Examples

# 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:

A = n A RT / V

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

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:

T = P A + P B

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

T = (nA + nB) * (RT / V)

Since n A + n B = n:

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 T depends only on the total number of moles of gas present , not on the nature of the gas molecules.

In general, the total pressure of a gas mixture is given by

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:

A / P T = (n A RT / V) / (nRT / V)

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

A / P T = n A / n

And since n = n A + n B (total moles = moles of gas A + moles of gas B):

A / P T = n A / (n A + n B )

A / P T = X A

Where 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:

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:

A = X A P T

And similarly, for gas B:

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

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:

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

T = 2.00 atm

Neon Partial Pressure:

Ne = 4.46 moles

Ne = 4.46 / 7.35 = 0.6068

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

Argon Partial Pressure:

Ar = 0.74 moles

Ar = 0.74 / 7.35 = 0.1007

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

Xenon Partial Pressure:

Xe = 2.15 moles

Xe = 2.15 / 7.35 = 0.2925

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:

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

T = 5.50 atm

Nitrogen Partial Pressure:

N2 = 5.05 moles

N2 = 5.05 / 9.03 = 0.5592

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

Methane Partial Pressure:

CH4 = 0.82 moles

CH4 = 0.82 / 9.03 = 0.0908

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

Carbon Dioxide Partial Pressure:

CO2 = 3.16 moles

CO2 = 3.16 / 9.03 = 0.3499

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:

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

T = 3.80 atm

Hydrogen Partial Pressure:

H = 10.20 moles

H = 10.20 / 16.76 = 0.6086

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

Carbon Monoxide Partial Pressure:

CO = 4.62 moles

CO = 4.62 / 16.76 = 0.2757

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

Helium Partial Pressure:

He = 1.94 moles

He = 1.94 / 16.76 = 0.1158

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:

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

T = 8.64 atm

Chlorine Partial Pressure:

Cl2 = 12 moles

Cl2 = 12 / 14.05 = 0.8541

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

Nitrogen Partial Pressure:

N2 = 1.90 moles

N2 = 1.90 / 14.05 = 0.1352

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

Partial Oxygen Pressure:

O2 = 0.15 moles

O2 = 0.15 / 14.05 = 0.0107

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:

T = 7.3794 atm + 1.1684 atm + 0.0922 atm = 8.64 atm