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# Dalton’s Law

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].

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

## Formula

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

Where

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.

Solution:

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.

Solution:

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

References