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×11th May 2019 @ 6 min read

Dalton's law is also known as the law of partial pressure or Gibbs-Dalton law (rarely). The law describes the relationship between the total pressure of a mixture of non-reacting ideal gases and the partial pressures of each individual component. Dalton's law is valid for ideal gases. The law is similar to Amagat's law of additive volumes.

In 1801 English scientist John Dalton discovered the law and published it in 1802. He is also known for its atomic theory.

The law states the total pressure exerted by a mixture of non-reacting ideal gases equals the sum of the partial pressure of each individual gas in the mixture.

where *P*_{mix} is the total pressure of the mixture and *P*_{1}, *P*_{2}, *P*_{3} … *P _{k}* are the partial pressures of individual components.

Consider a mixture of two gases: A and B. Let *P*_{mix} be the total pressure of gas A and gas B. The partial pressure of gas A and gas B is *P*_{A} and *P*_{B} respectively. As Dalton's law states the total pressure Pmix of the mixture is the addition of partial pressures of individual gases, we have the equation below.

Note: Partial pressure is a hypothetical pressure which is defined as a pressure exerted by that individual component if it alone occupies the same entire volume as that of the mixture at the same temperature.

As we can observe from the above figure, when gas A and gas B are mixed, the total pressure added up. If you have noticed, the volume *V* and the temperature *T* is the same for each gas and of the mixture; only pressures *P*_{A}, *P*_{B}, and *P*_{mix} and the number of moles (*n*_{A} , *n*_{A}, and *n*_{mix}) are added. This is very important because Dalton's law holds only at constant volume and temperature.

The above equation can be generalised as described below.

Consider a mixture of ideal gases of *k* components occupies volume *V* and at temperature *T*.

Let *n*_{1}, *n*_{2}, *n*_{3} … *n _{k}* be the moles of individual components.

The pressure of the ideal gas mixture at temperature *T* and volume *V* be *P*_{mix}(*T*, *V*, *n*_{1}, *n*_{2}, *n*_{3} … *n _{k}*).

And the partial pressures of individual components at the same temperature *T* and volume *V* be *P*_{1}(*T*, *V*, *n*_{1}), *P*_{2}(*T*, *V*, *n*_{2}), *P*_{3}(*T*, *V*, *n*_{3}) … *P _{k}*(

The above equation can be expressed in the mole fractions *x*_{1}, *x*_{2}, *x*_{3} … *x _{k}*.

Using the ideal gas equation,

Volume (*V*) and temperature (*T*) are the same for the pressure of an ideal gas mixture (*P*_{mix}) and partial pressures (*P _{i}*).

From the ideal gas eqaution,

Taking the ratio of the above equations,

where *x _{i}* is the mole fraction of the

This is the mole fraction version of Dalton's law.

Dalton's law is applicable only for ideal gases. The law holds good for real gases at low pressure, but at high pressure, it deviates significantly. The mixture of gases is non-reactive in nature. It is also assumed that the interaction among the molecules of each individual gas is the same as the molecules in the mixture.

Air consists of oxygen 20 %, nitrogen 79 %, water vapour 0.99 %, and carbon dioxide 0.01 %. The pressure of the air is 101.325 kPa. Calculate the partial pressure of each component present in the air using Dalton's law.

The total pressure of air, *P*_{mix} = 101.325 kPa = 101 325 Pa.

The mole fractions of oxygen (*x*_{O2}), nitrogen (*x*_{N2}), water vapour (*x*_{H2O}), and carbon dioxide (*x*_{CO2}) are 0.200, 0.790, 0.099, and 0.001.

The total pressure of air, *P*_{mix} = 101.325 kPa = 101 325 Pa.

The mole fractions of oxygen (*x*_{O2}), nitrogen (*x*_{N2}), water vapour (*x*_{H2O}), and carbon dioxide (*x*_{CO2}) are 0.200, 0.790, 0.099, and 0.001.

From Dalton's law of partial pressure,

Three gases hydrogen, oxygen, and methane are mixed in a container. The partial pressures of hydrogen, oxygen, and argon are 0.20bar, 0.32bar,and 0.01bar. Calculate the total pressure of the mixture.

From Dalton's law,

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Thanks for your response!

Mike

25th Jan 2021

25th Jan 2021

Thank you for this article. In Example one, the solution calculates N2 twice instead of H20.

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