# Dalton’s Law of Partial Pressure – Definition and Examples Dalton’s law of partial pressure states that the total pressure of a mixture of gases is the sum of their partial pressures.

Dalton’s law of partial pressure is an ideal gas law that states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. English scientist John Dalton observed the behavior of gases in 1801 and published the gas law in 1802. While Dalton’s law of partial pressures describes ideal gases, real gases follow the law under most conditions.

### Dalton’s Law Formula

The formula for Dalton’s law states that the pressure of a gas mixture is the sum of the partial pressures of its component gases:

PT = P1 + P2 + P3 + …

Here, PT is the total pressure of mixture and P1, P2, etc. are the partial pressures of the individual gases.

### Solving for Partial Pressure or Mole Fraction

Combining Dalton’s law with the ideal gas law makes it possible to solve for the partial pressure, mole fraction, or number of moles of a component of the gas mixture.

Pi = PT ( ni / nT )

Here, Pi is the partial pressure of an individual gas, PT is the total pressure of the mixture, ni is the number of moles of the gas, and nT is the total number of moles of all gases in the mixture.

You can solve for mole fraction, the pressure of a component or the total pressure, the volume of a component or the total volume, and the number of moles of a component and the total number of moles of gas:

Xi = Pi / PT = Vi / VT = ni / nT

Here, Xi is the mole fraction of a component (i) of a gas mixture, P is pressure, V is volume, and n is number of moles.

### Assumptions in Dalton’s Law of Partial Pressure

Dalton’s law assumes gases behave as ideal gases:

• The partial pressure of a gas is the pressure exerted by an individual component in a mixture of gases.
• Gas molecules follow the kinetic theory of gases. In other words, they behave as point masses with negligible volume that are widely separated from each other, are neither attracted nor repelled by one another, and have elastic collisions with each other and container walls.

Dalton’s law predicts gas behavior quite well, but it real gases deviate from the law as pressure increases. At high pressure, there is less space between gas molecules and interactions between them becomes more significant.

### Dalton’s Law Examples and Worked Problems

Here are examples showing how you use Dalton’s law of partial pressure:

#### Calculate Partial Pressure Using Dalton’s Law

For example, calculate the partial pressure of oxygen gas in a mixture of nitrogen, carbon dioxide, and oxygen. The mixtures has a total pressure of 150 kPa and the partial pressures of nitrogen and carbon dioxide are 100 kPa and 24 kPa, respectively.

This is a straightforward application of Dalton’s law:

PT = P1 + P2 + P3
Ptotal = Pnitrogen + Pcarbon dioxide + Poxygen
150 kPa = 100 kPa + 24 kPa + Poxygen
Poxygen = 150 kPa – 100 kPa – 24kPa
Poxygen = 26 kPa

Always check your work. Add up the partial pressures and make sure you get the proper total.

#### Calculate Mole Fraction Using Dalton’s Law

For example, find the mole fraction of oxygen in a mixture of hydrogen and oxygen gas. The total pressure of the mixture is 1.5 atm and the partial pressure of hydrogen is 1 atm.

Start with Dalton’s law and find the partial pressure of oxygen gas.

PT = P1 + P2
Ptotal = Phydrogen + Poxygen
1.5 atm = 1 atm + Poxygen
Poxygen = 1.5 atm – 1 atm
Poxygen = 0.5 atm

Next, apply the formula for mole fraction.

Xi = Pi / PT
Xoxygen = Poxygen/Ptotal
Xoxygen = 0.5/1.5 = 0.33

Note the mole fraction is a pure number. It does not matter what pressure units you use as long as they are the same in both the numerator and denominator of the fraction.

#### Combining the Ideal Gas Law and Dalton’s Law

Many Dalton’s law problems require some calculations using the ideal gas law. For example, find the partial pressures and total pressure of a mixture of nitrogen and oxygen gas. The mixture forms by combining a container of 24.0 L of nitrogen (N2) gas at 2 atm and a container of 12.0 L of oxygen (O2) gas at 2 atm. The container has a volume of 10.0 L. Both gases are at an absolute temperature of 273 K.

The problem gives the pressure (P), volume (V), and temperature (T) for the gases before forming the mixture, so apply the ideal gas law to find the number of moles (n) of each gas.

PV = nRT

Rearrange the ideal gas law and solve for the number of moles. Be sure you use the appropriate units for the ideal gas constant.

n = PV/RT

nN2 = (2 atm)(24.0 L)/(0.08206 atm·L/mol·K)(273 K) = 2.14 mol N2

nO2 = (2 atm)(12.0 L)/(0.08206 atm·L/mol·K)(273 K) = 1.07 mol O2

Next, find the partial pressures of each gas after they are mixed. The volume of the mixture is different from the starting volumes of the gases, so you know the pressure of the mixture is different from the initial pressures. This time, use the ideal gas law, but solve for pressure.

PV = nRT
P = nRT/V

PN2 = (2.14 mol) (0.08206 atm·L/mol·K)(273 K) / 10 L = 4.79 atm

PO2 = (1.07 mol) (0.08206 atm·L/mol·K)(273 K) / 10 L = 2.40 atm

The partial pressures of each gas in the mixture is higher than their initial pressures. This makes sense, since pressure is inversely proportional to volume.

Now, apply Dalton’s law and solve for the total pressure of the mixture.

PT = P1 + P2
PT = PN2 + PO2 = 4.79 atm + 2.40 atm = 7.19 atm

Since Dalton’s law and the ideal gas law both make the same assumptions about gas behavior, you get the same answer just plugging in the sum of the number of moles of gas into the ideal gas law.

PT = (nN2 + nO2)RT/V
PT = (2.14 mol + 1.07 mol) (0.08206 atm·L/mol·K)(273 K) / 10 L = 7.19 atm

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