Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures states that in a mixture of non-reacting gases, the total pressure exerted by the mixture is equal to the sum of the partial pressures of each gas.

Each gas in the mixture behaves independently, exerting pressure as if it alone occupied the entire volume at the given temperature.
This principle was first proposed by John Dalton in 1801.

1.0Mathematical Expression of Dalton’s Law

For a mixture containing gases A, B, and C:

$P_{\text{total }}=P_A+P_B+P_C$

Where:

• $(P_{\text{total}})$ = total pressure of the mixture
• $P_A,P_B,P_C$ = partial pressures of individual gases

For any gas i in the mixture:

$P_i=\frac{n_{i}RT}{V}$

Adding for all gases:

$P_{\text{total }}=\frac{\left(n_A+n_B+n_C\right)RT}{V}$

Thus proving Dalton’s Law, which states that the total pressure equals the sum of the partial pressures.

2.0Mole Fraction and Partial Pressure Relationship

The mole fraction of a gas in a mixture is defined as:

${\chi }_i=\frac{n_i}{n_{\text{total }}}$

And the partial pressure of a gas is related to its mole fraction by:

$P_i={\chi }_i\times P_{\text{total }}$

This equation shows that the contribution of each gas to the total pressure depends directly on its mole fraction.

Example

A gas mixture contains 1 mole of helium and 4 moles of argon at a total pressure of 5 atm.

$\begin{array}{c}{X}_{\mathrm{He}}=\frac{1}{1+4}=0.2\\ P_{\mathrm{He}}=0.2\times 5=1\mathrm{atm}\end{array}$

3.0Derivation of Dalton’s Law

Consider three gases (A, B, and C) present in a container of volume V and temperature T.
According to the ideal gas equation:

$P_{A}V=n_{A}RT,P_{B}V=n_{B}RT,P_{C}V=n_{C}RT$

Adding these equations:

$\left(P_A+P_B+P_C\right)V=\left(n_A+n_B+n_C\right)RT$

Hence, $P_{\text{total }}=P_A+P_B+P_C$

4.0Applications of Dalton’s Law

Gas Mixtures

Used to calculate the total pressure of a gas mixture in laboratory and industrial processes such as gas storage or fuel gas preparation.

Collection of Gases Over Water

When gases like O₂ or H₂ are collected over water, they contain water vapour.
Hence, the total pressure includes both the gas pressure and the vapour pressure of water:

$P_{gas}=P_{total}-H_{2}O$

Respiration and Blood Gases

Dalton’s Law explains partial pressures of O₂ and CO₂ in blood and tissues, crucial for understanding gas exchange in lungs.

Industrial and Laboratory Uses

• Predicting pressure in gas cylinders
• Determining fuel gas composition
• Calculations in vapor–liquid equilibrium systems

5.0Limitations of Dalton’s Law

1. Ideal Gas Behaviour:
   The law assumes gases behave ideally, which is not true at high pressure or low temperature, where intermolecular forces become significant.
2. Non-Reactive Mixtures:
   Dalton’s Law does not apply to reactive gases, such as H₂ and O₂, that combine to form new products.
3. Neglects Intermolecular Forces:
   It assumes gases do not interact with each other, which may lead to deviations in real systems.

6.0Problem-Solving Strategy

Follow these steps to solve numerical problems based on Dalton’s Law:

1. Identify Given Data:
   Note the number of moles, temperature, volume, and total pressure.
2. Apply Ideal Gas Law:
   Use ( PV = nRT ) to find missing parameters if needed.
3. Calculate Partial Pressures:
   Use ( $P_i=X_i\times P_{\text{total}}$ ).
4. Verify Total Pressure:
   Ensure that the sum of all partial pressures equals the total pressure.

7.0Solved Examples

Example 1: Mixture of Gases

A mixture contains 2 moles of O₂, 3 moles of N₂, and 1 mole of CO₂ at a total pressure of 6 atm.
Find the partial pressure of O₂.

Solution:
Total moles = 2 + 3 + 1 = 6
Mole fraction of O₂ $=\frac{2}{6}=\frac{1}{3}$

$P_{{\text{O}}_2}={\chi }_{{\text{O}}_2}\times P_{\text{total}}=\frac{1}{3}\times 6=2\text{ atm}$

Answer: Partial pressure of O₂ = 2 atm

Example 2: Gas Collected Over Water

A gas is collected over water at 298 K.
If total pressure = 780 mmHg and vapour pressure of water = 20 mmHg, find the pressure of the dry gas.

$P_{\text{gas}}=P_{\text{total}}-P_{{\text{H}}_2\text{O}}=780-20=760\text{ mmHg}$

Answer: Dry gas pressure = 760 mmHg

Example 3: Mixture of Hydrogen and Oxygen

A 5 L container holds 2 moles of H₂ and 3 moles of O₂ at 298 K. Find total and partial pressures.

Solution:
Total moles = 1 + 2 = 3

Mole fraction of O₂ $={\text{O}}_2=\frac{1}{3}$, Mole fraction of N₂ $={\text{N}}_2=\frac{2}{3}$

$P_{{\text{O}}_2}=\frac{1}{3}\times 9=3\text{ atm}$

$P_{{\text{N}}_2}=\frac{2}{3}\times 9=6\text{ atm}$

Answer: $P_{{\text{O}}_2}=3\text{ atm, }{P}_{{\text{N}}_2}=6\text{ atm}$

Example 4: Mole Fraction and Partial Pressure

A container has 1 mole of O₂ and 2 moles of N₂ at a total pressure of 9 atm. Find the partial pressures of O₂ and N₂.

Solution:
Total moles = 1 + 2 = 3

Mole fraction of O₂ $={\text{O}}_2=\frac{1}{3}$, Mole fraction of N₂ $={\text{N}}_2=\frac{2}{3}$

$P_{{\text{O}}_2}=\frac{1}{3}\times 9=3\text{ atm}$

$P_{{\text{N}}_2}=\frac{2}{3}\times 9=6\text{ atm}$

Answer: $P_{{\text{O}}_2}=3\text{ atm, }{P}_{{\text{N}}_2}=6\text{ atm}$

Answer: $(P_{O_2}=3,\text{atm},P_{N_2}=6,\text{atm})$