9.3 Dalton’s Law of Partial Pressure

Dalton’s Law

In a mixture of gases, each molecule acts independently of all the others, provided that the gases behave as ideal gases and do not interact with each other in any way. 

Dalton’s law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases.

Mathematically we can write,

Partial pressure of a gas can be calculated using mol fraction of a gas in a gas mixture and total pressure.

Figure 9.51 Mixture of Gases

Ref: commons.wikimedia.org/

Total pressure = 1 atm

For example: in the above diagram, total number of gas molecules =14. Number of A (blue) gas molecules= 7, number of B ( green) gas molecule= 7.

Mole fraction gas A= 7/14= 0.5

Mole fraction of B= 7/14= 0.5

If the total pressure of the gas molecule= 1 aim.

Partial pressure of A= 0.5*1= 0.5 atm

Partial pressure of B= 0.5*1= 0.5 atm

The following videdo might help you to understand the concept.

Dalton’s Law of Partial Pressures Explained

Mole Fractions of Gas Mixtures

The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction (X) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture (n_t):

The mole fraction is a dimensionless quantity between 0 and 1. If X_A = 1.0, then the sample is pure A, not a mixture. If X_A = 0, then no A is present in the mixture. The sum of the mole fractions of all the components present must equal 1.

To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas A to the total pressure of a gas mixture that contains A. We can use the ideal gas law to describe the pressures of both gas A and the mixture: P_A = n_ART/V and P_t = n_tRT/V. The ratio of the two is thus

Rearranging this equation gives

P_A = X_AP_t

That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties.  Please note that earth’s atmosphere is about 78% N₂, 21% O₂, and 0.9% Ar, with trace amounts of gases such as CO₂, H₂O, and others. This means that 78% of the particles present in the atmosphere are N₂; hence the mole fraction of N₂ is 78%/100% = 0.78. Similarly, the mole fractions of O₂ and Ar are 0.21 and 0.009, respectively. Using Equation 10.30, we therefore know that the partial pressure of N₂ is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of O₂ and Ar are 0.21 and 0.009 atm, respectively.

Expressing the composition of a gas mixture

Because most of the volume occupied by a gas consists of empty space, there is nothing to prevent two or more kinds of gases from occupying the same volume. Homogeneous mixtures of this kind are generally known as solutions, but it is customary to refer to them simply as gaseous mixtures.

We can specify the composition of gaseous mixtures in many different ways, but the most common ones are by volumes and by mole fractions.

Volume fractions in gas mixtures

From Avogadro’s Law we know that “equal volumes contain equal numbers of molecules”. This means that the volumes of gases, unlike those of solids and liquids, are strictly additive. So if a partitioned container has two volumes of gas A in one section and one volume of gas B in the other (both at the same temperature and pressure), and we remove the partition, the volume remains unchanged.

Figure 9.49 Mixture of Gases

We can specify the composition of this mixture by saying that the volume fraction of A is 2/3 and that of B is 1/3.

Volume fractions are often called partial volumes:

Note that we can employ partial volumes to specify the composition of a mixture even if it had never actually been made by combining the pure gases.

When we say that air, for example, is 21 percent oxygen and 78 percent nitrogen by volume, this is the same as saying that these same percentages of the molecules in air consist of O₂ and N₂. Similarly, in 1.0 mole of air, there is 0.21 mol of O₂ and 0.78 mol of N₂ (the other 0.1 mole consists of various trace gases, but is mostly neon.)

Note that you could never assume a similar equivalence with mixtures of liquids or solids, to which the E.V.E.N. principle does not apply.

Mole fractions in gas mixtures

These last two numbers (.21 and .78) also express the mole fractions of oxygen and nitrogen in air. Mole fraction means exactly what it says: the fraction of the molecules that consist of a specific substance. This is expressed algebraically by

so in the case of oxygen in the air, its mole fraction is

Don’t let this type of notation put you off! The summation sign Σ (Greek Sigma) simply means to add up the n’s (moles) of every gas. Thus if Gas A is the “i-th” substance as in the expression immediately above, the summation runs from i=1 through i=2.

Problem Example 7

A mixture of O₂ and nitrous oxide, N₂O, is sometimes used as a mild anesthetic in dental surgery. A certain mixture of these gases has a density of 1.482 g L^–1 at 25 and 0.980 atm. What was the mole-percent of N₂O in this mixture?

Solution: First, find the density the gas would have at STP:

The molar mass of the mixture is (1.65 g L^–1)(22.4 L mol^–1) = 37.0 g mol^–1. The molecular weights of O₂ and N₂ are 32 and 44, respectively. 37.0 is 5/12 of the difference between the molar masses of the two pure gases. Since the density of a gas mixture is directly proportional to its average molar mass, the mole fraction of the heavier gas in the mixture is also 5/12:

Problem Example 8

What is the mole fraction of carbon dioxide in a mixture consisting of equal masses of CO₂ (MW=44) and neon (MW=20.2)?

Solution: Assume any arbitrary mass, such as 100 g, find the equivalent numbers of moles of each gas, and then substitute into the definition of mole fraction:

n_CO2 = (100 g) ÷ (44 g mol^–1) = 2.3 mol

n_Ne= (100 g) ÷ (20.2 g mol^–1) = 4.9 mol

X_Ne = (2.3 mol) ÷ (2.3 mol + 4.9 mol) = 0.32

Question: A chemical engineer places a mixture of noble gases consisting of 5.50 g of He, 15.0 g of Ne and 35.0 g of Kr in a chamber at STP. Calculate the partial pressure of Kr gas.

Total pressure = 1 atm (STP)

Mols of He= 5.50/4.023= 1.367

Mols of Ne= 15.5/20.18= 0.7681

Mols of Kr= 35.5/83.798= 0.4237

Total mols= 1.367+0.7681+0.4237= 2.5588

Mol fraction of Kr= 0.4237/2.5588= 0.1656

Partial pressure of Kr= 0.1656*1= 0.166 atm.

***Dalton’s law of partial pressures

Figure 9.52 Dalton

The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the partial pressure, a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently.

The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas. Dalton himself stated this law in the simple and vivid way shown at the left.

The usual way of stating Dalton’s Law of Partial Pressures is

The total pressure of a gas is the sum
of the partial pressures of its components

We can express this algebraically as

or, equivalently

(If you feel a need to memorize these formulas, you probably don’t really understand Dalton’s Law!)

There is also a similar relationship based on volume fractions, known as Amagat’s law of partial volumes. It is exactly analogous to Dalton’s law, in that it states that the total volume of a mixture is just the sum of the partial volumes of its components. But there are two important differences: Amagat’s law holds only for ideal gases which must all be at the same temperature and pressure. Dalton’s law has neither of these restrictions.

Although Amagat’s law seems intuitively obvious, it sometimes proves useful in chemical engineering applications. We will make no use of it in this course.

Problem Example 9

Calculate the mass of each component present in a mixture of fluorine (MW and xenon (MW 131.3) contained in a 2.0-L flask. The partial pressure of Xe is 350 torr and the total pressure is 724 torr at 25°C.

Solution: From Dalton’s law, the partial pressure of F₂ is (724 – 350) = 374 torr:

The mole fractions are X_Xe = 350/724 = .48 and X_F2 = 374/724 = .52 . The total number of moles of gas is

The mass of Xe is (131.3 g mol^–1) × (.48 × .078 mol) = 4.9 g

Problem Example 10

Three flasks having different volumes and containing different gases at various pressures are connected by stopcocks as shown. When the stopcocks are opened,

a) What will be the pressure in the system?
b) Which gas will be most abundant in the mixture?

Assume that the temperature is uniform and that the volume of the connecting tubes is negligible.

Figure 9.53 Mixture of Gases

Solution: The trick here is to note that the total number of moles n_T and the temperature remain unchanged, so we can make use of Boyle’s law PV = constant. We will work out the details for CO₂ only, denoted by subscripts a.

For CO₂, P_aV_a = (2.13 atm)(1.50 L) = 3.19 L-atm.

Adding the PV products for each separate container, we obtain
ΣP_iV_i = 6.36 L-atm = n_T RT. We will call this sum P₁V₁.
After the stopcocks have been opened and the gases mix, the new conditions are denoted by P₂V₂.

From Boyle’s law, P₁V₁ = P₂V₂ = 6.36 L-atm. V₂ = ΣV_i = 4.50 L.
Solving for the final pressure P₂ we obtain (6.36 L-atm)/(4.50 L) = 1.41 atm.

For part (b), note that the number of moles of each gas is n = PV/RT. The mole fraction if any one gas is X_i = n_i /n_T . For CO₂, this works out to
(3.19/RT) / (6.36/RT) = 0.501. Because this exceeds 0.5, we know that this is the most abundant gas in the final mixture.

5  Some applications of Dalton’s law

Collecting gases over water

Figure 9.54 Collection of Gas over water

A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough, and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part.

The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure

confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H₂O. The partial pressure of H₂O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton’s Law to find the partial pressure of that gas.

Problem Example 11

Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of O₂ produced.

Solution:

From Dalton’s law, P_O2= P_total – P_H2O = 754 – 19.8 = 734 torr = .966 atm.

Scuba diving and Dalton’s law:

Figure 9.55 Scuba diving

Ref: commons.wikimedia.org/

                                     Figure 9.56 Atmospheric Pressure below sea level

Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O₂ is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure.

But at a total pressure of 2 atm, the partial pressure of O₂ in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the O₂ pressure of .8 atm would be sufficiently high to induce oxygen toxicity. For this reason, the air mixture in the pressurized tanks that scuba divers wear  must contain a smaller fraction of O₂. This can be achieved most simply by raising the nitrogen content, but high partial pressures of N₂ can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures.

The Chemistry of Scuba Diving

Diving physics and “fizzyology”

Scuba diving and hyperbaric medicine

What you should be able to do

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.

One mole of a gas occupies a volume of 22.4 L at STP (standard temperature and pressure, 273K, 1 atm = 103 kPa.)

The above fact allows us to relate the measurable property of the density of a gas to its molar mass.

The composition of a mixture of gases is commonly expressed in terms mole fractions; be sure you know how to calculate them.

Dalton’s Law of partial pressures says that every gas in a mixture acts independently, so the total pressure a gas exerts against the walls of a container is just the sum of the partial pressures of the individual components.

Practice Questions:

1. CO₂ is added to a container containing 2.5 atm of O₂ to give total pressure 4.0 atm of the gas. What is the partial pressure of CO₂ in the final mixture?

2. Blast furnaces give off many unpleasant and unhealthy gases. If the total air pressure is 0.99 atm, the partial pressure of carbon dioxide is 0.05 atm, and the partial pressure of hydrogen sulfide is 0.02 atm, what is the partial pressure of the remaining air?

3. Oxygen and chlorine gas are mixed in a container with partial pressures of 401 mmHg and 0.639 atm, respectively. What is the total pressure inside the container (in atm)?

4. If I place 3 moles of N2 and 4 moles of O2 in a 35 L container at a temperature of 250 C, what will the pressure (in atm) of the resulting mixture of gases be?

5. What’s the partial pressure of carbon dioxide in a container that holds 5 moles of carbon dioxide, 3 moles of nitrogen, and 1 mole of hydrogen and has a total pressure of 1.05 atm?

6. A container with two gases, helium and argon, is 30.0% by volume helium. Calculate the partial pressure of helium and argon if the total pressure inside the container is 4.00 atm.

7. Two flasks at the same temperature are joined by a glass tube with a stopcock. Flask A is a 4.0 L flask containing N2(g) at 2.0 atm, while flask B is a 10.0 L flask containing CO(g) at 1.4 atm. What is the final pressure in the flasks after the stopcock is opened? Hints: • Determine the final volume for the gases (easy!). • Find the final partial pressure for both N2 and CO individually. (P1V1 = P2V2) • Finally, use Dalton’s law to find the total final pressure.

8. Two flasks are connected with a stopcock. The first flask has a volume of 5 liters and contains nitrogen gas at a pressure of 0.75 atm. The second flask has a volume of 8 L and contains oxygen gas at a pressure of 1.25 atm. When the stopcock between the flasks is opened and the gases are free to mix, what will the pressure be in the resulting mixture?

9. Container A (with volume 1.23 L) contains a gas under 3.24 atm of pressure. Container B (with volume 0.93 L) contains a gas under 2.82 atm of pressure. Container C (with volume 1.42 L) contains a gas under 1.21 atm of pressure. If all of these gases are put into Container D (with volume 1.51 L), what is the pressure in Container D?