# Dalton’s Law of Partial Pressures Examples

We explain the Dalton’s Law of Partial Pressures examples. Dalton’s Law of Partial Pressures is a physical generality that explains gas mixtures. **Each component of the gas mixture exerts its own pressure that participates in the total pressure** of the mixture.

In cases such as the study of air pollution, the relationship of Pressure, Volume and Temperature of an air sample, which contains various gases, may be of interest. In this and all cases involving gas mixtures, the total gas pressure is related to the **Partial Pressures** , that is, the **pressures of the individual gaseous components of the mixture.**

In 1801, Dalton formulated a Law that today bears his name: **Dalton’s Law of Partial Pressures** , which describes that the Total Pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone.

## Description of Dalton’s Law

We consider the case in which two gases, A and B, are in a container of volume V. The pressure exerted by gas A, according to the ideal gas equation, is:

**P _{A} = n _{A} RT / V**

Where n _{A} is the number of moles of A present. Similarly, the pressure exerted by gas B is

**P _{B} = n _{B} RT / V**

In a mixture of gases A and B, the total pressure P _{T} is the result of the collisions of both types of molecules A and B, with the walls of the container. Therefore, according to Dalton’s law:

**P _{T} = P _{A} + P _{B}**

**P _{T} = (n _{A} RT / V) + (n _{B} RT / V)**

We regroup by putting RT / V as a common factor, and adding n _{A} and n _{B}

**P _{T} = (nA + nB) * (RT / V)**

Since n _{A} + n _{B} = n:

**P _{T} = nRT / V**

Where n, the total number of moles of gases present, is given by **n = n _{A} + n _{B}** , and P

_{A}and P

_{B}are the partial pressures of gases A and B, respectively. Thus, for a mixture of gases, the

**P**, not on the nature of the gas molecules.

_{T}depends only on the total number of moles of gas presentIn general, the total pressure of a gas mixture is given by

**P _{T} = P _{1} + P _{2} + P _{3} +…**

Where P _{1} , P _{2} , P _{3} ,… are the partial pressures of components 1, 2, 3,… To see how each partial pressure is related to the total pressure, the case of the mixture of two gases A and B. Dividing P _{A} by P _{T} gives:

**P _{A} / P _{T} = (n _{A} RT / V) / (nRT / V)**

Since RT / V is above and below in the relationship, they are eliminated, leaving:

**P _{A} / P _{T} = n _{A} / n**

And since **n = n _{A} + n _{B}** (total moles = moles of gas A + moles of gas B):

P _{A} / P _{T} = n _{A} / (n _{A} + n _{B} )

**P _{A} / P _{T} = X _{A}**

Where **X _{A}** is called

**Fraction Molar of the gas A**.

## Molar fraction

The Molar Fraction is a dimensionless Quantity (without units) that expresses the **ratio of the number of moles of a component with the number of moles of all** the components present.

In general, the mole fraction of component i in a mixture is given by:

**X _{i} = n _{i} / n _{T}**

Where ni and n _{T} are the number of moles of component i and the total number of moles present, respectively. The **mole fraction is always smaller than 1** , of course, since a fraction, by definition, is a part of 1.

It is possible to express the Partial Pressure of A as:

P _{A} = X _{A} P _{T}

And similarly, for gas B:

P _{B} = X _{B} P _{T}

The Sum of the Molar Fractions of a gas mixture must equal one. If only two components are present, then

X _{A} + X _{B} = (n _{A} / n) + (n _{B} / n) = 1

If a system contains more than two gases, the partial pressure of component i is related to the total pressure by:

**P _{i} = X _{i} P _{T}**

## Determination of Partial Pressures

A pressure gauge only measures the total pressure of a gas mixture. To obtain the partial pressures, it is necessary to know **the mole fractions of the components** , which could involve elaborate chemical analyzes.

The most direct method of measuring partial pressures is to **use a mass spectrometer** . The relative intensities of the signals of a mass spectrum are directly proportional to the quantities, and therefore to the mole fractions of the gases present.

With the values of the mole fractions and the total pressure, it is possible to calculate the partial pressures of the individual components.

## Examples of Partial Pressures

1.- A gaseous mixture contains 4.46 moles of Neon (Ne), 0.74 moles of Argon (Ar) and 2.15 moles of Xenon (Xe). Calculate the Partial Pressures of the gases if the Total Pressure is 2.00 atmospheres.

n = 4.46 + 0.74 + 2.15 = 7.35

**n = 7.35 moles**

**P _{T} = 2.00 atm**

**Neon Partial Pressure:**

n _{Ne} = 4.46 moles

X _{Ne} = 4.46 / 7.35 = 0.6068

P _{Ne} = X _{Ne} * P _{T} = (0.6068) * (2.00 atm) = **1.2136 atm**

**Argon Partial Pressure:**

n _{Ar} = 0.74 moles

X _{Ar} = 0.74 / 7.35 = 0.1007

P _{Ar} = X _{Ar} * P _{T} = (0.1007) * (2.00 atm) = **0.2014 atm**

**Xenon Partial Pressure:**

n _{Xe} = 2.15 moles

X _{Xe} = 2.15 / 7.35 = 0.2925

P _{Xe} = (0.2925) * (2.00 atm) = **0.5850 atm**

**Verification:**

The Total Pressure of 2.00 atm must be equal when adding all the partial pressures:

**P _{T} = 1.2136 atm + 0.2014 atm + 0.5850 atm = 2.00 atm**

2.- A gaseous mixture contains 5.05 moles of Nitrogen (N _{2} ), 0.82 moles of Methane (CH _{4} ) and 3.16 moles of Carbon Dioxide (CO _{2} ). Calculate the Partial Pressures of the gases if the Total Pressure is 5.50 atmospheres.

n = 5.05 + 0.82 + 3.16 = 9.03

**n = 9.03 moles**

**P _{T} = 5.50 atm**

**Nitrogen Partial Pressure:**

n _{N2} = 5.05 moles

X _{N2} = 5.05 / 9.03 = 0.5592

P _{N2} = X _{N2} * P _{T} = (0.5592) * (5.50 atm) = **3.0759 atm**

**Methane Partial Pressure:**

N _{CH4} = 0.82 moles

X _{CH4} = 0.82 / 9.03 = 0.0908

P _{CH4} = X _{CH4} * P _{T} = (0.0908) * (5.50 atm) = **0.4994 atm**

**Carbon Dioxide Partial Pressure:**

N _{CO2} = 3.16 moles

X _{CO2} = 3.16 / 9.03 = 0.3499

P _{CO2} = (0.3499) * (5.50 atm) = **1.9247 atm**

**Verification:**

The Total Pressure of 5.50 atm must be equal when adding all the partial pressures:

**P _{T} = 3.0759 atm + 0.4994 atm + 1.9247 atm = 5.50 atm**

3.- A gaseous mixture contains 10.20 moles of Hydrogen (H), 4.62 moles of Carbon Monoxide (CO) and 1.94 moles of Helium (He). Calculate the Partial Pressures of the gases if the Total Pressure is 3.80 atmospheres.

n = 10.20 + 4.62 + 1.94 = 16.76

**n = 16.76 moles**

**P _{T} = 3.80 atm**

**Hydrogen Partial Pressure:**

n _{H} = 10.20 moles

X _{H} = 10.20 / 16.76 = 0.6086

P _{H} = X _{H} * P _{T} = (0.6086) * (3.80 atm) = **2.3126 atm**

**Carbon Monoxide Partial Pressure:**

n _{CO} = 4.62 moles

X _{CO} = 4.62 / 16.76 = 0.2757

P _{CO} = X _{CO} * P _{T} = (0.2757) * (3.80 atm) = **1.0477 atm**

**Helium Partial Pressure:**

n _{He} = 1.94 moles

X _{He} = 1.94 / 16.76 = 0.1158

P _{He} = (0.1158) * (3.80 atm) = **0.44 atm**

**Verification:**

The Total Pressure of 2.00 atm must be equal when adding all the partial pressures:

**P _{T} = 2.3126 atm + 1.0477 atm + 0.44 atm = 3.80 atm**

4.- A gaseous mixture contains 12 moles of Chlorine (Cl _{2} ), 1.90 moles of Nitrogen (N _{2} ) and 0.15 moles of Oxygen (O _{2} ). Calculate the Partial Pressures of the gases if the Total Pressure is 8.64 atmospheres.

n = 12 + 1.90 + 0.15 =

**n = 14.05 moles**

**P _{T} = 8.64 atm**

**Chlorine Partial Pressure:**

N _{Cl2} = 12 moles

X _{Cl2} = 12 / 14.05 = 0.8541

P _{Cl2} = X _{Cl2} * P _{T} = (0.8541) * (8.64 atm) = **7.3794 atm**

**Nitrogen Partial Pressure:**

N _{N2} = 1.90 moles

X _{N2} = 1.90 / 14.05 = 0.1352

P _{N2} = X _{N2} * P _{T} = (0.1352) * (8.64 atm) = **1.1684 atm**

**Partial Oxygen Pressure:**

N _{O2} = 0.15 moles

X _{O2} = 0.15 / 14.05 = 0.0107

P _{O2} = (0.0107) * (8.64 atm) = **0.0922 atm**

**Verification:**

The Total Pressure of 8.64 atm must be equal when adding all the partial pressures:

**P _{T} = 7.3794 atm + 1.1684 atm + 0.0922 atm = 8.64 atm**