Chemistry Learner

It's all about Chemistry

Home / Physical Chemistry / Dalton’s Law

Dalton’s Law

Table Of Contents

Dalton’s law of partial pressure states that in a mixture of two or more non-reacting gases, the sum of the partial pressures of each gas is equal to the overall pressure of the gas mixture. A gas’s partial pressure is the pressure it would exert on the walls if it were the only gas in a container [1-4].

Dalton’s Law

English physicist and chemist John Dalton published the law in 1802.


For Dalton’s law to hold good, the following assumptions are made: [1-6]

  • The gas molecules occupy an insignificant volume compared to the container’s volume.
  • The gases are ideal, meaning there is no attraction among the molecules.
  • The pressure is not excessively high. Otherwise, the volume occupied by the molecules becomes significant compared to the free space between them, resulting in a deviation from Dalton’s law.

Based on these assumptions, one can calculate the total pressure of the gas mixture in terms of the partial pressure of the gases. Suppose a mixture contains n gases whose partial pressures are p1, p2, p3, …, pn. Then, the total pressure p of the mixture is given by

p = p1 + p2 + p3 + … + p4

p = Σpi

Partial Pressure from Mole Fraction

Dalton’s law can also be expressed in terms of mole fraction. The mole fraction X is the fraction of the gas present in the mixture. It is evaluated by taking the ratio of the number of moles (ni) of a given gas and the total number of moles (Σni) of all the gases.

 Xi = ni /(Σni)

Then, the partial pressure of the i-th gas is

pi = Xi p

Partial Pressure from Ideal Gas Equation

The partial pressure can be calculated from the ideal gas equation. Suppose ni moles of the i-th gas are present in the mixture. The partial pressure is given by

pi = niRT/V


T: Absolute temperature

V: Volume of the gas

Solved Problems

Problem 1: A mixture of oxygen and hydrogen exerts a total pressure of 2.3 atm on the walls of the container. If the partial pressure of hydrogen is 1.2 atm, find the mole fraction of oxygen in the mixture.


Given,    phydrogen = 1.2 atm             p = 2.3 atm        

Therefore, the partial pressure of oxygen is

poxygen = p – phydrogen = 2.3 atm – 1.2 atm = 1.1 atm

The mole fraction of oxygen can be found using the following formula.

poxtgen = Xoxygen p

=> Xoxygen = poxygen/p

=> Xoxygen = 1.1 atm/2.3 atm = 0.48

Problem 2: 14 L of oxygen gas at 2.4 atm and 18 L of nitrogen at 2.2 atm are added to a 12 L container at 300 K. Find the partial pressure of oxygen and nitrogen and then find the total pressure.


Given    Voxygen = 14 L       poxygen1 = 2.4 atm              Vnitrogen = 18 L     pnitrogen1 = 2.2 atm             T = 300 K

V = 12 L               R = 0.082 L-atm · mol-1 · K-1

The mole fraction of oxygen is

noxygen = (poxygen Voxygen)/(RT) = (2.4 atm · 14 L)/(0.082 L-atm · mol-1 · K-1 · 300 K) = 1.36 mol

nnitrogen = (pnitrogen Vnitrogen)/(RT) = (2.2 atm · 18 L)/(0.082 L-atm · mol-1 · K-1 · 300 K) = 1.61 mol

Therefore, the total number of moles is

n = noxygen + nnitrogen = 1.36 + 1.61 = 2.97 mol

The total pressure is

P = nRT/V = 2.97 mol · 0.082 L-atm · mol-1 · K-1 · 300 K/12 L = 6.09 atm

Therefore, the partial pressures are

poxygen = (noxygen/n) · P = (1.36 mol/2.97 mol) · 6.09 atm = 2.78 atm

pnitrogen = (nnitrogen/n) · P = (1.61 mol/2.97 mol) · 6.09 atm = 3.3 atm

Leave a Reply

Your email address will not be published.