g of nitrogen gas and 10 g of helium gas are placed together in a 10 L container at 25^{o}C.
Calculate the partial pressure of each gas and the total pressure of the gas mixture.

Calculate the moles (n) of each gas present: n = mass ÷ molar mass

nitrogen (N_{2(g)})

helium (He_{(g)})

mass (g)

10 g

10 g

molar mass (g mol^{-1})

2 x 14 = 28

4

n = mass ÷ molar mass

10 ÷ 28 = 0.4 mol

10 ÷ 4 = 2.5 mol

Calculate the total moles of gas present = 0.4 + 2.5 = 2.9 mol

Calculate the total gas pressure assuming ideal gas behaviour: PV = nRT P = n x R x T ÷ V
n = 2.9 mol
R = 8.314
T = 25^{o}C = 25 + 273 = 298 K
V = 10 L
P = 2.9 x 8.314 x 298 ÷ 10 = 718 kPa (7atm)

Partial pressure of nitrogen = n(N_{2}) ÷ n(total) x total pressure
Partial pressure of nitrogen = 0.4 ÷ 2.9 x 718 kpa = 99 kPa (0.9 atm)

Partial pressure of helium = n(He) ÷ n(total) x total pressure
Partial pressure of helium = 2.5 ÷ 2.9 x 718 = 619 kPa (6.1atm)

At 15^{o}C, 25 mL of neon at 101.3 kPa (1 atm) pressure and 75 mL of helium at 70.9 kPa (0.7 atm) pressure are both expanded into a 1 L sealed flask. Calculate the partial pressure of each gas and the total pressure of the gas mixture.

Since the temperature and moles of each gas is constant, the pressure exerted by each gas is inversely proportional to its volume (Boyle's Law).
P_{i}V_{i} = P_{f}V_{f} P_{f} = P_{i}V_{i} ÷ V_{f}

Partial pressure Neon = 101.3 kPa x 25 x 10^{-3}L ÷ 1 L = 2.5 kPa

Partial pressure Helium = 70.9 kPa x 75 x 10^{-3}L ÷ 1 L = 5.3 kPa

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