Dalton’s Law of Partial Pressure

Did you know the air we breathe from our environment contains not only oxygen but other gases too? In fact, oxygen is only 21% of the air, while the majority, around 78%, is nitrogen, and the rest are small amounts of carbon dioxide, argon, and other gases. But if the atmosphere is a mixture of so many gases, how come these gases do not mix? The answer lies in the theory of Dalton’s Law of Partial Pressure, introduced by the English chemist John Dalton in 1801. An important law of practical chemistry and our topic of discussion today. So let’s begin! 

1.0Dalton's Law Statement 

The Dalton’s Law of Partial Pressure states that, “In a combination of non-reacting gases, the overall pressure of the mixture is the same as the sum of the partial pressures of the component gases.”

The above statement is a crucial concept for understanding the working of different gases and is referred to as Dalton's Law Statement. It's an “assumption” that all gases behave separately and exert their own pressures as if each of them filled the container by itself.

What is Partial Pressure? 

Partial pressure is the pressure a specific gas in a mixture would have if it alone filled up the entire volume at the same temperature. It is a quantity of the separate contribution of the specific gas to the total pressure.

In practical terms, if a room is full of oxygen, nitrogen, and carbon dioxide, each of these gases adds a proportion of the sum of pressure you experience — that's their partial pressure.

2.0Dalton’s Law of Partial Pressure Formula

In mathematical terms, Dalton’s law of partial pressure can be expressed with the following formula: 

$P_{\text{Total }}={\sum }_{i=1}^n{p}_i$

Or, 

P_Total = P₁ + P₂ + P₃ + … + P_n 

Here, 

• P_Total = Total Pressure of the gas mixture 
• P₁, P₂, P₃, …, P_n = Partial Pressures of the individual gases in the mixture. 

Note: The above equation is applied to ideal gases only, gases that occupy negligible space and have no interactions with other gases. 

Mole Fraction and Partial Pressure: 

When studying a mixture of gases, the mole fraction of a gas is defined as the ratio of the number of moles of that specific gas to the total number of moles of all gases combined in the mixture. Mole fraction is a dimensionless quantity, and has values between 0 to 1. Mathematically, the Mole fraction is: 

$X_i=\frac{n_i}{n_{\text{total }}}$

Here, 

• X_i​ = Mole fraction of gas i
• n_i = Moles of gas i
• n_total = Total moles of all gases in the mixture

Now, according to Dalton’s law, the mole fraction can be expressed in terms of total pressure by the following relation: 

$\begin{array}{rl}& P_i=X_i\bullet P_{\text{total }}\\ & X_i=\frac{P_i}{P_{\text{total }}}\end{array}$

Hence, Partial Pressure Using Mole Fraction can be written as:  

$X_i=\frac{n_i}{n_{\text{total }}}=\frac{P_i}{P_{\text{total }}}$

Dalton’s Law and Vapour Pressure: 

When a gas is collected over water inside the container, water molecules can also vaporise and exert vapour pressure on the walls of the container. This leads to additional pressure more than the actual pressure of the dry gas. Hence, this actual pressure can be corrected using the equation: 

P_gas = P_Total – P_{Water vapour}

3.0Derivation of Dalton’s Law

Dalton’s Law of Partial Pressures can be derived using the equation for ideal gases, which is expressed as: 

PV = nRT 

Now, let’s assume a container is filled with two non-reacting ideal gases, say A and B. Let the temperature and pressure also remain constant for both gases. Hence, the ideal gas equation for gases A and B will be:  

For gas A: 

P_AV = n_ART 

$n_A=\frac{P_{A}V}{RT}$

For gas B: 

P_BV = n_BRT 

$n_B=\frac{P_{B}V}{RT}$

The total number of moles will be equal to the number of moles of individual gases: 

${\mathrm{n}}_{\text{Total }}=\frac{P_{A}V}{RT}+\frac{P_{B}V}{RT}$

Here, ${\mathrm{n}}_{\text{Total }}=\frac{P_{\text{Total }}V}{RT}$

Hence, the above equation can be rewritten as: 

$\begin{array}{rl}& \frac{P_{\text{Total }}V}{RT}=\frac{P_{A}V}{RT}+\frac{P_{B}V}{RT}\\ & \frac{P_{\text{Total }}V}{RT}=\frac{V}{RT}\left(P_A+P_B\right)\\ & P_{\text{Total }}=P_A+P_B\end{array}$

4.0Applications of Dalton’s Law 

Despite having several limitations, Dalton’s Law is applied to a large number of practical applications, some of which are: 

• Medical Application: Assists in measuring oxygen concentration in ventilators and anaesthesia supply.
• Scuba Diving: Maintains safe breathing gas mixtures under changing water pressures.
• Weather Forecasting: Employed by meteorologists to examine the atmospheric gas mix.
• Lab Gas Collection Over Water: Applied in labs to determine dry gas pressure.
• Chemical Industry: Assists in the design of reactors and in understanding gas-phase reactions.
• Aerospace & Aviation: Essential for life support systems in pressurised cabins.
• Research: Assists in the study of gas exchange in respiration and photosynthesis.

5.0Examples of Dalton’s Law

Problem 1: A mixture contains 4 g of hydrogen (H₂), 28 g of nitrogen (N₂), and 88 g of carbon dioxide (CO₂) in a 10 L vessel at 27°C. Calculate the total pressure of the mixture using Dalton’s Law. $(R=0.0821L\cdot atm/mol\cdot K)$

Solution: Firstly, convert the mass of each gas to moles: 

• $H_2=2g/mol=\frac{4}{2}=2\mathrm{mol}$
• $N_2=28g/mol=\frac{28}{28}=1\mathrm{mol}$
• $C{O}_2=44g/mol=\frac{88}{44}=2\mathrm{mol}$
• Total moles = 5 moles 

Now, using the law of ideal gas: 

P_TotalV = n_TotalRT

$\begin{array}{rl}& \begin{array}{rl}& P_{\text{Total }}\times 10=5\times 0.0821\times (273+27)\\ & P_{\text{Total }}=\frac{5\times 0.0821\times 300}{10}=\frac{123.15}{10}\end{array}\\ & P_{\text{Total }}=12.315\mathrm{atm}\end{array}$

Problem 2: A gas mixture contains 3 mol of methane (CH₄) and 7 mol of ethane (C₂H₆) at a total pressure of 800 mmHg. Find the mole fraction and partial pressure of each gas.

Solution: The total moles in the gas mixture = 3 + 7 = 10 mol 

Mole fraction for methane X_CH4 = 3/10 = 0.3 

Mole fraction for ethane X_C2H6 = 7/10 = 0.7 

Partial pressure of P_CH4 = 0.3 x 800 = 240 mmHg

Partial pressure of P_C2H6 = 0.7 x 800 = 560 mmHg

Problem 3: Hydrogen gas is produced by the reaction: 

$\mathrm{Zn}+2\mathrm{HCl}\to {\mathrm{ZnCl}}_2+{\mathrm{H}}_2$

The gas is collected over water in a 2 L container at 25°C. The vapour pressure of water at 25°C is 23.8 mmHg. Calculate the partial pressure of dry hydrogen gas. 

Solution: Converting mass of Zn to moles: 

Moles in $Zn=\frac{6.5}{65}=0.1\mathrm{mol}$

Since according to the equation, 1 mol of Zn gives 1 mol of H₂

Moles of H₂ produced will be 0.1 mol 

Using the law of ideal gases: 

$\begin{array}{rl}& {\mathrm{P}}_{\text{Total }}\mathrm{V}={\mathrm{n}}_{\text{Total }}\mathrm{RT}\\ & P_{\text{Total }}=\frac{0.1\times 0.0821\times 298}{2}=\frac{2.44858}{2}\\ & P_{\text{Total }}=1.2243\mathrm{atm}\text{ or }930.5\mathrm{mmHG}\end{array}$

Hence, the partial pressure of dry hydrogen: 

$\begin{array}{rl}{P}_{H_2}& =P_{\text{Total }}-P_{H_{2}O}\\ P_{H_2}& =930.5-23.8=906.7\mathrm{mmHg}\end{array}$