Dalton's Law and Ideal Gas Mixtures

Hello! Welcome back to our walk through thermodynamics.

In our last session, we mastered the Ideal Gas Law, our new best friend for understanding the state of a single, pure gas. We saw how pressure, volume, and temperature are all beautifully linked by the equation $P V = m R T$ . But, as we hinted at the end, the real world is rarely that pure. The air we breathe, the exhaust from a car engine, the gas in a scuba tank—these are all mixtures of different gases.

So, how do we handle a team of gases when we've only learned how to deal with individual players? That's our mission for today.

Our learning outcome is to determine the properties of an ideal gas mixture using Dalton's law of partial pressures. We'll start with a simple, intuitive idea of how mixed gases behave and build up to the engineering equations that let us analyze them precisely. By the end of our chat, you'll be able to look at a container full of mixed gases and confidently calculate the properties of the mixture as a whole, as well as the contribution of each individual gas.

Let's get started.

What Happens When Gases Mix?

Imagine our container of gas again. We know the pressure on the walls comes from billions of particles zipping around and colliding with them. Now, what if we have two different types of gas in the container, say, some energetic little helium atoms and some bigger, slower nitrogen molecules?

Both types of particles are in there, flying around and hitting the walls. Each gas contributes to the total pressure. The helium atoms create their own pressure, and the nitrogen molecules create their own. The total pressure you'd measure is simply the sum of the pressures created by each gas.

This simple, powerful idea is called Dalton's Law of Partial Pressures.

To get a great mental picture of this, watch the first part of this Crash Course video. It uses a fantastic analogy of adults and kindergartners at a party to explain how different gases in a mixture behave and contribute to the overall "bouncing" against the walls (pressure).

Partial Pressures & Vapor Pressure: Crash Course Chemistry #15

This video provides a fun and intuitive analogy to help visualize what happens when different gases are mixed in the same container.

Watch from the beginning to 1:48. Focus on the analogy of the two groups of people mixing and how their combined 'bouncing' adds up.

This leads us to the formal statement of the law:

Dalton's Law of Additive Pressures: The total pressure of a mixture of ideal gases is equal to the sum of the pressures that each gas would exert if it existed alone at the same mixture temperature and volume.

We call the pressure exerted by a single gas in the mixture its partial pressure. So, for a mixture of Gas A, Gas B, and Gas C:

$P_{total} = P_{A} + P_{B} + P_{C} + \dots$

Here, $P_{A}$ is the partial pressure of Gas A, $P_{B}$ is the partial pressure of Gas B, and so on.

Dalton's Law of Partial Pressures Illustration — This image shows the core idea of Dalton's Law. The pressure of the helium (112 atm) and the pressure of the neon (45 atm), when combined in the same volume, add up to the total pressure of the mixture (157 atm).

Describing the Mixture: Mass and Mole Fractions

To work with mixtures, we first need a clear way to describe their composition. If I have a tank of air, how much of it is nitrogen and how much is oxygen? We can describe this in two main ways.

1. Mass Fraction (mf)

This is the most straightforward way. It's simply the ratio of the mass of one component to the total mass of the mixture.

$Mass Fraction of component i (m f_{i}) = \frac{mass of component i}{total mass of mixture} = \frac{m _{i}}{m _{m}}$

For example, if we have a mixture of 7 kg of oxygen and 3 kg of nitrogen, the total mass is 10 kg.

The mass fraction of oxygen is $m f_{O_{2}} = 7 kg /10 kg = 0.7$ (or 70%).
The mass fraction of nitrogen is $m f_{N_{2}} = 3 kg /10 kg = 0.3$ (or 30%).

Notice that the sum of all mass fractions must always be 1.
$\sum m f_{i} = m f_{1} + m f_{2} + \dots = 1$

2. Mole Fraction (y)

This one is a little more abstract but incredibly important for Dalton's Law. Instead of comparing masses, we compare the number of molecules (or, more practically, the number of moles).

Remember, a mole is just a specific number of particles ( $6.022 \times 1 0^{23}$ ). The mole fraction is the ratio of the number of moles of one component to the total number of moles in the mixture. We use the symbol N to denote the number of moles.

$Mole Fraction of component i (y_{i}) = \frac{moles of component i}{total moles of mixture} = \frac{N _{i}}{N _{m}}$

Just like with mass fractions, the sum of all mole fractions must also be 1.
$\sum y_{i} = y_{1} + y_{2} + \dots = 1$

For a deeper dive into these definitions, the following article is an excellent, clear reference.

Dalton's law of partial pressure (article)

This article from Khan Academy formally defines partial pressure, Dalton's Law, and the concept of mole fraction, which connects everything together.

Read the 'Key points', 'Introduction', 'Ideal gases and partial pressure', and the two sections on 'Dalton's law of partial pressures'. Focus on how mole fraction (x in the article, but we'll use y) is defined and used.

The Key Link: Mole Fraction and Partial Pressure

So why is mole fraction so important? Because for an ideal gas mixture, a component's share of the total number of molecules is the same as its share of the total pressure.

Think about it: if 80% of the molecules in a box are nitrogen molecules, then it makes sense that they are responsible for 80% of the collisions with the wall, and therefore 80% of the total pressure.

This gives us the most useful form of Dalton's Law:

$P_{i} = y_{i} P_{m}$

Where:

$P_{i}$ is the partial pressure of component i.
$y_{i}$ is the mole fraction of component i.
$P_{m}$ is the total pressure of the mixture.

This relationship is incredibly powerful. If you know the composition of a gas mixture (its mole fractions) and its total pressure, you can immediately find the pressure exerted by each individual gas.

Test your understanding!

Question: Air is approximately 79% nitrogen and 21% oxygen by mole fraction. If you are in a room where the atmospheric pressure is 100 kPa, what is the partial pressure of oxygen?

Show answer

Using the formula $P_{i} = y_{i} P_{m}$ :

Mole fraction of oxygen, $y_{O_{2}} = 0.21$
Total mixture pressure, $P_{m} = 100 kPa$

Partial pressure of oxygen, $P_{O_{2}} = 0.21 \times 100 kPa = 21 kPa$ .

This 21 kPa is what your lungs actually use to absorb oxygen into your bloodstream!

Recap Break

So far, we've established that in a gas mixture:

The total pressure is the sum of the partial pressures of each gas (Dalton's Law).
We can describe the mixture's composition using mass fractions (based on weight) or mole fractions (based on number of molecules).
A gas's partial pressure is directly proportional to its mole fraction: $P_{i} = y_{i} P_{m}$ .

Now, how can we apply our trusty Ideal Gas Law, $P V = m R T$ , to the entire mixture? To do that, we need to find the "average" properties of the mixture.

Treating the Mixture as a Single Substance

It's convenient to treat a gas mixture, like air, as if it were a single "pseudo-pure" substance. This allows us to use the same familiar equations. To do this, we need to calculate two "apparent" or "average" properties for the mixture:

Apparent Molar Mass ( $M_{m}$ )
Apparent Gas Constant ( $R_{m}$ )

The apparent molar mass is the average molar mass of the mixture, weighted by the mole fractions. It's like calculating a weighted grade average. The components that have a higher mole fraction contribute more to the average.

$M_{m} = \sum y_{i} M_{i} = y_{1} M_{1} + y_{2} M_{2} + \dots$

Once we have the apparent molar mass for the mixture, we can find its apparent gas constant just like we did for a single gas:

$R_{m} = \frac{R _{u}}{M _{m}}$

Where $R_{u}$ is the universal gas constant, $8.314 kJ / (kmol \cdot K)$ .

With this apparent gas constant $R_{m}$ , we can now use the Ideal Gas Law for the entire mixture:

$P_{m} V_{m} = m_{m} R_{m} T_{m}$

This is fantastic! It means once we calculate the mixture's special gas constant, $R_{m}$ , we can forget that it's a mixture and treat it just like a single gas.

For a detailed walkthrough of these concepts and derivations, the following video is excellent. It covers everything from defining fractions to deriving the mixture properties.

Thermodynamics: Overview of ideal gas mixtures, Amagat's and Dalton's laws (42 of 51)

This lecture will guide us through the engineering approach to analyzing gas mixtures. It formalizes the definitions of mass and mole fractions and shows how to derive the apparent molar mass and gas constant for a mixture.

Watch these three segments: From 3:45 to 13:08 to understand the definitions of mass fraction, mole fraction, and mixture notation. From 13:08 to 17:15 to see how the apparent molar mass and mixture gas constant are derived. From 46:29 to 54:00 to see the formal derivation of the relationship between partial pressure and mole fraction from Dalton's Law.

Let's Solve Some Problems

Time to put all this theory into practice. We'll follow the same problem-solving strategy as last time.

Example 1: From Mass Composition to Mixture Properties

Problem: A gas mixture in a tank consists of 5 kg of Nitrogen ( $N_{2}$ ) and 3 kg of Carbon Dioxide ( $C O_{2}$ ).
(a) Find the mass fraction of each component.
(b) Find the mole fraction of each component.
(c) Find the apparent molar mass and apparent gas constant for the mixture.

Properties needed:

Molar Mass of $N_{2}$ , $M_{N_{2}} \approx 28.0 kg/kmol$
Molar Mass of $C O_{2}$ , $M_{C O_{2}} \approx 44.0 kg/kmol$

Solution:

(a) Mass Fractions:

Total mass, $m_{m} = m_{N_{2}} + m_{C O_{2}} = 5 kg + 3 kg = 8 kg$
$m f_{N_{2}} = \frac{5 kg}{8 kg} = 0.625$
$m f_{C O_{2}} = \frac{3 kg}{8 kg} = 0.375$
Check: $0.625 + 0.375 = 1.0$ . Perfect.

(b) Mole Fractions:
First, we need to find the number of moles (N) for each component using $N = m / M$ .

Moles of $N_{2}$ : $N_{N_{2}} = \frac{5 kg}{28.0 kg/kmol} = 0.1786 kmol$
Moles of $C O_{2}$ : $N_{C O_{2}} = \frac{3 kg}{44.0 kg/kmol} = 0.0682 kmol$
Total moles, $N_{m} = N_{N_{2}} + N_{C O_{2}} = 0.1786 + 0.0682 = 0.2468 kmol$

Now we can find the mole fractions (y).

$y_{N_{2}} = \frac{N _{N_{2}}}{N _{m}} = \frac{0.1786}{0.2468} = 0.7236$
$y_{C O_{2}} = \frac{N _{C O_{2}}}{N _{m}} = \frac{0.0682}{0.2468} = 0.2764$
Check: $0.7236 + 0.2764 \approx 1.0$ . Great.
Notice how Nitrogen is 62.5% of the mass but over 72% of the moles. This is because its molecules are lighter than CO₂'s.

(c) Apparent Molar Mass and Gas Constant:

Apparent Molar Mass, $M_{m}$ :
$M_{m} = \sum y_{i} M_{i} = (y_{N_{2}} \times M_{N_{2}}) + (y_{C O_{2}} \times M_{C O_{2}})$
$M_{m} = (0.7236 \times 28.0) + (0.2764 \times 44.0) = 20.26 + 12.16 = 32.42 kg/kmol$
A shortcut is to use $M_{m} = m_{m} / N_{m} = 8 kg /0.2468 kmol = 32.41 kg/kmol$ . Both methods work!
Apparent Gas Constant, $R_{m}$ :
$R_{m} = \frac{R _{u}}{M _{m}} = \frac{8.314 kJ/kmol \cdot K}{32.42 kg/kmol} = 0.2564 kJ/kg \cdot K$

Now we have a specific gas constant for this unique mixture, which we can use in $P V = m R T$ .

Example 2: Pressures in a Mixture

Problem: A 0.5 m³ rigid tank contains a mixture of 0.6 kg of Nitrogen ( $N_{2}$ ) and 0.4 kg of Oxygen ( $O_{2}$ ) at 300 K. Determine the partial pressure of each gas and the total pressure of the mixture.

Properties needed:

Gas constant for $N_{2}$ , $R_{N_{2}} = 0.2968 kJ/kg \cdot K$
Gas constant for $O_{2}$ , $R_{O_{2}} = 0.2598 kJ/kg \cdot K$

Solution:

Here's a cool thing about Dalton's Law. We can think of each gas occupying the entire volume all by itself. We can use the ideal gas law on each component to find its partial pressure.

For Nitrogen: It behaves as if it's the only gas in the 0.5 m³ tank.
$P_{N_{2}} V_{m} = m_{N_{2}} R_{N_{2}} T_{m}$
$P_{N_{2}} = \frac{m _{N_{2}} R _{N_{2}} T _{m}}{V _{m}} = \frac{( 0.6 kg ) ( 0.2968 kJ/kg \cdot K ) ( 300 K )}{0.5 m ^{3}}$
$P_{N_{2}} = \frac{53.424 kJ}{0.5 m ^{3}} = 106.8 kPa$
(Remember, a kJ is a kN·m, and a kPa is a kN/m², so the units work out perfectly!)
For Oxygen: It also behaves as if it's the only gas in the 0.5 m³ tank.
$P_{O_{2}} V_{m} = m_{O_{2}} R_{O_{2}} T_{m}$
$P_{O_{2}} = \frac{m _{O_{2}} R _{O_{2}} T _{m}}{V _{m}} = \frac{( 0.4 kg ) ( 0.2598 kJ/kg \cdot K ) ( 300 K )}{0.5 m ^{3}}$
$P_{O_{2}} = \frac{31.176 kJ}{0.5 m ^{3}} = 62.4 kPa$
Total Pressure:
According to Dalton's Law, the total pressure is the sum of the partial pressures.
$P_{m} = P_{N_{2}} + P_{O_{2}} = 106.8 kPa + 62.4 kPa = 169.2 kPa$

This approach is very intuitive. You just analyze each gas one by one and add up the results.

Conclusion

Fantastic session! We've successfully extended our knowledge from single gases to the much more common scenario of gas mixtures. You've learned the engineering language and tools to describe and analyze them.

Key Takeaways:

Dalton's Law is our guiding principle: The total pressure is the sum of the partial pressures ( $P_{m} = \sum P_{i}$ ).
A gas's partial pressure is the pressure it would exert if it were alone in the container.
The composition of a mixture can be described by mass fraction ( $m f_{i}$ ) or mole fraction ( $y_{i}$ ).
The most powerful tool is the link between partial pressure and mole fraction: $P_{i} = y_{i} P_{m}$ .
We can treat a mixture as a single substance by calculating its apparent molar mass ( $M_{m}$ ) and apparent gas constant ( $R_{m}$ ), which allows us to use $P_{m} V_{m} = m_{m} R_{m} T_{m}$ .

Preview of the Next Lesson

We have spent a lot of time now learning how to describe the state of a system—its pressure, temperature, volume, and composition. We're getting really good at taking a snapshot of a gas and understanding its properties.

But thermodynamics is the science of energy in motion. The really interesting stuff happens when things change—when a system goes from one state to another. This involves energy crossing the boundary of our system in the form of heat and work.

In our next lesson, we will introduce the single most important rule in all of thermodynamics: the First Law of Thermodynamics. It's the principle of energy conservation, and it is the foundation for analyzing every engine, power plant, and refrigerator in the world. Get ready to move from describing states to analyzing processes

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