## Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure states the total pressure exerted by a mixture of gases is equal to the sum of the partial pressure of each individual gas.

### Learning Objectives

Demonstrate an understanding of partial pressures and mole fractions.

### Key Takeaways

#### Key Points

- The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: [latex]P_{total}=P_1+P_2+…+\;P_n[/latex].
- The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.
- Boyle’s Law and the Ideal Gas Law tell us the total pressure of a mixture depends solely on the number of moles of gas, and not the kinds of molecules; Dalton’s Law allows us to calculate the total pressure in a system from each gas’ individual contribution.

#### Key Terms

**mole fraction**: number of moles of one particular gas divided by the total moles of gas in the mixture**Dalton’s Law of Partial Pressures**: the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of each individual gas; also known as Dalton’s Law of Partial Pressures

Because it is dependent solely the number of particles and not the identity of the gas, the Ideal Gas Equation applies just as well to mixtures of gases is does to pure gases. In fact, it was with a gas mixture—ordinary air—that Boyle, Gay-Lussac, and Charles performed their early experiments. The only new concept we need to deal with gas mixtures is partial pressure, a concept invented by the famous English chemist John Dalton (1766-1844). Dalton correctly reasoned that the low density and high compressibility of gases were indicative of the fact that they consisted mostly of empty space; from this, it Dalton concluded that when two or more different gases occupy the same volume, they behave entirely independently of one another.

Dalton’s Law (also called Dalton’s Law of Partial Pressures) states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases. Mathematically, this can be stated as follows:

[latex]{P}_{total} = {P}_{1}+{P}_{2}+…+\;{P}_{n}[/latex]

where P_{1}, P_{2} and P_{n} represent the partial pressures of each compound. It is assumed that the gases do not react with each other.

### Example

A 2.0 L container is pressurized with 0.25 atm of oxygen gas and 0.60 atm of nitrogen gas. What is the total pressure inside the container?

[latex]P_{total}=P_{\text{O}_2}+P_{\text{N}_2}=0.25+0.60=0.85\;\text{atm}[/latex]

The total pressure inside the contain is 0.85 atm.

### Calculating the Mole Fraction

The mole fraction is a way of expressing the relative proportion of one particular gas within a mixture of gases. We do this by dividing the number of moles of a particular gas *i* by the total number of moles in the mixture:

[latex]x_i=\frac{\text{number of moles }i}{\text{total number moles of gas}}[/latex]

### Example

A 3.0 L container contains 4 mol He, 2 mol Ne, and 1 mol Ar. What is the mole fraction of neon gas?

[latex]x_{\text{Ne}}=\frac{\text{number of moles Ne}}{\text{total number moles of gas}}=\frac{2}{4+2+1}=\frac{2}{7}[/latex]

The mole fraction of neon gas is 2/7 or 0.28.

### Calculating Partial Pressure

The partial pressure of one individual gas within the overall mixtures, *pi*, can be expressed as follows:

[latex]{P}_{i}={P}_{total}{x}_{i}[/latex]

where *xi* is the mole fraction.

### Example

A mixture of 2 mol H_{2} and 3 mol He exerts a total pressure of 3 atm. What is the partial pressure of He?

[latex]{P}_{\text{He}}={P}_{total}{x}_{\text{He}}=(3)\left(\frac{3}{5}\right)=\frac{9}{5}\text{atm}[/latex]

### Calculating Total Pressure

We know from Boyle’s Law that the total pressure of the mixture depends solely on the number of moles of gas, regardless of the types and amounts of gases in the mixture; the Ideal Gas Law reveals that the pressure exerted by a mole of molecules does not depend on the identity of those particular molecules; Dalton’s Law now allows us to calculate the total pressure in a system when we know each gas individual contribution.

### Example

Consider a container of fixed volume 25.0 L. We inject into that container 0.78 moles of N_{2} gas at 298 K. From the Ideal Gas Law, we can easily calculate the measured pressure of the nitrogen gas to be 0.763 atm.

We now take an identical container of fixed volume 25.0 L, and we inject into that container 0.22 moles of O_{2} gas at 298K. The measured pressure of the oxygen gas is 0.215 atm.

As a third measurement, we inject 0.22 moles of O_{2} gas at 298K into the first container, which already has 0.78 moles of N_{2}. (Note that the mixture of gases we have prepared is very similar to that of air. ) The measured pressure in this container is now found to be 0.975 atm.

Our data show that the total pressure of the mixture of N_{2} and O_{2} in the container is equal to the sum of the pressures of the N_{2} and O_{2} samples taken separately. We now define the partial pressure of each gas in the mixture to be the pressure of each gas as if it were the only gas present. Our measurements demonstrate that the partial pressure of N_{2 }as part of the gas PN_{2} is 0.763 atm, and the partial pressure of O_{2 }as part of the gas PO_{2}, is 0.215 atm.

## Collecting Gases Over Water

The amount of gas present can be determined by collecting a gas over water and applying Dalton’s Law.

### Learning Objectives

Apply Dalton’s Law to determine the partial pressure of a gas collected over water.

### Key Takeaways

#### Key Points

- The total pressure in an inverted tube can be determined by the height of the water displaced in the tube.
- When calculating the amount of gas collected, Dalton’s Law must be used to account for the presence of water vapor in the collecting bottle.

#### Key Terms

**pneumatic trough**: device used to collect a gas over water; the height of water displaced in the tube can be used to determine the total pressure inside the tube

Since gases have such small densities, it can be difficult to measure their mass. A common way to determine the amount of gas present is by collecting it over water and measuring the height of displaced water; this is accomplished by placing a tube into an inverted bottle, the opening of which is immersed in a larger container of water.

### The Pneumatic Trough

This arrangement is called a pneumatic trough, and it was widely used in the early days of chemistry. As the gas enters the bottle, it displaces the water and becomes trapped in the closed, upper part of the bottle. You can use this method to measure a pure gas (i.e. O_{2}) or the amount of gas produced from a reaction. The collected gas is not the only gas in the bottle, however; keep in mind that liquid water itself is always in equilibrium with its vapor phase, so the space at the top of the bottle is actually a mixture of two gases: the gas being collected, and gaseous H_{2}O. The partial pressure of H_{2}O is known as the vapor pressure of water and is dependent on the temperature. To determine the quantity of gas we have collected alone, we must subtract the vapor pressure of water from the total vapor pressure of the mixture.

### Calculating Gas Volume

### Example 1

O_{2} gas is collected in a pneumatic trough with a volume of 0.155 L until the height of the water inside the trough is equal to the height of the water outside the trough. The atmospheric pressure is 754 torr, and the temperature is 295 K. How many moles of oxygen are present in the trough? (At 295 K, the vapor pressure of water is 19.8 torr.)

The total pressure in the tube can be written using Dalton’s Law of Partial Pressures:

[latex]{P}_{total}={P}_{\text{H}_{2}\text{O}}+{P}_{\text{O}_{2}}[/latex]

Rearranging this in terms of [latex]P_{\text{O}_2}[/latex], we have:

[latex]{P}_{\text{O}_{2}}= {P}_{total} - {P}_{\text{H}_{2}\text{O}}[/latex]

Because the height of the water inside the tube is equal to the height of the water outside the tube, the total pressure inside the tube must be equal to the atmospheric pressure. With substitution, we have:

[latex]P_{\text{O}_2}=P_{total}-P_{\text{H}_2\text{O}}= 754 - 19.8 = 734\text{ torr} =.966\text{ atm}[/latex]

Next, we apply the Ideal Gas Law:

[latex]\begin{array}{Rcl}n&=&\frac{PV}{RT}\\{}&=&\frac {(.966\text{ atm})(.155L)}{(.082 \text{L}\cdot\text{atm}\cdot \text{mol}^{-1}\cdot \text{K}^{-1}) (295\text{K})}\\{}&=&.00619 \text{ mol O}_2\end{array}[/latex]

### Example 2

Oxygen gas generated in an experiment is collected at 25°C in a bottle inverted in a trough of water. The external laboratory pressure is 1.000 atm. When the water level in the originally full bottle has fallen to the level in the trough, the volume of collected gas is 1750 ml. How many moles of oxygen gas have been collected?

If the water levels inside and outside the bottle are the same, then the total pressure inside the bottle equals 1.000 atm; at 25°C, the vapor pressure of water (or the pressure of water vapor in equilibrium with the liquid) is 23.8 mm Hg or 0.0313 atm.

Therefore, the partial pressure of oxygen gas is 1.000 – 0.031, or 0.969 atm.

The mole fraction of oxygen gas in the bottle is 0.969 (not 1.000), and the partial pressure of oxygen also is 0.969 atm. The number of moles is: [latex]n=\frac{PV}{RT }=\frac{(.969 \text{ atm})( 1730\text{ cm}^{3})}{(82.054\text{ cm}^{3}\ K^{-1}\text{ mole}^{-1}\ )( \ 298 \text{ K})}[/latex]

n = 0.068 moles O_{2}