Gas Mixtures: Dalton and Amagat Models

Analyzing mixtures of gases is fundamental to countless engineering systems, from designing internal combustion engines and HVAC systems to modeling atmospheric processes and industrial reactors. To predict the behavior of these mixtures, engineers rely on two powerful, complementary models: Dalton's law of partial pressures and Amagat's law of partial volumes. These models provide the mathematical framework for calculating overall mixture properties—like pressure, volume, and composition—directly from the known behavior of the individual components, assuming ideal gas behavior. Mastering these concepts is essential for accurate thermodynamic analysis and design.

Defining the Ideal Gas Mixture

Before applying Dalton's or Amagat's laws, we must establish what we mean by an ideal gas mixture. This is a mixture where each component gas, as well as the mixture as a whole, obeys the ideal gas equation of state. Crucially, each gas behaves as if the others were not present; the molecules do not interact, and they occupy the entire volume available. The composition of a mixture is most precisely described using mole fractions. For a mixture with $k$ components, the mole fraction of component $i$ , denoted $y_{i}$ , is the ratio of the number of moles of $i$ ( $n_{i}$ ) to the total number of moles in the mixture ( $n_{m}$ ): $y_{i} = n_{i} / n_{m}$ . By definition, the sum of all mole fractions equals one: $\sum y_{i} = 1$ . Mass fraction is an alternative, but mole fractions are often more convenient for gas law calculations.

Dalton's Law of Partial Pressures

Dalton's law states that for a mixture of non-reacting ideal gases, the total pressure $P_{m}$ exerted by the mixture is equal to the sum of the partial pressures of each individual component. A component's partial pressure, $P_{i}$ , is the pressure that component would exert if it alone occupied the entire mixture volume $V_{m}$ at the mixture temperature $T_{m}$ .

Mathematically, Dalton's law is expressed as: $P_{m} = i = 1 \sum k P_{i}$

The relationship between partial pressure and mole fraction is direct and elegant. From the ideal gas law:

For component $i$ occupying the mixture volume: $P_{i} V_{m} = n_{i} R T_{m}$
For the mixture: $P_{m} V_{m} = n_{m} R T_{m}$

Dividing the first equation by the second, we find that the partial pressure is simply the mole fraction times the total pressure: $P_{i} = y_{i} P_{m}$

This is a cornerstone result. It means that if you know the total pressure and composition of a mixture, you can immediately find the pressure contribution of any component.

Example: A 100 L tank at 300 K contains a mixture of 2 mol O₂ and 8 mol N₂. What is the partial pressure of O₂ if the total pressure is 250 kPa?

Total moles, $n_{m} = 2 + 8 = 10$ mol.
Mole fraction of O₂, $y_{O_{2}} = 2/10 = 0.2$ .
Partial pressure of O₂, $P_{O_{2}} = y_{O_{2}} P_{m} = 0.2 * 250 = 50$ kPa.

Amagat's Law of Partial Volumes

Amagat's law (or Leduc's law) offers a volumetric perspective. It states that the total volume $V_{m}$ of a non-reacting ideal gas mixture is equal to the sum of the partial volumes of each component. A component's partial volume, $V_{i}$ , is the volume that component would occupy if it alone were at the mixture pressure $P_{m}$ and mixture temperature $T_{m}$ .

Mathematically, Amagat's law is: $V_{m} = i = 1 \sum k V_{i}$

Using the ideal gas law under the conditions of Amagat's definition:

For component $i$ at mixture pressure: $P_{m} V_{i} = n_{i} R T_{m}$
For the mixture: $P_{m} V_{m} = n_{m} R T_{m}$

Dividing again reveals that the partial volume is also proportional to the mole fraction: $V_{i} = y_{i} V_{m}$

This provides a powerful way to think about mixture composition. The partial volume is the volume "contributed" by a gas to the total mixture.

Example: Air at 101.3 kPa and 298 K can be approximated as 21% O₂ and 79% N₂ by volume. This volume percent is exactly the mole fraction because of Amagat's law. For 1 m³ of this air, the partial volume of O₂ is $V_{O_{2}} = 0.21 * 1 = 0.21$ m³. This means the oxygen in that cubic meter of air would occupy 0.21 m³ if isolated at the same 101.3 kPa and 298 K.

Comparing the Dalton and Amagat Models

While both models apply to ideal gas mixtures and yield the same results for properties like $P_{m}$ and $V_{m}$ , they conceptualize the mixture differently. The choice of model often depends on which set of conditions is more natural for a given problem.

Feature	Dalton's Model	Amagat's Model
Defining Condition	Each component occupies the mixture volume $V_{m}$ at $T_{m}$ .	Each component is at the mixture pressure $P_{m}$ and $T_{m}$ .
Key Property	Partial Pressure, $P_{i} = y_{i} P_{m}$	Partial Volume, $V_{i} = y_{i} V_{m}$
Useful For	Problems where volume and temperature are fixed (e.g., a rigid tank).	Problems where pressure and temperature are fixed (e.g., mixing streams at constant P & T).
Visual Analogy	All gases squeezed into the same box; their pressures add up.	Each gas gets its own compartment at the same pressure; their volumes add up.

The models are two sides of the same coin. You can derive one from the other using the ideal gas law. Their equivalence for ideal gases is a critical point of understanding.

Calculating Mixture Properties

The real power of these laws lies in determining mixture properties from component data. A common task is finding the apparent (or average) molar mass $M_{m}$ and specific gas constant $R_{m}$ of the mixture.

The mixture molar mass is the mole-fraction-weighted average of the component molar masses: $M_{m} = i = 1 \sum k y_{i} M_{i}$ Since the universal gas constant $R_{u}$ is the same for all gases, the mixture-specific gas constant is: $R_{m} = \frac{R _{u}}{M _{m}}$

With $R_{m}$ known, you can apply the ideal gas law to the mixture: $P_{m} V_{m} = m_{m} R_{m} T_{m}$ .

Worked Example: A gaseous mixture has a molar analysis of 60% CH₄, 30% C₂H₆, and 10% C₃H₈. The mixture is contained in a 2 m³ rigid tank at 350 K. The total pressure measured is 1500 kPa. Calculate: a) The partial volume of propane (C₃H₈) using Amagat's law. b) The partial pressure of ethane (C₂H₆) using Dalton's law. c) The mass of the mixture in the tank.

Step 1: Organize known data. Molar masses: $M_{C H_{4}} = 16$ , $M_{C_{2} H_{6}} = 30$ , $M_{C_{3} H_{8}} = 44$ kg/kmol. Mole fractions: $y_{C H_{4}} = 0.60$ , $y_{C_{2} H_{6}} = 0.30$ , $y_{C_{3} H_{8}} = 0.10$ . $V_{m} = 2$ m³, $T_{m} = 350$ K, $P_{m} = 1500$ kPa.

Step 2: Solve (a) Partial Volume of C₃H₈. Amagat's law: $V_{i} = y_{i} V_{m}$ . $V_{C_{3} H_{8}} = 0.10 * 2 = 0.2$ m³.

Step 3: Solve (b) Partial Pressure of C₂H₆. Dalton's law: $P_{i} = y_{i} P_{m}$ . $P_{C_{2} H_{6}} = 0.30 * 1500 = 450$ kPa.

Step 4: Solve (c) Mass of the mixture. First, find the mixture molar mass. $M_{m} = \sum y_{i} M_{i} = (0.60 * 16) + (0.30 * 30) + (0.10 * 44) = 9.6 + 9 + 4.4 = 23$ kg/kmol. The mixture specific gas constant: $R_{m} = R_{u} / M_{m} = 8.314/23 = 0.3615$ kJ/kg·K. Use the ideal gas law for the mixture to find mass $m_{m}$ : $P_{m} V_{m} = m_{m} R_{m} T_{m}$ $m_{m} = \frac{P _{m} V _{m}}{R _{m} T _{m}} = \frac{( 1500 ) ( 2 )}{( 0.3615 ) ( 350 )} = \frac{3000}{126.525} \approx 23.71$ kg.

Common Pitfalls

Applying the Laws to Non-Ideal Gases or Condensing Vapors: Dalton's and Amagat's laws are strictly for ideal gas behavior. They fail for real gases at high pressures or low temperatures, and they cannot correctly handle mixtures where a component may condense (like water vapor in air). For such cases, more complex equations of state or psychrometric charts are needed.

Confusing Partial Pressure with Vapor Pressure: Partial pressure is a system-dependent property based on the amount of gas present. Vapor pressure is an intrinsic, temperature-dependent property of a pure substance when liquid and vapor are in equilibrium. Do not assume a gas's partial pressure equals its vapor pressure unless the system is specifically at saturation.

Incorrectly Using Mass Fractions in Core Formulas: The simple relationships $P_{i} = y_{i} P_{m}$ and $V_{i} = y_{i} V_{m}$ use mole fractions. If you are given composition by mass (mass fraction $w_{i}$ ), you must first convert to mole fractions: $y_{i} = \frac{w _{i} / M _{i}}{\sum ( w _{i} / M _{i} )}$ .

Misidentifying the "Mixture" Conditions: Ensure you use the correct defining condition for each law. For a Dalton's law partial pressure calculation, the component is imagined at $(V_{m}, T_{m})$ . For an Amagat's law partial volume, the component is imagined at $(P_{m}, T_{m})$ . Mixing these conditions will lead to incorrect results.

Summary

Dalton's Law treats each gas in a mixture as if it alone occupies the total volume at the mixture temperature. The total pressure is the sum of these partial pressures, and each partial pressure is directly proportional to its mole fraction: $P_{i} = y_{i} P_{m}$ .
Amagat's Law treats each gas as if it exists at the total pressure and mixture temperature. The total volume is the sum of these partial volumes, and each partial volume is directly proportional to its mole fraction: $V_{i} = y_{i} V_{m}$ .
Both models are equivalent for ideal gas mixtures and provide the foundational link between mixture composition (mole fractions) and measurable properties like pressure and volume.
The choice between models is often a matter of convenience, depending on whether the problem involves fixed volume (favoring Dalton's) or fixed pressure (favoring Amagat's) conditions.
Accurate analysis requires vigilant checking of ideal gas assumptions and careful use of mole fractions, especially when composition data is provided by mass.

Gas Mixtures: Dalton and Amagat Models

Gas Mixtures: Dalton and Amagat Models

Defining the Ideal Gas Mixture

Dalton's Law of Partial Pressures

Amagat's Law of Partial Volumes

Comparing the Dalton and Amagat Models

Calculating Mixture Properties

Common Pitfalls

Summary

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