# 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 p_{1}, p_{2}, p_{3}, …, p_{n}. Then, the total pressure p of the mixture is given by

p = p_{1} + p_{2} + p_{3} + … + p_{4}

p = Σp_{i}

**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 (n_{i}) of a given gas and the total number of moles (Σn_{i}) of all the gases.

X_{i} = n_{i} /(Σn_{i})

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

p_{i} = X_{i} p

**Partial Pressure from Ideal Gas Equation**

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

p_{i} = n_{i}RT/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, p_{hydrogen} = 1.2 atm p = 2.3 atm

Therefore, the partial pressure of oxygen is

p_{oxygen} = p – p_{hydrogen} = 2.3 atm – 1.2 atm = 1.1 atm

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

p_{oxtgen} = X_{oxygen} p

=> X_{oxygen} = p_{oxygen}/p

=> X_{oxygen} = 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 V_{oxygen} = 14 L p_{oxygen}^{1} = 2.4 atm V_{nitrogen} = 18 L p_{nitrogen}^{1} = 2.2 atm T = 300 K

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

The mole fraction of oxygen is

n_{oxygen} = (p_{oxygen} V_{oxygen})/(RT) = (2.4 atm · 14 L)/(0.082 L-atm · mol^{-1} · K^{-1} · 300 K) = 1.36 mol

n_{nitrogen} = (p_{nitrogen} V_{nitrogen})/(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 = n_{oxygen} + n_{nitrogen} = 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

p_{oxygen} = (n_{oxygen}/n) · P = (1.36 mol/2.97 mol) · 6.09 atm = 2.78 atm

p_{nitrogen} = (n_{nitrogen}/n) · P = (1.61 mol/2.97 mol) · 6.09 atm = 3.3 atm